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|---|---|---|---|---|
https://chemistry.stackexchange.com/questions/90260/what-is-the-difference-between-a-radical-and-a-neutral-alone-atom
|
What is the difference between a radical and a neutral alone atom?
|
I've done a bit of googling but can't find a clear answer to this question. Is there any difference to e.g. an alone neutral chlorine atom and a chlorine radical?
| 2 |
[
[
"\nA chlorine radical and a chlorine atom are effectively the same thing. This does not mean that \"radical\" and \"atom\" are interchangeable as jargons. They are used in fairly particular contexts, namely\n\n\n* Saying an \"atom\" is a reference to an *instance* of a particular element. They are often discussed abstractly, in terms of their constituent parts (nuclei and electrons) without coming to discuss their surroundings or conditions within a system.\n* It is natural to say *molecules are made of atoms*, because they conserve the total number of nuclei and electrons with respect to the individual atoms, so we can rationalize them as bonded atoms. But saying molecules are made of radicals is weird.\n* Radicals can (very often) be polyatomic, like a methyl radical $\\ce{CH3.}$\n* A radical has an odd number of electrons at its outer shell, it is chemically unstable, although electrically neutral. Atoms can be neutral and chemically stable, as in the noble gases, which fully comply to the octet rule.\n\n\nThus an atom being a radical and vice-verse is more of a \"chemical coincidence\" than anything else, kind of like a hydrogen cation is a proton. But they are definitely not equivalent.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/90256/ozonolysis-reaction
|
Ozonolysis Reaction
|
Why does the starting compound not need an -OH group? I thought ozonolysis only produced carbonyls?
[](https://i.stack.imgur.com/cqRTY.png)
| -2 |
[
[
"\nThe product of ozonolysis depends on how you work up the intermediate ozonide. You are clearly familiar with the reductive workup (with dimethyl sulfide or PPh3) that gives two carbonyl compounds, aldehyde or ketone. But if you work up under oxidative conditions, as is the case here, you get carboxylic acids from mono-substituted double bonds and ketones from di-substituted.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/90243/xylan-vs-teflon-for-non-stick-food-pans
|
Xylan vs. Teflon for non-stick food pans? [closed]
|
**Closed**. This question is [opinion-based](/help/closed-questions). It is not currently accepting answers.
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I am looking at non-stick pans and some are Xylan® others, and most, are Teflon™.
* What are the chemical differences?
* What is better for cooking?
* What is better for health safety?
I do know that Teflon can be consumed without a problem, as it is never degraded, but at very high temperatures the fumes are dangerous.
| 1 |
[
[
"\nBoth Xylan and Teflon are chemically polytetrafluoroethylene. But a polymer just by name can mean a lot of different things if you consider the structure/arrangement, additives and impurities. These are highly process-dependent and, guess what, we do not know their chemistry for sure because they are industrially protected. For actual matters, both Xylan and Teflon are unsafe above 260 ⁰C (500 ⁰F), when they start to break down to toxic fluorine polymer fumes.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/90242/why-do-these-reactions-not-occur
|
Why do these reactions not occur?
|
The following reactions do not occur:
$$\ce{HCl(aq)}\ +\ \ce{Pb(NO3)4(aq)} \to$$
$$\ce{HCl(aq)}\ +\ \ce{Cu(NO3)2(aq)} \to$$
Is this because in both cases both reactants are entirely dissolved as their components ion?
$\ce{NO3}$ is a conjugate base of a strong acid so it is a very weak base so it doesn't participate in reactions, right?
| 0 |
[
[
"\nWhen put in water a mixture such as $\\ce{HCl}$ and $\\ce{Pb(NO3)4}$ goes to complete ionic dissociation. Lead nitrate is soluble in water and hydrochloric acid is a very strong acid. Nitric acid is also strong, which means those dissolved nitrate ions are not reversing back to the molecular form.\n\n\nTheir ions will be fairly stable in solution and will only combine in case a more stable compound is formed. You could come up with reactions in which some compounds will go out of solution (like $\\ce{Cl2}$), such as $\\ce{ 2HCl + HNO3 \\to HNO2 + Cl2 + H2O} $. Lead could also act as an oxidizing agent. But I would argue that if you consider the change in free energy or the net redox potential of those reactions, you'd find that they are quite unfavourable. If you want to calculate it itself, you could start from a reduction potential table like [this one](http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm).\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90240/hamiltonian-2nd-positional-derivative-analogous-to-acceleration
|
Hamiltonian 2nd positional derivative analogous to acceleration?
|
In Quantum Mechanics, we learn that the Hamiltonian operator for an electron confined to a 1-D space is:
[](https://i.stack.imgur.com/8b6Oq.png)
We learn in QM that many operators have analogous interpretations familiar to us from classical mechanics. My question is then: is there a classical mechanics interpretation for the second positional derivative? Is its physical significance at all related to the second positional derivative in classical mechanics (i.e. acceleration)?
| 0 |
[
[
"\nIt's related to kinetic energy\n$$T = \\frac{1}{2} mv^2 $$\n\n\nwhich can be rewritten in terms of momentum $p$\n$$ T = \\frac{p^2}{2m} $$\n\n\nReplacing $p$ with its quantum mechanical operator\n$$ p = -i\\hbar\\frac{\\partial}{\\partial x} $$\n\n\ngives\n$$\\hat{T} = -\\frac{\\hbar^2}{2m} \\frac{\\partial^2}{\\partial x^2} $$\n\n\nI'm not sure why you bring up acceleration. Acceleration $a$ is the second *time derivative* of position, $$a = \\frac{d^2 x}{d t^2}$$\n\n\n",
"4"
],
[
"\nIf you really want a classical physics comparison, a second spatial derivative comes up in physical problems in the context of diffusion systems. The 1-D heat equation for example\n\n\n$$\\frac{\\partial T}{\\partial t} = \\alpha \\frac{\\partial^2 T}{\\partial x^2}$$\n\n\nThis equation is fairly similar to the Shrodinger equation at a zero potential. You could even argue that the potential term is analogous to adding a source term to the equation, which would physically mean some form of heat generation within the system. \n\n\nThe only important difference is that $\\alpha$ is a real value in the heat equation, and thus the solution to it with proper boundary conditions leads to a vanishing time derivative at infinity: in fact, the equilibrium state. The wavefunction however does not travel to equilibrium but oscillates forever. This attribute can only be achieved by having an imaginary coefficient of the second-derivative term: remember $e^{-t}$ vanishes at infinity, but $e^{-it}$ corresponds to a sine wave according to Euler's formula. And that is what you get from Schrodinger's equation.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/90238/what-is-the-italian-acido-psammico-psammic-acid
|
What is the Italian ‘Acido Psammico’ (Psammic acid?)
|
I am reading an Italian detective novel — [*Lo stato delle anime* by Giorgio Todde](http://www.ciao.it/Lo_stato_delle_anime_Giorgio_Todde__Opinione_1104970) — in which someone is poisoned with ‘acido psammico’.
>
> Marini trova nello stomaco della vittima un ostia con all’interno dell’**acido psammico** (usato per conciare le pelli), l’acido ha ucciso la donna avvelenandola.
>
>
>
Which I think translates into English as:
>
> In the stomach of the victim Marini finds a communion wafer, inside of
> which is ‘psammic acid’ (used to tan hides): the acid had killed the
> woman by poisoning.
>
>
>
I have spent quite a time with dictionary and Google searches using the Italian, French (acide psammique) and English (psammitic acid comes up but is not defined) terms without success. The term ‘psammico’ seems to come up in a geological context, but infrequently and without clarifying things.
I should add that the novel is set in rural Sardinia in the 1890s and the implication is that shepherds would have access to it.
Does anyone know what the modern name and chemical composition of this is?
**Addendum: Further Information**
I am now reading further in the book (slowly, very slowly) and on p. 101 the protagonist (Marini) is in the capital city (Cagliari) and enters a chemist’s (pharmacist’s) shop to ask for ‘acido psammico’. He receives the following reply:
>
> La sabbiolina mortale?
>
>
>
In which ‘sabbiolina’ (not in my dictionaries) appears to be a diminutive related to ‘sabbia’ — sand. (‘mortale’ = ‘deadly, fatal’, of course.)
Consistent with this, further down the page we have:
>
> La sabbia velenosa…
>
>
>
“The poisonous sand…”
| 10 |
[
[
"\n(Edit: this answer turned indeed to be the right one)\n\n\nItalian *Psammico* - from the greek *Psámmos* for Sand - reads as \"sandy\". It can thus refer to a soil or the organisms living in or about sandy soils and waters, or to something derived from sands. \n\n\nWe as chemists know that sand is mostly silicates, so it seems difficult to think of a poison.\n\n\nHowever if we take \"sand\" as to mean \"from the soil, mineral, inorganic\" and combine this with the fact that the most historical and literary mineral poison is Arsenic, than I would suggest that the character has been killed by As2O3 or some other compound containing this element.\n\n\nArsenic and its compounds can (were) indeed use to tan skins and leathers, to preserve wood and timbers, and even in mummification processes.\n\n\nIn particular Wikipedia mentions cupper arsenochromate as a (modernly banned) ingredient in non-tannin based tanning. \n\n\nAt least in XIX century Italy, the word \"acid\" could have referred to non formally and/or non chemically acid compounds. Given the negative connotation of the word among those workers with no chemistry background, acid could have referred to any dangerous substance. Alternatively, it could have referred as it does today, to accepted traditional nomenclature or dehydrated forms, like in the case of tannic acid or in that of As2O3, respectively.\n\n\nMy answer remains speculative but I think has quite reasonable bases. The novel author in this case did not use any license, just found a more cryptic and archaic but \"original\" name for arsenic as poison.\n\n\n",
"10"
],
[
"\nActually, the meaning is: \n\n\n\n```\nused to tan leather.\n\n```\n\nThe acid used traditionally in tanneries is tannic acid, from there its name.\n\n\nTannic acid can be found in several plants one of which is the Sicilian Sumac (*Rhus coriaria*). This plant is also found in Sardegna (as per [Flora Italiana](http://luirig.altervista.org)), where the novel is placed.\nThe Italian name for this shrub is Sommacco siciliano.\nThe description of this plant is:\n\n\n\n```\npiccolo albero tipico di luoghi sassosi e rupestri, con fiori giallastri in \npannocchie, le cui foglie e corteccia, ricche di tannino, si adoperano nella \nconcia delle pelli (fam. Anacardiacee)\n\n```\n\n[the last part says: the leaves and bark, rich in tannin, are used in the tanning of leather.]\n\n\nFrom sommacco to psammicco there is not too much distance.\nBut I agree that it is speculative.\n\n\nTannic acid is toxic, but the oral LD50 in rats is 2.26±0.083 g/kg body weight, so too much to hide in an *ostia*. \nHowever, that can be a writers license.\n\n\nTannic acid is a complex polyphenol with structure:\n\n\n[](https://i.stack.imgur.com/dL6RQ.png)\n\n\n",
"10"
],
[
"\nThe term ‘acido psammico’ is from the Greek and literally means an acid derived from sand, as @Alchimista explains in his answer. He argues that it could be a term for arsenic, and, although I do not think there is strong evidence for this, I think it fair to accept his answer, especially as my own answer is speculative. However, *for the record*, here it is.\n\n\n\n> \n> I believe that *‘acido psammico’* is **not** a genuine Italian chemical (or psuedo-chemical) expression, nor a Sardinian dialect term, but a name invented by the author so that he could ascribe particular properties to the poison that suited his story.\n> \n> \n> \n\n\nMy reasons for thinking this is the term is used and recognized by a pharmacist, a doctor, a midwife, a priest and a professor of medicine. One would therefore expect it to have been in common use at the time and some record of it would have been found by an internet search or in the heavy-weight Italian dictionaries I have consulted. It is further understood by the daughter of a shepherd (although she, exceptionally had a formal education) in relation to tanning of hides, again suggesting wide general use not supported by any sources I have consulted.\n\n\nThe author of the book — Giorgio Todde — is still alive and there is an interview with him and a presentation by him on *YouTube*, so it may be possible to contact him and ask him directly. I will try to follow that avenue, with the help of colleagues with better written Italian, and update this answer if I have any successs.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90094/is-h2po4-stronger-than-hcn
|
Is H2PO4- stronger than HCN? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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Closed 5 years ago.
[Improve this question](/posts/90094/edit)
Wouldn't $\ce{HCN}$ be stronger, since $\ce{H2PO4-}$ predominantly acts as a base in aqueous solution?
| -3 |
[
[
"\nAs Zhe points out, dihydrogen phosphate ion [is not predominantly basic in water](http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf).\n\n\nBut the larger answer is that one thing can be both a stronger acid and a stronger base than another. For example, zinc hydroxide reacts with both acids and bases in water whereas carbon monoxide reacts with neither (at least under normal laboratory conditions). So zinc hydroxide is both a stronger acid and a stronger base than carbon monoxide.\n\n\nSimilarly any basicity in the dihydrogen phosphate ion does not rule out the possibility that it could also be an acid. But there is a limit to this, things that are both acidic and basic, called **amphoteric**, are generally weak in both directions.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/90085/how-to-control-relative-of-fe2-and-fe3
|
how to control relative of Fe2+ and Fe3+ [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
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hi how can I control relative of Fe2+ andFe3+ when we adding raw iron to acid solution To get the ratio 2:1
| -2 |
[
[
"\nMost acids aren't strong oxidisers and really only yield $\\ce{Fe^2+}$ when dissolving iron. The most common oxidising acid is nitric acid, $\\ce{HNO3}$, which does **fully** oxidise the iron to $\\ce{Fe^3+}$.\n\n\nIf you don't have access to nitric acid (now a restricted material in most European countries), you can dissolve the iron in hydrochloric acid, $\\ce{HCl}(aq)$, then oxidise the $\\ce{Fe^2+}$ subsequently with hydrogen peroxide $\\ce{H2O2}(aq)$ solution. Then boil that solution for a bit to get rid of any excess peroxide.\n\n\nYou can also buy ferric chloride solution, $\\ce{FeCl3}(aq)$, as [etching solution](https://www.ebay.co.uk/i/141742141841?chn=ps&adgroupid=49962971442&rlsatarget=pla-380178313640&abcId=1129946&adtype=pla&merchantid=107393898&poi=&googleloc=9046248&device=c&campaignid=974960578&crdt=0).\n\n\nNow how do you create a solution with a known $\\frac{\\ce{Fe^2+}}{\\ce{Fe^3+}}$ ratio?\n\n\nAdd a calculated amount of steel wool to a known quantity of the ferric solution. The following reaction takes place:\n\n\n$$\\ce{Fe^3+ + Fe -> 2Fe^2+}$$\n\n\nControl the concentration and quantity of ferric solution and the amount of steel wool used and you can obtain any $\\frac{\\ce{Fe^2+}}{\\ce{Fe^3+}}$ ratio. For example, if you set aside precisely half of your ferric solution, then add an excess of steel wool to the other part, filter of the unreacted iron (leave it to react overnight, e.g.) and mix both parts then your ratio will be $\\frac21$.\n\n\nThis how many instructables for ferrofluid are written. I've done this myself and it works well.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90076/whats-the-actual-or-official-definition-of-activation-energy
|
What's the actual or official definition of activation energy? [duplicate]
|
**This question already has an answer here**:
[How exactly is activation energy defined?](/questions/6486/how-exactly-is-activation-energy-defined)
(1 answer)
Closed 5 years ago.
In some sources I have seen that activation energy is the minimum energy required to cause a reaction. This implies that activation energy doesn't change with temperature (not considering very high changes).
In other sources it is the difference between the energy required to cause the reaction & the energy of the reactants. This implies that activation energy is temperature dependent since the energy of the reactants change with temperature. So which is correct?
Also another source (khan academy's guardian chemist) says that "The official definition of activation energy is a bit complicated and involves some calculus."
| 1 |
[
[
"\nActivation Energy is the minimum amount of energy needed extra than the average potential energy of the reactants to cause a reaction.And it does not change with the change of temperature, and for clarification the energy of the reactant molecules don't also change with the reaction. It is just the rate which changes with the reaction according to the arrhenius equation k=Ae^{-\\frac{Ea}{RT}}So,the fraction of molecules having energy higher than activation energy changes with change in temperature. I think the easier definition can be stated like this\n\n\n",
"-2"
]
] |
https://chemistry.stackexchange.com/questions/90073/simple-primitive-tetragonal-bravais-lattice
|
Simple/ Primitive Tetragonal Bravais Lattice [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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Are there any elements which exhibit the Simple(Primitive) Tetragonal Bravais Lattice?
| -1 |
[
[
"\nChances are there are no such elements. [This link](http://periodictable.com/Properties/A/CrystalStructure.html) seems to suggest so, but I wouldn't put too much trust in it. (To begin with, it claims only one crystal structure for each element, ignoring any polymorphs.) **So what?** This is not a fact of any consequence. It is about as (un)important as the knowledge that only one of the element names in English starts with \"K\", and none start with \"J\". Besides, both may change over time. New high-pressure crystal modifications of elements are discovered every now and then, and will be for a while, because no matter how far you reach, there is always a higher pressure.\n\n\nCome to think of it, there are *millions* of different crystal structures out there. Elemental compounds are just a very tiny minority. Surely there are examples of all Bravais lattices (not that it matters much).\n\n\nSo it goes.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/90063/is-it-possible-to-calculate-the-concentration-percentage-using-chromatography-pa
|
Is it possible to calculate the concentration percentage using chromatography paper?
|
I'm trying to calculate the concentrations of specific compounds using paper chromatography. Would I be able to calculate a rough estimate by comparing the size of each concentrated spot? If not is there another way to calculate this?
Thanks!
| 3 |
[
[
"\nYes it is possible. You must be thorough. To have a better resolution, place your mixture in two small points very close \"oo\" instead of one big \"o\". It is best to calibrate the TLC system using the compounds you are looking for (if available). Then measure not only the area of spots, but also their optical density.\n\n\n",
"1"
],
[
"\nYes, it can be done. \n\n\nI have done it on silica plates to compare the content of a natural product in different samples of plants. \nIt is , obviously, better to use HPLC, but in that case the sample preparation was very tedious so we decided to do it through TLC. \nThe content in the different samples was between 0 and 2%. Standards equivalent to a content of 0.5, 1, 1.5 and 2% we're prepared. \n\n\nWhat you need to have succes is to be very accurate in the quantity that you apply on the plate and to have a good developer. \nThe quantities applied must be all the same, preferably not too much because then the spots are very big and it is more difficult to compare them. \nOur product was developed with $\\ce{KMnO4}$ yielding an intense yellow spot on a purple background. \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90057/fock-operator-invariance-under-unitary-transformation
|
Fock operator invariance under unitary transformation
|
I know how to show that the Coulomb operator of the Fock operator is invariant under a unitary transformation of the orbitals, as on page 121 of Szabo and Ostlund, but the indices in my proof for the exchange operator are just not working. What might I be missing?
I have
\begin{align}
\sum\_i\hat{K}\_i'(1) &=\sum\_i\int dr\_2\chi\_i'^\*(2)\chi\_j'(2)/r\_{12} \\
&=\sum\_i\int dr\_2\sum\_kU\_{ki}^\*\chi\_k^\*(2)\sum\_lU\_{lj}\chi\_l(2)/r\_{12}\\
&=\sum\_{ikl}\int dr\_2U\_{ki}^\*U\_{lj}\chi\_k^\*(2)\chi\_l(2)/r\_{12}\\
&=\sum\_{i}\int dr\_2\chi\_i^\*(2)\chi\_i(2)/r\_{12}\\
&\neq\sum\_{i}\int dr\_2\chi\_i^\*(2)\chi\_j(2)/r\_{12}.
\end{align}
| 6 |
[
[
"\nThe exchange operator is not a operator by itself, it is only defined with the orbital it is working on: \n$%\n\\newcommand{\\ll}{\\left\\langle}\\newcommand{\\rr}{\\right\\rangle}\n\\newcommand{\\lb}{\\left|}\\newcommand{\\rb}{\\right|}\n\\newcommand{\\op}[1]{\\mathbf{#1}}$\n$$\\begin{align}\n && \\op{F}\\_i &= \\op{H}^\\mathrm{c} + \\sum\\_j (\\op{J}\\_j - \\op{K}\\_j),\\\\\n\\text{with}&& \n \\op{J}\\_j\\lb \\phi\\_i\\rr &=\n \\ll \\phi\\_j(\\op{x}\\_1) \\rb r\\_{12}^{-1} \n \\lb \\phi\\_j(\\op{x}\\_1) \\rr \\lb \\phi\\_i(\\op{x}\\_2) \\rr,\\\\\n\\text{and}&&\n \\op{K}\\_j\\lb \\phi\\_i\\rr &=\n \\ll \\phi\\_j(\\op{x}\\_1) \\rb r\\_{12}^{-1} \n \\lb \\phi\\_i(\\op{x}\\_1) \\rr \\lb \\phi\\_j(\\op{x}\\_2) \\rr.\n\\end{align}$$\n\n\nWhile in the Coulomb operator case the $\\lb \\phi\\_i\\rr$ doesn't do anything, so you can take it through the proof as a constant:\\*\n\\begin{align}\n \\sum\\_j\\op{J}\\_j\\lb \\phi\\_i\\rr \n &= \\sum\\_j \n \\ll \\phi\\_j(\\op{x}\\_1) \\rb r\\_{12}^{-1} \n \\lb \\phi\\_j(\\op{x}\\_1) \\rr \\lb \\phi\\_i(\\op{x}\\_2) \\rr\\\\\n &= \\sum\\_k \\sum\\_l \\sum\\_j U\\_{kj}^\\*U\\_{lj}\n \\ll \\phi\\_k(\\op{x}\\_1) \\rb r\\_{12}^{-1} \n \\lb \\phi\\_l(\\op{x}\\_1) \\rr \\lb \\phi\\_i(\\op{x}\\_2) \\rr\\\\\n &= \\sum\\_k \\sum\\_l \\delta\\_{kl}\n \\ll \\phi\\_k(\\op{x}\\_1) \\rb r\\_{12}^{-1} \n \\lb \\phi\\_l(\\op{x}\\_1) \\rr \\lb \\phi\\_i(\\op{x}\\_2) \\rr\\\\\n &= \\sum\\_k \n \\ll \\phi\\_k(\\op{x}\\_1) \\rb r\\_{12}^{-1} \n \\lb \\phi\\_k(\\op{x}\\_1) \\rr \\lb \\phi\\_i(\\op{x}\\_2) \\rr\\\\\n &= \\sum\\_k \\op{J}\\_k \\lb \\phi\\_i\\rr\n\\end{align}\n\n\nThat is not the case in the Exchange operator, as it switches the orbitals:\n\\begin{align}\n \\sum\\_j\\op{K}\\_j\\lb \\phi\\_i\\rr \n &= \\sum\\_j \n \\ll \\phi\\_j(\\op{x}\\_1) \\rb r\\_{12}^{-1} \n \\lb \\phi\\_i(\\op{x}\\_1) \\rr \\lb \\phi\\_j(\\op{x}\\_2) \\rr\\\\\n &= \\sum\\_k \\sum\\_l \\sum\\_j U\\_{kj}^\\*U\\_{lj}\n \\ll \\phi\\_k(\\op{x}\\_1) \\rb r\\_{12}^{-1} \n \\lb \\phi\\_i(\\op{x}\\_1) \\rr \\lb \\phi\\_l(\\op{x}\\_2) \\rr\\\\\n &= \\sum\\_k \\sum\\_l \\delta\\_{kl}\n \\ll \\phi\\_k(\\op{x}\\_1) \\rb r\\_{12}^{-1} \n \\lb \\phi\\_i(\\op{x}\\_1) \\rr \\lb \\phi\\_l(\\op{x}\\_2) \\rr\\\\\n &= \\sum\\_k \n \\ll \\phi\\_k(\\op{x}\\_1) \\rb r\\_{12}^{-1} \n \\lb \\phi\\_i(\\op{x}\\_1) \\rr \\lb \\phi\\_k(\\op{x}\\_2) \\rr\\\\\n &= \\sum\\_k \\op{K}\\_k \\lb \\phi\\_i\\rr\n\\end{align}\n\n\n\n\n---\n\n\n\\* Let \n$$\n\\mathbb{U}=\\left(\\begin{matrix}\n U\\_{11} & U\\_{12} & \\dots \\\\\n U\\_{21} & U\\_{22} & \\dots \\\\\n \\vdots & \\vdots & \\ddots\\\\\n \\end{matrix}\\right);\n\\mathbb{U}^\\dagger\\mathbb{U}=\\mathbb{E}\n\\implies U\\_{ij}^\\* U\\_{kl}= \\delta\\_{ij} = \\begin{cases}\n 1; & i = j = k = l \\\\ 0; & i \\neq j \\dots\\\\ \\end{cases}\n.$$\n\n\n",
"7"
]
] |
https://chemistry.stackexchange.com/questions/90056/racemization-of-ketones-in-presence-of-catalytic-acid
|
Racemization of ketones in presence of catalytic acid
|
**Question:**
Which of the following ketones racemise in aqueous solution containing acidic or basic impurities?
[](https://i.stack.imgur.com/xMYVW.png)
My answer:
A and B should racemise in aqueous solution, for the carbonyl groups in A and B contain alpha-hydrogens, leading to the possibility of enolisation, hence racemization. Compound C can't form an enol, hence shouldn't racemise.
It's very clear that compound C doesn't racemise, my confusion is about A and B.
My logic (stability of carbanion) says that A and B should both racemise, however, the answer I have is only B.
I agree that the carbanion formed by A would be less stable than that formed by B, but not so much so to ignore the possibility of enolisation and racemization.
Could someone please help? Is there anything obvious that I'm missing? or is it related to data? A good explanation would help. Thanks a lot.
| 7 |
[
[
"\nYou are right that the racemization occurs through the enol form. \n\n\nThe chiral center is on the carbon attached to the chlorine atom. In B, that carbon participates in the enol form, but in A not. Therefore, in A the enol is formed but there is no racemization. \n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/90055/overshooting-titration-effect-on-k
|
Overshooting titration effect on K
|
>
> a. Explain how overshooting the titration of the aqueous layer (ie. using too large a value for the volume of thiosulfate to react with the iodine and of triiodide ion) would affect the concentrations of iodide ion, iodine and of triiodide ion, and therefore of the calculated $K$.
>
>
>
>
**My thoughts regarding this question**
overshooting the titration will increase the concentrations of the iodide ion, iodine and triiodide ion
- According to me a change in $K$ comes only when the **temperature** is changed.
---
>
> b. Explain how using too large a value for the minimum volume of thiosulfate to react with the iodine (organic) in the mineral oil layer would affect the concentrations of iodide ion, iodine and triiodide ion, and therefore of the calculated $K$
>
>
>
I am unsure of the answer for this one too.
**My reasoning**
Using too large a value for the minimum volume of thiosulfate will increase the concentrations of the iodide ion, iodine and triiodide ion and nothing will happen to $K$ again.
---
Is my reasoning correct?
| 3 |
[
[
"\nYour reasoning is correct that $K$ is independent of concentration. But I think you are misreading the question: it wants you to tell how *you* will miscalculate *K* with respect to the correct value if you put in the wrong concentrations in the formula, that is, if there was an overshoot in your titration.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90041/can-you-make-deuterium-depleted-water-through-a-freezing-method
|
Can you make deuterium-depleted water through a freezing method?
|
Deuterium-depleted water (DDW) is sold commercially by some companies. It is apparently manufactured by fractional distillation, which leverages differences in boiling points of heavy and light water.
My question is this: can DDW be manufactured through a freezing method? Heavy water has a freezing point of 38.8 °F and normal water of 32 °F. If a jar of water were kept in a water bath of say a constant 35 °F, wouldn't the heavy water freeze and sink to the bottom of jar, allowing the DDW to be siphoned off of the top? I am surprised there appears to be no discussion of the merits or demerits of this idea anywhere on the internet.
| 2 |
[
[
"\nNot easily. For all practical purposes, $\\ce{D2O}$, $\\ce{DHO}$ and $\\ce{H2O}$ behave the same -- including complete miscibility. Though some small fraction of the liquid that freezes first would be enriched in $\\ce{D2O}$, and to a smaller extent in $\\ce{DHO}$, it would still be composed largely of $\\ce{H2O}$. \n\n\nAccording to [Wikipedia, the Girdler process](https://en.wikipedia.org/wiki/Girdler_sulfide_process) is the most energy efficient means of separation.\n\n\n",
"4"
],
[
"\nTo clarify, the objective is not to obtain heavy water. It is to obtain light water. So it is not an issue if the ice contains only some deuterium, but mostly light water, as the point is that the liquid water now has less deuterium than it did before. \n\n\nI actually just got a reply from a DDW manufacturer who said the following:\n\n\n\"Deuterated (\"heavy\") water has a higher freezing point than ordinary water, there is indeed a fractionation slightly above zero Celsius, but it does not mean that all heavy water will be frozen. \n\n\nBy holding the temperature at let’s say 1°C, you can get some ice that contains D2O or DHO in a higher concentration than in liquid because the freezing point for H2O is lower.\n\n\nThis way you can reduce the D-concentration of the water with 8-10 ppm in one step. If you wish to achieve further decrease you have to freeze this water in further steps.\n\n\nAccording to our knowledge, fractional distillation is the best way to produce DDW in large scale.\"\n\n\nThis same manufacturer says that home water distillers can at best reduce deuterium concentration by 1 or 2 PPM per pass, which is much less efficient than commercial evaporative methods. But an 8 to 10 drop in PPM from a freezing method (per pass) is considerable (if true), as the early clinical trials in this field suggest that even drops of say 20ppm are potentially clinically significant.\n\n\n",
"3"
],
[
"\nI would say. Diffusion rates play a role. The closer the surface of ice formation is to the freezing point of deuterium water. The purer the deuterium ice will be. The process will be slowed because of this but a purer form of deuterium depleted water would result. Since I have no experience this is speculation on my part. With the formation of ice some heat would be liberated and a natural form of convection would occur. \n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90039/what-happens-to-the-h-electron-during-acid-reaction
|
What happens to the H electron during acid reaction?
|
In this generic acid-base reaction:
$\ce{HX + B <=> X- + HB+}$
It's explained that the acid HX donates a proton H+ to its conjugate base HB+. What happens to the electron that was originally in the acid's H atom?
| 0 |
[
[
"\nThe electron belonging to elemental hydrogen forms part of a covalent bond with $\\ce{X}$. In this acid-base reaction the $\\ce{HX}$ bond breaks *heterolytically*, meaning that both electrons constituting the bond go to $\\ce{X-}$, hence the negative charge.\n\n\n[](https://i.stack.imgur.com/GKL9M.png)\n\n\nThe base then donates electrons to the proton, forming the electron deficient species $\\ce{HB+}$.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/90037/why-will-a-strong-base-neutralize-as-much-acid-as-a-weak-base
|
Why will a strong base neutralize as much acid as a weak base?
|
I don't understand how the volume of 0.200 M NaOH needed to neutralize 250.0 mL of 0.010 M HCl (0.0125 mL) is the same volume of 0.200 M NH3 that would be needed to neutralize 250.0 mL of 0.010 M HCl.
| 0 |
[
[
"\nWell it's because the strength of an acid or base is a measure of it's dissociation in water. But for the the reaction of an acid with a base the equilibrium lies completely to the right as the products (salt and water) are much more stable than the acid and base.\n\n\nYou're trying to compare the dissociation of a species in water with the equilibrium constant of an actual reaction. I understand why, but think of them as two different situations.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90030/why-is-bond-angle-of-h2o-greater-than-that-of-ph3
|
Why is the bond angle H-P-H smaller than H-N-H?
|
$\ce{N}$ & $\ce{P}$ are in the same group. Both $\ce{NH3}$ and $\ce{PH3}$ have one lone pair and according to VSEPR theory, both the central atoms are predicted to be $\ce{sp^3}$ hybridized.
But in spite of that, the bond angle in the former is $107^\circ$ while that in the latter is $92^\circ$. What is the cause of such a difference?
| 40 |
[
[
"\n**Starting point:** 2s orbitals are lower in energy than 2p orbitals.\n\n\nThe $\\ce{H-N-H}$ bond angle in ammonia is around 107 degrees. Therefore, the nitrogen atom in ammonia is roughly $\\ce{sp^3}$ hybridized and the 4 orbitals emanating from nitrogen (the orbitals used for the 3 bonds to hydrogen and for the lone pair of electrons to reside in) point generally towards the corners of a tetrahedron. \n\n\nIn the analogous case for phosphorus (phosphine, $\\ce{PH\\_3}$), the $\\ce{H-P-H}$ bond angle is 93.5 degrees. This angle indicates that the phosphorus atom is almost unhybridized (the bond angle would be 90 degrees if it were completely unhybridized). The 3 bonds from phosphorus to hydrogen roughly involve the three 3p orbitals on phosphorus and the phosphorus lone pair of electrons resides in the 3s orbital of phosphorus.\n\n\nSo the question becomes, why does the nitrogen atom in ammonia choose to hybridize, while the phosphorus atom in phosphine does not? Let's start by listing the factors that will stabilize or destabilize geometries in these compounds.\n\n\n**There are two choices for the central atom:**\n\n\n**remain unhybridized:** [$\\ce{p}$ orbital - H1s] bonds will form and they will be arranged 90 degrees with respect to one another. As a result, \n\n\n* substituents will be arranged closer together and destabilizing steric interactions will be increased\n* due to the absence of s-character in the $\\ce{X-H}$ bonds emanating from $\\ce{X}$, the electrons in these orbitals will be higher in energy\n* the lone pair electrons will be highly stabilized since they will reside in a low energy, pure s orbital\n\n\nor\n\n\n**hybridize:** [$\\ce{sp^3}$ orbital - H1s] bonds will form and they will be arranged 109 degrees with respect to one another. As a result, \n\n\n* steric interactions will be reduced because the tetrahedral orbital arrangement will space the attached substituents further apart\n* due to the presence of s-character in the $\\ce{X-H}$ bonds emanating from $\\ce{X}$, the electrons in these orbitals will be lower in energy\n* the lone pair electrons will be less stabilized since they will reside in a higher energy orbital that contains significant p-character\n\n\n**An example:**\n\n\nLet's now consider the example of $\\ce{NH\\_3}$ and $\\ce{NF\\_3}$. Fluorine is much more electronegative than hydrogen, therefore we would expect electron density in the $\\ce{N-F}$ bond to be shifted away from nitrogen towards fluorine. Because of this electron redistribution, the $\\ce{sp^3}$ orbital on nitrogen involved in this bond will contain less electron density. Consequently it will rehybridize - if there is less electron density in the orbital, there is less of a need for lower energy, electron stabilizing s-character in this orbital. The orbital will wind up with higher p-character and the s-character that has been \"saved\" can be used to stabilize other electrons (the lone pair!). Our prediction would be that $\\ce{NF\\_3}$ should have more p-character in its $\\ce{N-F}$ bonds than $\\ce{NH\\_3}$ has in its $\\ce{N-H}$ bonds. As a result we would expect the $\\ce{F-N-F}$ bond angle in $\\ce{NF\\_3}$ to be smaller than the $\\ce{H-N-H}$ bond angle in $\\ce{NH\\_3}$. Indeed the bond angle in $\\ce{NF\\_3}$ is 102 degrees compared to the 107 degrees observed in ammonia!\n\n\n**Back to our problem:**\n\n\nNitrogen (3.04) is more electronegative than phosphorus (2.19), which is about the same as hydrogen(2.2). In our $\\ce{X-H}$ bonds, we would therefore expect more electron density around the central atom when $\\ce{X~=~N}$ than when $\\ce{X~=~P}$. Using the same reasoning used in our example, we would then expect the $\\ce{N-H}$ bonds in ammonia to have higher s-character (and a larger $\\ce{H-N-H}$ angle) than the analogous bonds in phosphine, just as observed. The fact that phosphorus, being a second row element, has longer $\\ce{P-H}$ bonds (142 pm) than ammonia (102 pm) lessens steric problems in the unhybridized geometry and further lowers the energy of the unhybridized configuration for phosphine.\n\n\nIn the case of ammonia, the shorter $\\ce{N-H}$ bond lengths (increased steric interactions) and the increased electron density in the $\\ce{N-H}$ bonds makes the hybridized case the lowest energy. Whereas in the case of phosphine, steric interactions are of less consequence because of the longer bond lengths and the decreased electron density in the bonds around phosphorus make the energetics of the (nearly) unhybridized geometry more favorable.\n\n\n",
"42"
],
[
"\nFirst note, that hybridisation is a mathematical concept which can be applied to interpret a bonding situation. It has no physical meaning whatsoever. Instead it helps us to understand the direction of bonds better.\n\n\nSecond note, that the second period usually behaves quite differently from the remaining elements in a group. So in a way, ammonia behaves unnatural or anomalous. \n\n\n\n\n---\n\n\nIf you compare nitrogen with phosphorus, you will note, that the former is much smaller than the latter, i.e. van der Waals radii $r(\\ce{N})=155~\\mathrm{pm};\\ r(\\ce{P})=180~\\mathrm{pm}$ (ref. [wikipedia](http://en.wikipedia.org/wiki/Van_der_Waals_radius)), covalent radii $r(\\ce{N})=71~\\mathrm{pm};\\ r(\\ce{P})=107~\\mathrm{pm}$ (ref. [wikipedia](http://en.wikipedia.org/wiki/Covalent_radius)). Therefore also the orbitals in nitrogen are smaller, and $\\ce{s}$ and $\\ce{p}$ orbitals will occupy more of the same space than in phosphorus. As a result the $\\ce{N-H}$ bond distance will naturally also be shorter.\n\n\nA lone pair is usually most stable in an orbital that has high $\\ce{s}$ character. Bonds will most likely be formed with the higher lying $\\ce{p}$ orbitals. The orientation of these towards each other is exactly $90^\\circ$.\n\n\nIn ammonia this would lead to very close $\\ce{H\\cdots{}H}$ contacts, which are repulsive and therefore the hydrogen atoms are pushed away from each other. This is possible since in the second period the $\\ce{s-p}$ splitting is still very small and the nitrogen $\\ce{s}$ orbital is accessible for the hydrogen atoms. This will ultimately result in mixing $\\ce{s}$ and $\\ce{p}$ orbitals for nitrogen in the respective molecular orbitals. This phenomenon can be referred to as hybridisation - the linear combination of orbitals from the same atom. This term is therefore somewhat independent from its most common usage. \n\n\nIt is also very important to know, that the molecular wavefunction of a molecule has to reflect its overall symmetry. In this case it is $C\\_{3v}$, which means there is a threefold rotational axis and three vertical mirror planes (the axis is element of these planes). This gives also rise to degenerate orbitals. A canonical orbital picture has to reflect this property (BP86/cc-pVDZ; valence orbitals are ordered with increasing energy from left to right). \n\n \n\nNote that the lowest lying valence molecular orbital is formed only from $\\ce{s}$ orbitals (There is one additional $\\ce{1s^2-N}$ core orbital.) \n\nNow Natural Bond Orbital (NBO) Theory can be used to transform these delocalised molecular orbitals to a more common and familiar bonding picture, making use of atomic hybrid orbitals. This method is called localising orbitals, but it has the expense of losing the energy eigenvalue that may be assigned to canonical orbitals (NBO@BP86/cc-pVDZ; valence NBO cannot be ordered by energy levels). \n\n \n\nIn this theory you will find three equivalent $\\ce{N-H}$ bonds, that are composed of $32\\%~\\ce{1s-H}$ and $68\\%~\\ce{s^{$0.87$}p^3-N}\\approx\\ce{sp^3-N}$ orbitals. Note that the lone pair orbital at nitrogen has a slightly higher $\\ce{s}$ orbital contribution, i.e. $\\ce{s^{1.42}p^3-N}\\approx\\ce{sp^3-N}$.\n\n\nSo the thermodynamically most favoured angle is found to be $107^\\circ$ due to a compromise between optimal orbital overlap and least internuclear repulsion.\n\n\n\n\n---\n\n\nThe canonical bonding picture in phosphine is very similar to ammonia, only the orbitals are larger. Even in this case it would be wrong to assume, that there is no hybridisation present at all. However, the biggest contribution to the molecular orbitals stems from the $\\ce{p}$ orbitals at phosphorus. \n\n \n\nApplying the localisation scheme, one end up with a different bonding picture. Here are three equal $\\ce{P-H}$ bonds that are composed of $48\\%~\\ce{1s-H}$ and $52\\%~\\ce{s^{$0.5$}p^3-P}$ orbitals. The lone pair at phosphorus is composed of $57\\%\\ce{s} + 43\\%\\ce{p}$ orbitals. \n\n\n\n\n\n\n---\n\n\nOne can see the difference of the molecules also in their [inversion barrier](http://en.wikipedia.org/wiki/Nitrogen_inversion), while for ammonia the inversion is readily available at room temperature, $\\Delta E \\approx 6~\\mathrm{kcal/mol}$, it is very slow for phosphine, $\\Delta E \\approx 40~\\mathrm{kcal/mol}$.\n\n\nThis is mostly due to the fact, that the nitrogen hydrogen bonds have already a significant $\\ce{s}$ orbital contribution, which can be easily increase, to form the planar molecule with formally $\\ce{sp^2}$ hybrids.\n\n\n",
"15"
],
[
"\nSince Phosphorus has vacant 3d orbitals, the s-orbitals of hydrogen can directly overlap with the orbitals of Phosphorus. The procedure of bonding does not see much benefit in hybridisation as it is energetically expensive. Thus the hybridised character of the bonds is less dominant than in ammonia and the structure resembles the Lewis structure that we would expect instead of the tetreahedral (VSEPR) structure. This effect is called Drago's Rule and can also be observed in hydrogen Sulfide.\n\n\n",
"-1"
]
] |
https://chemistry.stackexchange.com/questions/89894/why-is-cyclopropylmethyl-radical-opening-so-fast
|
Why is cyclopropylmethyl radical opening so fast?
|
Ring opening of the cyclopropylmethyl radical is [reliable and fast](http://pubs.rsc.org/en/content/articlepdf/1993/CS/CS9932200347), and so has been used as a radical trap and a [radical clock](https://en.wikipedia.org/wiki/Radical_clock).
Could someone please explain what exactly makes this reaction so fast, besides the ring strain? Why is the anionic ring opening slower?
[](https://i.stack.imgur.com/fDPLp.png)
| 1 |
[] |
https://chemistry.stackexchange.com/questions/89892/transfer-of-odor-molecules-from-oil-to-water
|
Transfer of odor molecules from oil to water
|
I brew beer, I'm not a chemist. I want to add natural peanut aroma to a beer. The most common way this is being done by brewers is the addition of peanut powder, which is essentially the product left over after pressing most of the oil out of roasted peanuts. This product is preferable in brewing to whole peanuts because the introduction of significant amounts of oil into beer will affect desirable characteristics like the "head retention".
I don't like the idea of using this powder for various reasons. First, odor molecules like to be in oil, so pressing out the oil means pressing out a lot of the aroma. Second, by adding all that powder to the beer, the beer ends up with far too much body and the powder absorbs some of the beer which lowers the yield.
If you've ever tasted roasted peanut oil, you'd notice that it is intensely flavorful of roasted peanuts. What I'd like to do is extract odor molecules from the peanut oil and ultimately get them dissolved into a water based solution (a.k.a. beer). This is where I was looking for help/ideas.
I've mixed aromatic fats with ethanol/water solutions (like everclear) and I've found there to be a reasonable transfer of aroma into the ethanol/water. I don't see this working in this case though. The amount of odor molecules transferred to the solution would be insufficient when used to flavor beer. The alcohol content of the beer would end up too high by the time enough of the solution was added to get the desired aroma intensity.
| 3 |
[] |
https://chemistry.stackexchange.com/questions/89885/cant-nobelium-form-compounds-like-other-lanthanides-actinides-do
|
Can't nobelium form compounds like other lanthanides/actinides do?
|
Over the past week I've been browsing through [WebElements](https://www.webelements.com/nobelium/compounds.html) just to find out some interesting facts/properties about certain elements in the Periodic Table.
I've just come across the uncommon element [nobelium](https://www.webelements.com/nobelium/compounds.html) and it looks like it doesn't form compounds with anything despite being an actinide. For example, actinium can form Actinium Trichloride and promethium can form Promethium(III) Fluoride.
Is there a reason why nobelium is so 'special'?
| 5 |
[
[
"\n[Wikipedia](https://en.m.wikipedia.org/wiki/Nobelium) reports that nobelium does form compounds, but unlike other later actinides it could adopt a +2 oxidation state as well as +3, which complicates identifying specific compounds (e.g. nobelium reacts with chlorine but the specific product is not known). The +2 state is stabilized by the full shell configuration $[\\ce{Rn}]5f^{14}$. Experiments in aqueous solution seem to show a preference for the +2 oxidation state, and $\\ce{No}^{2+}$ has been coprecipitated into solids. But we have yet to definitively identify a neat nobelium compound in either specific oxidation state.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/89881/distinguishing-propan-2-ol-from-ethanol-using-ir-spectroscopy
|
Distinguishing propan-2-ol from ethanol using IR spectroscopy
|
A question requires me to distinguish between propan-2-ol and ethanol from IR spectra. To me, they both look nearly identical, even though I know both of them contain the same functional group and that IR spectra are only capable of identifying functional groups.
My question is, how do you distinguish the two, without using a mass spectrometer?
| 2 |
[
[
"\nHere is the spectrum for ethanol:\n\n\n[](https://i.stack.imgur.com/4jNUw.gif)\n\n\nHere is the spectrum for propan-2-ol:\n\n\n[](https://i.stack.imgur.com/OLRKc.gif)\n\n\nHere are the tabulated peak positions ($\\pu{cm^{-1}}$) and percent transmittance (%T) for the two. I've tried to match the appropriate peaks.\n\n\n$$\n\\begin{array}{cccc}%\n\\hline%\n\\text{ethanol position} & \\text{ethanol %T} & \\text{propan-2-ol position} & \\text{propan-2-ol %T} \\\\\n\\hline%\n3358 & 12 & 3346 & 9 \\\\\n & & 3334 & 9 \\\\\n2974 & 13 & 2972 & 5 \\\\\n2927 & 38 & 2933 & 19 \\\\\n & & 2907 & 23 \\\\\n2887 & 35 & 2884 & 20 \\\\\n & & 2722 & 60 \\\\\n & & 2669 & 60 \\\\\n & & 2521 & 74 \\\\\n & & 2408 & 79 \\\\\n & & 2387 & 81 \\\\\n & & 2198 & 84 \\\\\n & & 1903 & 85 \\\\\n & & 1766 & 84 \\\\\n1455 & 59 & 1467 & 24 \\\\\n & & 1409 & 29 \\\\\n1381 & 54 & 1379 & 16 \\\\\n & & 1368 & 17 \\\\\n1330 & 68 & 1341 & 28 \\\\\n1274 & 74 & 1309 & 23 \\\\\n & & 1162 & 10 \\\\\n1090 & 22 & 1130 & 9 \\\\\n1050 & 7 & 1110 & 15 \\\\\n & & \\color{red}{954} & 4 \\\\\n881 & 37 & 818 & 13 \\\\\n669 & 67 & 660 & 37 \\\\\n & & 654 & 37 \\\\\n & & 490 & 66 \\\\\n & & 436 & 68 \\\\\n\\hline%\n\\end{array}%\n$$\n\n\nOne way of differentiating between the two is that I don't believe the peak at 954 wavenumbers in propan-2-ol appears in ethanol. Another better way that doesn't require looking at the fingerprint region is to see that the hydrogen-bonding peaks (the broadest ones to the left) are red-shifted in propan-2-ol compared to ethanol.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89880/basicities-of-nitrogen-atoms-in-purine
|
Basicities of nitrogen atoms in purine
|
>
> Arrange the sites in purine in order of basicity.
>
>
> [](https://i.stack.imgur.com/5Gsgx.png)
>
>
>
I feel that **9 is the least basic** as nitrogen's lone pair is delocalised. However, I am unable to rank 1, 3, and 7 in a particular order. Is there a particular theory to decide basicity of sites?
| 11 |
[
[
"\nI don't think that question is trivial and except for what you have already deduced, I would not have an idea how to solve that without employing computational chemistry.\n\n\n\n\n\nThe problems already start that I suspect that the proton from position 9 readily exchanges to position 7 in any kind of polar solvent (and in absence why not with itself).\n\n\nWell from that point on forward, I'd say you need at least a lot of acid to protonate 9, as the π system is aromatic and will most likely not react first.\n\n\nAfter that you could probably deduce from the approximate bond angles which of the nitrogen would have the largest s- and p- character, then rank them from highest to lowest p contribution. The one with the most will probably be the HOMO and therefore react first. To be honest though, this is just guessing. \n\nI have absolutely no clue what the lesson with this exercise it, except for: \"They're pretty much the same.\" \n\n\nLet's have a closer look anyway. I have computed the neutral molecule at the DF-B97D3/def2-TZVPP level of theory, and used NBO6.0 for the partial charges:\n\n\n[](https://i.stack.imgur.com/I436J.jpg)\n\n\nAs expected, there is not much of a difference; I'd even go as far and say they are the same. Position 9 is a little bit more negative, because it has a hydrogen to draw from.\n\n\nNext up, let's have a look at the highest molecular orbital, as we would assume to protonate there:\n\n\n[](https://i.stack.imgur.com/SrIRI.jpg)\n\n\nAgain, there is only little to no difference. Or in numbers (contributions > 3%):\n\n\n\n```\nAlpha occ 31 OE=-0.217 is \nN1-p=0.3635 N1-s=0.0382\nN3-p=0.2551 \nN7-p=0.0667 \nC5-p=0.0563 \nC4-p=0.0473 \nC2-p=0.0452 \nC6-p=0.0382 \n\n```\n\nFrom this we would conclude that since `N1` has the largest contribution, the proton is most likely to go there. \n\n\nTherefore I also calculated all the protonated species, and their relative energies are:\n\\begin{array}{lr}\\hline\n\\text{Position} & \\Delta G(\\pu{298.15 K}, \\pu{1 atm})/(\\pu{kJ/mol})\\\\\\hline\n1 & 0.0 \\\\\n3 & 41.2 \\\\\n7 & 26.1 \\\\\n9 & 196.8 \\\\\\hline\n\\end{array}\n\n\nFrom this we conclude that the most likely position to be protonated is `N1`, followed closely by `N7`, and also `N3`. Off the charts is, as expected, `N9` as it breaks the aromaticity.\n\n\n**TL;DR**: Order of protonation $1 > 7 > 3 > 9$, calculated at DF-B97D3/def2-TZVPP.\n\n\n",
"10"
]
] |
https://chemistry.stackexchange.com/questions/89874/how-come-%e2%88%86g-%e2%89%a4-0-is-used-for-spontaneity-in-electrochemistry-not-%e2%88%86g-%e2%89%a4-wother
|
How come ∆G ≤ 0 is used for spontaneity in electrochemistry, not ∆G ≤ W(other)?
|
The spontaneity condition for a system at constant temperature and pressure in which the only type of work accomplished is of the $pV$ type can be expressed as:
$$\Delta G\le0\tag1$$
In case of other types of work (such as electrical, friction, etc.) there is an additional term and the previous equation becomes:
$$\Delta G\le W\_\text{other}\tag2$$
In case of an electrochemical process the maximum work is the electrical work and can be expressed as:
$$W\_\text{other}=-nFE\tag3$$
* **Question:** Why it is still said that the electrochemical reaction is spontaneous if the change in Gibbs free energy is lower than zero, thus using Equation $(1)$, even though the proper equation to be used is number $(2)$?
| 4 |
[
[
"\nMost likely the source **is wrong**, or it has purposfully omitted the correct equation to pseudosimplify a problem. It might be that an electrochemical reaction was under discussion, not a cell. A further possibility is pursued in the appendix. I recommend adding a source where you came upon the quoted exchange to avoid a straw man argument; even though in this case I have observed (and possessed) the misconception first hand.\n\n\n### One definition of spontaneity\n\n\nI define the true spontaneity condition as such when there is a tendency toward net 'something different' (chemical reaction, expansion etc) to establish state 2 instead of some original state 1. An example could be a chemical reaction where there is a higher reaction extent than at equilibrium, so a net reverse reaction takes place.\n\n\nYou are correct that the true spontaneity equation (from [Clausius's inequality](https://chemistry.stackexchange.com/a/84554/27806)) in the thermodynamic limit is, for net-constant pressure and temperature in the system,\n\n\n$$\\operatorname{d} G-\\delta w\\_\\pu{other}\\leq 0.\\tag1\\label1$$\n\n\nWhen electrochemical work is the only component besides expansion work, this implies (due to $|\\operatorname{d} n\\_\\pu{e}|=|\\nu\\_\\pu{e}|\\operatorname{d}\\xi$)\n\n\n$$\\Delta\\_\\pu{r} G\\_{T,P} +|\\nu\\_\\pu{e}|FE\\leq0\\tag2\\label2$$\n\n\nwhere $\\Delta\\_\\pu{r} = \\partial/\\partial \\xi$; here $\\xi$ is extent of the reaction. The Greek $|\\nu\\_\\pu{e}|$ signifies an absolute value of the stoichiometric coefficient of an electron in some half-reaction. Equation $\\eqref2$ assumes that only one net reaction occurs. The term $|\\nu\\_\\pu{e}|FE$\nshould be a good indicator of electrical work.\n\n\nIt might also be that they are discussing an electrochemical reaction, *not* the cell itself. We can have the process\n\n\n$$\\ce{Zn(sln) + Cu^2+(sln) -> Cu(sln) + Zn^2+(sln)}\\tag3\\label3$$\n\n\nwithout harnessing its electrical work via an external circuit. So $\\operatorname{d} G\\leq 0$ would hold for spontaneity.\n\n\n### Appendix: An alternative\n\n\nA different definition of spontaneity might be in play. Namely, the term *spontaneity* is also used to mean a large enough standard equilibrium constant (especially in biochemistry).\n\n\n$$RT\\ln \\frac1K = \\Delta\\_\\pu{r} G^\\circ\\_{T,P} \\tag4\\label4$$\n\n\nEquation $\\eqref4$ is technically a definition of the standard equilibrium constant. When non-negative absolute temperatures are quaranteed, the LHS of equation $\\eqref4$ will become non-positive for all $K$ big enough, *i.e.,* $K\\ge1$ . That implies\n\n\n$$\\Delta\\_\\pu{r} G^\\circ\\_{T,P} \\leq 0. \\tag5\\label5$$\n\n\nNote, however, that $K\\ge1$ doesn't necessarily imply much about the reaction balance itself because the equilibrium constant comprises of activities (not concentrations), and because the stoichiometric coefficients in the denominator may drown out the coefficients in the numerator. (Equation $\\eqref6$ assumes that surrounding fugacity is equal to an agreed fugacity in the standard state, denoted here and elsewhere by '$^\\circ$'.)\n\n\n$$K(\\pu{in solution, solvent \\ce{A}}) = \\left[a(\\ce{A})\\_\\pu{eq}\\right]^{\\nu\\_\\ce{A}}\\prod\\_i \\left[a(\\ce{B\\_$i$})\\_\\pu{eq}\\right]^{\\nu\\_i}. \\tag6\\label6$$\n\n\nReactants–products (other than solvent) are designated by $\\ce{B\\_$i$}$. But still, for $K$ big enough (2nd definition of spontaneity), equation $\\eqref5$ will hold by definition. Also keep in mind [$\\Delta\\_\\pu{r} G^\\circ\\_{T,P} \\neq \\Delta G$](https://chemistry.stackexchange.com/questions/41862/what-is-the-difference-between-%e2%88%86g-and-%e2%88%86g/41864#41864) (even their dimensions are different!). So the source is presumably still at fault for poor notation.\n\n\n",
"5"
],
[
"\nBy spontaneous we normally mean that the reaction is thermodynamically allowed WITHOUT forcing it (that is, without doing non-PV work from the outside). This is described by your eq. (1). \n\n\nA non-spontaneous reaction we can make thermodynamically allowed by adding enough work, this is your eq. (2). A typical chemistry lab example of the last is electrochemical splitting of water into oxygen and hydrogen gases by means of two electrodes and an external power supply.\n\n\nEq. (2) also tells the thermodynamic limit (i.e. maximum) electrical work one can possibly extract from the given electrochemical reaction, if it is spontaneous and fulfills eq. (1).\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89872/what-are-the-key-differences-between-raoult-s-law-and-henry-s-law
|
What are the key differences between Raoult’s Law and Henry’s Law
|
Raoult’s Law ($p\_A = \chi\_A \cdot p\_A^\circ$) becomes a special case of [Henry’s Law](https://en.wikipedia.org/wiki/Henry%27s_law#Henry's_law_volatility_constants_%7F'%22%60UNIQ--postMath-0000004E-QINU%60%22'%7F) ($p\_A = K\_H^{px} \cdot \chi\_A $) when $K\_H^{px} = p^\circ\_A$, but where do these laws differ from each other?
There’s a statement in my text book which I’m having trouble understanding.
The statement:
>
> As a real solution approaches the limit of infinite dilution its components behave more ideal. The solvent obeys Raoult's law whereas solute (minor component) obeys Henry's law for dilute solutions.
>
>
>
[](https://i.stack.imgur.com/UheRT.jpg)
| 13 |
[
[
"\nRaoult's law can be seen to be valid for ideal liquids. The assumption behind Raoult's law is that both the liquid phase and the vapour phase behave ideally. This means that the interaction between any two molecules in the liquid and the vapour have no interaction.\nThe Raolt's law just states the mole fraction of a component in the liquid phase is proportional to the mole fraction of the same component in the vapour phase. This is valid only for mixtures of ideal liquids. \nWhen you add a very small amount of impurity in a solvent (equivalent to infinite dilution) the interaction between the impurity and the solvent molecule is minuscule or technically infinitesimal. Therefore, the liquid still behaves ideally. Hence Raoult's law is applicable for the solvent.\n\n\nWhereas Henry's law is also defined for a ideal mixture only this time the assumption is that the mole fraction of the component is tending to zero. this means that when very few molecules a present in a vast space, the interaction between them is almost zero. This makes the system ideal. therefore, an infinitesimally small quantity of impurity does not feel any interaction with another molecule of the same impurity in the solvent. Therefore, Henry's law is used for the solute. That is why Henry's law constant have huge values. Huge values makes sure that even small amount of the solute is accounted for when calculating the mixture properties.\n\n\nFor further clarity, please refer to chapter 6 of 'Molecular Thermodynamics of Fluid phase equilibria' by John M.Prausnitz et. al.\n\n\n",
"6"
],
[
"\nRaoult's and Henry's laws are limiting laws, generally applicable when the solute concentration goes to zero. In this limit the vapor pressure of any component in the solution depends linearly on its mole fraction, implying the absence of solute-solute interactions.\n\n\nRaoult's law describes the dependence of the vapour pressure of a solvent as a function of its mole fraction $\\chi\\_1$:\n\n\n$$\\lim\\_{\\chi\\_1\\rightarrow1}\\left( \\frac{p}{\\chi\\_1}\\right) =p^\\ast$$\n\n\nwhere $p^\\ast$ is the vapour pressure of the pure solvent.\n\n\nHenry's law describes the dependence of the vapour pressure of a solute as a function of its concentration. In terms of mole fraction $\\chi\\_2$:\n\n\n$$\\lim\\_{\\chi\\_2\\rightarrow0}\\left( \\frac{p}{\\chi\\_2}\\right) =K$$\n\n\nFor a binary mixture of pure substances it can be shown that the laws are complementary: if one law holds for one component then the other law holds for the other component. It can also be shown that the laws imply that the solution satisfies other criteria for ideality, including zero enthalpy and volume of mixing. \n\n\n",
"6"
],
[
"\n**Compare and contrast**\n\n\nAs stated in the caption of the figure posted by the OP, Henry's law applies to the solute and Raoult's law applies to the solvent. If a solute is non-volatile (e.g. sucrose or sodium chloride), Raoult's law still works, but invoking Henry's law, while possible technically (by setting Henry's constant to zero), would be kind of pointless.\n\n\nAs you can see for the example shown in the picture posted by the OP, Henry's law describes the partial pressure of a component at very low concentration. On the other hand, Raoult's law describes the pressure of a component at a concentration close to pure liquid.\n\n\nRaoult's law is independent of the nature of the solute; the only parameter is the mole fraction of the solvent (or solute, if written as $\\Delta p = \\chi\\_\\mathrm{solute} \\cdot p^\\circ$). Henry's law, on the other hand, depends on the nature of both solute and solvent (i.e. the Henry's constant is different for different solvents).\n\n\n**Continuous transition from Henry's to Raoult's law**\n\n\nFor a binary system where both components are volatile, are liquid in the pure state and are miscible across the entire range, you can discuss the equivalence or non-equivalence of the two laws. The situation for the ethanol: water mixture is shown below:\n\n\n[](https://i.stack.imgur.com/MiRJO.gif) \n\n\nThe figure was taken from [R K Thomas's course page](http://rkt.chem.ox.ac.uk/lectures/liqsolns/regular_solutions.html), which has a great explanation of the entire topic. Note that the vapor pressure of either component increases linearly near 0% and 100% composition, but not with the same slope. In this particular case, both water and ethanol show positive deviation from Raoult's law at low concentration. There are also cases of [negative deviation](https://en.wikipedia.org/wiki/Raoult%27s_law#Negative_deviation) such as the chloroform: acetone system.\n\n\nThe case described by the OP, where components conform to Raoult's law for any composition, is realized for mixtures of the same molecule with different isotopic composition, e.g. ethanol with carbon-13 on the first or second carbon. To a good level of approximation, the boiling points of the two species are identical, and the interactions between identical or isotopically different molecules are the same, so Raoult's and Henry's law would give the same values. I don't know of a less trivial system (i.e. where the boiling points of the pure components differ) where that is the case. \n\n\n\n> \n> [quoted in OP's question] As a real solution approaches the limit of infinite dilution its components behave more ideal. The solvent obeys Raoult's law whereas solute (minor component) obeys Henry's law for dilute solutions.\n> \n> \n> \n\n\nI don't think this statement is true. Henry's law does not require ideal behavior (the solute: solvent interaction can be stronger or weaker than the solvent: solvent interactions), and the distinct Henry's law constants depending on solvent reflect that. For example, more oxygen will dissolve in water than in cyclohexane at the same partial pressure of oxygen. The range where Henry's law applies is called ideally diluted solution. Roult's law will apply for any solute, so it also does not require an ideal solution.\n\n\nIf the solution is ideal (no difference between solvent: solute and solvent: solvent interactions), the condition $K\\_H^{px} = p^\\circ\\_A$ mentioned in the OP's question becomes true.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89864/numerical-related-to-crystallisation-of-water-below-its-freezing-point
|
Numerical related to crystallisation of water below its freezing point
|
**Question:**
>
> There is $16\ \mathrm g$ of pure water in a container at temperature $-20\ \mathrm{^\circ C}$. A small piece of ice is added to start crystallization. Amount of water in container when temperature reaches to $0\ \mathrm{^\circ C}$ is: (assume specific heat of water below $0\ \mathrm{^\circ C}$ is $1\ \mathrm{cal\ g^{-1}\ ^\circ C^{-1}}$.)
>
>
>
**My attempt:**
I have read this: [Liquid water below freezing temperature](https://chemistry.stackexchange.com/questions/65784/liquid-water-below-freezing-temperature). So, I understand that the *small* piece of ice is only added to provide a *nucleation site* to promote the freezing of water.
But, all physical processes must follow the Law of Calorimetry, right? So, for water below $0$ degrees to freeze, it must first accept heat from somewhere, to reach zero degrees. But, the container is at an *even lower temperature*, so it won't provide water with any heat (negative temperature gradient). We have also not been given the mass of ice added, so we can't calculate the amount of heat the water will gain from the ice (the latter *is* at a high temperature than water). The system is isolated, so there is no other possible heat source...
I am apparently at a dead end in this problem. I hope I have put everything in detail and correctly. What is the logical mistake in my thinking? And what is the correct way to approach this problem? Thank you!
| 1 |
[
[
"\nAnother answer explains more generally how to interpret the problem. This completes that spoiler:\n\n\n\n> \n> If one can assume that the [enthalpy of fusion](https://en.wikipedia.org/wiki/Enthalpy_of_fusion) ($\\pu{334 J/g},~$[2](https://www.engineeringtoolbox.com/latent-heat-melting-solids-d_96.html)) is independent of temperature between the normal melting point ($\\pu{0^\\circ C}$) down to $\\pu{-20 ^\\circ C}$, then the heat released during freezing equals that absorbed during heating of the liquid to $\\pu{0^\\circ C}$ so that $$\\begin{align} \\pu{334 J/g} \\times m &= \\pu{20 K \\times 16 g \\times 4.18 J/g\\*K} \\\\ \\rightarrow m&= \\pu{4.0 g} \\end{align}$$ Note the important condition that the heat capacity of water remains constant and equal for liquid and solid over that temperature span. \n> \n> \n> \n\n\n",
"1"
],
[
"\nIn the question **Crystallization begins** after you add a small piece of ice \n.\n So it has to begin that's been given in the question. \n\n\nI don't know thermodynamically whether it's plausible or not, but it's a given in the question a logical alternative way to approach the question (according to me) is\n\n\n*I formatted it as a spoiler just in case you want to think a bit more or anybody else does*\n\n\n\n> \n> The temperature has increased because latent heat of fusion is being released i.e **ice being formed from water is releasing heat**. \n> \n> \n> \n\n\n\n\n---\n\n\n**PS**: I may be wrong\n\n\n",
"1"
],
[
"\n\n> \n> So, for water below 0 degrees to freeze, it must first accept heat from somewhere, to reach zero degrees.\n> \n> \n> \n\n\nNo, water can freeze at lower temperatures, see e.g. <https://www.youtube.com/watch?v=Fot3m7kyLn4>.\n\n\nAs it freezes, the exothermic process warms up the ice and the surrounding water. The difference in enthalpy between liquid and solid water is larger than the difference between liquid water at the two temperatures, so not all the water freezes. Instead, you end up with liquid and solid water at equilibrium once the water and the ice reach the normal freezing point (not because they are warming up from the environment, but as ice continuously forms).\n\n\nInstead of adding ice to start the process, you can also shake the container like in the video.\n\n\n",
"1"
],
[
"\nThe crystallization is an operation that delivers heat, and a good deal of heat. This is sufficient to heat the metastable water form -20°C to 0°C. \n\n\n",
"0"
],
[
"\nThe previous answers are good ones. I just want to add a (longwinded) metaphorical picture to help understanding the freezing of water at -20C. \n\n\nImagine a tower of bricks (2 x 4 x 8 inches each) about 20 high. It is metastable, but the tiniest push will topple it because gravity pulls each brick to the lowest level. This is a metaphor for the freezing of water at -20C, right? Yes, but there is a great dissimilarity, because the bricks fall down to a lower height, but the water temperature rises as it freezes. We all know that in reality, we can't reduce the temperature of water to -20C and have it be as stable as a pile of bricks 20 high. Then, too, freezing is a process that 99.9% of the time requires a reduction of temperature.\n\n\nNow let's try another metaphor: take 20 bricks of low-density foamed polystyrene and push them down into a large tank of water. You may be able to hold them down perfectly straight, or even balance a heavy lead brick on top of them to hold them down while they are semi-floating, but if the slightest imbalance occurs (like the ice-seed), the polystyrene bricks will all float up to the surface (~ ~ supercooled water will freeze) and the heat released will equal the energy required to push the polystyrene into the tank. The heat released as the polystyrene bricks rise is really from the water settling in a gravity field (~ ~ the heat of fusion that was added to the water to keep it from freezing while it was being cooled to -20C.). \n\n\nNow how can you cool water to -20C without removing the heat of fusion? In your imagination, separate the specific heat and the heat of fusion into different accounts. Take the heat out of the one and leave the other, but don't let the water know. This allows you to mentally reduce the temperature without allowing the solidification process to occur - it's like applying a \"negative catalyst\". Then when you remove the \"negative catalyst\" (i.e., add the ice-seed), some of the water freezes, releasing bottled-up heat of fusion to flow into some of the -20C water - the temperature rises. The result will be some ice and some water, and when the system comes to equilibrium, the temperature will have risen to 0C. \n\n\nAnd the process can be stated much more succinctly by an equation, but sometimes equations are too succinct!\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89863/several-carbonyl-groups-which-one-does-ethylene-glycol-protect
|
Several carbonyl groups, which one does ethylene glycol protect?
|
We know that ethylene glycol is used to protect carbonyl functional groups. But, consider this:
[](https://i.stack.imgur.com/iJEHe.png)
Here, we have *two* carbonyl groups, but *only one equivalent* of ethylene glycol. Obviously, *only one* carbonyl group will get protected in this case. But will it be the $\ce{C}^1$ or $\ce{C}^2$ carbonyl? And why?
| 1 |
[
[
"\nSeveral factors influence the overall rate of a reaction under various conditions. Among the crucial factors are:\n\n\n1. structural features of the carbonyl compound;\n2. the role of protons or other Lewis acids in activating the carbonyl group toward nucleophilic attack;\n3. the reactivity of the nucleophilic species and its influence on subsequent steps; and\n4. the stability of the tetrahedral intermediate and the extent to which it proceeds to product rather than reverting to starting material. *(Ref.1)*\n\n\nLimiting ourselves to current molecule, aldehydes are more reactive than ketones. Electron-withdrawing substituents render the carbonyl group more electrophilic.\n\n\n[](https://i.stack.imgur.com/C5Cjq.jpg)\n\n\n*Image credit: Organic Chemistry, Structure and Function Seventh Ed by Peter Vollhardt, University of California at Berkeley, Neil Schore, University of California at Davis*\n\n\nBecause aldehydes form acetals more readily than ketones, we can protect an aldehyde selectively in the presence of a ketone. The following example shows the reduction of a ketone in the presence of a more reactive aldehyde *(Ref.2)*:\n\n\n[](https://i.stack.imgur.com/48Gu1.jpg)\n\n\nWith reference to your question, C-2 is more reactive then C-1 carbonyl group for the above reasons.\n\n\n### References\n\n\n1. Advanced Organic Chemistry, Fifth Edition, Part A: Structure and Mechanisms by Francis A. Carey and Richard J. Sundberg, University of Virginia\n2. Organic Chemistry, Eighth Edition, by L. G. Wade Jr., Whitman College\n3. [Reactivity of Aldehydes & Ketones - Chem LibreTexts](https://chem.libretexts.org/?title=Core/Organic_Chemistry/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_%26_Ketones)\n\n\n",
"7"
]
] |
https://chemistry.stackexchange.com/questions/89860/calculate-the-cathode-electrode-potential-in-this-redox-reaction
|
Calculate the cathode electrode potential in this redox reaction
|
In this problem I'm asked to *calculate the electrode potential of* the **magnesium electrode** ($E\_{Mg^{2+}/Mg}$) given the red-ox reaction:$$\text{Mg(s)}+2\text{Ag}^+(10^{-2}\text{M})\leftrightarrow 2\text{Ag(s)}+\text{Mg}^{2+}(10^{-2}\text{M})$$
The information I'm given is:
1. Temperature equal to $278\,K$ (the standard).
2. The molar concentration of the $\text{Mg}^{2+}$ ions is of $0.01\,\text{M}$.
3. The standard reduction potential of the $Mg$ electrode is: $E^{º}\_{Mg^{2+}/Mg}=-2.34\,V$.
I know I have to use the [Nernst Equation](https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Nernst_Equation), but my solution and the book's answer are very different.
**My solution:**
$$E\_{Mg^{2+}/Mg}=E^{º}\_{Mg^{2+}/Mg}-\dfrac{0.059}{n}\,\color{#ff3300}{\mathbf{\log\frac{\left[\text{Mg}^{2+}\right]}{1}}}=\ldots=\boxed{-2.28\,V}$$
**My book's solution:**
$$E\_{Mg^{2+}/Mg}=E^{º}\_{Mg^{2+}/Mg}-\dfrac{0.059}{n}\,\color{#00ff00}{\mathbf{\log\frac{1}{\left[\text{Mg}^{2+}\right]}}}=\ldots=\boxed{-2.40\,V}$$
($n=2$ because 2 moles of electrons where exchanged per 1 mole of reaction)
---
I've colored the part where I go wrong (basically the logarithm). I don't know exactly why the book put the concentration in the denominator, since $\text{Mg}^{2+}$ is a product (not a reactant), therefore following what I studied, its concentration should be in the numerator.
Any help? Sorry for asking such an elementary question, but I'm still learning the basic stuff in Chemistry!
**Note**
I attach an image of the solution provided by the book
[](https://i.stack.imgur.com/VlwUh.png)
| 1 |
[
[
"\nZhe's answer is the best. Though, I will try to offer a different (mathematical) approach, which will hopefully enable you to see the answer yourself, or rather, verify that your book's solution is indeed correct.\n\n\nAccording to you, the standard *oxidation* potential of magnesium electrode would be $E\\_{Mg/Mg^{2+}}=E^{º}\\_{Mg/Mg^{2+}}-\\dfrac{0.059}{n}\\,\\mathbf{\\log\\frac{\\left[\\text{Mg}^{2+}\\right]}{1}}$. Agreed?\n\n\nAlso, note that $E\\_{Mg/Mg^{2+}}=-E\\_{Mg^{2+}/Mg}$ always holds. Agreed?\n\n\nThen, $E\\_{Mg^{2+}/Mg}=-E\\_{Mg/Mg^{2+}}$ $$=-(E^{º}\\_{Mg/Mg^{2+}}-\\dfrac{0.059}{n}\\,\\mathbf{\\log\\frac{\\left[\\text{Mg}^{2+}\\right]}{1}})$$ $$=(-E^{º}\\_{Mg/Mg^{2+}})-\\dfrac{0.059}{n}\\,\\cdot(-1)\\cdot\\mathbf{\\log\\frac{\\left[\\text{Mg}^{2+}\\right]}{1}}$$. \n$$=E^{º}\\_{Mg^{2+}/Mg}-\\dfrac{0.059}{n}\\,\\mathbf{\\log\\frac{1}{\\left[\\text{Mg}^{2+}\\right]}}$$. \n\n\nWhich is exactly the answer given by your book.\n\n\nI hope it helps!\n\n\n",
"2"
],
[
"\nThe equation that corresponds to the standard reduction potential is:\n\n\n$$\\ce{Mg^{2+} + 2e- -> Mg}$$\n\n\nNotice that this is not the actual half reaction in your total equation, where magnesium is oxidized. But the book is determining the *reduction* potential, so the magnesium ion is a reactant not a product. In the reaction quotient, the concentration/activity of magnesium ion should be in the denominator.\n\n\nI'm not sure I agree that we should choose the standard reduction potential as the electrode potential for the electrode where oxidation occurs, but at least the book is consistent.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89858/obtaining-benzene-from-toluene
|
Obtaining Benzene from toluene
|
What will be the reaction for producing benzene from toluene?
My attempt:
Toluene→Benzoic acid→Sodium benzoate→Benzene(decarboxylation) ...
Is there any other ways, like a direct reaction?
| 0 |
[
[
"\nThere is also a disproportionation reaction that [converts toluene to a mixture of benzene and xylene (Bawa et al)](http://www.insa.nic.in/writereaddata/UpLoadedFiles/PINSA/Vol41A_1975_1_Art06.pdf), also, see [wikipedia](https://en.wikipedia.org/wiki/Transalkylation). It has commercial applications (because the benzene/xylene mixture can often be sold for more than toluene) and can be [licensed](http://www.gtctech.com/technology-licensing/selective-toluene-conversion/). However, it is more of an industrial process than one you would use in the lab, as it is really driven by an exothermic dealkylation in vapor phase over zeolite catalyst.\n\n\nInterestingly, as zeolites act as a Lewis acid, this is strangely similar to Friedel-Crafts alkylation, which uses Lewis acid catalysis in liquid phase. The paper by Bawa et al discusses experiments in which the trans alkylation was done with Frieden-Crafts catalysts in liquid phase.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89855/preparation-of-di-phenyl-ester-from-phenol
|
Preparation of di-phenyl ester from phenol [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed 5 years ago.
[Improve this question](/posts/89855/edit)
How to create di-phenyl ester from phenol in laboratory. What would be the steps. I am asking this question as it can not be done by Williamson ether synthesis.
| -5 |
[
[
"\nDiphenyl ethers have been prepared from condensation in melt phase of potassium phenoxide and aryl bromides in the presence of Copper and Copper(I) Chloride - J. Chem. Eng. Data, 1978, **23** (2), 185–186\n\n\nUS patent 4564712 A also describes a similar process\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89854/pressure-volume-formula-question
|
Pressure-Volume Formula Question
|
I have a chemistry homework problem that states the following:
>
> $6.2$ L of $N\_2$ at $0.74$ bar is mixed with a $15.2$ L sample of $O\_2$ at $0.35$ bar. The gaseous mixture is placed in a container of volume $12.0$ L at the same temperature. What is the pressure of the system, in bar?
>
>
>
I answered the question and I **believe** I have the right answer, but I'm unsure as to whether it's a fluke or not. Here's my approach:
Use the formula $P\_1 \* V\_1 = P\_2 \* V\_2$.
$6.2 \* 74 = 21.4 \* P\_2$, $P\_2 = 21.4$ kPa.
Repeating the process for $O\_2$, I get $P\_2 = 46.26$ kPa.
I add the volumes together since the gases are now mixed, and I also add the pressures. Now that the mixture is placed into a $12.0 L$ container, we use the formula again:
$46.26 \* 21.4 = P\_2 \* 12$. $P\_2 = 83$ kPa. This is equal to 0.83 bar.
I'm still not 100% sure of my reasoning for this question, wondering if someone could verify my logic and provide perhaps a better way of doing this question that I might've overlooked.
Thank you!
| -2 |
[
[
"\nI arrived at the same conclusion, but I am not sure your approach is optimal, for efficiency and conceptual reasons. The exercise states nothing about where the two samples are mixed, and I think it is reasonable to simply mix them in the final container:\n$$\np\\_i \\cdot V\\_i / V\\_{\\text{final}} = p\\_{i,\\text{final}},\n$$\nwhere $i$ runs over $\\ce{O2}, \\ce{N2}$. Add the $p\\_{i,\\text{final}}$ to obtain total final pressure $p\\_\\text{final}$.\n\n\nThe actual calculation (I dropped the units) is thus: \n$$\n( 0.74 \\cdot 6.2 + 0.35 \\cdot 15.2) / 12 \\approx 0.83,\n$$\nconfirming your result.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89850/choice-of-lewis-acid-for-eas-reaction
|
Choice of Lewis Acid for EAS reaction
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Why is AlCl3 or FeCl3 might be preferred for chlorination reactions and AlBr3 or FeBr3 for bromination reactions? Why not use AlCl3 for bromination reactions?
Source: <https://www.masterorganicchemistry.com/2011/07/22/reagent-friday-aluminum-chloride-alcl3/>
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"\nHonestly speaking I think there won't be a difference as in clayden they have used $\\ce{AlCl3}$ for bromination\n\n\n\nIf I **had to give a reason** it would be that when the $\\ce{ AlCl3-Br2}$ complex is made there might be a teeny tiny chance that instead of bromine breaking away the chlorine breaks away. \n\n\n\nMy basis of saying that the probability of chlorine breaking away is low is because there's a **+tive charge on $\\ce{Br}$ in the complex**.Chlorine on the other hand is free from any such effects. There are no factors supporting the breakage of the $\\ce{Al-Cl}$ bond. So the chances of $\\ce{Br-Br}$ bond breaking are higher than the $\\ce{Al-Cl}$ bond. \n\n\nThe same reasoning can be applied for $\\ce{AlBr3}$ with $\\ce{Cl2}$\n\n\nThat's why in Master Organic they've written **preferred** not necessary.\n\n\n",
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"\nIn Electrophilic Aromatic Substitution reaction (EAS), the choice of catalyst may influence the yield of the desired product.\n\n\nFirst of all, note the following data:\n\n\n*Bond dissociation energy (in kJ/mol)* -\n\n\nAl–Br : 429.2\n\n\nAl–Cl : 502\n\n\n**As the bond dissociation energy for Al–Cl bond is greater, it must be stronger.**\n\n\nNow, consider using aluminium bromide as catalyst for an EAS chlorination.\nIn the first step, i.e. interaction of chlorine molecule with Lewis acid aluminium bromide, one can't possibly neglect the formation of bromonium ion which may act as electrophile - decreasing yield of chlorobenzene and giving unnecessary side product of bromobenzene. This possibility is supported by thermodynamics, as shown by the data mentioned above.\n\n\nTo make sure that **yield of product is maximum** and **no unnecessary side reactions** occur - aluminium chloride catalyst is employed for chlorination and aluminium bromide for bromination.\n\n\nHope this answers your question.\n\n\nP.S.\nCorrection to @AvnishKabaj's answer:\n*If I had to give a reason it would be that when the AlCl3-Br2 complex is made there might be a teeny tiny chance that instead of the bromine breaking away the chlorine breaks away.*\n\n\nAs per the bond energy data I have, [from this exhaustive table](https://www.google.co.in/url?sa=t&source=web&rct=j&url=https://notendur.hi.is/agust/rannsoknir/papers/2010-91-CRC-BDEs-Tables.pdf&ved=2ahUKEwiw19W_ronZAhWBKo8KHb-GAj0QFjAAegQIExAB&usg=AOvVaw34NLsVSCn1-7lpqgBhwL8e) ;it is easy to conclude that the bond strength of Al–Cl bond is greater than Al–Br bond. Hence, problems would arise when chlorination is done in presence of aluminium bromide catalyst. In the AlBr3-Cl2 complex, there'd be a *teeny tiny chance that instead of the chlorine breaking away, the bromine breaks away.* (based on thermodynamic favourability)\n\n\n",
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https://chemistry.stackexchange.com/questions/89847/hydrolysis-of-transition-metals-halides
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Hydrolysis of transition metals' halides?
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So, it's a relatively common known solubility rule that any hydroxide with a cation not in the first two groups is basically insoluble. So supposing we have a transition or post-transition metal $M$ and say it forms a charge of $n+$ (so it's cation is $\ce{M^{n+}}$) why would $\ce{MX\_n}$ (where $\ce{X}$ is a halide) remain in solution? Why doesn't the following reaction occur?
$$
\ce{MX\_n(aq)} + n\ \ce{H\_2O(l)} \rightarrow \ce{M(OH)\_n(s)} + n\ \ce{HX(aq)}
$$
My thought was that perhaps this reaction does occur, but just very slowly and not fast enough for the $\ce{M(OH)\_n}$ to precipitate in noticeable amounts. However, I don't really know and this is just a guess. Could someone explain to me why this doesn't happen?
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"\n$\\ce{MCl\\_n(aq)} + n\\ \\ce{H2O(l)} \\rightarrow \\ce{M(OH)n(s)} + n\\ \\ce{HCl(aq)}$ does tend to go to the right. \"Basically insoluble\" is not quantitative enough to generalize. If the hydroxide is very insoluble, the acid produced by hydrolysis may not keep it in solution. But if the hydroxide is even only slightly soluble, it may not precipitate out. The acidity of the solution is a subtle indicator that the reaction has shifted a little bit to the right. The reaction isn't slower, just not always so extreme.\n\n\nIf you put $\\ce{SiCl4}$ in water, it gives $\\ce{HCl}$ + silica gel because $\\ce{SiO2}$ (or \"$\\ce{Si(OH)4}$\") is so insoluble.\n\n\nIf you put $\\ce{MgCl2}$ in water, no precipitate appears because $\\ce{Mg(OH)2}$ is slightly soluble (~2 mg/L), and the $\\ce{HCl}$ produced by the hydrolysis reaction is acidic enough so that if a significant amount of $\\ce{Mg(OH)2}$ were produced, it would just redissolve. $\\ce{Mg(OH)2}$ gives a pH of about 10 in water, so at pH's lower than 10, $\\ce{Mg(OH)2}$ doesn't precipitate out. BTW, $\\ce{MgCl2}$ is slightly acidic because of the hydrolysis reaction.\n\n\n$\\ce{AlCl3}$ is complicated because of ionic complexes, but in water it hydrolyzes to give a very acidic solution and some aluminum hydroxide complexes or gels which are of varying solubility. $\\ce{AlCl3}$ has been used as a water treatment because it produces flocculent precipitates that carry down impurities.\n\n\n$\\ce{NiCl2}$ does not precipitate in water because $\\ce{Ni(OH)2}$ is even more soluble than $\\ce{Mg(OH)2}$. \n\n\n$\\ce{FeCl3}$ is quite acidic in water. You might expect the commercial 40% solution to be a cloudy product because of the production of acid, but it is a clear, deep red color. Oh, but when it is diluted and the pH rises because the acid is diluted, it precipitates in flocs like $\\ce{AlCl3}$. It is used to clarify water by precipitating dirt and dust.\n\n\n",
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https://chemistry.stackexchange.com/questions/89844/how-does-atomization-allow-droplets-to-evaporate-below-their-boiling-point-cool
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How does atomization allow droplets to evaporate below their boiling point, cooling the surrounding vapor to an even lower temperature? [closed]
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**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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I understand why the vapor is cooled in such a scenario, because of the heat required to change states, but I don't fully understand why the state change even occurs if below the BP. Why does it seem that in many cases, atomization almost forces evaporation, minimizing the role of temperature?
I understand that air can naturally hold a certain amount of water, but I've never learned how this "humidity" concept applies to other solvents. Is there some kind of formula I can use to estimate how much of a particular solvent a particular volume of air can hold at room temperature, without the solvent being at boiling point? Or really any sort of math model to understand how atomization assists in evaporation below boiling point?
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"\nEvery gaseous phase that is in contact with a liquid phase contains some vapor from the liquid phase.\n\n\nIf the liquid phase is pure and liquid and gaseous phase are in equilibrium, the partial pressure of the liquid is called [vapor pressure](https://en.wikipedia.org/wiki/Vapor_pressure).\n\n\nAs it turns out the vapor pressure not only depends on temperature but also on the shape of the surface. The [Kelvin equation](https://en.wikipedia.org/wiki/Kelvin_equation) shows that the vapor pressure for small droplets is greater than for larger droplets or plain surfaces.\n\n\nSo atomized liquids have higher vapor pressure, and more liquid can evaporate. It also shows that larger droplets grow on the cost of smaller ones.\n\n\nThe drop in temperature is due to the required [latent heat of vaporization](https://en.wikipedia.org/wiki/Enthalpy_of_vaporization), i.e. the energy required by the transition liquid to vapor.\n\n\n",
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https://chemistry.stackexchange.com/questions/89836/what-does-sine-mean
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What does "sine" mean?
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I see the suffix "sine" (seen/sin) a lot, adenosine, cytosine, lysine, tyrosine, etc. Most of where I hear it is in amino acid R groups, but it's usually only the prefix that is recognized as significant.
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"\nAccording to [Textbook of metabolism and metabolic disorders](https://books.google.com/books?id=4M9qAAAAMAAJ&q=%22suffix%20osine%22&dq=%22suffix%20osine%22&hl=en&sa=X&ved=0ahUKEwiA5oGcpYXZAhWBwVkKHUrqDmcQ6AEIKTAA) (1964): \n\n\n\n> \n> The ribosides derived from purines have the suffix -osine; those from pyrimidines, the suffix -idine. The corresponding deoxyribosides, with the exception of thymidine, do not have any such simple designation. Occasionally, instead of, for example, guanine-deoxyriboside, the term deoxy-guanosine is used. \n> \n> \n> \n\n\nBeyond the more common adenosine and guanosine, there are also: \n\n\n[inosine](https://pubchem.ncbi.nlm.nih.gov/compound/inosine#section=Top) and \n\n\n[xanthosine](https://pubchem.ncbi.nlm.nih.gov/compound/xanthosine#section=Top)\n\n\n",
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https://chemistry.stackexchange.com/questions/89835/synthesis-of-hydrogen-fluoride
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Synthesis of Hydrogen Fluoride
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At almost 300°C, $\ce{CaF2}$ reacts with sulfuric acid and this reaction produces $\ce{HF}$. Is there any way to make a relatively dilute (10-20%) hydrofluoric acid with $\ce{CaF2}$ as the source of fluoride ions without that much amount of heat?
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"\nIn desperation, for some unknown reason, I need 15% aqueous $\\ce{HF}$. What do I really need? A weak solution of $\\ce{H+}$, say about pH 2. And a solution of fluoride ions to wreak havoc on some unsuspecting metal or glass(Ooops, I may have given away the secret). \n\n\nMy subconscious suggests that I could get them from two different sources, like $\\ce{NaF}$ and $\\ce{HCl}$. Yeah, that'll work. \n\n\nMaybe. Maybe even with $\\ce{CaF2}$(The use of $\\ce{H2SO4}$ with the calcium salt suggests the reason was to precipitate the $\\ce{CaSO4}$ and flash off the $\\ce{HF}$, thereby separating, not just generating, the $\\ce{HF}$).\n\n\n",
"2"
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https://chemistry.stackexchange.com/questions/89833/why-is-mer-alq3-more-stable-than-fac-alq3
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Why is mer-AlQ3 more stable than fac-Alq3?
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Several research papers and books say that the meridional isomer of tris-(8-hydroxyquinoline) aluminium is thermodynamically more stable than the facial isomer.
What is the reason behind this stability?
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"\nThe stability of meridional isomer over facial isomer can be deduced by considering the interactions between atoms of the ligand 8-hydroxyquinoline.\n In considering the interactions if we take the repulsion between the electron rich atoms dominated in 90 degrees and at 180 degrees these interactions become negligible, we will find that in meridional isomer, there are dominating\ni) 2 N-N interactions\nii) 2 O-O interactions\niii) 8 N-O interactions.\n But if we look into the facial isomer of tris(8-hydroxyquinolinato)aluminium we will find there are dominating\ni) 3 N-N interactions\nii) 3 O-O interactions\niii) 6 N-O interactions.\n So in facial isomer two N-O interactions of mer isomer has turned into 1 N-N interaction and one O-O interaction. Due to high charge density on small sized O atoms O-O repulsions are much much greater than N-O interactions and thus the electronic repulsions are much more greater in fac isomer which decreases its stability than meridional isomer\n\n\n",
"1"
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] |
https://chemistry.stackexchange.com/questions/89831/analytical-gradient-of-the-hartree-fock-nuclear-electronic-repulsion-term
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Analytical gradient of the Hartree-Fock nuclear-electronic repulsion term
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The analytical gradient of the Hartree-Fock energy is given by (Szabo & Ostlund eq. C.16):
$$ \frac{\partial E}{\partial X\_{A}} = \sum\_{ij}{P\_{ji} \frac{\partial H\_{ij}}{\partial X\_{A}}} + \frac{1}{2} \sum\_{ijkl}{P\_{ji}P\_{kl} \frac{\partial (ij|kl)}{\partial X\_{A}}} - \sum\_{ij}{Q\_{ji} \frac{\partial S\_{ij}}{\partial X\_{A}}} + \frac{\partial V\_{NN}}{\partial X\_{A}} $$
Of course, the Born-Oppenheimer approximation means that the nuclear repulsion term is easy to handle, and the derivative is given as:
$$ \frac{\partial V\_{NN}}{\partial X\_{A}} = Z\_{A} \sum\_{B} \frac{Z\_{B} (X\_{B} - X\_{A})}{R\_{AB}^{3}}$$
The remaining terms require computing the derivatives of the Gaussian functions (depending upon the molecular integral evaluation scheme used). However, in Szabo & Ostlund the nuclear-electronic attraction term is provided as:
$$ \frac{\partial V\_{Ne}}{\partial X\_{A}} = -Z\_{A} \sum\_{i} \frac{X\_{i} - X\_{A}}{r\_{iA}^{3}}$$
In the Hartree-Fock equations, the molecular Hamiltonian is formed as $H = T + V\_{Ne}$. Presuming that the gradient of the molecular Hamiltonian is formed the same way, how can I treat this formulation of the analytical gradient with only a scalar value for the nuclear attraction term, when I must take into account the density matrix? Is this solution to the nuclear attraction gradient of no use in calculating the gradient of the energy?
>
> [1] A. Szabo and N. S. Ostlund, *Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory*, Dover Publications Inc., New York, pp. 441-442.
>
>
>
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"\nThere are a few points to discuss:\n\n\n* Since there are $3N$ possible $\\{X\\_{A}\\}$, each term where an $X\\_{A}$ appears will result in $3N$ matrix elements. In your first equation, that will be $S$, $T$, $V$, and the 2-electron contribution, which should really be rewritten into a Fock matrix term $F$ where the ket is precontracted with the density.\n* There is nothing wrong with this solution, though it is a little strange because it's in the MO basis, and we normally prefer to work in the AO basis to avoid any unnecessary transformations. I'm going to drop the $Ne$ from here on out since $V$ is understood in this context to only be the one-electron nuclear attraction term.\n* I think you're asking what happens to the density matrix. The derivative of $V$ expands into multiple parts, as we will see below. I will reference a bunch of equations that use $h$, where $\\left( h = H = H^{\\text{core}} \\right) \\equiv T\\_{e} + V\\_{Ne}$, and differentiating into a sum should hopefully be clear. What follows should work for any real-valued one-electron operator. I will use the book by Yamaguchi with liberties taken. In many places I use $p,q$ rather than $i,j$, as these indices run over all MOs, not just occupied ones.\n\n\n\n\n---\n\n\nExpanding $\\frac{\\partial V\\_{ij}}{\\partial X\\_{A}}$ using the product rule gives\n\n\n$$\n\\begin{align\\*}\n\\frac{\\partial V\\_{ij}}{\\partial X\\_{A}} &= \\frac{\\partial}{\\partial X\\_{A}} \\left( \\sum\\_{\\mu\\nu}^{\\text{AO}} C\\_{\\mu i} C\\_{\\nu j} V\\_{\\mu\\nu} \\right) \\tag{3.80} \\\\\n&= \\sum\\_{\\mu\\nu}^{\\text{AO}} \\left( \\frac{\\partial C\\_{\\mu i}}{\\partial X\\_{A}} C\\_{\\nu j} V\\_{\\mu\\nu} + C\\_{\\mu i} \\frac{\\partial C\\_{\\nu j}}{\\partial X\\_{A}} V\\_{\\mu\\nu} + C\\_{\\mu i} C\\_{\\nu j} \\frac{\\partial V\\_{\\mu\\nu}}{\\partial X\\_{A}} \\right), \\label{3.81}\\tag{3.81}\n\\end{align\\*}\n$$\n\n\nwhere the third/last term is the true AO integral derivative, and the first two terms, the MO coefficient derivatives, come from differentiating the density matrix.\n\n\n\n\n---\n\n\nThe simplest term to write out, but the most difficult to implement, is the AO integral derivative. I will use $\\mu,\\nu$ rather than $\\chi\\_{\\mu},\\chi\\_{\\nu}$, so they refer to both AO basis functions *and* their matrix indices.\n\n\n$$\n\\begin{align\\*}\n\\frac{\\partial V\\_{\\mu\\nu}}{\\partial X\\_{A}} &= \\frac{\\partial}{\\partial X\\_{A}} \\left< \\mu | \\hat{V} | \\nu \\right> \\tag{3.24} \\\\\n&= \\left< \\frac{\\partial \\mu}{\\partial X\\_{A}} | \\hat{V} | \\nu \\right> + \\left< \\mu | \\frac{\\partial \\hat{V}}{\\partial X\\_{A}} | \\nu \\right> + \\left< \\mu | \\hat{V} | \\frac{\\partial \\nu}{\\partial X\\_{A}} \\right> \\label{3.25}\\tag{3.25}\n\\end{align\\*}\n$$\n\n\nWhat you have presented,\n\n\n$$\n\\frac{\\partial \\hat{V}}{\\partial X\\_{A}} = -Z\\_{A} \\sum\\_{i} \\frac{X\\_{i} - X\\_{A}}{r\\_{iA}^{3}},\n$$\n\n\nis only the derivative of the operator. The basis function derivatives are also required, as the AOs in this case depend on the nuclear positions. Contrast this with an electric field perturbation, where only the derivative of the operator is required, or magnetic field derivatives, where basis functions *may* be perturbation-dependent if gauge-including atomic orbitals (GIAOs) are used. Also important is that the index $i$ here refers to an electron, *not* an occupied MO. This is an electric field integral.\n\n\n\n\n---\n\n\nNow for the derivative of the MO coefficients/density matrix. Clearly they don't appear in the final HF gradient expression. They disappear through the magic of Wigner's $2n + 1$ rule. From page 25:\n\n\n\n> \n> When the wavefunction is determined up to the $n$th order, the expectation value (electronic energy) of the the system is resolved, according to the results of perturbation theory, up to the $(2n+1)$st order. This principle is called Wigner's $2n+1$ theorem [29-31].\n> \n> \n> \n\n\nMore explicitly, we have the zeroth-order wavefunction, so we must be able to calculate the first-order correction to the energy. Worded differently, any first derivative of the energy must be easily calculated without differentiating MO coefficients, which is only required for second derivatives, such as the molecular Hessian or the dipole polarizability.\n\n\nFirst, rewrite the MO coefficient derivatives\n\n\n$$\n\\frac{\\partial C\\_{\\mu i}}{\\partial X\\_{A}} = \\sum\\_{m}^{\\text{MO}} U\\_{mi}^{X\\_{A}} C\\_{\\mu m}, \\tag{3.7}\n$$\n\n\nwhere the index $m$ runs over all occupied and unoccupied/virtual MOs. The same goes for $p,q$. I've replaced $a$ from the text with $X\\_{A}$, but this holds for any general perturbation (such as $\\lambda$, pick your favorite unused index). The key insight is that we can write the effect of a perturbation on the MO coefficients as the contraction of the unmodified MO coeffcients with a unitary matrix describing single-particle excitations from occupied to virtual MOs, as well as deexcitations from virtual to occupied MOs. In matrix form, this is\n\n\n$$\n\\mathbf{C}^{(X\\_{A})} = \\mathbf{C}^{(0)} \\left( \\mathbf{U}^{(X\\_{A})} \\right)^{T},\n$$\n\n\nwhere it is usually easiest to have the dimension of $\\mathbf{U}$ be $[N\\_{\\text{orb}}, N\\_{\\text{orb}}]$, with only the occ-virt and virt-occ blocks being non-zero. The problem now becomes solving for $\\mathbf{U}$ and eliminating it from the final gradient expression. I will walk through parts of section 4.3 to show how this is done.\n\n\nGiven the energy expression for an RHF wavefunction,\n\n\n$$\nE = 2 \\sum\\_{i}^{\\text{d.o.}} h\\_{ii} + \\sum\\_{ij}^{\\text{d.o.}} \\left[ 2(ii|jj) - (ij|ij) \\right], \\tag{4.1}\n$$\n\n\nthe first derivative with respect to $X\\_{A}$ is\n\n\n$$\n\\frac{\\partial E\\_{\\text{elec}}}{\\partial X\\_{A}} = 2 \\sum\\_{i}^{\\text{d.o.}} h\\_{ii}^{X\\_{A}} + \\sum\\_{ij}^{\\text{d.o.}} \\left[ 2(ii|jj)^{X\\_{A}} - (ij|ij)^{X\\_{A}} \\right] + 4 \\sum\\_{m}^{\\text{all}} \\sum\\_{i}^{\\text{d.o.}} U\\_{mi}^{X\\_{A}} F\\_{im}, \\label{4.16}\\tag{4.16}\n$$\n\n\nwhere the Fock matrix is defined as\n\n\n$$\n\\begin{align\\*}\nF\\_{pq} &= h\\_{pq} + \\sum\\_{k}^{\\text{d.o.}} \\left[ 2(pq|kk) - (pk|qk) \\right] \\tag{4.6} \\\\\n&= h\\_{pq} + 2J\\_{pq} - K\\_{pq},\n\\end{align\\*}\n$$\n\n\nand I've introduced the Coulomb and exchange matrices $\\mathbf{J}$ and $\\mathbf{K}$ as well. I skipped all the steps in expanding the first derivative, and $\\eqref{4.16}$ is what results upon collecting terms with $\\mathbf{U}$.\n\n\nUsing the RHF variational conditions, the Fock matrix from a converged calculation is diagonal in the MO basis, corresponding to the MO energies\n\n\n$$\nF\\_{pq} = \\delta\\_{pq} \\epsilon\\_{pq}, \\tag{4.7}\n$$\n\n\nso $\\eqref{4.16}$ simplifies to\n\n\n$$\n\\frac{\\partial E\\_{\\text{elec}}}{\\partial X\\_{A}} = 2 \\sum\\_{i}^{\\text{d.o.}} h\\_{ii}^{X\\_{A}} + \\sum\\_{ij}^{\\text{d.o.}} \\left[ 2(ii|jj)^{X\\_{A}} - (ij|ij)^{X\\_{A}} \\right] + 4 \\sum\\_{m}^{\\text{all}} \\sum\\_{i}^{\\text{d.o.}} U\\_{mi}^{X\\_{A}} \\epsilon\\_{im}, \\tag{4.17}\n$$\n\n\nwhich can be further simplified as\n\n\n$$\n\\frac{\\partial E\\_{\\text{elec}}}{\\partial X\\_{A}} = 2 \\sum\\_{i}^{\\text{d.o.}} h\\_{ii}^{X\\_{A}} + \\sum\\_{ij}^{\\text{d.o.}} \\left[ 2(ii|jj)^{X\\_{A}} - (ij|ij)^{X\\_{A}} \\right] + 4 \\sum\\_{i}^{\\text{d.o.}} U\\_{ii}^{X\\_{A}} \\epsilon\\_{ii}. \\tag{actual 4.17}\n$$\n\n\nNow we use one of the most important tricks in quantum chemistry. Given the orthonormality of the MOs,\n\n\n$$\nS\\_{pq} = \\delta\\_{pq}, \\tag{3.44}\n$$\n\n\nwe must have\n\n\n$$\n\\frac{\\partial S\\_{pq}}{\\partial X\\_{A}} \\overset{!}{=} 0. \\tag{3.45}\n$$\n\n\nThis is where not using general notation in the above is a bit confusing because it seems to conflict with your original expression, but remember that similar to $\\eqref{3.81}/\\eqref{3.25}$, this is in fact multiple terms: two for the basis functions, and two for the MO coefficients, giving\n\n\n$$\n\\begin{align\\*}\n\\frac{\\partial S\\_{pq}}{\\partial X\\_{A}} &= \\sum\\_{\\mu\\nu}^{\\text{AO}} C\\_{\\mu p} C\\_{\\mu q} \\frac{\\partial S\\_{\\mu\\nu}}{\\partial X\\_{A}} + \\sum\\_{m}^{\\text{all}} \\left( U\\_{mp}^{X\\_{A}} S\\_{mq} + U\\_{mq}^{X\\_{A}} S\\_{pm} \\right) \\tag{3.40 + 3.43} \\\\\n &= S\\_{pq}^{X\\_{A}} + \\sum\\_{m}^{\\text{all}} \\left( U\\_{mp}^{X\\_{A}} S\\_{mq} + U\\_{mq}^{X\\_{A}} S\\_{pm} \\right). \\tag{3.43}\n\\end{align\\*}\n$$\n\n\nThe sum over all MOs can be eliminated by reusing the orthonormality condition, so in the first term $m \\overset{!}{=} q$ and for the second term $m \\overset{!}{=} p$, and the overlap matrix in the MO basis is unity for those terms, giving\n\n\n$$\n\\frac{\\partial S\\_{pq}}{\\partial X\\_{A}} = S\\_{pq}^{X\\_{A}} + U\\_{qp}^{X\\_{A}} + U\\_{pq}^{X\\_{A}} \\overset{!}{=} 0. \\tag{3.46}\n$$\n\n\nRecognizing that we only need diagonal terms, this can be rewritten as\n\n\n$$\nU\\_{pp}^{X\\_{A}} = -\\frac{1}{2} S\\_{pp}^{X\\_{A}}, \\tag{4.20}\n$$\n\n\nwhich is then plugged back into the first derivative expression to give\n\n\n$$\n\\begin{align\\*}\n\\frac{\\partial E\\_{\\text{elec}}}{\\partial X\\_{A}} &= 2 \\sum\\_{i}^{\\text{d.o.}} h\\_{ii}^{X\\_{A}} + \\sum\\_{ij}^{\\text{d.o.}} \\left[ 2(ii|jj)^{X\\_{A}} - (ij|ij)^{X\\_{A}} \\right] + 4 \\sum\\_{i}^{\\text{d.o.}} \\left( -\\frac{1}{2} S\\_{ii}^{X\\_{A}} \\right) \\epsilon\\_{ii} \\\\\n &= 2 \\sum\\_{i}^{\\text{d.o.}} h\\_{ii}^{X\\_{A}} + \\sum\\_{ij}^{\\text{d.o.}} \\left[ 2(ii|jj)^{X\\_{A}} - (ij|ij)^{X\\_{A}} \\right] - 2 \\sum\\_{i}^{\\text{d.o.}} S\\_{ii}^{X\\_{A}} \\epsilon\\_{ii}. \\tag{4.21}\n\\end{align\\*}\n$$\n\n\nRewrite the last term\n\n\n$$\n\\begin{align\\*}\n\\sum\\_{i}^{\\text{d.o.}} S\\_{ii}^{X\\_{A}} \\epsilon\\_{ii} &= \\sum\\_{i}^{\\text{d.o.}} \\sum\\_{\\mu\\nu}^{\\text{AO}} C\\_{\\mu i} C\\_{\\mu i} \\frac{\\partial S\\_{\\mu\\nu}}{\\partial X\\_{A}} \\epsilon\\_{ii} \\\\\n &= \\sum\\_{i}^{\\text{d.o.}} \\sum\\_{\\mu\\nu}^{\\text{AO}} C\\_{\\mu i} C\\_{\\mu i} \\epsilon\\_{ii} \\frac{\\partial S\\_{\\mu\\nu}}{\\partial X\\_{A}} \\\\\n &= \\sum\\_{\\mu\\nu}^{\\text{AO}} W\\_{\\mu\\nu} \\frac{\\partial S\\_{\\mu\\nu}}{\\partial X\\_{A}} \\tag{4.24}\n\\end{align\\*}\n$$\n\n\nto use the energy-weighted density matrix $\\mathbf{W}$, which in your expression is called $\\mathbf{Q}$.\n\n\nAgain, the elimination of the $\\mathbf{U}$ matrix is one of the most important results in quantum chemistry, as it means the coupled-perturbed SCF equations do not need to be solved for first derivatives of SCF wavefunctions. This is why you do not see any density or MO coefficient derivatives in the gradient expression.\n\n\nReferences\n----------\n\n\n* [Yamaguchi, Yukio; Goddard, John D.; Osamura, Yoshihiro; Schaefer III, Henry F. *A New Dimension to Quantum Chemistry: Analytic Derivative Methods in Ab Initio Molecular Electronic Structure Theory (International Series of Monographs on Chemistry);* Oxford University Press: 1994.](https://isbnsearch.org/isbn/9780195070286)\n* [Epstein, Saul T. General Remainder Theorem. *J. Chem. Phys.* **1968**, *48*, 4725.](https://dx.doi.org/10.1063/1.1668053)\n* [Epstein, Saul T. Constraints and the $V^{2n+1}$ theorem. *Chem. Phys. Lett.* **1980**, *70*, 311.](https://dx.doi.org/10.1016/0009-2614(80)85340-1)\n\n\nIf you don't have access to the Yamaguchi book, the seminal article on HF derivatives contains a condensed version of this derivation, but be aware of typos.\n\n\n* [Pople, John A.; Raghavachari, Krishnan; Schlegel, H. Bernhard (Berny); Binkley, J. Stephen. Derivative Studies in Hartree-Fock and Møller-Plesset Theories. *Int. J. Quantum Chem.* **1979**, *16* (num. S13), 225-241.](https://dx.doi.org/10.1002/qua.560160825)\n\n\n",
"9"
],
[
"\nThe book doesn't explicitly list all the terms you need to evaluate $\\frac{\\partial H\\_{\\mu\\nu}^{core}}{\\partial X\\_A}$ (which I'll abbreviate as $\\partial H\\_{\\mu\\nu}^{core})$.\n\n\n$H\\_{\\mu\\nu}^{core} = \\langle \\mu |\\hat{T} + \\hat{V}\\_{eN}| \\nu \\rangle$ so\n\n\n$\\partial H\\_{\\mu\\nu}^{core} = \\langle \\partial \\mu |\\hat{T} + \\hat{V}\\_{eN}| \\nu \\rangle + \\langle \\mu | \\partial \\hat{V}\\_{eN}| \\nu \\rangle + \\langle \\mu |\\hat{T} + \\hat{V}\\_{eN}| \\partial \\nu \\rangle$\n\n\nwhere $\\partial \\hat{V}\\_{eN} = - Z\\_A \\sum\\_i \\frac{X\\_i-X\\_A}{r^3\\_{iA}}$\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/89816/whats-this-mystery-acid
|
What's this mystery acid? [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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Today, my dad told me about a cleaning product for membranes he bought a while ago. He told me he put his finger in it and it didn't hurt him at all, nor did it seem to damage any tissues.
But later that day, he brought the substance and checked it's pH with a digital pH tester. It showed that it was pH 0. after diluting it to a concentration of around 20%, he said the pH meter reads 0.68.
I googled "strong acid non corrosive" and it came up with carborane superacid ($\ce{H(CHB11Cl11)}$) as an answer.
Is there a specific answer or is there a number of possible answers?
And what tests can we carry out to identify this acid?
Edit: My dad kept the acid for 15 seconds before rinsing
| -1 |
[
[
"\nThis acid is actually not that concentrated. \nSuppose it were $\\ce{HCl}$, then:\n\n\n$\\frac{10^{-0.68}\\cdot 36.45}{0.20}\\frac{100}{1000}=4\\%\\ \\ce{HCl}$\n\n\nI am not going to try on my skin but it is not like the 37% which we have in the bottle. \n\n\nI guess that since it is for cleaning membranes, it is some chelant for the calcium with a bit of acid for the carbonate. \n\n\n**Edit**\n\n\nI could not resist...\n\n\nI had some 5% $\\ce{H2SO4}$ prepared for adjusting pH so I have tried it after all. \nSo, no more speculation. \n\n\n* 20 seconds with the tip of the finger in it, does not itch;\n* 20 seconds on the hand, does not itch either;\n* on the tongue, it has definitively an acid taste, but it does not burn. It is, however, too acidic to take it.\n\n\n**Warning** \n\n\nDo never ever touch chemicals, nor taste them.\nAlways work with proper attire, lab coat, gloves, goggles, etc. \n\n\n",
"6"
],
[
"\nI cannot tell you what acid is in the product. \n\n\nHowever burning and corrosion of skin and living tissues is not necessarily and directly due to fast acid/base reactions.\n\n\nIt involves reaction such as oxidation, nitration, dehydration, etc. or denaturing proteins.\n\n\nFor instance the damaging power of concentrated sulphuric acid resides mostly in its dehydrating and oxidizing power.\n\n\nAs such, the skin can get in touch with very acidic media without suffering immediate and permanent damage - and particularly so if the area is immediately and throughly rinsed.\n\n\nThis stated, I am not sure if such a low pH should cause an immediate hitching sensation. In my direct experience less than 15 seconds were enough to always start feeling hitches, independent of the nature and concentration of the acid handled.\n\n\nOn the other hand, we can exclude acid such as HCl, sulphuric acid, nitric acid, phosphoric acid. Plugging a finger on their solutions for 15 seconds will definitively be felt by the operator, unless very diluted, and burn its skin if conversely they are concentrated enough. \n\n\nFinal note. It is not a good idea to stick a finger in unknown chemicals as for some of them are very nasty, being them acid or alkali or *none* of the two.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89813/is-dacron-and-glyptal-the-same-polymer
|
Is Dacron and Glyptal the same polymer?
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I know Dacron or Terylene is a polymer of ethylene glycol and terephthalic acid . But is Glyptal also a name for Dacron? Glyptal has almost similar monomers.
| -1 |
[
[
"\nAccording to the [Polymer Properties Database](http://polymerdatabase.com/polymer%20classes/Alkyd%20Resin.html), Glyptal is a manufacturer of [alkyd](https://en.wikipedia.org/wiki/Alkyd) resins formed from polyhydric alcohols with polybasic acids. Certainly, they might make [polyethylene terephthalate](https://en.wikipedia.org/wiki/Polyethylene_terephthalate) as well.\n\n\n",
"1"
]
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https://chemistry.stackexchange.com/questions/89812/paint-removal-from-metal
|
Paint removal from metal [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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What chemical product works well to strip paint off to reveal bare metal in this instance a small tin (guessing to be aluminium or steel) with as good a finish as possible.
| -1 |
[
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"\nN-Methyl-2-pyrrolidone - typical paint stripper.\n\n\n",
"-2"
]
] |
https://chemistry.stackexchange.com/questions/89810/step-up-reactions-in-elementary-organic-chemistry
|
Step up reactions in elementary organic chemistry [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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What are some elementary (high school level) step up reactions? (in which the length of the carbon chain of an organic molecule is increased by one)
**Background:** I was looking for a way to methyl chloride to acetone (spoilers: I still haven't found one :( I could use Wurtz reaction to form ethane from methyl chloride, but I don't know of a way to add one more carbon to this alkane. I found [this question](https://chemistry.stackexchange.com/questions/23840/organic-chemistry-alkanes-step-up-reaction), but it only focuses on the ketene reagent's action on alkanes. I suppose there might be many more such "step up" reagents and their many more compounds which can be "stepped up". Hence, my question.
---
PS: A term related to "Step up" is "Step down", which observed in Hunsdiecker reaction and Hoffmann Bromamide reaction.
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[] |
https://chemistry.stackexchange.com/questions/89806/will-the-carboxylate-groups-of-the-citrate-anion-undergo-hydrogen-bonding
|
Will the carboxylate groups of the citrate anion undergo hydrogen bonding?
|
The citrate anion of trisodium citrate has three carboxylate groups. I am concerned with whether or not there will be hydrogen bonding in water, with hydroxyl groups of other ions or alcohols.
1. It has been suggested to me that the negative carboxylate oxygens won't hydrogen bond, due to resonance delocalisation of the charge. If it mattered whether or not the oxygen was partially charged, I would hypothesize that the oxygen is charged, because carbon has a lower electronegativity.
2. The oxygen should have two lone pairs, yet it has either been implied that they are not there, or for some reason that they will not interact in hydrogen bonding.
My question is: will the citrate anion hydrogen bond with other molecules or polymers with hydroxyl groups, in aqueous solution?
It has been made aware to me that there is an article that has done research to show that carboxylates are strong hydrogen bond acceptors in anhydrous solvents.
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[
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"\nAs the comments indicate, you don't need a formal charge on oxygen at all. \"Neutral\" oxygen atoms are usually negatively charged anyway just from bond polarity, as in water. That and the lone pairs on oxygen are quite enough to enable hydrogen bonds to protic hydrogen.\n\n\n",
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]
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https://chemistry.stackexchange.com/questions/89804/where-am-i-wrong-in-my-calculations
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Where am I wrong in my calculations? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
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I am revisiting the concept of mole and I decided to figure out random values (for instance: mass of 1 molecule of water) to see how much I understand the concept of moles and how it relates to amu. So here is how I started:
There is 1 mole of C atoms in 12g of C and 1 C atom is 12 amu. Therefore, there is 1 mole of 1 amu in 1g. That means there are 6.02e23 nucleons in 1 gram. So, 1 nucleon is 1g/6.02e23= 1.66e-24g. A molecule of water has 14 nucleons (6 neutrons and 6 protons from oxygen and 2 protons from both H); thus, 1.66e23 x 14 should be the g of a water molecule, which turns out to be 2.33e-23. However, this is not the correct mass of water molecule (2.992 x 10¯23 grams).
| -2 |
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"\nAs Ivan pointed out in the comments, I had the molar mass of water incorrect. By fixing the molar mass to 18g/mol, I get the correct answer.\n\n\n",
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https://chemistry.stackexchange.com/questions/89798/what-is-the-relation-between-rate-of-reaction-and-rate-of-consumption-of-reactan
|
What is the relation between Rate of reaction and rate of consumption of reactants?
|
>
> The rate of the following reaction is 0.300M/s. What is the relative rate of change of each species in the reaction?$$\ce{A + 3B -> 2C}$$
>
>
>
1. We are supposed to find A and B and I understand that we have to use $\frac{ [\delta]A}{[\delta]t}$ and then use a negative for the coefficient. But what I don't understand is, why for B we had to multiply -3 but not -1/3? And is it the same for A?
2. Also, I am confused on how I can differentiate between using these two equations of rates. The Delta of reactants/delta of time, versus the rate $v(nu)=k[A]m[B]n$
equation.
| 0 |
[
[
"\nFor your first part you're supposed to multiply it by -3 instead if $\\frac{-1}{3}$ as the **rate of reaction** is given for one mole. That is if $R$ is the rate of reaction it is defined as .\n$$\\frac{-d[A]}{dt} = \\frac{-d[B]}{3dt} = \\frac{d[C]}{2dt} = R$$\nSo the **rate of consumption** of B is $-3R$.\n\n\n\n\n---\n\n\nAs for your second part it's a tad unclear for me if you could further edit your question.I could take a shot at answering your second part as well\n\n\n",
"0"
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] |
https://chemistry.stackexchange.com/questions/89794/using-drip-extraction-how-much-caffeine-is-extracted-from-coffee-as-a-function
|
Using drip extraction, how much caffeine is extracted from coffee as a function of the amount of water used?
|
When using a drip coffee maker, there are basically two choices we make: the amount of water and the amount of coffee grounds.
If we use the same amount of coffee grounds but double the water, how much extra caffeine is in the resulting brew? Does all the available caffeine in the grounds get quickly liberated, so that there's not much of a difference between brewing with more water and diluting the brew with water afterwards? Or is it closer to the other extreme, where twice the water means twice the caffeine?
I saw on [wikipedia](https://en.wikipedia.org/wiki/Coffee_extraction#Strength) that only about 20% of the caffeine is extracted for an 'optimal brew,' but I wasn't sure if that means we can extrapolate and say that the remaining 80% can be extracted just as easily.
I have the same question about the amount of coffee grounds. I only care to consider the 'normal brewing regime', that is with water at ~80-90C, 3-6 cups of water, and 1/4-1/2 a cup of grounds.
Thanks!
| 3 |
[
[
"\nCaffeine has a solubility in water of 20% at 80 $^\\circ$C, while the coffee has between 1 and 4% caffeine. \nThat means that you have more than enough water to extract all the caffeine in the coffee. \nWith the quantities you have given, I estimate that you are using from 12 to 24 mL of water per gram of coffee (bulk density 0.25 g/mL). \n\n\nSo, it is not very likely that a change in volume will affect the total quantity of extracted caffeine much. Only the concentration per unit volume will be lower if more water is used. \nBut then, solid-liquid extraction is not that perfect and possibly you will extract some more caffeine with more water. The main property that influences the extraction is the size of the milled coffee. The smaller the better is the extraction. \n\n\nMany other components of coffee are extracted fairly fast. \nTherefore, here we do a percolation (expresso) which extracts everything with very less water. The concentration of caffeine per unit volume seems to be higher, but the dose of coffee is much smaller. \n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89790/regioselectivity-in-claisen-condensation-and-aldol-reaction
|
Regioselectivity in Claisen condensation and aldol reaction
|
I'm having a hard time determining regio-selectivity in those two reactions.
[](https://i.stack.imgur.com/yzD0A.png)
At the image above I've drawn one example for each on of them. My question is how do we choose where the deprotonation will occur. My approach was that the base will deprotonate the substrate in order to produce the most stable enolate. In the Claisen condensation, A should be the main product since the -CH2- group is present thus corresponding the mechanism. For B there will be no acidic hydrogens so it is not favoured, or at least that's what I thought.
Moving to the aldol condensation I can't figure out which product will be the main. Theoretically in C, the enolate is more stable than that of D, but the final unsaturated ketone isn't more stable in D?
Either way, I can't comprehend how can we determine what enolate will be produced each time.
| 6 |
[
[
"\nClaisen and aldol condensations are thermodynamically controlled,\\* so it is not a question of which enolate forms, it is a question of which product forms.\n\n\nIn your first example, product A can be deprotonated by ethoxide as there is still an acidic hydrogen between the two carbonyls. Product B, on the other hand, has a quaternary carbon between the two carbonyls and cannot be deprotonated.\n\n\nBecause the p*K*a of a 1,3-dicarbonyl (~10) is significantly smaller than the p*K*a of ethanol (~15),\\*\\* this deprotonation of A is almost complete and it therefore pulls the equilibrium position over to one which favours product A, simply by le Chatelier's principle. In fact if you somehow make product B via a different route and then treat it with ethoxide, you should exclusively get product A out of your flask, via retro-Claisen and Claisen condensation.\n\n\nFor the second example I would expect product D (tetrasubstituted alkene) to be preferentially formed over C (trisubstituted).\n\n\n\n\n---\n\n\n\\* Unless you use kinetic deprotonation conditions such as LDA. In this case you will then have to look at the stability of the transition states leading to enolate formation, but that's a story for another day.\n\n\n\\*\\* **Very** rough values.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/89789/comparing-the-h%c3%bcckel-and-extended-h%c3%bcckel-methods
|
Comparing the Hückel and extended Hückel methods
|
I'm very confused about the differences between these methods. From my textbook, it states that the Hückel method only takes into account the π bonding interactions, while the extended Hückel method takes into account all the valence electrons. But what exactly does this do?
To my knowledge, the s, p*z*, p*x*, and p*y* orbitals are all orthogonal to one another and wouldn't mix anyways. Are shielding effects the only thing accounted for by extended Hückel theory?
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[
[
"\nIn the simple Hückel method (SHM) the basis set is limited to p orbitals. This set is limited in a great extent to p*z* orbitals which constraints the molecular plane to be the *xy* plane. Basically, you are limited to planar molecules. \n\n The inclusion of all valence s and p orbitals in the extended Hückel method (EHM) naturally lifts the spacial constraints and you can work with non-planar molecules.\n\n\nThese are the two first differences between the two methods. The other points are:\n\n\n**SHM**:\n\n\n* Orbital energies are limited to same-atom interactions, adjacent-atom interactions while all other interactions are 0.\n* Fock matrix elements are not actually calculated.\n* Overlap integrals are limited to 1 or 0.\n\n\n**EHM**:\n\n\n* Orbital energies are calculated and vary smoothly with geometry.\n* Fock matrix elements are actually calculated.\n* Overlap integrals are actually calculated.\n\n\nYou can look up the derivation and steps for the implementation of these two methods in this book that I used as a reference:\n\n\nErrol G. Lewars; *Computational Chemistry, Introduction to the Theory and Applications of Molecular and Quantum Mechanics,* Second Edition;\nSpringer: 2011. DOI: [10.1007/978-90-481-3862-3](https://doi.org/10.1007/978-90-481-3862-3)\n\n\n",
"11"
],
[
"\nPentavalentcarbon brought up an interesting point that I thought I could elaborate on: How are the matrix elements of this simplified Hamiltonian obtained? I'll basically just summarize the relevant parts of this Chem-Libre section on the [Extended Hückel method](https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al.)/10%3A_Theories_of_Electronic_Molecular_Structure/10.06%3A_Semi-Empirical_Methods%3A_Extended_H%C3%BCckel)\n\n\nThis is where the empirical part of this method comes into play. If you have the overlap matrix elements $S\\_{ij}$, then the elements of the Hamiltonian $H\\_{ij}$ can obtained using experimentally obtained values of a given orbital's ionization potential. The diagonal elements are simply taken to the negative of the ionization potential of the corresponding atomic orbital, $$H\\_{ii}=-\\mathrm{IP}$$ The off-diagonal elements take only slightly more effort with $$H\\_{ij}=\\frac{K}{2}(H\\_{ii}+H\\_{jj})S\\_{ij} \\text{ , } i\\not=j$$ $K$ is just a proportionality constant and a commonly used value is $1.75$, based on a study by Hoffman on the orbital energies of ethane. \n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89781/diels-alder-reaction-question
|
Diels-Alder reaction question
|
In class, we learned that Diels-Alder reactions only occurs with s-cis conformations and that conformations like those in cyclopentadiene are permanently locked in a cis-transformation and that is why they are so reaction. We also learned that the reaction proceeds at a higher rate with an electro-withdrawing substituent (eg. carbonyl) . Why then is a molecule like the one in the figure not reactive while a similar structure, a cyclopentadienone, is?[](https://i.stack.imgur.com/D1mEd.png)
| -1 |
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"\nYou need to draw this structure in a 3D visualizer program, eg, Avogadro. While this structure looks similar to cyclopentadiene, in 3D it looks vastly different.\n[](https://i.stack.imgur.com/c7xFd.jpg)\n\n\n",
"-1"
]
] |
https://chemistry.stackexchange.com/questions/89769/what-is-150-degree-test-kerosene
|
What is 150 degree test kerosene?
|
I saw an old ad for kerosene in which it was advertised at "150 degree test".
What does that mean?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/89768/how-do-we-indicate-a-volume-fraction-like-0-05-in-procedure-descriptions
|
How do we indicate a volume fraction like 0.05 in procedure descriptions?
|
I'm translating a Russian biotech text, and there's this sentence with a number without measurement units:
>
> The contents of the wells were discarded; the wells were washed three times with 200 µL of Buffer containing **0.05 of Tween 20**.
>
>
>
I was told that this figure indicates a volume fraction, and is dimensionless, hence no units after it. Will this be understood by the native speaker of English as-is, without any units, as in my quote?
Or is it better to write it down as:
>
> The contents of the wells were discarded; the wells were washed three times with 200 µL of Buffer containing **5% (v/v) of Tween 20**.
>
>
>
Or maybe like this:
>
> The contents of the wells were discarded; the wells were washed three times with 200 µL of Buffer containing **5 vol% of Tween 20**.
>
>
>
Is there a way to retain **0.05** and somehow indicate that this is a volume fraction?
| 1 |
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"\nI think it would be wise to include units, because how would you distinguish between mass and volume fractions? Options 2 and 3 that you present are both seen in the scientific literature and text books. My opinion is that option 3 is the least ambiguous way to present both the numeric value and the context (i.e. the units).\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89766/sodium-carbonate-saturation-curve
|
Sodium Carbonate saturation curve [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
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Where can I find the saturation curve for Na2CO3 (Sodium Carbonate) versus temperature?
Thanks,
ECD40
| -3 |
[
[
"\nWikipedia has a table in the article on sodium carbonate. You could plot the numbers or Google it. I got a neat curve (attached below)but you have to do a little figuring to get the temperatures. Or you could Google images for sodium carbonate solubility to get other curves.\n\n\nWashing soda is the decahydrate; soda ash is anhydrous.\n\n\nDecahydrate:\n7 g/100 mL (0 °C)\n16.4 g/100 mL (15 °C)\n34.07 g/100 mL (27.8 °C)\n\n\nHeptahydrate:\n48.69 g/100 mL (34.8 °C)\n\n\nMonohydrate:\n50.31 g/100 mL (29.9 °C)\n48.1 g/100 mL (41.9 °C)\n45.62 g/100 mL (60 °C)\n43.6 g/100 mL (100 °C)\n\n\n[](https://i.stack.imgur.com/7EBFm.jpg)\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89763/mechanism-for-reaction-of-an-allene-with-the-grignard-reagentrmgx
|
Mechanism for reaction of an allene with the Grignard Reagent(RMgX)
|
How does the Grignard Reagent($\ce{RMgX}$) react with an allene $\ce{CH\_2=C=CH\_2}$ and what are the products formed?
This is exactly what I tried to do.
[](https://i.stack.imgur.com/9xb3N.png)
Any help will be appreciated.
| 3 |
[] |
https://chemistry.stackexchange.com/questions/89762/is-a-beaker-reaction-vessel-like-the-one-pictured-available-commercially
|
Is a beaker/reaction vessel like the one pictured available commercially?
|
We're doing photocatalysis with the catalyst fixed to a 2" Si wafer, washing an aqueous solution (methylene blue, the concentration varies but in the picture I believe it's 0.1 mMol/l) over the top in beakers that look like this
[](https://i.stack.imgur.com/U3msk.jpg)
(the lid is the heatsink for the 365 nm UV LED that drives the reaction; that part is pretty much optimised).
The few we have are custom made, and we need some more. I can't find a source of anything comparable, but I'm only a physicist who doesn't necessarily know what's available or what terms to search for.
It needs two hose barbs as near the bottom as possible, and a flat base wide enough to lift a 2" wafer out (i.e. about a 60 mm minimum internal diameter).
They're also visible in this [timelapse video of the reaction](https://www.youtube.com/watch?v=spnoti9n50M).
If such a thing exists, what's it called? Or is there something we could repurpose? Both the cost and the lead time of custom glassware are a bit high.
| 5 |
[
[
"\nA beaker / reactor of such type would be rather expensive as well. If design and elegance do not matter you can easily adapt every suitably sized beaker by holding two pipes at the levels at which the hose barbs are supposed to be. Perhaps you choose a beaker a bit wider so there will be no turbulence or differences of sort as compared to what you are using.\n\n\nAny flask or other container, combined with two small tubes and a couple of stands will effectively let you save money and time. Perhaps you must get space for the inlet and outlet through the led holder lid.\n\n\n",
"2"
],
[
"\nI don't think these are UV LED. Most likely vanilla blue LED. You can buy strips of blue LED light and wrap it around a 500 mL beaker. Run your reaction in a second vessel inside that larger beaker. or else, you can use a aluminium sheet, cut vertical perforations and wrap the LED strips around it. Place the photocage around your reaction vessel.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89759/setting-to-basis-set-for-electron-transfer-in-type-ii-cds-tio2-heterostructre-fo
|
Setting to basis set for electron transfer in type II CdS/TiO2 heterostructre for PYXAID package
|
I am studying electron injection from $\ce{CdS}$ into $\ce{TiO2}$ using NAMD approach. Now I'm using pyxaid package.
Would be possible someone kindly help me with construction of basis set for this process?
30 bands selected for Hamiltonian construction and from this selected band set number and active space of [6-13] has been made. 7 is the valence band and 8 to 13 are conduction states.
For initial electronic state I selected -11,-12,-13 (one of which could be $\ce{CdS}$'s lumo) as I found from some preliminary decoherence rate calculations (12 decoherence more rapidly than 9-12).
```
params["active_space"] = [6,7,8,9,10,11,12,13]
# Generate basis states
params["states"] = []
params["states"].append(["ES1",[6,-6,7,-8]])
params["states"].append(["ES2",[6,-6,7,-9]])
params["states"].append(["ES2",[6,-6,7,-10]])
params["states"].append(["ES2",[6,-6,7,-11]])
params["states"].append(["ES2",[6,-6,7,-12]])
params["states"].append(["ES2",[6,-6,7,-13]])
# Initial conditions
nmicrost = len(params["states"])
ic = []
i = 0
while i<100:
j = 3 # started from -11,-12,-13 one of which I believe belongs to CdS (donor semiconductor)
while j<nmicrost:
ic.append([2*i,j])
j = j + 1
i = i + 1
params["iconds"] = ic
```
| 2 |
[] |
https://chemistry.stackexchange.com/questions/89751/how-to-choose-alcohol-solvent-for-organic-extraction
|
How to choose alcohol solvent for organic extraction?
|
Assuming I have an organic substance that is known to be "soluble in alcohol" (e.g. [capsaicin](https://en.wikipedia.org/wiki/Capsaicin)) I want to extract, *which* alcohol should/can I use? I know that both ethanol and methanol are commonly used, but what about:
* isopropanol
* glycerol
* n-Butanol
* 1,4-Butanediol
* tert-Butyl alcohol
* 1-Pentanol
* tert-Amyl alcohol
In general, which alcohols are suitable/unsuitable for organic compound extraction (e.g. which should be the best solvent in terms of amount of solute dissolved in given volume of solvent), and why? Is there any *a priori* way to determine e.g. if isopropanol extraction will be OK for a given "soluble in alcohol" organic substance?
---
Note: I am aware of [this discussion](https://www.researchgate.net/post/Which_is_the_best_solvent_for_herbal_extraction), but it doesn't cover the specific nature of alcohol extraction.
| 1 |
[
[
"\nThe optimal solvent will extract the greatest amount of desired product and the least amount of difficult-to-remove substances.\n\n\nThe extraction is only the first step; separation from solvent and other extractives follows. Try one, then another solvent to see if there is a difference. Maybe number of carbons or hydroxyls is important; you can also mix alcohols, and/or add water. \n\n\nCapsaicin has a phenolic group; perhaps it would be extracted from its source with alkali (NaHCO3 or Na2CO3).\n[](https://i.stack.imgur.com/DWSP2.jpg)\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89747/thermodynamic-stability-of-meta-xylene-over-ortho-and-para-isomers
|
Thermodynamic stability of meta-xylene over ortho- and para-isomers
|
When talking about the example of the alkylation of toluene by chloromethane in the presence of $\ce{AlCl3}$, Hepworth, Waring and Waring (2002) mentioned that:
>
> At room temperature, a mixture of 1,2-dimethylbenzene and 1,4-dimethylbenzene results, but at 80 °C the product is mainly 1,3-dimethylbenzene. In fact, heating either of the 1,2- or 1,4-isomers in the presence of $\ce {AlCl\_3}$/$\ce {HCl}$ results in rearrangement to the more stable 1,3-dimethylbenzene.
>
>
>
This leads me to question: Why would the 1,3-isomer be more thermodynamically stable than the 1,2- and 1,4-isomers? Clearly, there must be an activation energy barrier to overcome as heat is required for the conversion. The methyl group being an *ortho/para*-director would lead to the formation of 1,2- and 1,4- isomers. Thus, wouldn't these isomers also be the more thermodynamically stable isomers?
**Reference**
Hepworth, J. D., Waring, D. R., & Waring, M. J. (2002). Aromatic Chemistry. United Kingdom: The Royal Society of Chemistry.
| 23 |
[
[
"\nThe thermodynamic stability of *m*-xylene over *o*-xylene or *p*-xylene can be deduced by considering the hyperconjugative effect. More specifically, resonance forms with a positive charge on methyl-substituted carbons are more important than resonance forms with a negative charge on them, as the methyl group stabilises the positively charged carbon by hyperconjugation.\n\n\nIn the *o*- or *p*- isomers, the hyperconjugation of one methyl group produces negative charge at ortho and para position where another methyl group resides. This is highly destabilising. But, in the *m*-isomer, hyperconjugation does not destabilise the structure and both methyl groups' hyperconjugation effects act in a cooperative way with each other.\n\n\n[](https://i.stack.imgur.com/Dws5H.png)\n\n\n",
"16"
]
] |
https://chemistry.stackexchange.com/questions/89746/what-is-the-lewis-structure-of-ozonide-ion
|
What is the Lewis structure of Ozonide ion? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed 5 years ago.
[Improve this question](/posts/89746/edit)
What will be the Lewis structure of the ozonide ion, including formal charges?
Do all oxygen atoms have filled octate?
| 0 |
[
[
"\nFrom a [cursory e-book search](https://books.google.co.in/books?id=L-1K9HmKmUUC&pg=PA354&lpg=PA354&dq=ozonide%20ion%20structure&source=bl&ots=zXmBZQRXr-&sig=IEQPRusbE8ZUH9RZoKSJ-lJV8Co&hl=en&sa=X&ved=0ahUKEwjX4tTW6__YAhUdSY8KHTNmCRwQ6AEIRjAH#v=onepage&q=ozonide%20ion%20structure&f=false): \n\n\n\n> \n> Ozonide ion is paramagnetic and is shown to have magnetic\n> susceptibility. Bond length is 135 Å and bond angle is 108°. The\n> structure is:\n> \n> \n> [](https://i.stack.imgur.com/GySVu.png)\n> \n> \n> \n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89745/what-exactly-is-an-electron-sponge
|
What exactly is an "electron-sponge"?
|
What exactly an "electron-sponge" [behavior/action/property/system] nickname is, and what makes a material an "electron-sponge" (preferably, quantitatively)?
From what I found, it's typically a molecular cluster or a metal complex which is capable of undergoing multiple oxidation and reduction waves, or is able of accepting several electrons without losing their chemical identity. However, this definition seems to be rather qualitative and quite vague. Also, it would be interesting to find out who and when coined this term.
| 7 |
[] |
https://chemistry.stackexchange.com/questions/89744/calculating-compressibility-factor-from-the-van-der-waals-gas-equation
|
Calculating Compressibility factor from the Van der Waals' Gas equation
|
So this problem has been bugging me for a long time. According to [Wikipedia](https://en.m.wikipedia.org/wiki/Compressibility_factor) the compressibility factor $Z$ is defined as the ratio of the volume occupied by a real gas divided by the volume occupied by an ideal gas at the same temperature and pressure.
$$Z = \frac{V\_\text{real}}{V\_\text{ideal}}$$
Now the ideal gas equation is
$$P\_\text{ideal} V\_\text{ideal} = nRT$$
Now from here we can substitute the value of $V\_\text{ideal}$ into the $Z$ expression to get $$Z = \frac{P V\_\pu{m}}{RT}$$
Where $V\_\pu{m}$ is the molar volume of the gas and $P$ is the value of $P\_\pu{ideal}$ that was in the ideal gas equation.
OK so far so good.
*Deriving the compressibility factor value from the van der Waals' gas equation:*
We know for real gases the pressure exerted by them is always less than the pressure exerted by an ideal gas so we correct the pressure by the term
$an^2/V^2$
$$P\_{\text{ideal}}= P\_{\text{real}} + \frac{an^2}{V^2}$$
And the volume occupied by the gas molecules can be written as
$$ V\_{\text{ideal}} = V\_{\text{real}} - nb$$
On substituting the above values of $P\_\text{ideal}$ and $V\_\text{ideal}$ in the ideal gas equation, we get our van der Waals' Gas equation
$$\left(P+\frac{an^2}{V^2}\right)(V-nb) = nRT$$
Please note that in the above equation $P$ is $P\_\text{real}$ *and not* $P\_\text{ideal}$. Now I'm going a step further to calculate the value of $Z$ from this equation by shifting the terms and I got this $$Z = \frac{P V\_\pu{m}}{RT} = \frac{V\_\pu{m}}{V\_\pu{m} - b} - \frac{a}{RTV\_\pu{m}}$$
Now here's the problem that I'm facing: In the above expression the value of $P$ is not the $P\_\text{ideal}$ but it is the value of $P\_\pu{real}$ that we had in the van der Waals' gas equation. Doesn't that contradict the whole definition of Compressibility factor that we had learned in the first place? How this expression for $Z$ can be correct?
Thanks for reading.
| 7 |
[
[
"\n$$Z = \\frac{V^{\\mathrm{r}}\\_{\\mathrm{m}}}{V^{\\circ}\\_{\\mathrm{m}}}\\tag{1}\\label{1}$$\n\n\n$V^{\\mathrm{r}}\\_{\\mathrm{m}} = \\text{Volume of 1 mol real gas.}$\n\n\n$V^{\\circ}\\_{\\mathrm{m}} = \\text{Volume of 1 mol perfect gas.}$\n\n\nThe van der Waals equation is $$ \\left( P^{\\text{r}} + \\frac{an^2}{(V^{\\text{r}})^2}\\right) (V^{\\text{r}} - nb ) = nRT \\tag2\\label{2}$$\n\n\nNow, consider you have a container containing a $1~\\mathrm{mol}$ real gas. You know its pressure $P^{\\text{r}}$, volume $V^{\\text{r}}$ and temperature $T$ and you wish to find the compressibility factor of this gas. \n\n\nSo, for calculating $Z$, we know the real volume of the gas, i.e $V^{\\mathrm{r}}$ and now we need to calculate $V^{\\circ}$ which is *the volume it should have occupied if it behaved like a perfect gas,* i.e. it obeyed the perfect gas law.\n\n\nThus, $$V^{\\circ} = \\frac{RT}{P^{\\text{r}}}$$\n\n\nTherefore, substituting this value of $V^{\\circ}$ in $\\eqref{1}$ $$Z = \\frac{V^{\\mathrm{r}} P^{\\text{r}}}{RT}\\tag3\\label{3}$$\n\n\nFrom $\\eqref{2}$, by rearranging the terms, we get $$P^{\\text{r}} = \\frac{RT}{(V^{\\text{r}} - b)} - \\frac{a}{(V^{\\text{r}})^2} \\tag4\\label{4}$$\n\n\nSubstituting this value of $P^{\\text{r}}$ in $\\eqref{3}$, we get $$Z = \\frac{V^{\\text{r}}}{(V^{\\text{r}} - b)} - \\frac{a}{V^{\\text{r}}(RT)} \\tag5\\label{5}$$\n\n\n",
"6"
],
[
"\nFrom van der Waals' equation: $$(p + \\frac{an^2}{V^2})(V - nb) = nRT\\\\\n\\color{brown}{p} = \\frac{nRT}{V - nb} - \\frac{an^2}{V^ 2}$$\n\n\nCompressibility factor is the ratio of volume of real gas to ideal gas:\n$$Z= \\frac{V}{V\\_{\\text{ideal}}} = \\frac{\\color{brown}{p}V}{nRT} = \\frac{V}{V - nb} - \\frac{an}{VRT}$$\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89742/making-para-dibromobenzene
|
Making para-dibromobenzene [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed 5 years ago.
[Improve this question](/posts/89742/edit)
What steps do I need to take to make the compound in the picture from just a benzene ring? Thank you.
[](https://i.stack.imgur.com/U5p29.png)
| -3 |
[
[
"\nAccording to US Patent 3062899A p-dibromobenzene is the major product of the bromination of benzene with 2 equivalents of bromine at ambient temperature. With catalytic aluminium trihalide this remains the outcome below 40C.\n\n\n",
"3"
],
[
"\nAccording to this [forum](http://www.chemicalforums.com/index.php?topic=29556.0):\n\n\n\n> \n> To benzene, (add) $\\ce{Br2}$ and $\\ce{HOAc}$ to get bromobenzene.\n> Then (add) $\\ce{ZnBr2}$ to get p-bromobenzene as the major isomer. Or\n> from aniline, NBS and $\\ce{HOAc}$ to get p-bromoaniline. Then\n> diazotonization of aniline with $\\ce{NaNO2}$ and subtitution with\n> $\\ce{NaBr}$ to get p-bromobenzene.\n> \n> \n> \n\n\nAs the source is a forum, I found it highly unreliable but then I found [a page](https://erowid.org/archive/rhodium/chemistry/bromobenzene.html) which shows the process of synthesis of bromobenzene. It says that the para isomer is formed as a minor by-product and there is also a note on the isolation and purification of the para isomer:\n\n\n\n> \n> .....Now distill the crude bromobenzene slowly, rejecting the fraction\n> boiling up to 150°C, and collecting that of bp 150-160°C. Yield,\n> 28-29g (about 19ml). **A small quantity of crude p-dibromobenzene\n> remains in the flask.**\n> \n> \n> [....]\n> \n> \n> The p-dibromobenzene formed as a by-product in the above reaction\n> usually solidifies when the undistilled residue obtained in the first\n> distillation is chilled. It may be isolated by adding about 10ml of\n> methylated spirit and some activated charcoal to the flask, boiling\n> for a few minutes, and filtering hot. On cooling the filtrate in\n> ice-water, crystals of p-dibromobenzene, mp 89°C separate:\n> recrystallise a second time if necessary to obtain colourless\n> crystals.\n> \n> \n> \n\n\nThere is also [a patent](https://www.google.com/patents/US3062899)(@Waylander) which says that p-bromobenzene can be prepared by using bromine and aluminum trihalide as catalyst.\n\n\n\n> \n> In general, the bromination of benzene with two moles of bromine per\n> mole of benzene at ambient temperature produces predominantly\n> p-dibromobenzene. If the bromination is conducted in the presence of\n> an aluminun trihalide catalyst at low temperature, e.g., below about\n> 40 C., the para-dibromo isomer is still the principal product.\n> \n> \n> \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89736/how-to-find-the-molar-mass-of-an-iron-compound-based-on-its-mass-and-mole-fracti
|
How to find the molar mass of an iron compound based on its mass and mole fraction in the compound?
|
I'm trying to solve this question:
>
> We have a molecule composed of $3$ iron atoms, and $4$ atoms of another element. We are given the following information: it has $\pu{2.36 g}$ of iron for $\pu{3.26 g}$ of molecule.
>
>
>
I want to find the molar mass of the compound, I have tried so far:
$$ m = \pu{3.26 g} = \pu{0.00326 kg}$$
Since it has $3$ atoms of $\ce{Fe}$ and $4$ atoms of an unknown substance, therefore:
$$3 + 4 = 7~\text{atoms},\\
\pu{1 mol} = \pu{6.022\* 10^23 atoms}\\
\frac{7}{\pu{6.022\* 10^23}} = \pu{1.16 \* 10^-23}$$
As we know: $M = m / n,$ I tried to divide $0.00326$ by $\pu{1.16 \* 10^-23}$ and I obtained $\pu{2.79429 \* 10^19}$, but the correct answer is $\pu{231.43 g/mol}$.
What I have done wrong?
| 1 |
[
[
"\nLet $\\ce Z$ denote the unknown element.\nThe total amount of iron atom in the compound is,\n\n\n$$n(\\ce{Fe})=\\frac{\\pu{2.36g}}{\\pu{55.845 g\\cdot mol^{-1}}}=\\pu{0.0423 mol},$$\n\n\nThe molecule comprises of 3 iron atoms and 4 other atoms, so the amount of the unknown atom is,\n\n\n$$n(\\ce Z)=\\pu{0.0423 mol}\\times \\frac43=\\pu{0.0564 mol},$$\n\n\nThe molar mass of $\\ce Z$ is,\n\n\n$$M(\\ce Z)=\\frac{\\pu{3.26g}-\\pu{2.36g}}{\\pu{0.0564 mol}}=\\pu{15.97 g\\cdot mol^{-1}}$$\n\n\nSo $\\ce Z$ is $\\ce{O}$. The molar mass is\n$$M(\\ce{Fe3O4})=\\pu{55.845 g\\cdot mol^{-1}}\\times 3 + \\pu{15.97 g\\cdot mol^{-1}}\\times 4=\\pu{231.43 g}.$$\n\n\n",
"5"
],
[
"\nBefore I start explaining it, I would say that the data provided in the question and the correct answer corresponds to the compound $\\ce{Fe3O4}$, whose molar mass is $\\pu{231.43 g}$. But, I am not proceeding with this known fact rather I'm considering only the given data and the molar mass of iron, $\\pu{55.845 g}$. \n\n\nIn $\\pu{1 mol}$ of the compound, $\\ce{Fe3X4}$, there are $\\pu{3 mol}$ of $\\ce{Fe}$ and $\\pu{4 mol}$ of the unknown element $\\ce{X}$.\nLet molar mass of the compound be $M$ grams.\nThere is $3\\times\\pu{55.845 g}$ of $\\ce{Fe}$ in $M$ grams of the compound.\n\n\nGiven, in $\\pu{3.26 g}$ of the compound, $\\pu{2.36 g}$ of $\\ce{Fe}$ is present.\nTherefore, $\\pu{1 g}$ of $\\ce{Fe}$ is present in $\\pu{(3.26 / 2.36) g}$ of the compound.\nTherefore, $3\\times\\pu{55.845 g}$ of $\\ce{Fe}$ is present in\n$(3.26 / 2.36) \\times 3 \\times \\pu{55.845 g}$ of the compound.\n\n\nTherefore, $M = (3.26 / 2.36) \\times 3 \\times 55.845 = 231.43$, the molar mass is $\\pu{231.43g/mol}$.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89728/best-settings-for-electrolysis-of-water
|
Best Settings for Electrolysis of water
|
I want to electrolyze water as a personal project, but I have found conflicting power recommendations on google. Some sources recommended a higher voltage, claiming that higher amps will just destroy the electrodes, while others insist a higher amperage is required, and extra voltage will be wasted heating up the water. Unable to find one absolute result, I'm turning here- assuming not a crazy amount of amps or volts, what is a good ratio to get the best results? (an explanation of the answer would be appreciated.)
| 2 |
[
[
"\nIdeally, it takes 1.23 volts to split water. Less voltage than this will give essentially no current, no electrolysis. If you want more H2 + O2, you raise the potential. (Of course, you use an electrolyte like H2SO4 or NaOH with electrodes that will not corrode. Platinum is OK, also gold. Oh, yeah, also stainless steel for ordinary people.) If the surface becomes covered with bubbles of gas, the effective area of the electrode is reduced, so the current will drop. So you raise the voltage, and bubbles come pouring off. Then the electrolysis uses more energy (volts x amps). 1 amp for 96,500 seconds (27 hours) will give you 1 gram of hydrogen and 8 grams of oxygen. The product is dependent on the current, and the current is dependent on how high you raise the voltage.\n\n\nIt's not as slow as watching grass grow, but even at furious bubbling, it's surprising how slowly the volumes build up.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89726/acetic-acid-odor-while-testing-co2-sensor-can-it-be-easily-removed-is-there-a
|
Acetic acid odor while testing CO2 sensor; can it be easily removed, is there a better way?
|
A friend asked me to help test a hobby-quality $\ce{CO2}$ sensor [MG-811](http://datasheet.octopart.com/27929-Parallax-datasheet-13538405.pdf) (updated link). It uses a solid electrolyte and should provide a sensitivity of something like -40 mV per decade. I though I'd just use a small amount of acetic acid + baking soda to make roughly 4,000 ppm in a large container, and compare to "fresh air" which should be roughly 400 ppm, or at least within the [350 to 600 ppm range](https://earthscience.stackexchange.com/q/10145/6031), as a very rough initial test.
But then I noticed/remembered (duh!) that the acid smells quite strong. Are these simply molecules of $\ce{CH3COOH}$ that are evaporating, or are they bringing along molecules of water in some kind of cluster?
I realized that they could potentially contaminate the sensor, so I'd like to know what I'm up against in terms of removing them from the gas before immersing the sensor.
I've tried to look up the vapor pressure of acetic acid, and it seems to be quite large, about 10 Torr at 20C. See for example [here](https://en.wikipedia.org/wiki/Acetic_acid_(data_page)) and [here](http://ddbonline.ddbst.de/AntoineCalculation/AntoineCalculationCGI.exe?component=Acetic+acid). So I'm thinking of using some kind of ad-hoc cold-trap to try to remove it. Might there be another way to do this without using special laboratory equipment? Would passing the gas over dry $\ce{NaHCO3}$ or a solution of it help to "getter" the remaining acetic acid vapor?
**Or**, is there a *better way* to produce a controlled amount of $\ce{CO2}$ to begin with?
---
[](https://i.stack.imgur.com/KN1gS.png)
| 3 |
[
[
"\nYou could use a less volatile acid, like oxalic. But I think the reaction with NaHCO3 probably gives off a mist as the reaction bubbles, and that might be why the odor is so strong.\n\n\nHow about using seltzer water (or Perrier? or Diet Coke!) as the source of CO2? I estimate that the pressure of CO2 is about 45 psi in the unopened bottle, and if you attach a balloon to a freshly opened bottle and let it outgas slowly, the balloon will fill up and be essentially pure CO2. If a latex balloon is unsuitable or too hard to handle, a mylar balloon ($1 at the Dollar Store) will work just as well. The gas will be saturated with water vapor. \n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89724/colorimetry-at-home
|
Colorimetry at home?
|
Is there a reasonably priced option for colorimetry at home? Is this even a good option to accomplish my goal?
I'd like to roughly determine the concentration of a chemical in an ethanol extract from a natural product. It has a definite color. I would be able to buy small quantities of the purified product from a chemical supplier. Based on my recollections of college chemistry, I should be able to make a solution of known concentration and compare them with a colorimeter to determine how much I've extracted. Would this be reasonable, or should I be looking at another way of doing this?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/89720/rule-of-thumb-of-how-much-vacuum-does-lab-glass-stand
|
Rule of thumb of how much vacuum does lab glass stand?
|
Sometimes I have to apply vacuum to all kind of different borosilicate glass equipment like:
* round bottom flasks
* liebig condensers
* fractional distillation apparatus
* spiral condensers
* chromatography column
* and so on...
While I know that a thick walled round bottom flask will pretty much stand any vacuum my vacuum pump can produce, I am not so sure for something as fragile as a spiral condenser.
Is there a kind of rule of thumb how much mBar thin walled glass equipment like spiral condensers or chromatography columns can stand without risk of implosion?
| 1 |
[
[
"\nThe basic rule is that the stress on an evacuated glass vessel is pretty much the same at 0.01 mbar (oil pump), 1 mbar (membrane) or 25 mbar (water jet) absolute internal pressure. 0.99999, 0.999 or 0.975 bar $\\Delta p$ to the outside make very little difference if the thing goes *pop*.\n\n\nAnd basically everything made out of borosilicate *and* with no flat surface is safe. Condensors definitely, they're not delicate at all. It's basic procedure to make them oxygen-free by evacuating.\n\n\nBe careful with chromatography columns and flasks in the > 1 liter range. They're principally safe to be evacuated, too, but then are also a *bomb*, armed and ready to drop. If it goes off, you *better not* stand next to it. The danger from evacuated vessels is more or less proportional to their volume.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89715/what-acid-can-dissolve-thorium-dioxide
|
What acid can dissolve thorium dioxide?
|
In one of my projects (which is about alpha-voltaic cells, don't worry, I know that thorium isn't a strong alpha source, it's only for measurements) I would need to get water soluble thorium compound from a small sample of thorium dioxide for purification with selective reactions and further measurements. Would any mineral acid work? I found some sources and publications on the internet by google searching but the biggest amount of information I could get is that
>
> "...Several hot acids can solve it..."
>
>
>
Which is not nearly as accurate as I would need it because I don't want to spend the budget of the project on expensive acids which doesn't even work instead of better equipments. (Their price is from their necessary purity)
| 8 |
[
[
"\nThorium dioxide can be dissolved in the mixture of hydrofluoric acid and hydrochloric acid, as described by Kluger and Lieser in 1981.1\n\n\nProceed with caution since hot condensed acid are known to be highly corrosive and hydrofluoric acid is also [highly toxic and penetrating](https://en.wikipedia.org/wiki/Hydrofluoric_acid#Health_and_safety).\n\n\n\n\n---\n\n\n1. Kluge, E.; Lieser, K. H. Dissolution of ThO2 in HCl/HF-mixtures. *Radiochim. Acta* **1981,** *29* (2–3), 167–168. [DOI: 10.1524/ract.1981.29.23.167](https://doi.org/10.1524/ract.1981.29.23.167).\n\n\n",
"11"
],
[
"\n$\\color{red}{WARNING:DO \\ NOT \\ ATTEMPT \\ TO \\ REPEAT \\ THE \\\\ FOLLOWING \\ EXPERIMENTS \\ WITHOUT \\ EXPERIENCE \\\\\n IN \\ WORKING \\ WITH \\ HF \\ OR \\ WITHOUT \\ THE \\ PROPER \\\\\nWORK \\ ENVIRONMENT \\ AND \\ PERMISSIONS\\ !}$ \n \n In the procedure of refining used nuclear fuel pellets which are typically consisting of $(Th,U)O\\_{2}$ with 1-4% $UO\\_{2}$ content The most common acid mixtures which are used are $HCl/HF,HNO\\_{3}/HF,H\\_{2}SO\\_{4}/HF.$ If the Thorium oxide was previously heated above $1000°C$ It makes the process much more slower. The least dangerous way to obtain a water soluble thorium compound is to dissolve the Thorium oxide in a solution of $8-13 \\ mol\\cdot l^{-1} \\ \\ HNO\\_{3}$ and $0.02-0.05 \\ mol\\cdot l^{-1} \\ \\ F^{-}$ at temperatures of $50-70°C$ (The fluoride ion concentration can be achieved by adding any fluoride salts which will produce it or by adding hydrofluoric acid to it). This method does not needs the acids to be at the temperature of boiling.The produced Thorium nitrate is much more soluble in water than it's halides so this is a good procedure from this side too.\n\n\n\n\n---\n\n\nReference: [DISSOLUTION OF $(Th,U)O\\_{2}$ IN NITRIC ACID - HYDROFLUORIC ACID SOLUTIONS](https://www.ipen.br/biblioteca/rel/R39145.pdf)\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89713/how-can-i-interpolate-between-two-densities-at-different-temperatures
|
How can I interpolate between two densities at different temperatures?
|
I'm trying to find the temperature of methanol ($\ce{CH3OH}$) when its density is equal to $\pu{780 kg m-3}$.
I know that when $T = \pu{30^\circ C}$ the density is equal to $\pu{783 kg m-3}$, and when $T = \pu{40^\circ C}$ the density is equal to $\pu{774 kg m-3}$.
Is there any formula or a trick to find the temperature corresponding to an intermediate density? Likewise is it possible to find the density at a temperature intermediate between the two?
| 2 |
[
[
"\nYou could use tabulated literature data and then interpolate between the data points. For example, the following plot was created using data from NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9. It shows the liquid saturation line as well as the vapour saturation line.\n\n\n[](https://i.stack.imgur.com/T5Nzz.png)\n\n\nIn this case, interpolation for a liquid density of $780.00\\ \\mathrm{kg/m^3}$ yields a temperature of $31.643\\ \\mathrm{^\\circ C}$.\n\n\nBy way of comparison, the calculated liquid density is $781.55\\ \\mathrm{kg/m^3}$ for a temperature of $30.000\\ \\mathrm{^\\circ C}$, and $772.10\\ \\mathrm{kg/m^3}$ for a temperature of $40.000\\ \\mathrm{^\\circ C}$.\n\n\n",
"7"
]
] |
https://chemistry.stackexchange.com/questions/89711/explaining-stereochemistry-in-a-sigmatropic-1-3-alkyl-migration
|
Explaining stereochemistry in a sigmatropic 1,3-alkyl migration
|
I have quite a bit of difficulty in explaining sigmatropic rearrangement reactions. Consider the following reaction:
[![[1,3]-sigmatropic rearrangement reaction](https://i.stack.imgur.com/9hw5A.png)](https://i.stack.imgur.com/9hw5A.png)
How do I explain the formation of the shown product? When simply moving the electron pair in my head, I come to the other enantiomer, where the $\ce{-CH3}$ group shows in our direction.
I've repeated all my notes on sigmatropic rearrangements but unfortunately, I don't get it. I suppose that I should treat this like an alkyl migration. The latter should happen over a suprafacial intermediate state as we have an uneven number of electron pairs (1) and we have thermal conditions. This results in a suprafacial reaction. Knowing this, I went on and added the HOMO and LUMO orbitals and postulated intermediate steps. Note: Yes, I am aware, that pericyclic reactions happen in a concerted fashion, but somehow I need to see, why the groups on the bridging carbon have this conformation.
[](https://i.stack.imgur.com/mTVKH.png)
So, let's go on to my question...
>
> Is my proposed explanation okay? If not, please provide some information. I really have trouble with this kind of reactions.
>
>
>
| 8 |
[
[
"\nProblem #1 is that if you interchange the methyl and hydrogen groups, you get diastereomers, not enantiomers. But this is a relatively minor point. I assume your question is about the diastereoselectivity, not enantioselectivity (as ron rightly pointed out there is no enantioselectivity in this reaction).\n\n\n[](https://i.stack.imgur.com/P0WIz.png)\n\n\n\n\n---\n\n\nThe analysis you have drawn is somewhat similar to a Fukui frontier molecular orbital analysis. In that picture, for a reaction to be thermally allowed, you need to check for constructive bonding overlap between the HOMO and the LUMO, within geometrical reason. This is achieved by connecting lobes of \"like\" shading: i.e. shaded lobe with shaded lobe, and unshaded with unshaded.\n\n\nYou don't necessarily have to draw it stepwise (although I gather it's your own way of doing it). You can simply draw both interactions in the starting material, like this. The bottom interaction is not an issue. The top interaction, however, is an issue.\n\n\n[](https://i.stack.imgur.com/Nyryp.jpg)\n\n\nProblem #2 is that you tried to connect the large unshaded lobe of the σ\\* orbital with the white lobe of the π orbital. That is not geometrically reasonable: if you connect these two lobes, then you need to somehow end up with the new σ bond being formed underneath the ring. In other words, you would have to have an antarafacial migration of the alkyl group.\n\n\nProblem #3 is that you have concluded that the reaction must be suprafacial based on electron counting. Note that in this context, the *reaction* being suprafacial means that the migrating alkyl group stays on the same face of the ring. In fact, the reaction must occur in a suprafacial manner simply because of geometrical constraints: there's simply no way that the alkyl group is going to migrate to the opposite side. Technically it could, but you would have to break so many bonds in the transition state that it would not be possible.\n\n\n[Note that this has a different meaning from the suprafacial or antarafacial *components* in a Woodward–Hoffmann analysis.]\n\n\nSo what is the solution? Well, because the reaction is suprafacial, you need to use the shaded lobe of the π HOMO, which is on the top face of the ring – the same face as the migrating alkyl group. So then for constructive bonding you need to also use the shaded lobe of the σ\\* LUMO.\n\n\n[](https://i.stack.imgur.com/eWYUW.jpg)\n\n\nNow one might argue that just like how the unshaded lobe of the π HOMO was geometrically inaccessible, so should the shaded lobe of the σ\\* LUMO. But we already know that the C–C σ bond is going to break in the course of the reaction, so you're going to go through a pathway whereby the bond behind it (with an arrow pointing towards it) can rotate such that the shaded lobe of the σ\\* can overlap with the shaded lobe of the π\\* orbital. In fact, that's not too different from what you drew. The transition state is pretty similar to the intermediate in your stepwise reaction – it just has a few more dotted lines.\n\n\n[](https://i.stack.imgur.com/ZaBJk.jpg)\n\n\nIf you now \"join\" the shaded lobes then you can see how the correct stereochemistry arises.\n\n\n\n\n---\n\n\nIn general my preferred way of analysing pericyclic reactions is to use Woodward and Hoffmann's original rules. Here, there is no shading of lobes: you connect the lobes within geometrical reason, then count the number of $4q+2$ suprafacial *components*, and the number of $4r$ antarafacial *components*. If the total is odd, the process is thermally allowed; if it is even, the process is thermally forbidden.\n\n\nI am not going to explain this approach in detail - it is too long for here and it is adequately covered in textbooks which you should have access to - but I will upload the two diagrams I drew:\n\n\n[](https://i.stack.imgur.com/JS0Sd.jpg)\n\n\n[](https://i.stack.imgur.com/W3mxE.jpg)\n\n\n",
"7"
]
] |
https://chemistry.stackexchange.com/questions/89708/how-does-the-hexaaquacopper-ii-form
|
Why do copper (II) complexes contain so many valence electrons?
|
Consider tetraaminecopper (II) ion. $\ce{[Cu(NH3)4]^2+}$ or $\ce{[Cu(NH3)4(H2O)2]^2+}$.
The copper(II) ion has the electron configuration $\mathrm{[Ar]\ 3d^9}$. How do the 4 electron pairs from the ammonia ligands form the coordination bond with the copper ion, if the 3d subshell is almost entirely occupied?
| 9 |
[
[
"\nI'll address $\\ce{[Cu(NH3)4(H2O)2]++}$, but the reason is similar for $\\ce{[Cu(NH3)4]++}$\n\n\nThe 9 d-electrons in the complex are in non-bonding and antibonding orbitals. \n\n\nSee pages 6-9 of [Ligand Field and Molecular Orbital Theory](http://www.chem.mun.ca/homes/cmkhome/Chemistry%203211%20Pt3.pdf). \n\n\n\n> \n> the ligand σ orbitals are derived from AOs with energies much lower than those of the metal d-orbitals. Thus, the six bonding MOs of the complex are mostly ligand-orbital in character. These six bonding orbitals can accommodate the 12 electrons provided by the six ligand lone pairs. Therefore, the electrons provided by the ligands are largely confined to the ligands in the complex.\n> \n> \n> \n\n\nSo, to a large degree, the 12 electrons from the ligands are in bonding orbitals, but the 9 d-electrons are in nonbonding or antibonding orbitals. \n\n\n\n> \n> The d electrons of the metal centre...occupy the lower energy, non-bonding metal $t\\_{2g}$ set and the higher energy, antibonding $e\\_g$ set. Therefore, the metal based d electrons tend to reside largely on the metal atom. In summary, the frontier orbitals of the complex are the non-bonding, entirely metal based $t\\_{2g}$ orbitals and the antibonding $e\\_g$ orbitals, which are mainly metal in character anyway. \n> \n> \n> \n\n\n",
"8"
],
[
"\nThis is best explained using a molecular orbital scheme of a typical octahedral complex. Now, this is assuming $O\\_\\mathrm{h}$ symmetry which is not correct for $\\ce{[Cu(NH3)4(H2O)2]}$; the actual symmetry of the complex is $D\\_\\mathrm{4h}$. However, it is close enough for discussion of why the complex can form, and we can even use a gedankenexperiment to find out the MO scheme with the correct $D\\_\\mathrm{4h}$ symmetry. The scheme also ignores π interactions which are present but not important in ammine complexes.\n\n\n[](https://i.stack.imgur.com/GKvJQ.png) \n\n**Figure 1:** Octahedral $\\ce{[ML6]}$ complex with no π interactions. Metal orbitals on the left-hand side; ligand orbitals on the right-hand side. Molecular orbitals in the centre. Image copied from [this answer](https://chemistry.stackexchange.com/a/38339/7475) and originally taken from [Professor Klüfers’ internet scriptum to his coordination chemistry course](http://www.cup.lmu.de/ac/kluefers/homepage/L_kc.html).\n\n\nThe first thing we notice is that the metal’s $\\mathrm{t\\_{2g}}$ orbitals (that would be $\\mathrm{d}\\_{xy}, \\mathrm{d}\\_{xz}$ and $\\mathrm{d}\\_{yz}$) do not take part in metal-ligand interactions; they are nonbonding. This explains why complexes with populated $\\mathrm{t\\_{2g}}$ orbitals are often stable and favourable. Next, we see that the six σ-symmetric ligand group orbitals transform as $\\mathrm{a\\_{1g} + e\\_g + t\\_{1u}}$ which is exactly the transformation of the metal’s remaining d-orbitals, its 4s and three 4p-orbitals. Thus, these are all destabilised while the ligands’ orbitals are lowered. This is the net gain of ligand bonding.\n\n\nStill, the metal’s $\\mathrm{e\\_g}$ orbitals are still antibonding, i.e. $\\mathrm{e\\_g^\\*}$. Electrons in antibonding orbitals are generally unfavourable. But if we take a $\\mathrm{d^9}$ case and calculate the bond order:\n\n\n$$\\begin{multline}\\mathrm{B.O.} = \\frac{n(\\text{bonding electrons}) - n(\\text{antibonding electrons})}{2 \\times n(\\text{bonds})}=\\\\\n\\frac{\\mathrm{(a\\_{1g} + e\\_g + t\\_{1u}) - e\\_g^\\*}}{2\\times 6}\n= \\frac{12 - 3}{2\\times 6} = \\frac{9}{12} = \\frac34\n\\end{multline}$$\n\n\nA bond order of $\\frac34$ is larger than $0$ and thus favourable. Thus, having the bond is better than not having it. The MO explanation is that the destabilisation of the populated orbitals is less than the stabilisation of populated orbitals: most of the stabilisation is offset by destabilising unpopulated orbitals, not populated ones.\n\n\nAnd what changes if we remove symmetry by going from octahedral $O\\_\\mathrm{h}$ to Jahn-Teller distorted $D\\_\\mathrm{4h}$? Well, what happens is that ligands are removed on the $z$-axis, so simply spoken anything containing $z$ contribution will be stabilised. That includes one of the two $\\mathrm{e\\_g^\\*}$ orbitals as well as two of the three $\\mathrm{t\\_{2g}}$ orbitals. Similarly, the ligand orbitals that used to be $\\mathrm{t\\_{1u}}$ are split up into a set of two and a single one, depending on their principal direction. These will no longer be degenerate. The same applies to the metal’s p-orbitals, which are also $\\mathrm{t\\_{1u}^\\*}$. However, the basic picture remains the same; stabilised orbitals are more populated than destabilised ones for a net gain in the overall system.\n\n\n",
"8"
]
] |
https://chemistry.stackexchange.com/questions/89705/can-i-use-a-primary-amine-to-substitute-deprotect-a-tosylate-at-high-yields
|
Can I use a primary amine to substitute (deprotect) a tosylate at high yields?
|
I am trying to use tosyl chloride to allow for a easy substitution of the primary alcohol on glucose then use a BOC-diamine to substitute the tosylate to form a primary amine with a short alkyl chain. But I am having trouble finding any methods in the literature that use primary or secondary amines as the nucleophile for the substitution reaction. Most use relatively stronger nuc like sodium azide or sodium ethoxide.
Is a primary amine not a strong enough nuc? And if so what methods/reagents should I use to obtain a relatively high yield?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/89697/can-carbon-monoxide-from-a-floor-heater-in-a-poorly-ventilated-room-eventually-b
|
Can carbon monoxide from a floor-heater in a poorly ventilated room eventually be ignited by the fire and explode?
|
I have often wondered about carbon monoxide accidents caused by free-standing heaters burning in a poorly ventilated room causing fatalities. I am aware that CO is explosive between 12 and 75% volume air and wonder why it doesn't eventually ignite from the flames or glow present when the CO level reaches 12%? (This never seems to happen). Or does the decreasing oxygen content in the air due to the combustion cause the fire to die out before the critical level of 12% is reached?
| 4 |
[
[
"\nAnother possibility: the \"carbon monoxide\" formed by combustion is not pure enough CO to get to flammability (12% total CO) in air. The gas could be mostly carbon dioxide but still enough CO to poison you.\n\n\n",
"1"
],
[
"\nI know from experience that carbon monoxide/air has a very low fame velocity. I think that air/carbon monoxide mixtures would make a very poor fuel air explosive.\n\n\nI think however that by the time you get to the point of having sufficient carbon monoxide in a room to allow a flame front to start to travel from one side of the room to the other that unless you are wearing self containing breathing apparatus you will be dead of carbon monoxide poisoning.\n\n\nJon Custer is very right, if we consider methane as the fuel then we can write the equation.\n\n\n$\\ce{2CH4 + 3O2 -> 2CO + 4H2O}$\n\n\nThus to reach 11% carbon monoxide in a sealed room with a methane flame burning under very special and rich conditions to only form water and carbon monoxide. I think that all the oxygen in the room will be used up long before we reach 11% $\\ce{CO}$ in the air. I think that the highest $\\ce{CO}$ concentration we could get in the room would be 6.67% by volume. I think that as the oxygen level in the room will be so low that even with additional $\\ce{CO}$ added to the room air then the room gas/air mixture will be unable to explode.\n\n\nThe only time I imagine you could get a carbon monoxide/air mixture to \"explode\" would be if a carbon monoxide cylinder was to leak and then an ignition source was to be added.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89690/lab-methods-for-determination-of-phosphoric-acid-concentration
|
Lab Methods for Determination of Phosphoric Acid Concentration
|
Dear stack exchange,
For my next lab session we have been tasked to come up with two different methods for determining the concentration of an unknown sample of phosphoric acid.
>
> Design at least two different experimental methods to determine the
> concentration of the phosphoric acid solution using the materials and
> equipment provided. (NB: two different volumetric titrations are not two
> different methods.)
>
>
>
For reference, the "materials and equipment provided" are (along with the standard burettes, pipettes, etc.):
>
> A sample of the phosphoric acid solution
>
> A selection of pH indicators:
>
> - Bromocresol green (range 3.8-5.4)
>
> - Phenolphthalein (range 8.0-10.0)
>
> - Thymolphthalein (range 9.3-10.5)
>
> A standardised NaOH (approximately 0.1 M) solution
>
> A pH meter
>
>
>
One obvious method is an acid-base titration, which is fine, but I have no idea what to do as a second method since we cannot do two different volumetric titrations - perhaps there is some finnicky definition of "volumetric" which means there is another type of titration we can do? It seems as though it has to be something along those lines based on the provided equipment.
Many thanks!
| -1 |
[
[
"\nSpecific gravity would be a useful technique; perhaps not as accurate as titration. Weigh 50 mL of the unknown and check vs tables in the CRC Handbook.\n\n\nDo a reaction: add weighed amounts of NaHCO3 (slowly, carefully) until the foaming stops. I think this will just neutralize 2 of the 3 acidic protons on H3PO4. Knowing how much NaHCO3 reacts, you can determine how much H3PO4 there was.\n\n\nEvaporation to \"dryness\": in a porcelain dish, weigh a decent amount of the unknown and heat at about 110C for an hour or so. The liquid at the end will be highly concentrated, but perhaps not 100% H3PO4. Check with a known concentration of H3PO4.\n\n\nIf you can get the apparatus, you could check conductivity. You could even do it yourself (roughly) with a multimeter and stainless steel wire (of course, platinum wire is preferable).\n\n\nYou could try to determine viscosity; check vs known standards. Very rough for analysis.\n\n\nYou could offer the lab director $10 for the correct answer.\n\n\n",
"0"
],
[
"\nYou can add a single dose of NaOH to a sample of the acid and measure the pH with the pH meter. You can easily calculate the charge on a mole of phosphate at a given pH and as the moles of NaOH you added is equal to the moles of phosphate multiplied by the charge change per mole you can get the number of moles of phosphate.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89689/for-two-compounds-that-are-symmetrical-why-does-the-more-compact-one-have-a-hig
|
For two compounds that are symmetrical, why does the more compact one have a higher melting point?
|
In this post, I refer to two molecules B (Di-tert-butyl ether) and C (Dibutyl ether).
```
Wikipedia: Di-tert-butyl ether (B)
b.p. 107.2 C
m.p. - 61 C
```
```
ChemSpider: Dibutyl ether (C)
b.p. 142 C
m.p. - 96 C
```
I understand that Compound C has a higher boiling point than compound B, because of its greater surface area and therefore stronger intermolecular VDWs.
I don't understand, however, why Compound B has a higher melting point than compound C. A lecturer told me that this is due to compound B being more compact and symmetrical. I think that I understand the concept that some structures pack into a crystalline lattice better than others, such that the structure of the packing influences the strength of intermolecular forces within the solid, therefore affecting the melting point.
However, it seems to me that both B and C above are equally symmetrical in at least one conformation. Beyond that, C has conformations of greater surface area than B, which should further strengthen its intermolecular bonds in a solid state. I would expect the melting point of C to be higher than B; yet this is not the case.
I feel like the answer may lie in the fact that C has many more conformers than B. Even so, I can't seem to draw a link anywhere. Shouldn't the solid formed take the most stable crystalline structure, much like how water tends to just one of many crystal structures?
| 4 |
[
[
"\nThe terms used by your teacher are quite in line with at least a qualitative explanation. \n\n\nI really suspect that there are various ways to express compactness and sort of \"sphericity\" of molecules. I am not a specialist so forgive me for being imprecise on this.\n\n\nHowever, if we forget a rigorous classification in symmetry group, a potato is more spherical symmetrical than a banana.\n\n\nIn such a sense the ramified isomer is definitely more symmetric and \"round\" than its linear counterpart. \n\n\nThis is even more evident if all conformers are taken into account: in the linear molecule a chain end can even turn toward the center while the opposite being extended.\n\n\nNow this does not necessarily explain the observed behaviour as for the more spherical molecule can pack easily but at expense of VdW forces that depend on molecular surface.\n\n\nHowever, it relates to the fact that the linear isomer has a number of conformers much higher than the ramified one (rotation about 4 bonds versus 1 for each side of the ether)\\*\\*\\* and results in the first having a bigger conformational entropic term.\n\n\nAlthough the crystals will generally contain one isomer, the crystal formation take place at expense of the intramolecular forces. As such, *when keeping entalpy terms comparable*, bigger is the entropy term then less stronger is the binding.\n\n\n$\\Delta G = \\Delta H - T\\Delta S$,\n\n\nFor comparable enthalpies and very different conformational entropy terms, than the melting occurs at lower T for the entropy rich molecule, i.e. that one for wich more conformations become accessible. \n\n\nThis obviously is not the case for the liquid to gas transition, as the conformational S term is similar in both phases and negligible faced to the gas entropic content. \n\n\n\\*\\*\\*Conformers due to the rotation around the bonds in the terbutyl groups do not really alter the shape of the molecules (connection with the compactness mentioned above). More rigorously at each side we should have (calling x the number of conformers around a bond) that the number of conformations is circa $x^4$ and about $3x^2$ for the linear and the branched molecules, respectively.\n\n\nAdditional note: conformational entropic terms are crucial to many phenomena in polymers and proteins (solubility, crystallization, protein folding/unfolding, etc).\nA treatment of this requires statistics that I have forgotten. So take my numerical estimated about conformers with a grain of salt. I am satisfied if I did convey the message. \n\n\nAdd. Note 2. Another useful example is, for instance, a stiff and insoluble chain such as polythiophene.\n\n\nIt is insoluble as for electron delocalisation makes it flat and rigid. Once side substituted with alkyl chains, the entropic term brought in by the flexible chain is the main responsible for the acquired solubility. \n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89687/stoichiometry-of-incomplete-combustion
|
Stoichiometry of incomplete combustion
|
>
> Methane is burned with air in a continuous steady-state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide and water.
>
>
> The feed to the reactor contains: $\pu{7.8 mole \% }\ce{CH4}$, $\pu{19.4 mole \% } \ce{O2}$, $\pu{72.8 mole \% } \ce{N2}$. The conversion of methane is $90\%$ and the gas leaving the reactor contains a ratio of $\pu{8 mol}\,\ce{CO2}/\pu{1 mol}\,\ce{CO}$.
>
>
>
My attempt:
* $\ce{2 CH4 + 3 O2 -> 2 CO + 4 H2O}$
* $\ce{CH4 + 2 O2 —> CO2 + 2 H2O}$
Both of these reactions have $90\%$ conversion.
From the ratio, can I say that approximately $88.88\%$ of the methane in the feed produces $\ce{CO2}$ while $11.11\%$ produces $\ce{CO}$?
| 0 |
[
[
"\nComplicated little rascal!\nWhat we know: \n1) The N2 is inert. \n2) The oxygen is in excess. \n3) 90% of the 7.8 mole % CH4 reacts; this is 7.02 mole % CH4. \n4) This gives 8 moles CO2 per mole of CO, or 88.89% CO2 plus 11.11% CO. \n\n\nThe resultant output contains CO2: 7.02 mole % of the original CH4 x 88.89% = 6.24 mole % CO2.\n\n\nThe resultant output contains CO: 7.02 mole % of the original CH4 x 11.11% = 0.78 mole % CO.\n\n\nThe resultant output also contains 10% unreacted CH4. The 90% conversion refers to the starting material, not to the products. (Yield % would refer to one product.)\n\n\nSo the stoichiometry is 80% of the methane goes to CO2, 10% goes to CO, and 10% is unreacted.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89682/reversed-trends-in-stability-of-transition-metal-ions-with-respect-to-oxidation
|
Reversed trends in stability of transition metal ions with respect to oxidation state
|
Why is titanium more stable with a lower oxidation state, but vanadium more stable with a larger oxidation state?
More specifically, why is $\ce{Ti^{+}}$ more stable than $\ce{Ti^{3+}}$, but $\ce{V^{2+}}$ is less stable than $\ce{V^{4+}}$ in aqueous solution?
Both $\ce{Ti^{+}}$ ($\mathrm{3d^{2}4s^{1}}$) and $\ce{V^{2+}}$ ($\mathrm{3d^{3}4s^{0}}$) have a half filled $t^3\_{\mathrm{2g}}$ orbital configuration which imparts stability during the formation of aqua complexes, assuming the $\mathrm{4s}$ electron of $\ce{Ti^{+}}$ can go to the $\mathrm{3d}$ sub-shell.
Is this discrepancy related to the charge/size ratio, which is larger in $\ce{Ti^{3+}}$ than in $\ce{Ti^+}$?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/89676/why-does-ammonium-chloride-form-white-crystals
|
Why does ammonium chloride form white crystals? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed 5 years ago.
[Improve this question](/posts/89676/edit)
Why does ammonium chloride ($\ce{NH4Cl}$) form white crystals on top of a test tube when heated?
| -2 |
[
[
"\nI suppose you are not asking why crystalline $\\ce{NH4Cl}$ is white.\n\n\nAmmonium chloride decomposes upon heat\n$$\\ce{NH4Cl(s) ->[\\Delta] NH3(g) + HCl(g)}$$\n\n\nThe top of the test tube is cold, and they recombine to generate the original salt.\n$$\\ce{NH3(g) + HCl(g) -> NH4Cl(s)}$$\n\n\nThis is the white crystal you see.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89673/when-two-negatively-charged-molecules-are-hydrogen-bonding-with-each-other-why
|
When two negatively charged molecules are hydrogen bonding with each other, why do the charges increase?
|
### Context
[EDIT: As some commentators have suggested, I should conclude that the energy minimize feature is not intended for the purpose of Hydrogen-bonding, and primarily for refining the modelling of molecules.]
Using the Chem3D automatic Huckel charge calculator, I was able to calculate the charges of the atoms in the citrate and borate ion.
Individually, the borate ion's oxygen atoms have a small negative charge, i.e. -0.07. Individually, the citrate ion's oxygen atoms have a large negative charge, i.e. -0.56.
I use the energy minimize function to make the two hydrogen bond. I then do the charge calculation, and I found that some of the citrate's oxygens' charges increased to ~-0.67, while some of them decreased, and the borate ion's increased to ~-0.5. The hydrogen atoms do increase positive charge.
[citrate ion vs borate ion](https://i.stack.imgur.com/Dqvli.png)
----------------------------------------------------------------
I actually noticed another particular feature, the oxygen atom of the hydroxyl involved in both hbond donating and accepting has a positive charge now. The other charges remain as I've explained above. The highlighted yellow atom was just an oxygen atom that I clicked.
### Question
For my own explanation, I would suggest that the hydrogen bonding is the cause of the charges being induced and increased. Yet, due to the size of the molecules, only two or four atoms seem to be involved in the hydrogen bonding. Then, why should the *OTHER* atoms still increase their charge? The atoms not involved in hydrogen bonding between the two molecules?
Would it be a feature of the calculator, or would this be the scenario in real life?
I am interested in how different ions are compatible with borate in terms of hydrogen bonding and screening, based on their charges. This will aid my experiments.
| 2 |
[] |
https://chemistry.stackexchange.com/questions/89669/is-it-possible-to-make-synthetic-oil-without-fossil-fuel
|
Is it possible to make synthetic oil without fossil fuel? [duplicate]
|
**This question already has answers here**:
[How can we produce gasoline from electricity without fossil fuels?](/questions/39948/how-can-we-produce-gasoline-from-electricity-without-fossil-fuels)
(2 answers)
Closed 5 years ago.
If human ran out of crude oil and all other fossil fuel, are we still able to create synthetic oil? I've searched online about the raw material of synthetic oil, but I'm still not sure if none of the raw materials is from crude oil.
Also is there any synthetic oil with viscosity less than 10cSt?
| 0 |
[
[
"\nYes, it is being theorized and it's being done: \n\n\nExample A:\n<https://www.researchgate.net/publication/264810861_Engineering_cyanobacteria_for_production_of_a_cyclic_hydrocarbon_from_CO2_and_H2O>\n\n\nExample B:\n<https://www.sciencedirect.com/science/article/pii/S2211926416304982>\n\n\nExample B is more theoretical with economic considerations, and example A is more methodical. In these papers, we see cyanobacteria being used to make the aromatic hydrocarbon limonene with the limonene synthase gene derived from a plant. \n\n\nAs for can we make synthetic oil that is chemically proportional to fossil fuel, Probably not ... but by adjusting the ratios of aromatic hydrocarbons, we can measure some useful properties like the heat value of the fuel for which limonene has a heat value very close diesel fuel and the density is close as well as you can see if you read the aforementioned Life Cycle Analysis paper. If you want to explore the chemical composition of fossil fuel then you need to do some spectroscopy work.\n\n\nBut fossil fuel isn't perfect. There is residual incombustible material, and perhaps by adjusting the proportion of combustible aromatic hydrocarbons in fossil fuel maybe we can get more energy from fossil fuel. This is also a function of the funding that these research projects get and also industrial support which is also a function of money ... which is why you need high yield in these reactions, hence genetically engineered cyanobacteria. \n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89667/reactions-of-differents-halohydrins-with-tert-butoxide
|
Reactions of differents halohydrins with tert-butoxide
|
I have two compounds (1*R*,2*S*)-3,3-dimethyl-2-bromocyclohexanol and (1*R*,2*R*)-3,3-dimethyl-2-bromocyclohexanol, both of them are treated with $\ce{KO^tBu}$ (a strong but bulky base) to form two different isomers of $\ce{C8H14O}$ which is not an alcohol.
I though that (1*R*,2*R*)-3,3-dimethyl-2-bromocyclohexanol is going to form an epoxide due the configuration of the halogen and the hydroxyl (although I don't know if it is required an specific configuration) and (1*R*,2*S*)-3,3-dimethyl-2-bromocyclohexanol is going to have a E2 reaction due steric hindrance but that will end with an alcohol anyway.
| 0 |
[
[
"\nChair conformation **RR-1** (shown as its alkoxides) is in equilibrium with conformation **RR-2**, which is the unfavorable partner owing to the strong 1,3-diaxial Me/alkoxide interaction. It is this conformation that leads to the epoxide **1**. This route is possible if the rate of formation is much greater than the rate constant for **RR-1** --> aldehyde **2**. **RR-1** has concentration on its side and anti-periplanar alignment of bonds for ring contraction. I prefer ring contraction but this does not preclude epoxide formation. \n \n\nThe **RS** isomer has two chair conformations, **RS-1** and **RS-2**. Again, **RS-2** is an unfavorable conformation (*vide supra*) but has proper bond alignment for ring contraction. Conformation **RS-1** has the necessary bond alignment for E2 elimination to form cyclohexanone **4** via the enol **3**. \n \n\nA priori, it is not easy to predict with certainty what the product distribution will be. The epoxide can only arise from the **RR** isomer while aldehyde **2** can arise from either bromohydrin. Organic chemistry is an experimental science. \n\n\n[](https://i.stack.imgur.com/2Be3e.jpg)\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89655/relative-reactivities-of-alkyl-halides-in-an-e2-reaction
|
Relative reactivities of alkyl halides in an E2 reaction
|
What is the exact order of reactivity of alkyl halides in an E2 reaction?
Organic Chemistry by Paula Yurikanis Bruice says that the order is $3^\circ>2^\circ>1^\circ$. And I think this order is correct because the transition state of the tertiary alkyl halide will be better stabilized by more hyperconjugating structures as compared to the secondary and primary halides.
But my teacher made me write that the order is $3^\circ<2^\circ<1^\circ$ and I even saw the this order in a book.
So I'm kind of confused about the correct order in E2 mechanism.
Any help will be appreciated.
| 3 |
[
[
"\nI believe you might be mixing up E1 and E2 elimination reactions. In the first case the reaction first proceeds by formation of a carbocation from the alkyl halide and then a proton extractio. In this case, the order of reactivity is indeed $3^∘>2^∘>1^∘$ as such is the order of carbocation stability - in order the E1 reaction to proceed the carbocation must be formed.\n\n\nBut the situation is completely different in an E2 elimination. In this case the mechanism is the abstraction of the proton that is oriented *anti* to the halide with the simultaneous formation of the double bond and halide leaving the molecule as anion. That means that in this case the stability of carbocation plays no role and the significant factor are the sterical considerations. If the carbon atoms around the halide make it hard for the base to get to the *anti* proton then the reaction is less likely to proceed *via* E2 mechanism.\n\n\nBut bear in mind that usually both E1 and E2 elimination can take place and because of that tertiary halides would usually undergo elimination more readily than primary halides - just not *via* E2 mechanism.\n\n\nMore details on the E2 mechanism, with pictures: [Master Organic Chemistry](https://www.masterorganicchemistry.com/2012/09/27/the-e2-mechanism/)\n\n\n",
"2"
],
[
"\nIn E1 and E2 both are same reactivity order of alkyl halide because both of reaction 3^• alkyl halide give easily product because after elimination reaction on 3^•alkyle halide. alkene is attached more alpha hydrogen which is more stable.so reactivity order in E1and E2 both are same 3^•>2^•>1^and is also same in SN1 and but different in SN2\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89654/how-can-chlorine-have-10-electrons-around-it-in-chlorine-trifluoride
|
How can chlorine have 10 electrons around it in chlorine trifluoride?
|
I'm relatively new to chemistry, and while doing some exercises about $\ce{ClF\_3}$ I tried to draw where the electrons are. I checked Google and got this:
[](https://i.stack.imgur.com/xyoUJ.jpg)
How is this structure possible? From what I've learned from high school every element wants 8 valence electrons. From the figure, $\ce{F}$ has 8, but $\ce{Cl}$ has 10.
| 2 |
[
[
"\nInterhalogens consist of one $\\ce{X-F}$ pair whith a regular two electron bond, and pairs of flourine atoms taking over one free electron pair of X (chlorine, bromine, iodine).\n\n\nWhat forms are three-centre four-electron bonds (one pair of electrons from the chlorine, two single electrons from two fluorine atoms). That's why interhalogens always have an uneven number of fluorine atoms (except for the dimer $\\ce{Cl\\_2F\\_6}$).\n\n\n<https://en.wikipedia.org/wiki/Three-center_four-electron_bond>\n\n\nIf this is still octet rule is debatable, or at least not directly obvious (see the wp article). Let's just say that fluorine is a bad, bad guy who has things his way.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89652/assignment-of-the-ir-spectrum-of-2-4-pentanedione-acetylacetone
|
Assignment of the IR spectrum of 2,4-pentanedione (acetylacetone)
|
The IR spectrum is shown here:
[](https://i.stack.imgur.com/TisND.jpg)
My assignments so far:
* The 1728 and 1708 stretches correspond to $\ce{C=O}$ stretches in the keto form - the symmetric/asymmetric stretches respectively.
* The 3003 stretches are typical $\ce{C-H}$ stretches.
* The 1606 stretch corresponds to the $\ce{C=C}$ in the enol form. However, hidden under this peak there is also the $\ce{C=O}$ stretch in the enol form, which is lowered by conjugation to the $\ce{C=C}$ and the $\ce{O}$ of the $\ce{-OH}$ respectively.
I have two questions about the spectrum, however:
1. Why is there no $\ce{-OH}$ peak? Even if the enol is hydrogen bonded to itself, there should still be an $\ce{-OH}$ peak in the $\pu{3400 cm^{-1}}$ range?
2. Does the fact that the carbonyl peak in the enol form is hydrogen bonded change its shift? Assuming the only contribution to the lowering of the $\ce{C=O}$ stretching frequency is conjugation to the one $\ce{C=C}$ bond, this would lower the peak to ~$\pu{1655 cm^{-1}}$: Surely this would be seen as a peak in its own right? It seems to be shifted even lower to be fully hidden by the 1606 peak.
| 6 |
[
[
"\nOH str for beta diketones is tabulated from 3200 to 2400 1/cm. The shift is explained by resonance of the hydrogen bonded structure. \n\n\nIts absorption features is small or unseen as for\n\n\n* commonly the ketonic form is highly dominant; i.e. we won't expect to see OH str in the acetaldehyde spectrum, even\n* owing to the above mentioned resonance, the dipole change associated to symmetric OH str is null, and it is minimal for the AS one.\n\n\nThe symmetry of 2,4-pentanedione makes the last point even more important.\n\n\nConcerning the second part. The fact that the O of the carbonyl of the enoform is H bonded does further shift down the related CO str. Moreover, in addition to resonance with the C=C, note that again there will be resonance with the CO in beta as above, and this *further* shift the frequency to lower wn.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89646/d3d-point-group-matrices
|
D3d point group matrices
|
I am trying to construct all the matrices for the point group D3d operations. I find that the matrices of E, i, S6, inverse of S6, C3, and inverse of C3 are easy, but I do not know how to construct the matrices of the three C2s and the three dihedral reflections. When I construct these matrices, they do not look like those of usual reflections or rotations. I cannot find any resources about finding these more complicated matrices. Any help would be appreciated.
| 2 |
[
[
"\nOne of the usual setup for the coordination system would be that $x$ axis coincident with a two-fold axis. If we use this setup, we can easily write out the matrix for rotation around this 2-fold axis:\n$$\\begin{bmatrix}\n1&0&0\\\\\n0&-1&0\\\\\n0&0&-1\n\\end{bmatrix}.$$\n\n\nThen you can combine this with the transform matrix of the 3-fold axis to get the transform matrices of the other two 2-fold axes. That's just matrix multiplication.\n\n\nIn this setup, $yz$ is one of the reflection plane and the corresponding matrix is\n$$\\begin{bmatrix}\n-1&0&0\\\\\n0&1&0\\\\\n0&0&1\n\\end{bmatrix}.$$\n\n\nNote that this is just the combination of the previous matrix with inversion. You can apply the 3-fold axis rotation to it (once again, matrix multiplication) to obtain the matrix with regard to other two reflection planes.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89644/stability-of-carbanion
|
Stability of carbanion
|
I am confused about how one can determine the stability of carbanions using pKa-values. I know that the stability of carbanions can be determined by the inductive effect, hybridization of the charged-bearing atom and resonance, but when searching about how pKa can do so I don't find anything. The lower the pKa-value, the stronger the acid and hence, the weaker the conjugate base, does this have anything to do with the stability since weak bases are unreactive and thus more stable?
| 2 |
[
[
"\nAs commented by SteffX, less the $\\mathrm{p}K\\_\\mathrm{a}$, more stronger the acid. More stronger acid means that the conjugate base or the anion formed after donating the proton is more stable compared to the carbanion of an acid of a higher $\\mathrm{p}K\\_\\mathrm{a}$ value. So, if the anion formed is a carbanion, then lower the $\\mathrm{p}K\\_\\mathrm{a}$, the stable the carbanion is. \n\n\n",
"2"
],
[
"\nThe stability of every molecule is related to the spreading of charges: the more you can share a charge among atoms, the more stable the molecule is.\n\n\nFor a carbanion, every group WITHDRAWING electrons to the carbanion center will stabilize it (i.e. the negative charge will be shared by several atoms)\n\n\nFor a carbocation, every group GIVING electrons to the carbocation will stabilize it (i.e. the localized positive charge will be compensated for... at least partially)\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89637/niosh-7300-icp-oes-calculation
|
NIOSH 7300 - ICP OES calculation
|
need some help with NIOSH 7300 method <https://www.cdc.gov/niosh/docs/2003-154/pdfs/7300.pdf>
I have two 37mm cellulose filter paper, one is my filter blank while other is sample. The sample has 80L of natural gas passed through and I plan to test for Mg , final concentration in mg/kg.
Both are digested and top up to 10 ml with dilution acid.
ICP OES Result:
Filter Blank : Mg = 0.022 mg/L , Pb = 0.002 mg/L
Sample : Mg = 0.038 mg/L , Pb = 0.006 mg/L
---
(From Method)
Given the formula is
Solution concentrations for the sample, Cs(µg/mL),
Filter blank , Cb(µg/mL)
Volumes of sample, Vs(mL), and Filter blank , Vb(mL),
calculate the concentration,C (mg/m3), of each element in the air volume sampled, V (L):
C= (CsVs - CbVb)/V = mg/m3
NOTE: µg/L / mg/m3
---
My calculation,
[(0.038 mg/L x 0.01 L) - (0.022 mg/L X 0.01 L)]/ 80L
= 0.00016 mg/ 80L
Edit:
According to the composition result, density is = 17 gmol-1 ,
and I calculate this under STP.
= 0.00016 mg / [(80/22.4)mol \* (0.017kg/mol)]
= 0.00263 mg/Kg
is this correct?
| 2 |
[] |
https://chemistry.stackexchange.com/questions/89634/which-is-more-stable-out-of-the-given-carbocation
|
Which is more stable out of the given carbocation?
|
>
> Which is more stable?
>
>
> 1. $\ce{^+C(CH3)3}$
> 2. $\ce{^+C(CD3)3}$
>
>
>
I have read in Solomon's and Fryhle that the +I effect of D is more than that of H but here in this problem it is given that $\ce{(CH3)3C+}$ is more stable.
How is this possible? Since, deuterium enriches the electron density over the central carbon resulting in the diminishing of a positive charge eventually making the second structure more stable.
Second explanation can be on the basis of hyperconjugation which says that $\ce{-CH3}$ is better in showing hyperconjugation than $\ce{-CD3}$, but why is this so?
| 3 |
[
[
"\n$\\ce{C-D}$ bonds are stronger. Hyperconjugation is thus more easily seen with hydrogen rather than deuterium. \n **Hyperconjugation> Inductive effects** at stabilization.I believe you can reason the rest out by yourself. \n To see why $\\ce{C-D}$ bonds are stronger. \n \n\n[See this](https://chemistry.stackexchange.com/questions/70128/why-is-a-c-d-bond-stronger-than-a-c-h-bond) \n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89633/ionization-energy-of-beryllium
|
Ionization energy of beryllium
|
Suppose I have $\ce{Be^3+}$. What would be its 4th ionization energy?
By trying to solve the issue I saw that its a "hydrogen-like" atom – means that beryllium left with only $1$ electron in his valance shell. Hence, I can use the Bohr atom model to solve the problem.
Is my guess right? If so, what values I need to plug in the equation and why so?
| 10 |
[
[
"\nIn terms of Bohr model ionization potential $E\\_\\mathrm{i}$ is the work $A\\_\\mathrm{i}$ on eliminating an electron in vacuum from its current non-excited orbital level to infinity:\n\n\n$$E\\_\\mathrm{i} = \\frac{A\\_\\mathrm{i}}{e}$$\n$$A\\_\\mathrm{i} = h\\nu = \\frac{hc}{\\lambda}$$\n\n\nUnknown wavelength $\\lambda$ can be determined from the [Rydberg formula](https://en.wikipedia.org/wiki/Rydberg_formula):\n\n\n$$\\frac{1}{\\lambda} = R\\_\\infty Z^2\\left(\\frac{1}{n\\_1^2} - \\frac{1}{n\\_2^2}\\right)$$\n\n\nso that final equation for ionization energy looks like this:\n\n\n$$E\\_\\mathrm{i} = \\frac{hc}{e}R\\_\\infty Z^2\\left(\\frac{1}{n\\_1^2} - \\frac{1}{n\\_2^2}\\right)$$\n\n\nSince we are determining 4th ionization energy of beryllium ($Z = 4$), $n\\_1 = 1$ and $n\\_2 = \\infty$:\n\n\n$$E\\_\\mathrm{i}^\\mathrm{IV} = \\frac{\\pu{6.63e-34 m^2 kg s-1}\\cdot\\pu{3e8 m s-1}}{\\pu{1.602e-19 C}}\\cdot\\pu{10973732 m-1}\\cdot 4^2\\left(\\frac{1}{1^2} - \\frac{1}{\\infty^2}\\right) = \\pu{217.86 eV}$$\n\n\nThis value is in a good agreement with the one listed in *CRC Handbook of Chemistry and Physics* [1, p. 10-204]: $E\\_\\mathrm{i}^\\mathrm{IV}(\\ce{Be}) = \\pu{217.71865 eV}$.\n\n\n### References\n\n\n1. Haynes, W. M.; Lide, D. R.; Bruno, T. J. *CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data*, 97th ed.; CRC Press, **2016**. ISBN 978-1-4987-5429-3.\n\n\n",
"12"
]
] |
https://chemistry.stackexchange.com/questions/89630/ph-value-of-distilled-pure-water
|
pH value of distilled (pure) water [duplicate]
|
**This question already has an answer here**:
[What is exactly the pH value of distilled water?](/questions/86948/what-is-exactly-the-ph-value-of-distilled-water)
(1 answer)
Closed 5 years ago.
What is exactly the pH value of distilled water?
which is correct?
* Neutral(not acidic not alkaline)pH 7?
* pH 6?
| -2 |
[
[
"\nDistilled water usually has a pH of 7. However, pH might change with temperature. See [this answer](https://chemistry.stackexchange.com/a/39610/48829) for more detail.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89621/what-is-the-product-of-this-reaction-and-mechanism
|
what is the product of this reaction and mechanism?
|
![picture from national chemistry Olympiad]
[](https://i.stack.imgur.com/yEOtk.png)
I cant get the mechanism.
Answer shows there will be formation of keto group.
But how?
As I can understand the $\ce{H^+}$ is going to attack the $\ce{O}$ and can remove the $\ce{CH3}$ groups forming two adjacent alcohols which may join to form keton.
I'm not sure.
| -1 |
[
[
"\nThe mechanism to convert acetal to ketone in acid based hydrolysis is in following steps-\n\n\n* Protonation of (1) to make OCH3 a good leaving group forming (2)\n* O lone pair allows LG to leave forming oxonium ion and alcohol(3)\n* Water nucleophile attacks C of C=O (3) to give (4)\n* Base removes proton to give (5).\n* Reprotonates (5) to make OCH3 a better leaving group(6).\n* CH3OH leaves to give (7)\n* Base deproronates to give ketone (8)\n\n\n**Note :In this scheme, the base, B:, could be R-O-R, R-OH, H2O or the conjugate base of the acid catalyst**\n\n\n[](https://i.stack.imgur.com/cLwtY.png)\n\n\n**References:**\n\n\n* <http://www.chem.ucalgary.ca/courses/353/exams/3513/353w05/353fin05me.html>\n* <https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/aldket1.htm>\n\n\n",
"2"
],
[
"\nYes, the oxigen is protonated, water enters, methanol leaves and another $\\ce{H^+}$ is released.\nThen the oxigen of the other $\\ce{CH3O}$ group is protonated, the keto group is formed while the methanol leaves and $\\ce{H^+}$ is released.\n\n\nEdit:\n\n\nThe first part results in:\n\n\n[](https://i.stack.imgur.com/IdFnH.png)\n\n\nand $\\ce{CH3OH}$.\n\n\n[](https://i.stack.imgur.com/t6h6n.gif)\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89620/conversion-of-z-matrix-to-cartesian-coordinates
|
Conversion of Z-matrix to Cartesian coordinates
|
To begin with, I wrote a script that gets Cartesian coordinates of molecule as input in the below. These are $x,y,z$ coordinates of H2O2 molecule.
```
1 O -1.7529 -0.5188 1.3324
2 O -0.4737 -0.1091 0.7774
3 H -2.2902 0.1678 0.8933
4 H 0.0636 -0.7957 1.2164
```
Then, the script constructs a Z-matrix with them, like this:
```
Z-mat :
O
O 1 1.45335189476
H 1 0.976176039452 2 96.5694760083
H 2 0.976131061897 1 96.5720363573 3 -179.995395182
```
Now I need to perform the reverse operation and use this Z-matrix as input and define $x,y,z$ coordinates for each atom. This is converting a Z-matrix to Cartesian coordinates.
My question is after setting first atom as 0,0,0
```
1 O 0 0 0
```
and the second one as 0,0,(distance from first) to put it on the z-axis
```
2 O 0 0 1.45335189476
```
How should I treat the 3rd and 4th atoms?
The 3rd atom must have coordinates that are something like this if read correctly:
```
3 H 0 distance*sin(angle) z2+distance*cos(angle)
```
Taking `z2` as the z-coordinate of atom 2. I am not sure if I should calculate this as `z2 + distance.cos(angle)` or `z2 - distance.cos(angle)` and what it depends on, if both are possible.
For the 4th atom, I use spherical coordinates
```
r, theta, phi = (0.976, 96.572, -179.995)
```
to calculate $x,y,z$ values from the formulas
```
x = r * sin(theta) * cos(phi)
y = r * sin(theta) * sin(phi)
z = r * cos(theta)
```
If there aren't any mistakes up to this point, how I will calculate Cartesian coordinates of the 4th atom using these $x,y,z$ values?
---
While searching I found [TMPChem's work](https://github.com/tmpchem/computational_chemistry/blob/b44ee6d2e1d2ec203811140b732cdb38f5164d76/scripts/geometry_analysis/zmat2xyz.py) on GitHub and it does exactly what I want. However, in his work, there is a mathematical part that I don't understand:
```
# get local axis system from 3 coordinates
def get_local_axes(coords1, coords2, coords3):
u21 = get_u12(coords1, coords2) #calculating vector between that points 1-2
u23 = get_u12(coords2, coords3) #calculating vector between that points 2-3
if (abs(get_udp(u21, u23)) >= 1.0):
print('\nError: Co-linear atoms in an internal coordinate definition')
sys.exit()
u23c21 = get_ucp(u23, u21) # unit cross product
u21c23c21 = get_ucp(u21, u23c21) # unit cross product
z = u21
y = u21c23c21
x = get_ucp(y, z)
local_axes = [x, y, z]
return local_axes
```
What is "getting local axis system from 3 coordinates"?
Here is some context of how this function is used:
```
bond_vector = get_bond_vector(atom.rval, atom.aval, atom.tval)
disp_vector = np.array(np.dot(bond_vector, self.atoms[i].local_axes))
for p in range(3):
atom.coords[p] = self.atoms[atom.rnum].coords[p] + disp_vector[p]
```
The bond vector definition is here:
```
def get_bond_vector(r, a, t):
x = r * math.sin(a) * math.sin(t)
y = r * math.sin(a) * math.cos(t)
z = r * math.cos(a)
bond_vector = [x, y, z]
return bond_vector
```
Again, the only part I don't understand is `# get local axis system from 3 coordinates`. What is that function doing?
| 7 |
[
[
"\nI will try to answer the question in bold from a mathematical perspective.\n\n\nBasically, the first two lines in `get_local_axes` are to build the vectors $\\vec r\\_{12}$ and $\\vec r\\_{23}$. The next line builds the cross product $$\\frac{\\vec r\\_{23}\\times\\vec r\\_{12}}{|\\vec r\\_{23}\\times\\vec r\\_{12}|},$$\n\n\nThe next vector is\n$$\\frac{\\vec r\\_{12}\\times(\\vec r\\_{23}\\times\\vec r\\_{12})}{|\\vec r\\_{12}\\times(\\vec r\\_{23}\\times\\vec r\\_{12})|},$$\n\n\nThis is a vector perpendicular to $\\vec r\\_{12}$, and lies in the plane where $\\vec r\\_{12}$ and $\\vec r\\_{23}$ lie.\n\n\nThe next lines are defining the three axes of the system. $\\vec e\\_z$ is just $\\vec r\\_{12}$ normalized, $\\vec e\\_y$ is chosen so that $yz$ is the plane where the three points (atoms) locate. and $\\vec e\\_x$ is chosen so that $(\\vec e\\_x, \\vec e\\_y, \\vec e\\_z)$ forms a orthonormal set and are the [versors](https://en.wikipedia.org/wiki/Versor_(physics)) of a right-handed coordinate system. This system is the local axis system.\n\n\nSo this is the meaning of \"getting local axis system from 3 coordinates\".\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/89618/can-h-act-as-a-strong-br%c3%b8nsted-base-like-ch3
|
Can H– act as a strong Brønsted base like CH3–?
|
In the reduction of carboxylic acids with $\ce {LiAlH\_4}$ we get an alcohol. But isn't there a possibility of the hydride donor abstracting $\ce {H^{+}}$ from the acid to form $\ce {H\_2}$?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/89613/can-a-molecule-be-neither-gerade-nor-ungerade
|
Can a molecule be neither gerade nor ungerade?
|
Or can any molecular orbit always be written a linear combination of gerade and ungerade basis states?
| 3 |
[
[
"\n1. g/u is not a property of a molecule; I assume you meant molecular orbital.\n2. All functions can be written as a linear combination of even + odd functions: there is a short explanation on [Wikipedia](https://en.wikipedia.org/wiki/Even_and_odd_functions#Other_algebraic_properties). Basically, you have a function $f(x)$; now define \n\n\n$$\\begin{align}\nf\\_\\mathrm e(x) &= \\frac{1}{2}[f(x) + f(-x)] \\\\\nf\\_\\mathrm o(x) &= \\frac{1}{2}[f(x) - f(-x)] \\\\\n\\end{align}$$\n\n\nIt is clear from the definition of parity that $f\\_\\mathrm e$ is even (since $f\\_\\mathrm e(x) = f\\_\\mathrm e(-x)$) and likewise that $f\\_\\mathrm o$ is odd. Now since $f(x) = f\\_\\mathrm e(x) + f\\_\\mathrm o(x)$, we have shown that any function $f(x)$ can be expressed as a linear combination of even and odd functions.\n\n\nIn three dimensions where you have $\\psi(x,y,z)$ you simply need to define\n\n\n$$\\begin{align}\n\\psi\\_\\mathrm g(x) &= \\frac{1}{2}[\\psi(x,y,z) + \\psi(-x,-y,-z)] \\\\\n\\psi\\_\\mathrm u(x) &= \\frac{1}{2}[\\psi(x,y,z) - \\psi(-x,-y,-z)] \\\\\n\\end{align}$$\n\n\nand analogously to above one can see that every function $\\psi(x,y,z)$ can be expressed as the sum of a gerade component $\\psi\\_\\mathrm g$ and an ungerade component $\\psi\\_\\mathrm u$. These components don't have any physical meaning (unless $\\psi$ itself is either g or u, in which case one component is simply $\\psi$ and the other is zero), but that's somewhat beside the point.\n\n\nIf a molecule does not possess a centre of inversion then its (canonical) molecular orbitals will not possess g/u symmetry.\n\n\nSo the answer is yes, and yes. There is no \"or\".\n\n\n",
"8"
]
] |
https://chemistry.stackexchange.com/questions/89608/comparison-of-stability-of-cyclic-imines
|
Comparison of stability of cyclic imines [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed 5 years ago.
[Improve this question](/posts/89608/edit)
[](https://i.stack.imgur.com/HWT5l.png)
I am looking for the reason for the lowest stability of cyclohexane-1,3,5-triimine (D).
I was not able to find any explanation using resonance or aromaticity.
All help is appreciated.
| 0 |
[
[
"\nThe cyclohexane-1,3,5-triimine tautomer is benzene-1,3,5-triamine, which is aromatic. This makes this trienamine form much more stable than any of the other possible tautomeric forms.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89604/why-does-pyrolysis-activate-carbon
|
Why does pyrolysis activate carbon?
|
I am doing a project that involves activating carbon. The method my team is using is simply to heat it within the range of 400-500°C, and then testing for activity by using a dye.
Why does simply heating it (without any additional reactants like zinc chloride and so on) activate it?
| 2 |
[] |
https://chemistry.stackexchange.com/questions/89603/why-is-cr-unstable
|
Why is Cr+ unstable?
|
We know that half-filled and fully filled orbitals are highly stable. In ground state $\ce{Cr}$ has a $\ce{3d^5\! 4s^1}$ configuration. Therefore, the elctronic configuration of $\ce{Cr+}$ should have $\ce{3d^5\! 4s^0}$ state. The half-filled $\ce{3d}$ orbital should make the ion stable.
However, we don't see many $\ce{Cr(I)}$ compounds. There are one or two rare examples, but other than them, most compounds of $\ce{Cr}$ are $\ce{Cr(II)}$, $\ce{Cr(III)}$ and so on. Why does this happen? Why $\ce{Cr+}$ is unstable?
[If you find a reference, add it to your answer. As I am a student, it will help me immensely. But if you don't have a reference, don't hesitate to post an answer. Getting an answer is the priority.]
| 3 |
[] |
https://chemistry.stackexchange.com/questions/89601/orientation-in-benzene-rings-with-more-than-one-substituent
|
Orientation in benzene rings with more than one substituent
|
Why is it that the chlorination of 1-chloro-3-nitrobenzene (**18**) gives 1,4-dichloro-2-nitrobenzene (**19**) as the major product, 1,2-dichloro-3-nitrobenzene (**20**) as a minor product and 1,2-dichloro-4-nitrobenzene (**21**) is not formed at all?
[](https://i.stack.imgur.com/plMuW.png)
I know $\ce{-NO2}$ is a deactivating, meta-directing group and hence it decreases the electron density at positions 2, 4, 6 while position 5 is unaffected. On the other hand, $\ce{-Cl}$ is a deactivating, o-p directing group and hence it directs the incoming group to positions 2,4 or 6. Taking into account the fact that $\ce{-NO2}$ is a more powerful deactivating group than $\ce{-Cl}$, the incoming $\ce{Cl+}$ should be attached at position 5 as it has more electron-density than 6,2 or 4 position. But the product with $\ce{Cl}$ at position 5 is not formed. Why so? I am unable to rationalise this.
**Source**: page 583 of "March's Advanced Organic Chemistry, 7th ed." under the heading "Orientation in benzene rings with more than one substituent".
| 7 |
[
[
"\nI quote March's Advanced Organic Chemistry, Reactions, Mechanisms and Structure, 6th edition, page 666.\n\n\n\n> \n> Some substituents have a pair of electrons (usually unshared) that may be contributed\n> toward the ring. The three arenium ions would then look like this:[](https://i.stack.imgur.com/yR9Lp.jpg)\n> \n> \n> The stability of these two ions is increased by the extra form not only because it is another canonical form, but because it is more stable than the others and makes a greater contribution to the hybrid. Every atom (except of course hydrogen) in these forms (C and D) has a complete octet, while all the other forms have one carbon atom with a sextet. No corresponding form can be drawn for the meta isomer.The inclusion of this form in the hybrid lowers the energy , but also because it spreads the positive charge over a larger area—out onto the group Z. **Groups with a pair of electrons (e.g., as the halogens) to contribute would be expected, then, in the absence of field effects, not only to direct ortho and para, but also to activate these positions for electrophilic attack.** (my emphasis)\n> \n> \n> \n\n\nIn your example, if chlorine electrophile attacks at C-4 or C-6, the resulting intermediate is stabilized by an extra resonating structure, lowering the energy of this intermediate.Hence electrophillic aromatic substitution takes place.\n\n\n[](https://i.stack.imgur.com/qKYCw.png)\n\n\n[](https://i.stack.imgur.com/e1i46.png) \n\n\nIf chlorine electrophile attacks at C-5 , the resulting intermediate is not stabilized compared to attack at C-6 .Hence the resulting electrophillic aromatic substitution is not seen.\n\n\n[](https://i.stack.imgur.com/E2Ndr.png) \n\n\nThis is the possible reason that the product with Cl at position 5 is not formed.\n\n\n\n\n---\n\n\nQuoting from [chemistry.msu.edu](https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/benzrx1.htm#benz1):\n\n\n\n> \n> The influence a substituent exerts on the reactivity of a benzene ring may be explained by the interaction of two effects:\n> \n> \n> The first is the inductive effect of the substituent. Most elements other than metals and carbon have a significantly greater electronegativity than hydrogen. Consequently, substituents in which nitrogen, oxygen and halogen atoms form sigma-bonds to the aromatic ring exert an inductive electron withdrawal, which deactivates the ring.\n> \n> \n> The second effect is the result of conjugation of a substituent function with the aromatic ring. This conjugative interaction facilitates electron pair donation or withdrawal, to or from the benzene ring, in a manner different from the inductive shift. If the atom bonded to the ring has one or more non-bonding valence shell electron pairs, as do nitrogen, oxygen and the halogens, electrons may flow into the aromatic ring by p-π conjugation (resonance).\n> [](https://i.stack.imgur.com/qTqVM.jpg)\n> \n> \n> The second factor that becomes important in reactions of substituted benzenes concerns the site at which electrophilic substitution occurs.\n> \n> \n> [](https://i.stack.imgur.com/X4TNR.jpg)\n> \n> \n> The exact influence of a given substituent is best seen by looking at its interactions with the delocalized positive charge on the benzenonium intermediates generated by bonding to the electrophile at each of the three substitution sites.\n> \n> \n> \n\n\n[](https://i.stack.imgur.com/fUPEC.jpg)\n\n\nTherefore chlorine is an ortho para directing group with minor meta product.\n\n\n**References:**\n\n\n* Advanced Organic Chemistry 5th edition Part A: Structure and Mechanisms, Francis A. Carey and Richard J. Sundberg\n* March's Advanced Organic Chemistry, Reactions, Mechanisms and Structure, 6th edition\n\n\n",
"-1"
]
] |
https://chemistry.stackexchange.com/questions/89596/what-is-the-iupac-name-for-ch3ch2cooag
|
What is the iupac name for ch3ch2cooag? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed 5 years ago.
[Improve this question](/posts/89596/edit)
Like ch3coo is acetate group and is named accordingly,how to name this one.
| -2 |
[
[
"\nStart with the acid $\\ce{CH\\_3CH\\_2COOH}$. This is clearly the one and only simple carboxylic acid derived from an unsubstituted propane chain, so it's\n\n\n\"Propan-\" from the propane chain\n\n\nplus \"-oic acid\" from the carboxyl group (we don't need a locant because there is only one place to fit that group)\n\n\n= propanoic acid\n\n\nThen the silver salt is silver propanoate, the \"-ic acid\" part becomes \"-ate\" just like inorganic salts. \n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89595/bond-order-of-h2o-and-nh3-using-group-theory-to-construct-mos
|
Bond order of H2O and NH3 using group theory to construct MOs
|
The bond order of H2O is 2, and that of NH3 is 3, which makes sense, when considering the number of bonds they have.
However, if using basic group theory, you get the MO diagrams as presented in this paper:
<http://www.scielo.br/pdf/qn/v35n7/v35n7a32.pdf>
Upon examination, we see that in the case of H2O, there are two orbitals that are strongly bonding, one weakly bonding orbital, and one non-bonding orbital. If we then tried to apply the bonding order equation to this MO diagram, would we not end up with a bond order of 3 - despite the a1 orbital being only weakly bonding, its occupation still stabilizes the molecule relative to its constituent atoms. Is it the case that the concept of bonding order breaks down using these kinds of MO diagrams?
A further example is given using the NH3 MO diagram. Here, they calculate the bond order as 3, ignoring the fact that the NH3 a1 orbital is weakly bonding. If we were to in theory calculate the MO diagram for NH3 in a trigonal planar geometry (same as for BH3), we would also get the bond order as 3. Of course, the occupied a1 orbital is weakly bonding in the trigonal pyrimidal NH3 geometry, leading to a preference over the trigonal planar geometry, but why is this not reflected in the bond order?
| 2 |
[
[
"\nIt depends on your definition of bond order. If you are using the simple formula \n\n\n$$\\text{Bond order} = \\frac{\\text{Number of bonding electrons} - \\text{Number of antibonding electrons}}{2}$$\n\n\nthen, yes, it won't be able to capture the subtleties that you describe. After all, when you use that, you're categorising electrons into bonding vs antibonding in a black and white manner. So it's really no surprise that it doesn't reflect \"weakly bonding\" MOs.\n\n\nYou gave the example of water. Well, for that weakly bonding MO, you need to make a choice: is that considered a bonding MO, leading to a bond order of 1.5 per O–H bond in water (doesn't make much sense)? Or is it considered a nonbonding MO, leading to a bond order of 1 (makes more sense but then you have this problem you brought up)? Or is it somewhere in between, leading to a bond order between 1 and 1.5 – and crucially, if this is the route you want to go down, how do you quantify the bonding character of the MO?\n\n\nI don't have any experience using different definitions of bond order, but if you used one of these, I would not be surprised to see an O–H bond order that deviates ever so slightly from 1.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89580/is-there-a-way-to-quantify-the-surface-area-or-simply-the-shape-of-organic-com
|
Is there a way to quantify the 'surface area' or simply the shape of organic compounds?
|
I wish to write an investigation into the effects of the shape of organic compounds and the location of the functional groups on the enthalpy of combustion of the organic compounds. Thus, I wonder if there is a way of quantifying these variables, as databases I have looked at do not provide much of an answer.
| 0 |
[
[
"\nIt is possible to compute the surface area of a molecules, and to compute the surface area of each individual atom in a molecule. These areas depend on the conformation of the molecule. An example of a software that does this is MSMS (<http://mgltools.scripps.edu/packages/MSMS>).\n\n\nWith respect to the shape, it is possible to categorize molecules in a triangular diagram describing similarity with \"rods\", \"disks\" and \"spheres\" (e.g. <http://pubs.acs.org/doi/10.1021/ci300415d>). Details about this analysis were published here: <http://pubs.acs.org/doi/10.1021/ci025599w>.\n\n\n",
"1"
]
] |
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