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I have a question about 1H NMR splitting of 3-methyl-1-butanol.
[![enter image description here][1]][1]
The spectrum looks as the following.
[![enter image description here][2]][2]
[1]: https://i.stack.imgur.com/mZYMz.png
[2]: https://i.stack.imgur.com/xvteM.png
The assignment of the NMR spectrum is the following.
Assign. Shift(ppm)
A 3.673
B 1.66
C 1.57
D 1.49
E 0.922
So based on the results, A hydrogen is splitted into triplet. Is this splitted by D hydrogen? If so, why not C hydrogen on the alcohol group?
The results are retrieved from the following link: https://www.chemicalbook.com/SpectrumEN_123-51-3_1HNMR.htm |
I actually have a (or many) big issue(s) with the quote:
> The central C-atom is in an sp<sup>2</sup> hybridized state, for which the carbocations have planar geometry. The p<sub>$z$</sub>-AO remains empty.
The authors here have clearly scrambled up their reasoning, making carbocations seem as something they are definitely not. Suffice to say (tl;dr) **the above statement cannot be true.** Let's get a few points straight before moving on to a more complex examples.
1. *The p orbital remains empty.*
We know that s orbitals ($\ell=0$) of the same principle quantum number $n$ have a lower energy than the corresponding p orbitals ($\ell=1$). It is therefore (almost) always energetically more favourable to occupy orbitals with as much s character as possible.
2. *The coordination is planar.*
Ideally one (any) of the p orbitals will remain completely unoccupied. Because of symmetry considerations, a planar arrangement of ligands around the central atom virtually ensures that. The planar coordination is a result of a favourable electronic state. Obviously there will be other interactions at play, but in a first approximation the above is always true.
(Also note that I am avoiding the word geometry, because that should rather be reserved for the whole molecule.)
3. *Orbitals are hybridised, not atoms.*
There is no such thing as an *"hybridised state"*. There might be an atom of which the wave function can be described with hybrid orbitals. The colloquial phrase *"the carbon is sp<sup>3</sup> hybridised"*, which is especially popular with organic chemists, is a garbage simplification.
4. *Valence Bond Theory is not a simplification; a.k.a. Bent's rule.*
The description with sp<sup>$n$</sup> orbitals is a relic of the very, very first days of VB theory. Nowadays this theory has well evolved past these rigid kinds of descriptions. Essentially, allowing $n\in\mathbb{R}$ produces better descriptions and a better agreement with experimental data. (Read more: https://chemistry.stackexchange.com/q/15620/4945 https://chemistry.stackexchange.com/q/15671/4945)
5. **Hybridisation is a mathematical description.**
We would be completely fine without hybridisation. We choose to use hybrid orbitals, because they (in most cases) represent the geometry of molecules in a much easier view than the very generic canonical orbitals.
Unfortunately, hybrid orbitals became a tool of prediction in organic chemistry textbooks because they are so temptingly easy to understand. As a result many things get explained in this way where it would not the least be necessary. Often leading to wrong conclusions, other times being right only by coincidence (right for the wrong reasons).
6. *Carbocations are nothing trivial.*
It took [a couple of years](https://en.wikipedia.org/wiki/Carbocation#History) for the theory to be accepted and then confirmed by experiments, showing that there is nothing easy to fathom. In terms of electronic stability, only occupied orbitals count. Molecular entities will always adopt the lowest lying electronic state in the the optimal geometry.
---
Just because of Bent's rule it is only logical to assume that carbocations in general may differ significantly from the often taught 3×sp<sup>2</sup> + p hybridisation scheme. In principle, only carbocations of the form $\ce{^+CR3}$ are symmetric enough to have this scheme. This already starts to break down with $\ce{R{ = }CH3}$ because of hyperconjugation. In first approximation, however, the convenient model holds. Just keep the limitations in mind.
With all of that we can go to your specific questions. All of your examples are what we often refer to non-classical carbocations. You may now ask yourself: https://chemistry.stackexchange.com/q/31732/4945 I therefore recommend reading the linked Q&A before continuing. ([Importance of such cations.](https://chemistry.stackexchange.com/q/33398/4945) Shameless self-promotion.)
I personally dislike the terminology and the definition in the [gold book](http://goldbook.iupac.org/html/N/N04185.html), as I find it a little reactionary, but we're stuck with it, there is no use in complaining.
> **nonclassical carbocation**
A carbocation the ground state of which has delocalized (bridged) bonding π- or σ-electrons. (N.B. Allylic and benzylic carbocations are not considered nonclassical.)
Note for the remaining part of the answer I am keeping things short as I am just summarising stuff from two sources on our network: (1)https://chemistry.stackexchange.com/q/38093/4945 (2) https://chemistry.stackexchange.com/q/49648/4945
1. **Phenyl cation/ Aryl carbocation**
In this case we have a cationic carbon which is already planar. Therefore the necessary change would be to adopt a linear coordination. This obviously is restricted by the cyclic backbone.
[![Benzene and the Phenyl cation][1]][1]
Technically this is not a non-classical carbocation according to the definition (or is it?), which is one of the reasons why I don't like this definition in the first place.
A true non-classical version with a bridging proton is not a stable stationary point on DF-BP86/def2-SVP.
While the bridging $C_\mathrm{5v}$ symmetric $\ce{^+C(CH)5}$ is a stationary point, it is about $\pu{145 kJ mol-1}$ higher in energy.
2. **Vinyl cation**
*tl;TL;DR;dr:* More recent work indicates that the bridged form of the vinyl cation with is slightly more stable (by about 1-3 kcal/mol).
[![Classical vs non-classical vinyl cation][2]][2]
3. **Ethynylene Carbocation**
*tl;dr:* The linear $\ce{HCC+}$ is not a stationary point at DF-BP86/def2-SVP. The stable structure is an almost three-membered ring, which is best thought of as a protonated dicarbon.
[![etynyl carbocation][3]][3]
---
## Conclusion (?!)
Throw out the restrictive thinking of hybridisation. It is almost always useless when it comes to carbocations (best case scenario) or even gives you the completely wrong ideas. Always remember that orbitals can be described hybridised, but not atoms, and that hybridisation itself is never a fixed deal.
Always keep in mind that the smallest molecular entities do the weirdest things, with the most complicated bonding situations.
Stay open-minded.
[1]: https://i.stack.imgur.com/rVzhN.jpg
[2]: https://i.stack.imgur.com/tKUPm.png
[3]: https://i.stack.imgur.com/hfvWJm.jpg |
Gefitinib has been shown to be an effective tyrosine kinase inhibitor in a fraction (~$10$%) of non-small cell lung cancer patients.
These patients are characterized as having a mutation (usually a deletion or missense) in the _EGFR_ gene affecting the ATP-binding site.
I have two questions:
-- How does gefitinib impede the proliferation and apoptosis pathway? Does it in effect "ace out" the binding of EGF? If this is correct, how does it accomplish that? If not, I would appreciate a brief explanation as to what the successful mechanism is.
-- It seems that (similar to imatinib, another tyrosine kinase used to treat CML) successfully treated patients tend to relapse after a period of time. Why is this?
|
I intend to do a kinetic study of simple alcohol catalytic dehydrogenation reactions in the gas phase. I want to start with simple power law kinetics using Keq to account for the reversibility of the reactions. Would the equation below, knowing the Gibbs free energy of reaction, be valid?
[![enter image description here][1]][1]
I came across a paper that used this equation to calculate Keq to build a kinetic model of the ethanol-to-butadiene reaction. However, other papers I read, i.e. on the methanol-to-diethyl ether reaction, have used complex equations derived from empirical observations to calculate Keq. Why use one approach over the other?
[1]: https://i.stack.imgur.com/lknjG.jpg |
Is Keq at a given T derived from ΔGr° valid for doing simple kinetic modelling? |
I intend to do a kinetic study of simple alcohol catalytic dehydrogenation reactions in the gas phase. I want to start with simple power law kinetics using $K_\mathrm{eq}$ to account for the reversibility of the reactions. Would the equation below, knowing the Gibbs free energy of reaction, be valid?
$$Δ_\mathrm{r}G^\circ = -RT\ln K_\mathrm{eq}$$
I came across a paper that used this equation to calculate $K_\mathrm{eq}$ to build a kinetic model of the ethanol-to-butadiene reaction. However, other papers I read, i.e. on the methanol-to-diethyl ether reaction, have used complex equations derived from empirical observations to calculate $K_\mathrm{eq}.$ Why use one approach over the other? |
Is equilibrium constant at a given temperature derived from Gibbs free energy of reaction valid for doing simple kinetic modelling? |
Why is alpha-Furoic acid less acidic than acetic acid?
Alpha furoic acid's conjugate base is aromatic and hence is more stable. Then why acetic acid is more acidic than alpha furoic acid? |
We have two solutions:
- Solution 1 is $\ce{HCOOH}$, its concentration is $c_1 = \pu{10^-2 mol/l}$, its volume is $V_1 = \pu{50 ml}$, and its $\mathrm{pH}_1 = 2.9$.
- Solution 2 is $\ce{CH3COOH}$, its concentration is $c_2 = \pu{10-2 mol/l}$, its volume is $V_2 = \pu{50 ml}$, and its $\mathrm{pH}_2 = 3.4$.
How would be the equation and the ICE table, and what is the $\mathrm{pH}$ of the mixture of these two solutions?
I used numbers to differentiate the solutions, and to understand how it works, but for no other reason.
I tried to do $\mathrm{pH}=\frac{1}{2}\left(\mathrm{pH}_1+\mathrm{pH}_2\right)$, but it’s trivial and doesn't work. |
α-Furoic acid's conjugate base is aromatic and hence is more stable. Then why acetic acid is more acidic than α-furoic acid? |
On a recent quiz, we were asked the following question:
> $$\ce{A(aq) + B(aq) <=> C(aq)}$$
> At $\pu{298 K}$, the equilibrium amounts of each of the component are as follows: $[\ce{A}] = \pu{0.125 M}$, $[\ce{B}] = \pu{0.400 M}$, $[\ce{C}] = \pu{1.200 M}$. At $\pu{298 K}$, the system is perturbed by increasing the concentration of $\ce{B}$ by adding $X$ amount. When equilibrium is re-established, the concentration of $[\ce{C}] = \pu{1.250 M}$. The volume of the system remains unchanged. How much $X$ (in $\pu{M}$) was added to the system?
The correct answer for this is $\pu{0.344 M}$, but I'm struggling to understand the setup for how the ICE table would work, since my original construction would mean the $X$ value is just $\pu{0.05 M}$:
\begin{array}{lccc}
\text{Stage} & \ce{A (aq)} & \ce{B (aq)} & \ce{C (aq)} \\\hline
\text{Initial} & \pu{0.125 M} & \pu{0.4 M} & \pu{1.2 M} \\
\text{Change} & -X & +X & +X \\
\text{Equilibrium} & \pu{0.125 M} - X & \pu{0.4 M} +X & \pu{1.25 M} \\
\end{array}
But this does not seem to be right. Have I set up my initial values incorrectly? Any help would be appreciated.
<!-- For reference only
**Picture of Question:**
[![The Problem in Question][1]][1]
[1]: https://i.stack.imgur.com/Qcpxz.png
--> |
How to solve for the amount the equilibrium was disturbed with an ICE table? |
What is the largest, noncrystalline, non-polymer inorganic molecule? |
When light is passed through a collidal solution, due to Tyndall scattering, the path of light appears to be a bright cone. Why does it have to form only a conical shape?
The conditions for Tyndall effect are that the diameter of dispersed particles should not be much smaller than wavelength of light used and refractive index of dispersed phase and dispersed medium should differ greatly.
This is perhaps because colloidal particles scatter light in all directions. Normal refraction shouldn't cause such a shape.
Is there any logical reson for the conical shape? |
Why does Tyndall scaterring produce conical shape? |
On my organic chemistry exam there was a skeletal formula that had, supposedly 1 methyl group at each end of the skeletal formula and 1 somewhere in the middle. But is it correct if you draw a formula like that and then put those methyl groups on each end? I feel like it's very misleading because I thought that those were just the ends of the main chain, so NOT methyl groups. I lost a lot of marks on that question because of this confusion and I just really want to know if it's even correct to write it that way.
[![Skeletal formula from my exam question][1]][1]
[1]: https://i.stack.imgur.com/jkOIn.png |
Is it correct to put a methyl group on each end of a skeletal formula? |
When light is passed through a collidal solution, due to Tyndall scattering, the path of light appears to be a bright cone. Why does it have to form only a conical shape?
The conditions for Tyndall effect are that the diameter of dispersed particles should not be much smaller than wavelength of light used and refractive index of dispersed phase and dispersed medium should differ greatly.
This is perhaps because colloidal particles scatter light in all directions. Normal refraction shouldn't cause such a shape.
Is there any logical reson for the conical shape?
[![Tyndall Cone][1]][1]
Source: NCERT class 12
[1]: https://i.stack.imgur.com/YHQFR.jpg |
It is simple.
Light always travels in a straight line and if a lens is present it would diverge or converge ( in a straight line ) accordingly.
**The question is why no other shape like maybe a hemisphere or a paraboloid ?**
Now let me ask the question, have you seen such thing?
[Check out this diagram to make everything clear.][1]
If the beam of light increases in width ( diverging beam) linearly with distance. It is because all the particles scatter equally in all directions ***and some of them reach your eyes and this reveals the shape of the beam.*** The adjacent particles does the same and so on.
Therefore you can also see the light getting more and more diffused linearly as more and more light is scattered.
It also depends on the refractive index of the medium where the incident light ray will either move closer to the normal or move further. Again this would lead to a conically diverging beam.
[1]: https://images.app.goo.gl/d3pbcSLFMKjyumB17 |
On my organic chemistry exam there was a skeletal formula that had, supposedly one methyl group at each end of the skeletal formula and one somewhere in the middle.
[![Skeletal formula from my exam question][1]][1]
But is it correct if you draw a formula like that and then put those methyl groups on each end? I feel like it's very misleading because I thought that those were just the ends of the main chain, so *not* methyl groups. I lost a lot of marks on that question because of this confusion and I just really want to know if it's even correct to write it that way.
[1]: https://i.stack.imgur.com/jkOIn.png |
It is simple.
Light always travels in a straight line and if a lens is present it would diverge or converge ( in a straight line ) accordingly.
**The question is why no other shape like maybe a hemisphere or a paraboloid ?**
Now let me ask the question, have you seen such thing?
[Check out this diagram to make everything clear.][1]
If the beam of light increases in width ( diverging beam) linearly with distance. It is because all the particles scatter equally in all directions ***and some of them reach your eyes and this reveals the shape of the beam.*** The adjacent particles does the same and so on.
Therefore you can also see the light getting more and more diffused linearly as more and more light is scattered.
It also depends on the refractive index of the medium where the incident light ray will either move closer to the normal or move further. Again this would lead to a conically diverging beam.
**EDIT**
>***This is to be noted that the colloidal solution does not affect the shape of the beam of light ( converging or diverging )***.
This is also evident from your diagram.
From your question I feel that you think that the colloidal solution is responsible for the cone shape, *it is not so*. *It is simply the geometry of the light rays*
*What the solution does is scatter the light in all directions and makes the "beam" visible*. A straight beam of light in perfect vacuum would not be visible as a stalk of light ( that we see when light enters a dark room thought a window )
[1]: https://images.app.goo.gl/d3pbcSLFMKjyumB17 |
I am making a simple restricted HF code using the Python interface of Psi4. I am now evaluating convergence by tracking the change in the sum of orbital energies, but I want to do this in a better way. It is common to use the fact that at self-consistency, the Fock and density matrices commute
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} - \mathbf{DF} = \mathbf{0}
$$
However, the above expression is only valid in MO basis, while F and D in my code are computed in AO basis. So I need to derive an equivalent expression in AO basis. I am quite sure the correct expression is
$$
[\mathbf{F}, \mathbf{D}]^{\text{AO}} = \mathbf{FDS} - \mathbf{SDF}
$$
as this is equal to zero to within $1\times10^{-14}$. But how to derive this?
Derivation
--------------------
An arbitrary molecular orbital $\phi_i$ is expanded in atomic orbital basis functions
$$
\phi_i = \sum_\alpha C_{\alpha i} \chi _{\alpha i}
$$
Acting the commutator on $\phi_i$ and expanding it to AO basis yields
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha i} - \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha i}
$$
Since we know the solution contains the overlap matrix $\mathbf{S}$, lets look at the definition
$$
\mathbf{S}_{ij} = \langle \chi_i(\mathbf{r}) \vert \chi_j(\mathbf{r}) \rangle = \int d\mathbf{r} \chi^*(\mathbf{r})\chi(\mathbf{r})
$$
Since this must be part of our expression, it seems to me a good approach is to multiply from the left by $\chi_{\beta i}$ (dropping the $\mathbf{r}$ dependence from now on) and integrating over $\mathbf{r}$
$$
[\mathbf{F}, \mathbf{D}] = \int \chi^*_{\beta i} \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha i} - \int \chi^*_{\beta i} \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha i}
$$
from in braket notation becomes
$$
[\mathbf{F}, \mathbf{D}] = \langle \chi_{\beta i} \vert \mathbf{FD} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha i} \rangle - \langle \chi_{\beta i} \vert \mathbf{DF} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha i} \rangle
$$
At this point I am not sure what to do - or if am I even on the right track. |
>How would be the equation and the ICE table, and what is the pH of the mixture of these two solutions?
There are three equations:
$$\ce{HCOOH <=> H+ + HCOO-}$$
$$\ce{CH3COOH <=> H+ + CH3COO-}$$
$$\ce{H2O <=> H+ + OH-}$$
Hydroxide is a minor species, so you can neglect it. As I explain below, at the pH of the mixture, the acetate ion is also a minor species, so you can first neglect the dissociation of acetic acid, and then correct for it later.
I would not use an ICE table in this situation, at least not one that describes all three reactions.
**My solution**
First, you can determine the equilibrium constants for each acid from the pH values of their aqueous solutions (before mixing). The corresponding pKa values are 3.74 and 4.78 for formic and acetic acid, respectively.
After mixing, the concentration of total formic acid and total acetic acid drop by a factor of 2 (mutual dilution). Formic acid is the stronger acid, and has a pH of 3.06 at this concentration (down from 2.9 because of dilution).
At pH 3.06, acetic acid dissociates very little (the difference between pH and pKa is more than 1.5, vs. less than 0.7 for formic acid). As a first approximation, we can say that the pH is determined by the stronger acid.
To figure out how much the presence of acetic acid does influence the pH, we can calculate how much it would dissociate at pH 3.06, and correct the pH accordingly. Our new pH will be more acidic, so we assumed a too acidic pH in our correction, overshooting a bit. After a couple of iterations of this, the pH comes out as 3.04.
So the biggest effect of adding the acetic acid comes from diluting the formic acid (pH would go from 2.9 to 3.06 just by adding water). A secondary effect comes from acetic acid dissociating a tiny bit, giving a pH of 3.04.
Here is the spreadsheet used for the calculations, along with the formulas used (to dampen the overshooting effect, the pH value used in each step is the average of the updated pH and the previous pH):
[![enter image description here][1]][1]
[![enter image description here][2]][2]
>I tried to do pH=12(pH1+pH2), but it’s trivial and doesn't work.
To convince yourself once and for all that does not work, consider two example:
1. Mixing 5 mL of 1 M HCl with 6 mL of 1 M NaOH (starting pH values are 0 and 14, resulting pH is roughly 13).
2. Mixing 10 ml of 1 M HCl with 90 mL of water (starting pH values are 0 and 7, resulting pH is 1)
[1]: https://i.stack.imgur.com/4QoQW.jpg
[2]: https://i.stack.imgur.com/XKUoi.jpg |
I am making a simple restricted HF code using the Python interface of Psi4. I am now evaluating convergence by tracking the change in the sum of orbital energies, but I want to do this in a better way. It is common to use the fact that at self-consistency, the Fock and density matrices commute
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} - \mathbf{DF} = \mathbf{0}
$$
However, the above expression is only valid in MO basis, while F and D in my code are computed in AO basis. So I need to derive an equivalent expression in AO basis. I am quite sure the correct expression is
$$
[\mathbf{F}, \mathbf{D}]^{\text{AO}} = \mathbf{FDS} - \mathbf{SDF}
$$
as this is equal to zero to within $1\times10^{-14}$. But how to derive this?
Derivation
--------------------
An arbitrary molecular orbital $\phi_i$ is expanded in atomic orbital basis functions
$$
\phi_i = \sum_\alpha C_{\alpha i} \chi _{\alpha}
$$
Acting the commutator on $\phi_i$ and expanding it to AO basis yields
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha} - \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$
Since we know the solution contains the overlap matrix $\mathbf{S}$, lets look at the definition
$$
\mathbf{S}_{ij} = \langle \chi_i(\mathbf{r}) \vert \chi_j(\mathbf{r}) \rangle = \int d\mathbf{r} \chi^*(\mathbf{r})\chi(\mathbf{r})
$$
Since this must be part of our expression, it seems to me a good approach is to multiply from the left by $\chi_{\beta i}$ (dropping the $\mathbf{r}$ dependence from now on) and integrating over $\mathbf{r}$
$$
[\mathbf{F}, \mathbf{D}] = \int \chi^*_{\beta} \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha} - \int \chi^*_{\beta} \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$
from in braket notation becomes
$$
[\mathbf{F}, \mathbf{D}] = \langle \chi_{\beta} \vert \mathbf{FD} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle - \langle \chi_{\beta} \vert \mathbf{DF} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle
$$
At this point I am not sure what to do - or if am I even on the right track. I can see that we have the "pieces" that make up the overlap matrix, but I don't know how to put them together. The Fock and density matrices are Hermitian matrices, and so perhaps they can be shuffled around the braket in some ways. But what to do with the coefficients $C_{\alpha i}$? |
>How would be the equation and the ICE table, and what is the pH of the mixture of these two solutions?
There are three equations:
$$\ce{HCOOH <=> H+ + HCOO-}$$
$$\ce{CH3COOH <=> H+ + CH3COO-}$$
$$\ce{H2O <=> H+ + OH-}$$
Hydroxide is a (very) minor species, so you can neglect it. As I explain below, at the pH of the mixture, the acetate ion is also a minor species, so you can first neglect the dissociation of acetic acid, and then correct for it later.
I would not use an ICE table in this situation, at least not one that describes all three reactions.
**My solution**
First, you can determine the equilibrium constants for each acid from the pH values of their aqueous solutions (before mixing). The corresponding pKa values are 3.74 and 4.78 for formic and acetic acid, respectively.
After mixing, the concentration of total formic acid and total acetic acid drop by a factor of 2 (mutual dilution). Formic acid is the stronger acid, and has a pH of 3.06 at this concentration (down from 2.9 because of dilution).
At pH 3.06, acetic acid dissociates very little (the difference between pH and pKa is more than 1.5, vs. less than 0.7 for formic acid). As a first approximation, we can say that the pH is determined by the stronger acid.
To figure out how much the presence of acetic acid does influence the pH, we can calculate how much it would dissociate at pH 3.06, and correct the pH accordingly. Our new pH will be more acidic, so we assumed a too acidic pH in our correction, overshooting a bit. After a couple of iterations of this, the pH comes out as 3.04.
So the biggest effect of adding the acetic acid comes from diluting the formic acid (pH would go from 2.9 to 3.06 just by adding water). A secondary effect comes from acetic acid dissociating a tiny bit, giving a pH of 3.04.
Here is the spreadsheet used for the calculations, along with the formulas used (to dampen the overshooting effect, the pH value used in each step is the average of the updated pH and the previous pH):
[![enter image description here][1]][1]
[![enter image description here][2]][2]
>I tried to do pH=12(pH1+pH2), but it’s trivial and doesn't work.
To convince yourself once and for all that does not work, consider two example:
1. Mixing 5 mL of 1 M HCl with 6 mL of 1 M NaOH (starting pH values are 0 and 14, resulting pH is roughly 13).
2. Mixing 10 ml of 1 M HCl with 90 mL of water (starting pH values are 0 and 7, resulting pH is 1)
[1]: https://i.stack.imgur.com/4QoQW.jpg
[2]: https://i.stack.imgur.com/XKUoi.jpg |
>How would be the equation and the ICE table, and what is the pH of the mixture of these two solutions?
There are three equations:
$$\ce{HCOOH <=> H+ + HCOO-}$$
$$\ce{CH3COOH <=> H+ + CH3COO-}$$
$$\ce{H2O <=> H+ + OH-}$$
Hydroxide is a (very) minor species, so you can neglect it. As I explain below, at the pH of the mixture, the acetate ion is also a minor species, so you can first neglect the dissociation of acetic acid, and then correct for it later.
I would not use an ICE table in this situation, at least not one that describes all three reactions.
**My strategy to determine the pH of the mixture**
First, you can determine the equilibrium constants for each acid from the pH values of their aqueous solutions (before mixing). The corresponding pKa values are 3.74 and 4.78 for formic and acetic acid, respectively.
After mixing, the concentration of total formic acid and total acetic acid drop by a factor of 2 (mutual dilution). Formic acid is the stronger acid, and would have a pH of 3.06 at this concentration in the absence of acetic acid (down from 2.9 because of dilution).
At pH 3.06, acetic acid dissociates very little (the difference between pH and pKa is more than 1.5, vs. less than 0.7 for formic acid). As a first approximation, we can say that the pH is determined by the stronger acid.
To figure out how much the presence of acetic acid does influence the pH, we can calculate how much it would dissociate at pH 3.06, and correct the pH accordingly. Our new pH will be more acidic, so we assumed a too acidic pH in our correction, overshooting a bit. After a couple of iterations of this, the pH comes out as 3.04.
So the biggest effect of adding the acetic acid comes from diluting the formic acid (pH would go from 2.9 to 3.06 just by adding water). A secondary effect comes from acetic acid dissociating a tiny bit, giving a pH of 3.04.
Here is the spreadsheet used for the calculations, along with the formulas used (to dampen the overshooting effect, the pH value used in each step is the average of the updated pH and the previous pH):
[![enter image description here][1]][1]
[![enter image description here][2]][2]
>I tried to do pH=12(pH1+pH2), but it’s trivial and doesn't work.
To convince yourself once and for all that does not work, consider two example:
1. Mixing 5 mL of 1 M HCl with 6 mL of 1 M NaOH (starting pH values are 0 and 14, resulting pH is roughly 13).
2. Mixing 10 ml of 1 M HCl with 90 mL of water (starting pH values are 0 and 7, resulting pH is 1)
[1]: https://i.stack.imgur.com/4QoQW.jpg
[2]: https://i.stack.imgur.com/XKUoi.jpg |
I am making a simple restricted HF code using the Python interface of Psi4. I am now evaluating convergence by tracking the change in the sum of orbital energies, but I want to do this in a better way. It is common to use the fact that at self-consistency, the Fock and density matrices commute
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} - \mathbf{DF} = \mathbf{0}
$$
However, the above expression is only valid in MO basis, while F and D in my code are computed in AO basis. So I need to derive an equivalent expression in AO basis. I am quite sure the correct expression is
$$
[\mathbf{F}, \mathbf{D}]^{\text{AO}} = \mathbf{FDS} - \mathbf{SDF}
$$
as this is equal to zero to within $1\times10^{-14}$. But how to derive this?
Derivation
--------------------
An arbitrary molecular orbital $\phi_i$ is expanded in atomic orbital basis functions
$$
\phi_i = \sum_\alpha C_{\alpha i} \chi _{\alpha}
$$
Acting the commutator on $\phi_i$ and expanding it to AO basis yields
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha} - \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$
Since we know the solution contains the overlap matrix $\mathbf{S}$, lets look at the definition
$$
\mathbf{S}_{ij} = \langle \chi_i(\mathbf{r}) \vert \chi_j(\mathbf{r}) \rangle = \int d\mathbf{r} \chi^*(\mathbf{r})\chi(\mathbf{r})
$$
Since this must be part of our expression, it seems to me a good approach is to multiply from the left by $\sum_\beta C_{\beta i}^* \chi_\beta^*$ (dropping the $\mathbf{r}$ dependence from now on) and integrating over $\mathbf{r}$
$$
[\mathbf{F}, \mathbf{D}] = \int \sum_\beta C_{\beta i}^* \chi_\beta^* \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha} - \int \sum_\beta C_{\beta i}^* \chi_\beta^* \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$
from in braket notation becomes
$$
[\mathbf{F}, \mathbf{D}] = \langle \sum_\beta C_{\beta i}^* \chi_\beta^* \vert \mathbf{FD} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle - \langle \sum_\beta C_{\beta i}^* \chi_\beta^* \vert \mathbf{DF} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle
$$
At this point I am not sure what to do - or if am I even on the right track. I can see that we have the "pieces" that make up the overlap matrix, but I don't know how to put them together. |
I am making a simple restricted HF code using the Python interface of Psi4. I am now evaluating convergence by tracking the change in the sum of orbital energies, but I want to do this in a better way. It is common to use the fact that at self-consistency, the Fock and density matrices commute
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} - \mathbf{DF} = \mathbf{0}
$$
However, the above expression is only valid in MO basis, while F and D in my code are computed in AO basis. So I need to derive an equivalent expression in AO basis. I am quite sure the correct expression is
$$
[\mathbf{F}, \mathbf{D}]^{\text{AO}} = \mathbf{FDS} - \mathbf{SDF}
$$
as this is equal to zero to within $1\times10^{-14}$. But how to derive this?
Derivation
--------------------
An arbitrary molecular orbital $\phi_i$ is expanded in atomic orbital basis functions
$$
\phi_i = \sum_\alpha C_{\alpha i} \chi _{\alpha}
$$
Acting the commutator on $\phi_i$ and expanding it to AO basis yields
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha} - \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$
Since we know the solution contains the overlap matrix $\mathbf{S}$, lets look at the definition
$$
\mathbf{S}_{ij} = \langle \chi_i(\mathbf{r}) \vert \chi_j(\mathbf{r}) \rangle = \int d\mathbf{r} \chi^*(\mathbf{r})\chi(\mathbf{r})
$$
Since this must be part of our expression, it seems to me a good approach is to multiply from the left by $\sum_\beta C_{\beta i}^* \chi_\beta^*$ (dropping the $\mathbf{r}$ dependence from now on) and integrating over $\mathbf{r}$
$$
[\mathbf{F}, \mathbf{D}] = \int \sum_\beta C_{\beta i}^* \chi_\beta^* \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha} - \int \sum_\beta C_{\beta i}^* \chi_\beta^* \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$
from in braket notation becomes
$$
[\mathbf{F}, \mathbf{D}] = \langle \sum_\beta C_{\beta i}^* \chi_\beta^* \vert \mathbf{FD} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle - \langle \sum_\beta C_{\beta i}^* \chi_\beta^* \vert \mathbf{DF} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle
$$
At this point I am not sure what to do - or if am I even on the right track. I can see that we have the "pieces" that make up the overlap matrix, but I don't know how to put them together. Further, due to the orthonormality of the MOs, then I can imagine that the summation terms only survive when $\alpha = \beta$. But I'm not sure how to derive this properly. |
I am making a simple restricted HF code using the Python interface of Psi4. I am now evaluating convergence by tracking the change in the sum of orbital energies, but I want to do this in a better way. It is common to use the fact that at self-consistency, the Fock and density matrices commute
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} - \mathbf{DF} = \mathbf{0}
$$
However, the above expression is only valid in MO basis, while F and D in my code are computed in AO basis. So I need to derive an equivalent expression in AO basis. I am quite sure the correct expression is
$$
[\mathbf{F}, \mathbf{D}]^{\text{AO}} = \mathbf{FDS} - \mathbf{SDF}
$$
as this is equal to zero to within $1\times10^{-14}$. But how to derive this?
Derivation
--------------------
An arbitrary molecular orbital $\phi_i$ is expanded in atomic orbital basis functions
$$
\phi_i = \sum_\alpha C_{\alpha i} \chi _{\alpha}
$$
Acting the commutator on $\phi_i$ and expanding it to AO basis yields
$$
[\mathbf{F}, \mathbf{D}] = \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha} - \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$
Since we know the solution contains the overlap matrix $\mathbf{S}$, lets look at the definition
$$
\mathbf{S}_{ij} = \langle \chi_i(\mathbf{r}) \vert \chi_j(\mathbf{r}) \rangle = \int d\mathbf{r} \chi^*(\mathbf{r})\chi(\mathbf{r})
$$
Since this must be part of our expression, it seems to me a good approach is to multiply from the left by $\sum_\beta C_{\beta i}^* \chi_\beta^*$ (dropping the $\mathbf{r}$ dependence from now on) and integrating over $\mathbf{r}$
$$
[\mathbf{F}, \mathbf{D}] = \int \sum_\beta C_{\beta i}^* \chi_\beta^* \mathbf{FD} \sum_\alpha C_{\alpha i} \chi_{\alpha} - \int \sum_\beta C_{\beta i}^* \chi_\beta^* \mathbf{DF} \sum_\alpha C_{\alpha i} \chi_{\alpha}
$$
from in braket notation becomes
$$
[\mathbf{F}, \mathbf{D}] = \langle \sum_\beta C_{\beta i} \chi_\beta \vert \mathbf{FD} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle - \langle \sum_\beta C_{\beta i} \chi_\beta \vert \mathbf{DF} \vert \sum_\alpha C_{\alpha i} \chi_{\alpha} \rangle
$$
At this point I am not sure what to do - or if am I even on the right track. I can see that we have the "pieces" that make up the overlap matrix, but I don't know how to put them together. Further, due to the orthonormality of the MOs, then I can imagine that the summation terms only survive when $\alpha = \beta$. But I'm not sure how to derive this properly. |
Yes, HNO2 is unstable, however, [per Wikipedia](https://en.wikipedia.org/wiki/Dinitrogen_trioxide) the gaseous anhydride N2O3 is described as a:
>deep blue solid[1] is one of the simple nitrogen oxides. It forms upon mixing equal parts of nitric oxide and nitrogen dioxide and cooling the mixture below −21 °C (−6 °F):[2]
>$\ce{NO + NO2 ⇌ N2O3}$
>Dinitrogen trioxide is only isolable at low temperatures, i.e. in the liquid and solid phases. At higher temperatures the equilibrium favors the constituent gases...
And further:
>Nitrite salts are sometimes produced by adding N2O3 to solutions of bases:
>$\ce{N2O3 + 2 NaOH → 2 NaNO2 + H2O}$
So, equivalently, one may be able to employ stable stored cold N2O3 and not work with in situ formed nitrous acid.
So, on the question, "Why is nitrous acid prepared in situ?", so one possible answer is that it is actually an elective (not to employ N2O3), and not always a requirement.
|
Yes, HNO2 is unstable, however, [per Wikipedia](https://en.wikipedia.org/wiki/Dinitrogen_trioxide) the gaseous anhydride N2O3 is described as a:
>deep blue solid[1] is one of the simple nitrogen oxides. It forms upon mixing equal parts of nitric oxide and nitrogen dioxide and cooling the mixture below −21 °C (−6 °F):[2]
>$\ce{NO + NO2 ⇌ N2O3}$
>Dinitrogen trioxide is only isolable at low temperatures, i.e. in the liquid and solid phases. At higher temperatures the equilibrium favors the constituent gases...
And further:
>Nitrite salts are sometimes produced by adding N2O3 to solutions of bases:
>$\ce{N2O3 + 2 NaOH → 2 NaNO2 + H2O}$
So, equivalently, one may be able to employ stable stored cold N2O3 and not work with in situ formed nitrous acid.
So, on the question, "Why is nitrous acid prepared in situ?", one possible answer is that it is actually an elective (not to employ N2O3), and not always a requirement.
|
I want to ask a question about the C=C streching modes of Salicylamide.
I was presented with the following spectra and asked to comment on the bonds causing the peaks.
[![enter image description here][1]][1]
I identified the molecule from this and its NMR spectra to be salicylamide.
[![Diagram of Salicylamide][2]][2]
Here is my understanding of the IR spectra (solid state, ATR).
> - $3391.66 cm ^{-1}$ is the asymmetric stretch of the $\ce{N-H}$ in the amide.
> - $3184.80 cm^{-1}$ is the symmetric stretch of the $\ce{N-H}$ in the amide.
From previous understanding, I was aware that the asymmetric stretch requires more energy, and hence a higher wavenumber than the symmetric stretch.
> - $2736.27 cm^{-1}$ is most possibly the $\ce{O-H}$ peak of the phenol. I attributed this to a few reasons, possibly due to hydrogen bonding or due to the dissociation of the $\ce{O-H}$ bond, which is driven to the right hand side of the equilibrium by the presence of the EWG amide group.
> - $1672.98 cm^{-1}$ is due to the $\ce{C=O}$ stretch of the amide.
> - $1628.09$ and $1589.49 cm^{-1}$ are due to $\ce{C=C}$ aromatic stretches.
I tried to find information to help me describe the $\ce{C=C}$ stretches (i.e. what orientation, any dipole arrows) but I failed to find any information to support my argument.
I recall in ferrocene and acetalferrocene, the *ring breathing* effect is observed, whereby ferrocene has fewer IR peaks than acetalferrocene due to a lack of a net dipole moment being non-zero, and I wondered whether there was a similar vibrational mode to apply for these two $\ce{C=C}$ peaks, but I was not sure.
What are the directions of the C=C stretches in the molecule of salicylamide in the range $1580 - 1630 cm^{-1}$?
[1]: https://i.stack.imgur.com/bSrZi.png
[2]: https://i.stack.imgur.com/w4Hwq.png |
>How would be the equation and the ICE table, and what is the pH of the mixture of these two solutions?
There are three equations:
$$\ce{HCOOH <=> H+ + HCOO-}$$
$$\ce{CH3COOH <=> H+ + CH3COO-}$$
$$\ce{H2O <=> H+ + OH-}$$
Hydroxide is a (very) minor species, so you can neglect it. As I explain below, at the pH of the mixture, the acetate ion is also a minor species, so you can first neglect the dissociation of acetic acid, and then correct for it later.
I would not use an ICE table in this situation, at least not one that describes all three reactions.
**My strategy to determine the pH of the mixture**
First, you can determine the equilibrium constants for each acid from the pH values of their aqueous solutions (before mixing). The corresponding pKa values are 3.74 and 4.78 for formic and acetic acid, respectively.
After mixing, the concentration of total formic acid and total acetic acid drop by a factor of 2 (mutual dilution). Formic acid is the stronger acid, and would have a pH of 3.06 at this concentration in the absence of acetic acid (less acidic than pH = 2.9 because of dilution).
At pH 3.06, acetic acid dissociates very little (the difference between pH and pKa is more than 1.5, vs. less than 0.7 for formic acid). As a first approximation, we can say that the pH is determined by the stronger acid.
To figure out how much the presence of acetic acid does influence the pH, we can calculate how much it would dissociate at pH 3.06, and correct the pH accordingly. Our new pH will be more acidic, so we assumed a too acidic pH in our correction, overshooting a bit. After a couple of iterations of this, the pH comes out as 3.04.
So the biggest effect of adding the acetic acid comes from diluting the formic acid (pH would go from 2.9 to 3.06 just by adding water). A secondary effect comes from acetic acid dissociating a tiny bit, giving a pH of 3.04.
Here are results from the spreadsheet used for the calculations, along with the formulas used (to dampen the overshooting effect, the pH value used in each step is the average of the updated pH and the previous pH):
[![enter image description here][1]][1]
[![enter image description here][2]][2]
>I tried to do pH=12(pH1+pH2), but it’s trivial and doesn't work.
To convince yourself once and for all that pH averaging does not work, consider these two extreme examples:
1. Mixing 5 mL of 1 M HCl with 6 mL of 1 M NaOH (starting pH values are 0 and 14, resulting pH is roughly 13).
2. Mixing 10 ml of 1 M HCl with 90 mL of water (starting pH values are 0 and 7, resulting pH is 1)
[1]: https://i.stack.imgur.com/4QoQW.jpg
[2]: https://i.stack.imgur.com/XKUoi.jpg |
>How would be the equation and the ICE table, and what is the pH of the mixture of these two solutions?
There are three equations:
$$\ce{HCOOH <=> H+ + HCOO-}$$
$$\ce{CH3COOH <=> H+ + CH3COO-}$$
$$\ce{H2O <=> H+ + OH-}$$
Hydroxide is a (very) minor species, so you can neglect it. As I explain below, at the pH of the mixture, the acetate ion is also a minor species, so you can first neglect the dissociation of acetic acid, and then correct for it later.
I would not use an ICE table in this situation, at least not one that describes all three reactions.
**My strategy to determine the pH of the mixture**
First, you can determine the equilibrium constants for each acid from the pH values of their aqueous solutions (before mixing). The corresponding pKa values are 3.74 and 4.78 for formic and acetic acid, respectively.
After mixing, the concentration of total formic acid and total acetic acid drop by a factor of 2 (mutual dilution). Formic acid is the stronger acid, and would have a pH of 3.06 at this concentration in the absence of acetic acid (less acidic than pH = 2.9 because of dilution).
At pH 3.06, acetic acid dissociates very little (the difference between pH and pKa is more than 1.5, vs. less than 0.7 for formic acid). As a first approximation, we can say that the pH is determined by the stronger acid.
To figure out how much the presence of acetic acid does influence the pH, we can calculate how much it would dissociate at pH 3.06, and correct the pH accordingly. Our new pH will be more acidic, so we assumed a too acidic pH in our correction, overshooting a bit. After a couple of iterations of this, the pH comes out as 3.04.
So the biggest effect of adding the acetic acid comes from diluting the formic acid (pH would go from 2.9 to 3.06 just by adding water). A secondary effect comes from acetic acid dissociating a tiny bit, giving a pH of 3.04.
Here are results from the spreadsheet used for the calculations, along with the formulas used (to dampen the overshooting effect, the pH value used in each step is the average of the updated pH and the previous pH):
[![enter image description here][1]][1]
[![enter image description here][2]][2]
>I tried to do $\mathrm{pH}=\frac{1}{2}\left(\mathrm{pH}_1+\mathrm{pH}_2\right)$, but it’s trivial and doesn't work.
To convince yourself once and for all that pH averaging does not work, consider these two extreme examples:
1. Mixing 5 mL of 1 M HCl with 6 mL of 1 M NaOH (starting pH values are 0 and 14, resulting pH is roughly 13).
2. Mixing 10 ml of 1 M HCl with 90 mL of water (starting pH values are 0 and 7, resulting pH is 1)
[1]: https://i.stack.imgur.com/4QoQW.jpg
[2]: https://i.stack.imgur.com/XKUoi.jpg |
## Below a more general approach.
Suppose that we have two weak acids $\ce{HA}$ and $\ce{HB}$.
The initial concentrations are $C^0_\ce{HA}$ and $C^0_\ce{HB}$, and their constants are $K_{\mathrm{a},\ce{(HA)}}$ and $K_{\mathrm{a},\ce{(HB)}}$.
Suppose yet that volumes, $V_\ce{HA}$ and $V_\ce{HB}$, are additives.
So we have:
1. *Reactions*
\begin{align}
\ce{HA + H2O &<=> H3O+ + A-} &
K_{\mathrm{a},(\ce{HA})}
&= \frac{\ce{[H3O+][A-]}}{\ce{[HA]}}
\tag{1}\label{eq:KAcidHA}\\
\ce{HB + H2O &<=> H3O+ + B-} &
K_{\mathrm{a},(\ce{HB})}
&=\frac{\ce{[H3O+][B-]}}{\ce{[HB]}}
\tag{2}\label{eq:KAcidHB}\\
\ce{2 H2O &<=> H3O+ + OH-} &
K_\mathrm{w} &= \ce{[H3O+][OH-]}\tag{3}\label{eq:KWater}
\end{align}
2. *Mass balance*
\begin{align}
C_\ce{HA} &= \frac{C^0_\ce{HA} V_\ce{HA}}{V_\ce{HA} + V_\ce{HB}}
&&=\ce{[HA] + [A-]}\tag{4}\label{eq:MassBalanceHA}\\
C_\ce{HB} &= \frac{C^0_\ce{HB} V_\ce{HB}}{V_\ce{HA} + V_\ce{HB}}
&&=\ce{[HB] + [B-]}\tag{5}\label{eq:MassBalanceHB}
\end{align}
3. *Charge balance*
$$\ce{[H3O+] = [OH-] + [A-] + [B-]}\tag{6}\label{eq:ChargeBalance}$$
Replacing \eqref{eq:KAcidHA}–\eqref{eq:MassBalanceHB} equations on \eqref{eq:ChargeBalance}, we have:
$$
\ce{[H3O+]} =
\frac{K_\mathrm{w}}{\ce{[H3O+]}}
+\frac{C_\ce{HA} K_{\mathrm{a},(\ce{HA})}}{\ce{[H3O+]}
+ K_{\mathrm{a},(\ce{HA})}}
+ \frac{C_\ce{HB} K_{\mathrm{a},(\ce{HB})}}{\ce{[H3O+]}
+ K_{\mathrm{a},(\ce{HB})}}
\tag{7}\label{eq:GeneralEquation}$$
or as polynomial
\begin{align}
\begin{split}
&\ce{[H3O+]}^4\\
+&\ce{[H3O+]}^3 (K_{\mathrm{a},(\ce{HA})}
+ K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}^2 \left[
K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
- ( C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB}K_{\mathrm{a},(\ce{HB})} )
- K_\mathrm{w} \right]\\
-&\ce{[H3O+]} \left[
(C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
+ K_\ce{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})}) \right]\\
-& K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w}\\
=&\ 0
\end{split}\tag{8}\label{eq:GeneralPol}
\end{align}
This single equation will exactly solve any equilibrium problem involving the mixture of any two monoprotic acids, in any concentration (as long as they're not much higher than about $\pu{1 mol L-1}$) and any volume. Depending of $K_\mathrm{a}$ values, we can yet obtain a simpler version.
The \eqref{eq:GeneralPol} equation can simplified considering that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} K_\mathrm{w} \ll 1$.
\begin{align}
\begin{split}
&\ce{[H3O+]}^3\\
+&\ce{[H3O+]}^2 (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})\\
+&\ce{[H3O+]}\left[
K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
- (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})})
- K_\mathrm{w} \right]\\
-&\left[
(C_\ce{HA} + C_\ce{HB}) K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})}
+ K_\mathrm{w} (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
\right]\\
=&\ 0
\end{split}\tag{9}\label{eq:GeneralPolSimp1}
\end{align}
The \eqref{eq:GeneralPolSimp1} equation can simplified considering that $K_{\mathrm{a},(\ce{HA})} K_{\mathrm{a},(\ce{HB})} \ll 1$ and disregarding the autoionization of water.
\begin{align}
\ce{[H3O+]}^2 + \ce{[H3O+]} (K_{\mathrm{a},(\ce{HA})}
+ K_{\mathrm{a},(\ce{HB})})
- (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})})
= 0\tag{10}\label{eq:GeneralPolSimp2}
\end{align}
The \eqref{eq:GeneralPolSimp2} equation can be solved as usual.
$$\ce{[H3O+]} =
\frac{
- (K{_\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
+ \sqrt{
(K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
+ 4 (C_\ce{HA} K_{\mathrm{a},(\ce{HA})}
+ C_\ce{HB} K_{\mathrm{a},(\ce{HB})})}
}{2}$$
Or using the initial concentrations
$$\ce{[H3O+]} =
\frac{
- (K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})
+ \sqrt{
(K_{\mathrm{a},(\ce{HA})} + K_{\mathrm{a},(\ce{HB})})^2
+ 4\left(%\displaystyle
\frac{C^0_\ce{HA} V_\ce{HA} K_{\ce{a},(\ce{HA})}
}{V_\ce{HA} + V_\ce{HB}}
+ \frac{C^0_\ce{HB} V_\ce{HB} K_{\ce{a},(\ce{HB})}
}{V_\ce{HA} + V_\ce{HB}}
\right)}
}{2}$$
Replacing $C^0_\ce{HA}=\pu{0.01 mol L-1}$, $C^0_\ce{HB}=\pu{0.01 mol L-1}$, $V_\ce{HA}=\pu{0.050 L}$ and $V_\ce{HB}=\pu{0.050 L}$, and using $\text{p}K_\ce{a}=3.75$ for formic acid and $\text{p}K_\ce{a}=4.756$ for acetic acid<sup>\[1\]</sup>, we have
$$\ce{pH}=3.05.$$
**Reference**
1. [Haynes, W. M.; Lide, D. R.; Bruno, T. J., *CRC Handbook of Chemistry and Physics*. 97 ed.; CRC Press: 2017; pp. 5–88][1].
[1]: https://books.google.com.br/books?id=VVezDAAAQBAJ&lpg=PP1&hl=pt-BR&pg=SA5-PA88#v=onepage&q&f=false |
Since -OH and —CH3 are -o,p directing groups, shouldn’t the major product be ***3-Methyl-2-nitrophenol***? Why not?[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/awAnv.jpg |
The following question comes from *General Chemistry: Principles and Modern Applications, 11th ed.* by Pettruci.
> **Example 15-33.** A mixture consisting of 0.150 mol H<sub>2</sub> and 0.150 mol I<sub>2</sub> is brought to equilibrium at 445 °C, in a 3.25 L flask. What are the equilibrium amounts of H<sub>2</sub>, I<sub>2</sub>, and HI?
$$\ce{H2(g) + I2(g) <=> 2HI(g)}\qquad K_c=50.2\text{ at }\pu{445 °C}$$
First, I found the concentrations of H<sub>2</sub> and I<sub>2</sub>:
$$
\begin{aligned}
\phantom{}[\ce{H2}]=[\ce{I2}]&=\frac{\pu{0.150 mol}}{\pu{3.25 L}} \\
&=\pu{0.0461538 M}
\end{aligned}
$$
Setting up the ICE table:
$$
\begin{array}{|l|c c c c c|}
\hline
&\ce{H2}&+&\ce{I2}&\rightleftharpoons&\ce{2HI} \\
\hline
\text{Initial}&0.0461538&&0.0461538&&0 \\
\text{Change}&-x&&-x&&+2x \\
\text{Equilibrium}&0.0461538-x&&0.0461538-x&&2x \\
\hline
\end{array}
$$
Then,
$$
K_c=50.2=\frac{(2x)^2}{(0.0461538-x)(0.0461538-x)}.
$$
Most solutions I've seen for similar problems involve taking the square root of both sides of this equation. However, they only take the principal root, effectively ignoring a second valid solution. Here, I'll do the proper solution for finding $x$:
$$
\begin{aligned}
\\
50.2&=\left(\frac{2x}{0.0461538-x}\right)^2 \\
\sqrt{50.2}&=\pm\frac{2x}{0.0461538-x}
\end{aligned} \\
\begin{aligned}
7.0852&=\frac{2x}{0.0461538-x}&&\text{or}&7.0852&=-\frac{2x}{0.0461538-x} \\
0.327009-7.0852x&=2x&&&0.327009-7.0852x&=-2x \\
0.327009&=9.0852x&&&0.327009&=5.0852x \\
x&=0.0359936&&&x&=0.064306
\end{aligned}
$$
As you can see, there are two positive (and therefore, valid) solutions for $x$. However, only one of the solutions results in a positive $[\ce{H2}]_\mathrm{eq}$:
$$
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}=0.0461538-x
\end{aligned}
\\
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}&=0.0461538-0.0359936&&\text{or}&[\ce{H2}]_\mathrm{eq}&=0.0461538-0.064306 \\
&=\pu{0.0101602 M}&&&&=\pu{-0.0181522 M}
\end{aligned}
$$
And $x=0.0359936$ does indeed lead to the correct answer according to the textbook.
My question is: was it just a coincidence that I was able to eliminate one of the possible solutions for $x$? In other words, is there a scenario where I could end up with two possible $[\ce{H2}]_\mathrm{eq}$s? If not, what forbids this? |
Can ICE tables give two solutions? What does this mean physically? |
The following question comes from *General Chemistry: Principles and Modern Applications, 11th ed.* by Pettruci.
> **Example 15-33.** A mixture consisting of 0.150 mol H<sub>2</sub> and 0.150 mol I<sub>2</sub> is brought to equilibrium at 445 °C, in a 3.25 L flask. What are the equilibrium amounts of H<sub>2</sub>, I<sub>2</sub>, and HI?
$$\ce{H2(g) + I2(g) <=> 2HI(g)}\qquad K_c=50.2\text{ at }445\,\mathrm{^\circ C}$$
First, I found the concentrations of H<sub>2</sub> and I<sub>2</sub>:
$$
\begin{aligned}
\phantom{}[\ce{H2}]=[\ce{I2}]&=\frac{0.150\text{ mol}}{3.25\text{ L}} \\
&=0.0461538\text{ M}
\end{aligned}
$$
Setting up the ICE table:
$$
\begin{array}{|l|c c c c c|}
\hline
&\ce{H2}&+&\ce{I2}&\rightleftharpoons&\ce{2HI} \\
\hline
\text{Initial}&0.0461538&&0.0461538&&0 \\
\text{Change}&-x&&-x&&+2x \\
\text{Equilibrium}&0.0461538-x&&0.0461538-x&&2x \\
\hline
\end{array}
$$
Then,
$$
K_c=50.2=\frac{(2x)^2}{(0.0461538-x)(0.0461538-x)}.
$$
Most solutions I've seen for similar problems involve taking the square root of both sides of this equation. However, they only take the principal root, effectively ignoring a second valid solution. Here, I'll do the proper solution for finding $x$:
$$
\begin{aligned}
\\
50.2&=\left(\frac{2x}{0.0461538-x}\right)^2 \\
\sqrt{50.2}&=\pm\frac{2x}{0.0461538-x}
\end{aligned} \\
\begin{aligned}
7.0852&=\frac{2x}{0.0461538-x}&&\text{or}&7.0852&=-\frac{2x}{0.0461538-x} \\
0.327009-7.0852x&=2x&&&0.327009-7.0852x&=-2x \\
0.327009&=9.0852x&&&0.327009&=5.0852x \\
x&=0.0359936&&&x&=0.064306
\end{aligned}
$$
As you can see, there are two positive (and therefore, valid) solutions for $x$. However, only one of the solutions results in a positive $[\ce{H2}]_\mathrm{eq}$:
$$
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}=0.0461538-x
\end{aligned}
\\
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}&=0.0461538-0.0359936&&\text{or}&[\ce{H2}]_\mathrm{eq}&=0.0461538-0.064306 \\
&=0.0101602\text{ M}&&&&=-0.0181522\text{ M}
\end{aligned}
$$
And $x=0.0359936$ does indeed lead to the correct answer according to the textbook.
My question is: was it just a coincidence that I was able to eliminate one of the possible solutions for $x$? In other words, is there a scenario where I could end up with two possible $[\ce{H2}]_\mathrm{eq}$s? If not, what forbids this? |
An ICE table is moderately complicated, so it ends up seeming a bit mysterious when you get two solutions. But it's really not that mysterious at all. In fact, this sort of thing is pretty common, since it happens anytime you have two variables related to each other through a square. Let's use a very simple example to illustrate this.
Suppose you are given the following problem:
The area of a square is $25\, \mathrm m^2$. What is the length of each side?
To solve this, we use $\mathrm{A} = \mathrm l^2$, where A is the area, and l is the length of each side. Hence:
$$l = \sqrt{25} =\pm 5$$
Now clearly a side can't have a negative length—it's non-physical. So we choose $\mathrm l = +5 \mathrm \,m$ instead of $\mathrm l = -5 \mathrm \,m$. Nothing mysterious about that. I.e., what you were encountering was merely a property of squares, not a property specific to ICE tables.
|
When an electron is excited to a singlet state, must it flip its spin state to enter the triplet state and flip once more to the ground state? |
Compounds containing an oxime functional group, such as 2-PAM and Obidoxime, are used in the treatment of poisoning with organophosphorus compounds. These oximes reactivate the enzyme acetylcholinesterase, which has been inhibited by the OPC.
However, some organophosphorus compounds are resistant to re-activation by oximes such as 2-PAM and Obidoxime for reasons not related to denaturation of the enzyme (Aging), but are responsive to some other oximes. For instance, methylfluorophosphonylcholine and related compounds such as GV are more susceptible to treatment by HI-6 than Obidoxime. Why?
Both of these compounds have fluorine act as a leaving group and have an amino-chain that would occupy the anionic site of the enzyme, so I presume that their resistance to reactivation by older oximes has to do with steric effects, though I have not been able to find research confirming or denying this. Nevertheless, why are some oxime compounds, especially the larger ones such as HI-6 and Hlo-7, better reactivators of AChE than others, especially if some OPCs' resistance to reactivation actually is due to steric effects?
Sources:
https://pubmed.ncbi.nlm.nih.gov/9106387-therapeutic-efficacy-of-obidoxime-or-hi-6-with-atropine-against-intoxication-with-some-nerve-agents-in-mice/
https://link.springer.com/article/10.1007%2Fs00204-003-0533-0
https://www.sciencedirect.com/science/article/abs/pii/0006295259900589?via%3Dihub |
I understand the Jablonski diagram in that it has intersystem crossing from the singlet state to the triplet state, but how many spin switches are necessary to complete to phosphores? When an electron is excited to a singlet state, must it flip its spin state to enter the triplet state and flip once more to the ground state? |
The following question comes from *General Chemistry: Principles and Modern Applications, 11th ed.* by Pettruci.
> **Example 15-33.** A mixture consisting of 0.150 mol H<sub>2</sub> and 0.150 mol I<sub>2</sub> is brought to equilibrium at 445 °C, in a 3.25 L flask. What are the equilibrium amounts of H<sub>2</sub>, I<sub>2</sub>, and HI?
$$\ce{H2(g) + I2(g) <=> 2HI(g)}\qquad K_c=50.2\text{ at }445\,\mathrm{^\circ C}$$
First, I found the concentrations of H<sub>2</sub> and I<sub>2</sub>:
$$
\begin{aligned}
\phantom{}[\ce{H2}]=[\ce{I2}]&=\frac{0.150\text{ mol}}{3.25\text{ L}} \\
&=0.0461538\text{ M}
\end{aligned}
$$
Setting up the ICE table:
$$
\begin{array}{|l|c c c c c|}
\hline
&\ce{H2}&+&\ce{I2}&\rightleftharpoons&\ce{2HI} \\
\hline
\text{Initial}&0.0461538&&0.0461538&&0 \\
\text{Change}&-x&&-x&&+2x \\
\text{Equilibrium}&0.0461538-x&&0.0461538-x&&2x \\
\hline
\end{array}
$$
Then,
$$
K_c=50.2=\frac{(2x)^2}{(0.0461538-x)(0.0461538-x)}.
$$
Most solutions I've seen for similar problems involve taking the square root of both sides of this equation. However, they only take the principal root, effectively ignoring a second valid solution. Here, I'll do the proper solution for finding $x$:
$$
\begin{aligned}
\\
50.2&=\left(\frac{2x}{0.0461538-x}\right)^2 \\
\sqrt{50.2}&=\pm\frac{2x}{0.0461538-x}
\end{aligned} \\
\begin{aligned}
7.0852&=\frac{2x}{0.0461538-x}&&\text{or}&7.0852&=-\frac{2x}{0.0461538-x} \\
0.327009-7.0852x&=2x&&&0.327009-7.0852x&=-2x \\
0.327009&=9.0852x&&&0.327009&=5.0852x \\
x&=0.0359936&&&x&=0.064306
\end{aligned}
$$
As you can see, there are two positive (and therefore, valid) solutions for $x$. However, only one of the solutions results in a positive $[\ce{H2}]_\mathrm{eq}$:
$$
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}=0.0461538-x
\end{aligned}
\\
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}&=0.0461538-0.0359936&&\text{or}&[\ce{H2}]_\mathrm{eq}&=0.0461538-0.064306 \\
&=0.0101602\text{ M}&&&&=-0.0181522\text{ M}
\end{aligned}
$$
And $x=0.0359936$ does indeed lead to the correct answer according to the textbook.
My question is: was it just a coincidence that I was able to eliminate one of the possible solutions for $x$? In other words, is there a scenario where I could end up with two possible $[\ce{H2}]_\mathrm{eq}$s? If not, what forbids this?
**Edit:** To reiterate, I am aware that one of the solutions is to be eliminated for being "non-physical". My question is whether it is possible for both solutions to be physical. |
Compounds containing an oxime functional group, such as 2-PAM and Obidoxime, are used in the treatment of poisoning with organophosphorus compounds. These oximes reactivate the enzyme acetylcholinesterase, which has been inhibited by the OPC.
However, some organophosphorus compounds are resistant to re-activation by oximes such as 2-PAM and Obidoxime for reasons not related to denaturation of the enzyme (Aging), but are responsive to some other oximes. For instance, methylfluorophosphonylcholine and related compounds such as GV are more susceptible to treatment by HI-6 than to Obidoxime. Why?
Both of these compounds have fluorine act as a leaving group and have an amino-chain that would occupy the anionic site of the enzyme, so I presume that their resistance to reactivation by older oximes has to do with steric effects, though I have not been able to find research confirming or denying this. Nevertheless, why are some oxime compounds, especially the larger ones such as HI-6 and Hlo-7, better reactivators of AChE than others, especially if some OPCs' resistance to reactivation actually is due to steric effects?
Sources:
https://pubmed.ncbi.nlm.nih.gov/9106387-therapeutic-efficacy-of-obidoxime-or-hi-6-with-atropine-against-intoxication-with-some-nerve-agents-in-mice/
https://link.springer.com/article/10.1007%2Fs00204-003-0533-0
https://www.sciencedirect.com/science/article/abs/pii/0006295259900589?via%3Dihub |
All organic matter seems to be inflammable when dried up. Petroleum, Alcohol, Wood, and Coal burns readily. Can I call it a general rule? Are there known exceptions? What makes them the exceptions? |
Do all anhydrous organic matter burn? |
An ICE table is moderately complicated, so it ends up seeming a bit mysterious when you get two solutions. But it's really not that mysterious at all. In fact, this sort of thing is pretty common, since it happens anytime you have two variables related to each other through a square. Let's use a very simple example to illustrate this.
Suppose you are given the following problem:
The area of a square is $\pu{25 m2}$. What is the length of each side?
To solve this, we use $A = l^2$, where $A$ is the area, and $l$ is the length of each side. Hence:
$$l = \sqrt{\pu{25 m2}} = \pu{\pm 5 m}$$
Now clearly a side can't have a negative length — it's non-physical. So we choose $l = \pu{+5 m}$ instead of $l = \pu{-5 m}.$ Nothing mysterious about that. I.e., what you were encountering was merely a property of squares, not a property specific to ICE tables.
|
I am currently studying the textbook *Mass Spectrometry*, third edition, by Jürgen H. Gross. Chapter **2.4.3 Bond Dissociation Energies and Heats of Formation** says the following:
> Great efforts have been made to generate accurate and reliable ion thermochemistry data. Once such data is available, it can be employed to elucidate fragmentation mechanisms and in addition, it is useful for obtaining some background on the energetic dimensions in mass spectrometry.
> Heats of formation of neutral molecules, $\Delta H_{f(RH)}$, can be obtained from combustion data with high accuracy. Bond dissociation energies can either be derived for *homolytic bond dissociation*
> $$\ce{R - H \rightarrow R^. + H^., \Delta H_{Dhom}} \tag{2.12}$$
> or *heterolytic bond dissociation*
> $$\ce{R - H \rightarrow R^+ + H^-, \Delta H_{Dhet}} \tag{2.13}$$
> The homolytic bond dissociation enthalpies give the energy needed to cleave a bond of the neutral molecule which, in the gas phase is in its vibrational and electronic ground states, to obtain a pair of radicals which also are not in excited modes. Homolytic bond dissociation enthalpies are in the range of $3-5 \ eV$ (table 2.2). The heterolytic bond dissociation energies apply for the case of creating a cation and an anion in their ground states from the neutral precursor which means that these values include the amount of energy needed for charge separation; they are in the order of $10-13 \ eV$ (Table 2.3). Due to the significantly altered bonding situation, breaking of a bond is clearly less demanding in molecular ions than in neutral molecules.
I think this is good explanation. However, since this is my first exposure to these concepts, I'm interested in a higher-level explanation. Specifically,
1. What does "heats of formation of neutral molecules, $\Delta H_{f(RH)}$", mean in the context of mass spectrometry, and why is it relevant?
2. What is the relevance of homolytic vs heterolytic bond dissociation in mass spectrometry?
The Wikipedia articles for these two concepts, [1. Standard Enthalpy of Formation](https://en.wikipedia.org/wiki/Standard_enthalpy_of_formation) and [2. Bond Dissociation Energy: Homolytic Versus Heterolytic_Dissociation](https://en.wikipedia.org/wiki/Bond-dissociation_energy#Homolytic_versus_heterolytic_dissociation), offer more general explanations, but I'm interested in understanding in a general explanation *in the context of mass spectrometry*.
I would greatly appreciate it if some of the more knowledgeable contributors would please take the time to clarify these concepts. |
From NCERT Class Chemistry Part 2:
> **Example 11.5**
> Write the structures of the major products expected from the following reactions:
> (a) Mononitration of 3-methylphenol
> (b) Dinitration of 3-methylphenol
> (c) Mononitration of phenyl methanoate.
> **Solution**
>The combined influence of $\ce{-OH}$ and $\ce{-CH3}$ groups determine the position of the incoming group.
> [![nitration products][1]][1]
Since $\ce{-OH}$ and $\ce{-CH3}$ are *o*,*p*-directing groups, shouldn’t the major product be **3-methyl-2-nitrophenol**? Why not?
[1]: https://i.stack.imgur.com/6xkdK.png |
For a substructure search I would like to search for structures containing unfused benzyl.
The idea was to explicitely add hydrogen. But apparently this does not give the expected results.
So, I assume either my understanding or expectations are wrong or I'm using RDKit not properly.
I just learnt (https://chemistry.stackexchange.com/q/128440) that I better use SMARTS expressions with `Chem.MolToSmarts()` where I get the expected results.
However, if I have much more complex search structures, how can I get from the SMILES to the SMARTS? Do I have always to figure out the SMARTS code myself? Is there maybe a way just drawing a molecule and add explicit hydrogen and search with this somehow? I guess it should be pretty clear (at least to a human) what I want to find (or not find) with structure 2. Are there maybe any tools or code which can assist here?
[![enter image description here][1]][1]
**Code:**
from rdkit import Chem
smiles_strings = '''CC1=CC=CC=C1
CC1=C([H])C([H])=C([H])C([H])=C1[H]
C1(C2=CC=CC=C2)=CC=CC=C1
C12=C(CC3=CC=CC=C3C2)C=CC=C1
C12=CC=CC=C1C=C3C(C=CC=C3)=C2
C12=C(C(C(C=C3)=CC=C4)=C4C=C2)C3=CC=C1
'''
smiles_list = smiles_strings.splitlines()
def search_structure(pattern):
found = []
for idx,smiles in enumerate(smiles_list):
m = Chem.MolFromSmiles(smiles)
if m.HasSubstructMatch(pattern):
found.append(idx+1)
print("Structures found: {}".format(found))
smiles_1 = smiles_list[0]
pattern_1 = Chem.MolFromSmiles(smiles_1)
smiles_2a = smiles_list[1]
pattern_2a = Chem.MolFromSmiles(smiles_2a)
smiles_2b = Chem.MolToSmiles(pattern_2a)
pattern_2b = Chem.MolFromSmiles(smiles_2b)
smiles_2c = smiles_2b
pattern_2c = Chem.MolFromSmarts(smiles_2c)
smarts_3 = '[cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]'
pattern_3 = Chem.MolFromSmarts(smarts_3)
print("\nSMILES 1 : {}\n# my expectation: [1, 2, 3, 4, 5, 6]".format(smiles_1))
search_structure(pattern_1)
print("\nSMILES 2a: {}\n# my expectation: [1, 2, 3]".format(smiles_2a))
search_structure(pattern_2a)
print("\nSMILES 2b: {} (Mol from SMILES)\n# my expectation: [1, 2, 3]".format(smiles_2b))
search_structure(pattern_2b)
print("\nSMILES 2c: {} (Mol from SMARTS)\n# my expectation: [1, 2, 3]".format(smiles_2c))
search_structure(pattern_2c)
print("\nSMARTS 3 : {}\n# my expectation: [1, 2, 3]".format(smarts_3))
search_structure(pattern_3)
**Result:**
SMILES 1 : CC1=CC=CC=C1
# my expectation: [1, 2, 3, 4, 5, 6]
Structures found: [1, 2, 3, 4]
SMILES 2a: CC1=C([H])C([H])=C([H])C([H])=C1[H]
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3, 4]
SMILES 2b: Cc1ccccc1 (Mol from SMILES)
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3, 4]
SMILES 2c: Cc1ccccc1 (Mol from SMARTS)
# my expectation: [1, 2, 3]
Structures found: [1, 2, 4]
SMARTS 3 : [cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3]
[1]: https://i.stack.imgur.com/iUM91.png |
RDKit: how to search substructures with explicit hydrogen? |
Why does attaching an $\ce{-NO2}$ group at the *ortho* position of benzoic acid cause $\ce{-COOH}$ to become perpendicular to the plane of the benzene ring? Why doesn't the nitro group get pushed out while $\ce{-COOH}$ stays on the plane?
Also, in the case of two $\ce{-NO2}$ groups attached on the *ortho* and *meta* positions of benzoic acid, both the nitro groups become perpendicular to the benzene ring while $\ce{-COOH}$ remains at an angle to the plane. Why doesn't it become perpendicular this time?
Here is what I referred to for the structure of 2-nitrobenzoic acid :https://pubchem.ncbi.nlm.nih.gov/compound/2-Nitrobenzoic-acid#section=3D-Conformer&fullscreen=true
Here is what I referred to for the structure of 2,3-dinitrobenzoic acid :https://pubchem.ncbi.nlm.nih.gov/compound/2,3-diNitrobenzoic-acid#section=3D-Conformer&fullscreen=true
If you feel these structures arent correct please do give a reference to where I can find the right structure. :) |
Compounds containing an oxime functional group, such as [2-PAM](https://en.wikipedia.org/wiki/Pralidoxime) and [obidoxime](https://en.wikipedia.org/wiki/Obidoxime), are used in the treatment of poisoning with organophosphorus compounds. These oximes reactivate the enzyme acetylcholinesterase, which has been inhibited by the OPC.
However, some organophosphorus compounds are resistant to re-activation by oximes such as 2-PAM and obidoxime for reasons not related to denaturation of the enzyme (aging), but are responsive to some other oximes. For instance, methylfluorophosphonylcholine and related compounds such as GV are more susceptible to treatment by HI-6 than to obidoxime. Why?
Both of these compounds have fluorine act as a leaving group and have an amino-chain that would occupy the anionic site of the enzyme, so I presume that their resistance to reactivation by older oximes has to do with steric effects, though I have not been able to find research confirming or denying this. Nevertheless, why are some oxime compounds, especially the larger ones such as HI-6 and Hlo-7, better reactivators of AChE than others, especially if some OPCs' resistance to reactivation actually is due to steric effects?
### References
1. Kassa, J.; Bajgar, J. Therapeutic Efficacy of Obidoxime or HI-6 with Atropine against Intoxication with Some Nerve Agents in Mice. *Acta Medica (Hradec Kralove)* **1996**, 39 (1), 27–30. PMID: [9106387](https://www.ncbi.nlm.nih.gov/pubmed/9106387).
2. Worek, F.; Thiermann, H.; Szinicz, L. Reactivation and Aging Kinetics of Human Acetylcholinesterase Inhibited by Organophosphonylcholines. *Arch Toxicol* **2004**, 78 (4), 212–217. DOI: [10.1007/s00204-003-0533-0](https://doi.org/10.1007/s00204-003-0533-0).
3. Fredriksson, T.; Tibbling, G. Reversal of Effects on the Rat Nerve-Diaphragm Preparation Produced by Methylfluorophosphorylcholines. *Biochemical Pharmacology* **1959**, 2 (1), 63–64. DOI: [10.1016/0006-2952(59)90058-9](https://doi.org/10.1016/0006-2952(59)90058-9). |
Compounds containing an oxime functional group, such as [2-PAM](https://en.wikipedia.org/wiki/Pralidoxime) and [obidoxime](https://en.wikipedia.org/wiki/Obidoxime), are used in the treatment of poisoning with organophosphorus compounds. These oximes reactivate the enzyme acetylcholinesterase, which has been inhibited by the OPC.
However, some organophosphorus compounds are resistant to re-activation by oximes such as 2-PAM and obidoxime for reasons not related to denaturation of the enzyme (aging), but are responsive to some other oximes. For instance, methylfluorophosphonylcholine and related compounds such as GV are more susceptible to treatment by [HI-6](https://en.wikipedia.org/wiki/Asoxime_chloride) than to obidoxime. Why?
Both of these compounds have fluorine act as a leaving group and have an amino-chain that would occupy the anionic site of the enzyme, so I presume that their resistance to reactivation by older oximes has to do with steric effects, though I have not been able to find research confirming or denying this. Nevertheless, why are some oxime compounds, especially the larger ones such as HI-6 and [Hlo-7](https://pubchem.ncbi.nlm.nih.gov/compound/HLo-7-dimethanesulfonate), better reactivators of AChE than others, especially if some OPCs' resistance to reactivation actually is due to steric effects?
### References
1. Kassa, J.; Bajgar, J. Therapeutic Efficacy of Obidoxime or HI-6 with Atropine against Intoxication with Some Nerve Agents in Mice. *Acta Medica (Hradec Kralove)* **1996**, 39 (1), 27–30. PMID: [9106387](https://www.ncbi.nlm.nih.gov/pubmed/9106387).
2. Worek, F.; Thiermann, H.; Szinicz, L. Reactivation and Aging Kinetics of Human Acetylcholinesterase Inhibited by Organophosphonylcholines. *Arch Toxicol* **2004**, 78 (4), 212–217. DOI: [10.1007/s00204-003-0533-0](https://doi.org/10.1007/s00204-003-0533-0).
3. Fredriksson, T.; Tibbling, G. Reversal of Effects on the Rat Nerve-Diaphragm Preparation Produced by Methylfluorophosphorylcholines. *Biochemical Pharmacology* **1959**, 2 (1), 63–64. DOI: [10.1016/0006-2952(59)90058-9](https://doi.org/10.1016/0006-2952(59)90058-9). |
Why ZnS can be seen as making solution alkaline? Explain with reactions |
Why Normal modes of vibration are important when compared to translational and rotation modes. And why the frequencies for rotational and translational modes are close to zero? |
I am currently studying the textbook *Mass Spectrometry*, third edition, by Jürgen H. Gross. Chapter **2.4.3 Bond Dissociation Energies and Heats of Formation** says the following:
> **Energetics of $\ce{H}^\bullet$ loss from $\ce{CH^{+ \bullet}}$** The minimum energy needed to form a $\ce{CH_3^+}$ ion and a hydrogen radical from the methane molecular ion can be estimated from the heat of reaction, $\ce{\Delta H_r}$, of this process. According to Fig 2.6, $\ce{\Delta H_r} = \ce{AE_{(CH_3^+)}} - \ce{IE_{(CH_4)}}$. In order to calculate the missing $\ce{AE_{(CH_3^+)}}$ we use the tabulated values of $\ce{\Delta H_{f(H^\bullet)}} = 218.0 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_3^+)}} = 1093 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_4)}} = -74.9 \ \text{kJ mol}^{-1}$, and $\ce{IE_{(CH_4)}} = 12.6 \ \text{eV} = 1216 \ \text{kJ mol}^{-1}$. First, the heat of formation of the methane molecular ion is determined based on the experimental value of $\ce{IE_{(CH_4)}}$:
> $$\ce{\Delta H}_{f(CH_4^{+\bullet})} = \ce{\Delta H_{f(CH_4)}} + \ce{IE_{(CH_4)}} \tag{2.14}$$
> $$\ce{\Delta H}_{f(CH_4^{+\bullet})} = -74.9 \ \text{kJ mol}^{-1} + 1216 \ \text{kJ mol}^{-1} = 1141.1 \ \text{kJ mol}^{-1}$$
> Then, the heat of formation of the products is calculated from:
> $$\ce{\Delta H_{f(prod)}} = \ce{\Delta H_{f(CH_3^+)}} + \ce{\Delta H_{f(H^\bullet)}} \tag{2.15}$$
> $$\ce{\Delta H_{f(prod)}} = 1093 \ \text{kJ mol}^{-1} + 218 \ \text{kJ mol}^{-1} = 1311 \ \text{kJ mol}^{-1}$$
> Now, the heat of reaction is obtained from the difference
> $$\ce{\Delta H_r} = \ce{\Delta H_{f(prod)}} - \ce{\Delta H}_{f(CH_4^{+\bullet})} \tag{2.16}$$
> $$\ce{\Delta H_r} = 1311 \ \text{kJ mol}^{-1} - 1141.1 \ \text{kJ mol}^{-1} = 169.9 \ \text{kJ mol}^{-1}$$
> The value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 14.35 \ \text{eV}$, which is in good agreement with published values of about $14.3 \ \text{eV}$ (Fig. 2.7).
> [![enter image description here][1]][1]
Note: The amount of energy needed to be transferred to the neutral $\ce{M}$ to allow for the detection of the fragment ion $m_1^+$ is called the *appearance energy* ($AE$) of that fragment ion.
> [![enter image description here][2]][2]
> [![enter image description here][3]][3]
> [![enter image description here][4]][4]
> [![enter image description here][5]][5]
Taking the values from Fig. 2.7 and using [this][6] calculator, we get that $1311 \ \text{kJ mol}^{-1} = 13.588 \ \text{eV}$ and $1386 \ \text{kJ mol}^{-1} = 14.365 \ \text{eV}$.
Unless I have made a mistake or am misunderstanding something, these values do not agree with what the author has presented. It seems that, if we take the values from Fig 2.7, the value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 13.588 \ \text{eV}$, which implies that the published value should be around $14.365 \ \text{eV}$.
So has the author made a small mistake here? Or am I misunderstanding something?
I would greatly appreciate it if people would please take the time to review this.
[1]: https://i.stack.imgur.com/Y7HcS.png
[2]: https://i.stack.imgur.com/4zBn5.png
[3]: https://i.stack.imgur.com/HgOVR.png
[4]: https://i.stack.imgur.com/6gFxS.png
[5]: https://i.stack.imgur.com/9x6fP.png
[6]: http://www.colby.edu/chemistry/PChem/Hartree.html |
I am currently studying the textbook *Mass Spectrometry*, third edition, by Jürgen H. Gross. Chapter **2.4.3 Bond Dissociation Energies and Heats of Formation** says the following:
> **Energetics of $\ce{H}^\bullet$ loss from $\ce{CH^{+ \bullet}}$** The minimum energy needed to form a $\ce{CH_3^+}$ ion and a hydrogen radical from the methane molecular ion can be estimated from the heat of reaction, $\ce{\Delta H_r}$, of this process. According to Fig 2.6, $\ce{\Delta H_r} = \ce{AE_{(CH_3^+)}} - \ce{IE_{(CH_4)}}$. In order to calculate the missing $\ce{AE_{(CH_3^+)}}$ we use the tabulated values of $\ce{\Delta H_{f(H^\bullet)}} = 218.0 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_3^+)}} = 1093 \ \text{kJ mol}^{-1}$, $\ce{\Delta H_{f(CH_4)}} = -74.9 \ \text{kJ mol}^{-1}$, and $\ce{IE_{(CH_4)}} = 12.6 \ \text{eV} = 1216 \ \text{kJ mol}^{-1}$. First, the heat of formation of the methane molecular ion is determined based on the experimental value of $\ce{IE_{(CH_4)}}$:
> $$\ce{\Delta H}_{f(CH_4^{+\bullet})} = \ce{\Delta H_{f(CH_4)}} + \ce{IE_{(CH_4)}} \tag{2.14}$$
> $$\ce{\Delta H}_{f(CH_4^{+\bullet})} = -74.9 \ \text{kJ mol}^{-1} + 1216 \ \text{kJ mol}^{-1} = 1141.1 \ \text{kJ mol}^{-1}$$
> Then, the heat of formation of the products is calculated from:
> $$\ce{\Delta H_{f(prod)}} = \ce{\Delta H_{f(CH_3^+)}} + \ce{\Delta H_{f(H^\bullet)}} \tag{2.15}$$
> $$\ce{\Delta H_{f(prod)}} = 1093 \ \text{kJ mol}^{-1} + 218 \ \text{kJ mol}^{-1} = 1311 \ \text{kJ mol}^{-1}$$
> Now, the heat of reaction is obtained from the difference
> $$\ce{\Delta H_r} = \ce{\Delta H_{f(prod)}} - \ce{\Delta H}_{f(CH_4^{+\bullet})} \tag{2.16}$$
> $$\ce{\Delta H_r} = 1311 \ \text{kJ mol}^{-1} - 1141.1 \ \text{kJ mol}^{-1} = 169.9 \ \text{kJ mol}^{-1}$$
> The value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 14.35 \ \text{eV}$, which is in good agreement with published values of about $14.3 \ \text{eV}$ (Fig. 2.7).
> [![enter image description here][1]][1]
**Note**: The amount of energy needed to be transferred to the neutral $\ce{M}$ to allow for the detection of the fragment ion $m_1^+$ is called the *appearance energy* ($AE$) of that fragment ion.
> [![enter image description here][2]][2]
> [![enter image description here][3]][3]
> [![enter image description here][4]][4]
> [![enter image description here][5]][5]
Taking the values from Fig. 2.7 and using [this][6] calculator, we get that $1311 \ \text{kJ mol}^{-1} = 13.588 \ \text{eV}$ and $1386 \ \text{kJ mol}^{-1} = 14.365 \ \text{eV}$.
Unless I have made a mistake or am misunderstanding something, these values do not agree with what the author has presented. It seems that, if we take the values from Fig 2.7, the value of $169.9 \ \text{kJ mol}^{-1}$ ($1.75 \ \text{eV}$) corresponds to $\ce{AE_{CH_3^+}} = 13.588 \ \text{eV}$, which implies that the published value should be around $14.365 \ \text{eV}$.
So has the author made a small mistake here? Or am I misunderstanding something?
I would greatly appreciate it if people would please take the time to review this.
[1]: https://i.stack.imgur.com/Y7HcS.png
[2]: https://i.stack.imgur.com/4zBn5.png
[3]: https://i.stack.imgur.com/HgOVR.png
[4]: https://i.stack.imgur.com/6gFxS.png
[5]: https://i.stack.imgur.com/9x6fP.png
[6]: http://www.colby.edu/chemistry/PChem/Hartree.html |
> $$\ce{CF_3I + NaOH(aq) -> Major Product}$$
The Major product, according to me, should be **ditrifluorometyl ether $\ce{CF3-O-CF3}$**. The hydrogen on the $\ce{-OH}$ group is very acidic and hence reacts with $\ce{CF3I}$ to give a stable compound according to the following reaction mechanism:
[![Reaction][1]][1]
However, the textbook gives the following reaction mechanism and claims that the product is **trifluoromethane $\ce{CF_3}$**
[![Textbook reaction][2]][2]
Which one of these is correct?
[1]: https://i.stack.imgur.com/cUlWl.png
[2]: https://i.stack.imgur.com/kfbrd.png |
I came upon the following statement in a true/false question:
> Glycine can be esterified by an alcohol in the presence of inorganic acid
**Is this statement correct?**
Glycine contains the basic $\ce{-NH2}$ group, which would get protonated in the presence of inorganic acid. As a consequence, would the esterification process still be carried out on the $\ce{-COOH}$ group? [This][1] paper says that the reaction needs tetrachloromethylsilane as a catalyst and not an inorganic acid.
[1]: https://www.mdpi.com/1420-3049/13/5/1111 |
> $$\ce{CF_3I + NaOH(aq) -> Major Product}$$
The Major product, according to me, should be **ditrifluorometyl ether $\ce{CF3-O-CF3}$**. The hydrogen on the $\ce{-OH}$ group is very acidic and hence reacts with $\ce{CF3I}$ to give a stable compound according to the following reaction mechanism:
[![Reaction][1]][1]
However, the textbook gives the following reaction mechanism and claims that the product is **trifluoromethane $\ce{CHF_3}$**
[![Textbook reaction][2]][2]
Which one of these is correct?
[1]: https://i.stack.imgur.com/cUlWl.png
[2]: https://i.stack.imgur.com/kfbrd.png |
Burning is a non-scientific term, it implies appearance of a luminous chemical reaction in a gaseous phase. It is commonly called a flame. It does not necessarily imply the presence of carbon or even oxygen. Boron and hydrogen compounds can burn in air, similarly, sodium burns in the presence of chlorine.
You can rather ask, can I oxidize carbon containing materials to carbon dioxide or carbon monoxide in the presence of oxygen? The general answer is yes. You already mentioned the examples of wood, sugar, coal etc. which are carbon rich materials. They will burn with a luminous flame. The carbon in inorganic materials such as refractory carbides, like silicon carbide, can also be oxidized that but only under extreme conditions.
IUPAC acknowledges that the "The boundaries between ‘organic’ and ‘inorganic’ compounds are blurred." So there is no need to create an arbitrary division.
|
> $$\ce{CF_3I + NaOH(aq) -> Major Product}$$
The Major product, according to me, should be **ditrifluoromethyl ether $\ce{CF3-O-CF3}$**. The hydrogen on the $\ce{-OH}$ group is quite acidic and hence reacts with $\ce{CF3I}$ to give a stable compound according to the following reaction mechanism:
[![Reaction][1]][1]
However, the textbook gives the following reaction mechanism and claims that the product is **trifluoromethane $\ce{CHF_3}$**
[![Textbook reaction][2]][2]
Which one of these is correct?
[1]: https://i.stack.imgur.com/cUlWl.png
[2]: https://i.stack.imgur.com/kfbrd.png |
I came upon the following statement in a true/false question:
> Glycine can be esterified by an alcohol in the presence of inorganic acid
**Is this statement correct?**
Glycine contains the basic $\ce{-NH2}$ group, which would get protonated in the presence of inorganic acid. As a consequence, would the esterification process still be carried out on the $\ce{-COOH}$ group? [This][1] paper says that the reaction needs trimethylchlorosilane as a catalyst and not an inorganic acid.
[1]: https://www.mdpi.com/1420-3049/13/5/1111 |
**TL;DR**: It's **2** (2,2'‐dibromo‐6,6'‐diiodo‐1,1'‐biphenyl) and maybe even **3** (2,2'‐diiodo‐1,1'‐biphenyl; 2,6‐dimethyl‐1,1'‐biphenyl) at higher temperatures.
### Introduction
According to the IUPAC gold book [**chirality**](https://doi.org/10.1351/goldbook.C01058) is defined the as the geometric property of a rigid object of being non-superposable on its mirror image. Orr in other words, such an object has neither a mirror plane, nor a centre of inversion.
For examples, please see this Q&A: [What is the perfect definition for chirality?](https://chemistry.stackexchange.com/q/59124/4945)
Asymmetric substituted biphenyl derivatives may possess axial chirality. This is quite well explained in [Matthew's answer](https://chemistry.stackexchange.com/a/124020/4945) (at the time of writing, I think he has the numbering wrong though). This arises (mainly) due to hindered rotation, see [**atropisomers** in the IUPAC gold book](https://doi.org/10.1351/goldbook.A00511) and on [Wikipedia](https://en.wikipedia.org/wiki/Atropisomer).
### Analysis
Based on this, we can immediately rule out biphenyl **4** (2,6‐dimethyl‐1,1'‐biphenyl). I have simulated the structure<sup>\[1\]</sup> and the equilibrium structure is close to <i>C</i><sub>2v</sub>.
[<img src="https://i.stack.imgur.com/jp51Z.png" width="400" alt="C2V highlighted in 2,6‐dimethyl‐1,1'‐biphenyl">][1]
The remaining compounds will have chiral conformers as they are asymmetric substituted. The first compound **1** (3‐iodo‐3'‐nitro‐1,1'‐biphenyl) has no (<i>C</i><sub>1</sub>) symmetry. The equilibrium structures of the second compound **2** (2,2'‐dibromo‐6,6'‐diiodo‐1,1'‐biphenyl), as well as the third compound **3** (2,2'‐diiodo‐1,1'‐biphenyl; 2,6‐dimethyl‐1,1'‐biphenyl), have <i>C</i><sub>2</sub> symmetry. These are potential candidates for atropisomerism.
The question now becomes, which one is (most) hindered. I have (very crudely) simulated the interconversion barriers.<sup>\[2\]</sup> As a reference, I have also included 1,1'-biphenyl as **0**. The barrier is calculated in a zeroth order approximation as the difference of the conformer(s) and the highest energy image.
\begin{array}{cr}
\text{Structure} & \Delta E /\pu{kJ mol-1}\\\hline
\mathbf{0} & 11.0 \\
\mathbf{1} & 9.7 \\
\mathbf{2} & 208.4 \\
\mathbf{3} & 107.6 \\
\mathbf{4} & \text{n.a.} \\
Note that I have only considered one conversion path, which actually only would make a difference for **3** anyway.
We can immediately spot that there is not much hindrance in **1**, as its value is in the same ballpark as **0**. The value for the latter is approximately the same as what Henry Rzepa calculated.<sup>\[3\]</sup> For the cinephiles, here is the animated minimum energy path (MEP):
[![NEB MEP interconversion of 1][2]][2]
We can also deduce that **3** is more hindered, but conversion is probably still doable at room temperature. Without further analysis, I'd make an educated guess that the large size of iodine is the reason for the higher barrier. In the local minima the structures are probably stabilised by non-covalent interactions, while during the transition Pauli repulsion will have a significant effect.
When you cool down the system enough, you should be able to differentiate the two forms.
[![NEB MEP interconversion of 3][3]][3]
The highest barrier is for **2**, which is not really surprising given the bulky halogens used. During the transition the molecule has to distort significantly as the MEP will show.
[![NEB MEP interconversion of 2][4]][4]
---
### Notes & References
1. Optimisation has been done on the semiempirical GFN2-xTB level of theory. (a) [Mulliken Center for Theoretical Chemistry](https://www.chemie.uni-bonn.de/pctc/mulliken-center/software/xtb/xtb), [xTB on Github](https://github.com/grimme-lab/xtb/), [xTB documentation](https://xtb-docs.readthedocs.io/en/latest/contents.html), xTB: [*J. Chem. Theory Comput.* **2017,** *13* (5), 1989–2009](https://doi.org/10.1021/acs.jctc.7b00118) & [*J. Chem. Theory Comput.* **2019,** *15* (3), 1652–1671](https://doi.org/10.1021/acs.jctc.8b01176).
Cartesian coordinates (Xmol format) of the optimised structure:
```
28
energy: -37.106373072108 gnorm: 0.000617993871 xtb: 6.3.0 (preview)
C -0.74457039092978 0.00668380904848 0.00403847586541
C -1.43960105448658 -0.45330956062940 1.12306879090173
C -1.43111456476918 0.48819737048977 -1.11130113065019
C -2.82681693607605 -0.42604266071721 1.11315113205938
C -2.81837054760201 0.50359494897240 -1.09421940563873
C -3.51401551801468 0.04941173994657 0.01133914771101
H -3.37205771113385 -0.78013743294762 1.97649654009736
H -3.35684427561920 0.87533913913887 -1.95437486971859
H -4.59411676716931 0.06680911498157 0.01451573070096
C 0.73741240721740 -0.01399221137562 -0.00121303974220
C 1.45737190356414 1.07903119590108 0.46615300935510
C 1.42452698329909 -1.12468405637333 -0.47649552340433
H 0.86686558778550 -1.97662540315982 -0.84058815377695
H 0.92547790883778 1.94454881021216 0.83667570046312
C 2.80773675401329 -1.14118822737596 -0.48485609074287
C 2.84061966747731 1.06166601900200 0.45746804395492
H 3.33358058669268 -2.00908593002598 -0.85653458093421
H 3.39191821123738 1.91650859033419 0.82242788728035
C 3.51831774061953 -0.04807951153356 -0.01828710942118
H 4.59840909035030 -0.06111117644652 -0.02569905851317
C -0.68780013266629 -0.96774724332108 2.31428913997988
H -1.37323200644906 -1.27392883680533 3.10034826852651
H -0.06975635021791 -1.82092520510799 2.03621620918938
H -0.02338556354040 -0.19893824461916 2.70781448086228
C -0.66947839003714 0.98202574672165 -2.30497384579965
H -0.01808999107620 1.80968104054958 -2.02509754296330
H -0.03719586612218 0.19173089684959 -2.70896585204466
H -1.34904789687459 1.32012824269556 -3.08395328163034
```
2. The interconversion has been calculated with the same level of theory using the Nudged Elastic Band (NEB) algorithm of ORCA 4.2.0 ([Forum](https://orcaforum.kofo.mpg.de/app.php/portal), [*WIREs Comput. Mol. Sci.* **2018,** *8* (1)](https://doi.org/10.1002/wcms.1327)). The path had a total of 15 images, here is the example file for **2** along with the necessary coordinate files. If you want to try this at home, on my laptop (old!) with a single core this calculation took about 2 hours and 40 minutes.
```
! XTB2 NEB
%neb
NEB_End_XYZFile "conformer_end.xyz"
Nimages 13
end
*xyzfile 0 1 conformer_start.xyz
```
Xmol `conformer_start.xyz` of conformer 1:
```
22
energy: -44.541438926695 gnorm: 0.000273788204 xtb: 6.3.0 (preview)
C -0.73842516239767 0.12099397705083 0.10023099347275
C -1.44934703887101 1.30959596420843 -0.04635714487514
C -1.46459032906802 -1.05364356156319 0.26275562729962
C -2.83260221816166 1.33338660213393 -0.03856888366719
Br -0.49031781314024 2.94358437512547 -0.26534599360856
C -2.84794544579427 -1.04806937194567 0.27338021399222
I -0.43874303211246 -2.89099212545445 0.49464849107241
C -3.52725076437943 0.14798638986577 0.12187031074955
H -3.35400213047419 2.27002355888684 -0.15598934111452
H -3.39039633102590 -1.97145652424781 0.40014772926329
H -4.60579519489111 0.15562863225057 0.12966905294877
C 0.73854134684734 0.12608798707818 0.09393840630342
C 1.44940095781461 0.26239315529407 1.28380896054353
C 1.46477927365802 0.00836392380816 -1.08597026657536
I 0.43777759512819 -0.19754357173773 -2.92573037522205
Br 0.49031834581031 0.43262730661288 2.92353161238781
C 2.84816021024059 0.01980313175890 -1.08292737332063
C 2.83266002230666 0.27532688568945 1.30522345586451
H 3.39064596265727 -0.07352515350963 -2.01027791369359
H 3.35396295523884 0.38108640783580 2.24326316270063
C 3.52740515329100 0.15308581626132 0.11533500559347
H 4.60593238792661 0.16262139848447 0.12098160378166
```
Xmol `conformer_end.xyz` of conformer 2:
```
22
energy: -44.541436788329 gnorm: 0.000524902200 xtb: 6.3.0 (preview)
C -0.73832361282032 0.11983310306066 0.10840520776248
C -1.44801951621468 1.30797897222752 -0.04711380834348
C -1.46589375900469 -1.05507032640145 0.26341702181708
C -2.83136671292522 1.33298447781511 -0.04044389495310
Br -0.48672217710205 2.94010480951284 -0.27007089450699
C -2.84922363862597 -1.04840911658639 0.27193924049683
I -0.44386752640863 -2.89430038039744 0.49812818908251
C -3.52726058220375 0.14843604873707 0.12049840516296
H -3.35201572570858 2.26977601051251 -0.16025533523296
H -3.39258284243005 -1.97170525823195 0.39578778183960
H -4.60581188092237 0.15713306581713 0.12812396988730
C 0.73861053874909 0.11915254649018 0.09757065282745
C 1.43899785537134 -0.05963599109061 -1.09283010773567
C 1.47533465683736 0.30394299799110 1.26237171334068
I 0.46538312669624 0.58600628876877 3.10142029802479
Br 0.46500728057852 -0.30932233235425 -2.71351652270371
C 2.85860590331322 0.30295779336713 1.24612034762903
C 2.82215760585763 -0.06398015847435 -1.12722135693164
H 3.40910925948393 0.44613646389996 2.16234539469840
H 3.33549654196190 -0.20778602755665 -2.06467923531197
C 3.52731730663299 0.11776860585707 0.04879749941162
H 4.60580925392358 0.11518160422810 0.03305180289059
```
3. On ωB97XD/6-31G(d,p) about $\pu{3 kcal mol-1}$, which is in real units $\pu{12.5 kJ mol-1}$. Henry Rzepa: [*Conformational analysis of biphenyls: an upside-down view*](https://www.ch.imperial.ac.uk/rzepa/blog/?p=1856)
[1]: https://i.stack.imgur.com/jp51Z.png
[2]: https://i.stack.imgur.com/ZhZOM.gif
[3]: https://i.stack.imgur.com/2CcLL.gif
[4]: https://i.stack.imgur.com/xOePl.gif |
In pig iron, from where does carbon in it comes from? |
How do I convert purity percentage to grams? |
**TL;DR**: It's **2** (2,2'‐dibromo‐6,6'‐diiodo‐1,1'‐biphenyl) and maybe even **3** (2,2'‐diiodo‐1,1'‐biphenyl; 2,6‐dimethyl‐1,1'‐biphenyl) at higher temperatures.
### Introduction
According to the IUPAC gold book [**chirality**](https://doi.org/10.1351/goldbook.C01058) is defined the as the geometric property of a rigid object of being non-superposable on its mirror image. Orr in other words, such an object has neither a mirror plane, nor a centre of inversion.
For examples, please see this Q&A: [What is the perfect definition for chirality?](https://chemistry.stackexchange.com/q/59124/4945)
Asymmetric substituted biphenyl derivatives may possess axial chirality. This is quite well explained in [Matthew's answer](https://chemistry.stackexchange.com/a/124020/4945) (at the time of writing, I think he has the numbering wrong though). This arises (mainly) due to hindered rotation, see [**atropisomers** in the IUPAC gold book](https://doi.org/10.1351/goldbook.A00511) and on [Wikipedia](https://en.wikipedia.org/wiki/Atropisomer).
### Analysis
Based on this, we can immediately rule out biphenyl **4** (2,6‐dimethyl‐1,1'‐biphenyl). I have simulated the structure<sup>\[1\]</sup> and the equilibrium structure is close to <i>C</i><sub>2v</sub>.
[<img src="https://i.stack.imgur.com/jp51Z.png" width="400" alt="C2V highlighted in 2,6‐dimethyl‐1,1'‐biphenyl">][1]
The remaining compounds will have chiral conformers as they are asymmetric substituted. The first compound **1** (3‐iodo‐3'‐nitro‐1,1'‐biphenyl) has no (<i>C</i><sub>1</sub>) symmetry. The equilibrium structures of the second compound **2** (2,2'‐dibromo‐6,6'‐diiodo‐1,1'‐biphenyl), as well as the third compound **3** (2,2'‐diiodo‐1,1'‐biphenyl; 2,6‐dimethyl‐1,1'‐biphenyl), have <i>C</i><sub>2</sub> symmetry. These are potential candidates for atropisomerism.
The question now becomes, which one is (most) hindered. I have (very crudely) simulated the interconversion barriers.<sup>\[2\]</sup> As a reference, I have also included 1,1'-biphenyl as **0**. The barrier is calculated in a zeroth order approximation as the difference of the conformer(s) and the highest energy image.
\begin{array}{cr}
\text{Structure} & \Delta E /\pu{kJ mol-1}\\\hline
\mathbf{0} & 11.0 \\
\mathbf{1} & 9.7 \\
\mathbf{2} & 208.4 \\
\mathbf{3} & 107.6 \\
\mathbf{4} & \text{n.a.} \\\hline
\end{array}
Note that I have only considered one conversion path, which actually only would make a difference for **3** anyway.
We can immediately spot that there is not much hindrance in **1**, as its value is in the same ballpark as **0**. The value for the latter is approximately the same as what Henry Rzepa calculated.<sup>\[3,4\]</sup> For the cinephiles, here is the animated minimum energy path (MEP):
[![NEB MEP interconversion of 1][2]][2]
We can also deduce that **3** is more hindered, but conversion is probably still doable at room temperature. Without further analysis, I'd make an educated guess that the large size of iodine is the reason for the higher barrier. In the local minima the structures are probably stabilised by non-covalent interactions, while during the transition Pauli repulsion will have a significant effect.
When you cool down the system enough, you should be able to differentiate the two forms.
[![NEB MEP interconversion of 3][3]][3]
The highest barrier is for **2**, which is not really surprising given the bulky halogens used. During the transition the molecule has to distort significantly as the MEP will show.
[![NEB MEP interconversion of 2][4]][4]
---
### Notes & References
1. Optimisation has been done on the semiempirical GFN2-xTB level of theory. (a) [Mulliken Center for Theoretical Chemistry](https://www.chemie.uni-bonn.de/pctc/mulliken-center/software/xtb/xtb), [xTB on Github](https://github.com/grimme-lab/xtb/), [xTB documentation](https://xtb-docs.readthedocs.io/en/latest/contents.html), xTB: [*J. Chem. Theory Comput.* **2017,** *13* (5), 1989–2009](https://doi.org/10.1021/acs.jctc.7b00118) & [*J. Chem. Theory Comput.* **2019,** *15* (3), 1652–1671](https://doi.org/10.1021/acs.jctc.8b01176).
Cartesian coordinates (Xmol format) of the optimised structure:
```
28
energy: -37.106373072108 gnorm: 0.000617993871 xtb: 6.3.0 (preview)
C -0.74457039092978 0.00668380904848 0.00403847586541
C -1.43960105448658 -0.45330956062940 1.12306879090173
C -1.43111456476918 0.48819737048977 -1.11130113065019
C -2.82681693607605 -0.42604266071721 1.11315113205938
C -2.81837054760201 0.50359494897240 -1.09421940563873
C -3.51401551801468 0.04941173994657 0.01133914771101
H -3.37205771113385 -0.78013743294762 1.97649654009736
H -3.35684427561920 0.87533913913887 -1.95437486971859
H -4.59411676716931 0.06680911498157 0.01451573070096
C 0.73741240721740 -0.01399221137562 -0.00121303974220
C 1.45737190356414 1.07903119590108 0.46615300935510
C 1.42452698329909 -1.12468405637333 -0.47649552340433
H 0.86686558778550 -1.97662540315982 -0.84058815377695
H 0.92547790883778 1.94454881021216 0.83667570046312
C 2.80773675401329 -1.14118822737596 -0.48485609074287
C 2.84061966747731 1.06166601900200 0.45746804395492
H 3.33358058669268 -2.00908593002598 -0.85653458093421
H 3.39191821123738 1.91650859033419 0.82242788728035
C 3.51831774061953 -0.04807951153356 -0.01828710942118
H 4.59840909035030 -0.06111117644652 -0.02569905851317
C -0.68780013266629 -0.96774724332108 2.31428913997988
H -1.37323200644906 -1.27392883680533 3.10034826852651
H -0.06975635021791 -1.82092520510799 2.03621620918938
H -0.02338556354040 -0.19893824461916 2.70781448086228
C -0.66947839003714 0.98202574672165 -2.30497384579965
H -0.01808999107620 1.80968104054958 -2.02509754296330
H -0.03719586612218 0.19173089684959 -2.70896585204466
H -1.34904789687459 1.32012824269556 -3.08395328163034
```
2. The interconversion has been calculated with the same level of theory using the Nudged Elastic Band (NEB) algorithm of ORCA 4.2.0 ([Forum](https://orcaforum.kofo.mpg.de/app.php/portal), [*WIREs Comput. Mol. Sci.* **2018,** *8* (1)](https://doi.org/10.1002/wcms.1327)). The path had a total of 15 images, here is the example file for **2** along with the necessary coordinate files. If you want to try this at home, on my laptop (old!) with a single core this calculation took about 2 hours and 40 minutes.
```
! XTB2 NEB
%neb
NEB_End_XYZFile "conformer_end.xyz"
Nimages 13
end
*xyzfile 0 1 conformer_start.xyz
```
Xmol `conformer_start.xyz` of conformer 1:
```
22
energy: -44.541438926695 gnorm: 0.000273788204 xtb: 6.3.0 (preview)
C -0.73842516239767 0.12099397705083 0.10023099347275
C -1.44934703887101 1.30959596420843 -0.04635714487514
C -1.46459032906802 -1.05364356156319 0.26275562729962
C -2.83260221816166 1.33338660213393 -0.03856888366719
Br -0.49031781314024 2.94358437512547 -0.26534599360856
C -2.84794544579427 -1.04806937194567 0.27338021399222
I -0.43874303211246 -2.89099212545445 0.49464849107241
C -3.52725076437943 0.14798638986577 0.12187031074955
H -3.35400213047419 2.27002355888684 -0.15598934111452
H -3.39039633102590 -1.97145652424781 0.40014772926329
H -4.60579519489111 0.15562863225057 0.12966905294877
C 0.73854134684734 0.12608798707818 0.09393840630342
C 1.44940095781461 0.26239315529407 1.28380896054353
C 1.46477927365802 0.00836392380816 -1.08597026657536
I 0.43777759512819 -0.19754357173773 -2.92573037522205
Br 0.49031834581031 0.43262730661288 2.92353161238781
C 2.84816021024059 0.01980313175890 -1.08292737332063
C 2.83266002230666 0.27532688568945 1.30522345586451
H 3.39064596265727 -0.07352515350963 -2.01027791369359
H 3.35396295523884 0.38108640783580 2.24326316270063
C 3.52740515329100 0.15308581626132 0.11533500559347
H 4.60593238792661 0.16262139848447 0.12098160378166
```
Xmol `conformer_end.xyz` of conformer 2:
```
22
energy: -44.541436788329 gnorm: 0.000524902200 xtb: 6.3.0 (preview)
C -0.73832361282032 0.11983310306066 0.10840520776248
C -1.44801951621468 1.30797897222752 -0.04711380834348
C -1.46589375900469 -1.05507032640145 0.26341702181708
C -2.83136671292522 1.33298447781511 -0.04044389495310
Br -0.48672217710205 2.94010480951284 -0.27007089450699
C -2.84922363862597 -1.04840911658639 0.27193924049683
I -0.44386752640863 -2.89430038039744 0.49812818908251
C -3.52726058220375 0.14843604873707 0.12049840516296
H -3.35201572570858 2.26977601051251 -0.16025533523296
H -3.39258284243005 -1.97170525823195 0.39578778183960
H -4.60581188092237 0.15713306581713 0.12812396988730
C 0.73861053874909 0.11915254649018 0.09757065282745
C 1.43899785537134 -0.05963599109061 -1.09283010773567
C 1.47533465683736 0.30394299799110 1.26237171334068
I 0.46538312669624 0.58600628876877 3.10142029802479
Br 0.46500728057852 -0.30932233235425 -2.71351652270371
C 2.85860590331322 0.30295779336713 1.24612034762903
C 2.82215760585763 -0.06398015847435 -1.12722135693164
H 3.40910925948393 0.44613646389996 2.16234539469840
H 3.33549654196190 -0.20778602755665 -2.06467923531197
C 3.52731730663299 0.11776860585707 0.04879749941162
H 4.60580925392358 0.11518160422810 0.03305180289059
```
3. On ωB97XD/6-31G(d,p) about $\pu{3 kcal mol-1}$, which is in real units $\pu{12.5 kJ mol-1}$. Henry Rzepa: [*Conformational analysis of biphenyls: an upside-down view*](https://www.ch.imperial.ac.uk/rzepa/blog/?p=1856)
4. Possibly relevant dissertation by David Vonlanthen *Biphenyl-Cyclophanes: The Molecular Control over the Conductivity of Single-Molecule Junctions* (in German): [pdf at researchgate](https://www.researchgate.net/publication/283056096_Biphenyl-Cyclophanes_The_Molecular_Control_over_the_Conductivity_of_Single-Molecule_Junctions). He cites the value for 1,1'-biphenyl as $\pu{2-3 kcal mol-1}$.
[1]: https://i.stack.imgur.com/jp51Z.png
[2]: https://i.stack.imgur.com/ZhZOM.gif
[3]: https://i.stack.imgur.com/2CcLL.gif
[4]: https://i.stack.imgur.com/xOePl.gif |
Acetic Acid + K2CrO4(aq) + Ba(NO3)2 (aq) —> ? |
What are the products of these 3 reactants? Acetic Acid + K2CrO4 + Ba(NO3)2 |
What are the products of these 3 reactants? Acetic Acid + K2CrO4(aq) + Ba(NO3)2(aq) |
>There are several examples where the negative inductive effect of a substituent gets increased when a hydrogen atom on that substituent is replaced by an alkyl group.
>Some particular examples:
\begin{align}
\ce{-NR3^+} &> \ce{-NH3^+}& \ce{-OR} &> \ce{-OH}
\end{align}
About comparing $\ce{-NR3^+}$ and $\ce{-NH3^+}$, at first, the electron-donating inductive effect of the methyl groups (+I) makes us feel that the electron-withdrawing inductive effect trend should be the *opposite*. Neither, could I find any reliable sources with an explanation. So, I came up with a few of my own.
# Part I: The Bent's Rule Interpretation (More Reliable)
To analyse the inductive effect, let's take the groups attached to a hydrocarbon chain and then see.
[![enter image description here][1]][1]
The $\ce{C-H}$ bond orbitals will have more s-character than in the $\ce{N-C}$ bond orbitals because there is more electron-density in the former due to the greater electronegativity of carbon over hydrogen and nitrogen over carbon, which requires the presence of more, lower-energy s-character in them to lower the system's overall energy. Thus, an implication of this would be the increase of p-character in the three $\ce{N-C_{Me}}$ bond-orbitals and the subsequent decrease of p-character in the $\ce{N-C_{chain}}$. This results in an increase of electronegativity of $\ce{N}$, thus the (-I) trend is:
$$\ce{-NR3^+} > \ce{NH3^+}$$
# Part II: I don't know what to call this (Need help with this)
Let's consider the examples of phenol and anisole. It is considered that phenol donates more electron density to the benzene ring than anisole because it exists in equilibrium with the phenoxide ion, which is a far-more electron-donating group, despite anisole's $\ce{-OCH3}$ group being a better electron-donating group.
If we apply this to our case, we can assume that $\ce{-NH3^+}$ may be in equilibrium with $\ce{-NH2}$ which makes it a better (+I) group and poorer (-I) group, while $\ce{-NMe3^+}$ can't deprotonate itself and establish an equilibrium like that, making it a better (-I) group.
(somebody please help here)
#Part III: Should we really consider $\ce{-CH3}$ as a (+I) group here? (<strike>Just thinking</strike>, very reliable<sup>*</sup>)
It is considered that the inductive effect of $\ce{-CH3}$ is mainly attributed to its hyperconjugative effect. However, hyperconjugation requires an empty p orbital carrying a positive charge into which it donates its $\sigma_{\ce{C-H}}$ bond-electron density. However, nitrogen's orbitals are sp³ hybridized here, with no empty orbital carrying the charge as such. So, I think (not really sure<sup>*</sup>), we should only consider carbon's greater electronegativity over hydrogen, thus making $\ce{-NMe3^+}$ more inductively electron-withdrawing than $\ce{-NH3^+}$. You can also apply this to $\ce{-OR}$ and $\ce{-OH}$.
<sup>*</sup> Just saw [Jan's answer](https://chemistry.stackexchange.com/questions/65106/) in the middle of my writing process. That makes this a valid point. Great answer, a must check-out.
[1]: https://i.stack.imgur.com/hpFOI.png |
For a substructure search I would like to search for structures containing unfused benzyl.
The idea was to explicitly add hydrogen. But apparently this does not give the expected results.
So, I assume either my understanding or expectations are wrong or I'm using RDKit not properly.
I just learnt (https://chemistry.stackexchange.com/q/128440) that I better use SMARTS expressions with `Chem.MolToSmarts()` where I get the expected results.
However, if I have much more complex search structures, how can I get from the SMILES to the SMARTS? Do I have always to figure out the SMARTS code myself? Is there maybe a way just drawing a molecule and add explicit hydrogen and search with this somehow? I guess it should be pretty clear (at least to a human) what I want to find (or not find) with structure 2. Are there maybe any tools or code which can assist here?
[![enter image description here][1]][1]
**Code:**
``` lang-python
from rdkit import Chem
smiles_strings = '''CC1=CC=CC=C1
CC1=C([H])C([H])=C([H])C([H])=C1[H]
C1(C2=CC=CC=C2)=CC=CC=C1
C12=C(CC3=CC=CC=C3C2)C=CC=C1
C12=CC=CC=C1C=C3C(C=CC=C3)=C2
C12=C(C(C(C=C3)=CC=C4)=C4C=C2)C3=CC=C1
'''
smiles_list = smiles_strings.splitlines()
def search_structure(pattern):
found = []
for idx,smiles in enumerate(smiles_list):
m = Chem.MolFromSmiles(smiles)
if m.HasSubstructMatch(pattern):
found.append(idx+1)
print("Structures found: {}".format(found))
smiles_1 = smiles_list[0]
pattern_1 = Chem.MolFromSmiles(smiles_1)
smiles_2a = smiles_list[1]
pattern_2a = Chem.MolFromSmiles(smiles_2a)
smiles_2b = Chem.MolToSmiles(pattern_2a)
pattern_2b = Chem.MolFromSmiles(smiles_2b)
smiles_2c = smiles_2b
pattern_2c = Chem.MolFromSmarts(smiles_2c)
smarts_3 = '[cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]'
pattern_3 = Chem.MolFromSmarts(smarts_3)
print("\nSMILES 1 : {}\n# my expectation: [1, 2, 3, 4, 5, 6]".format(smiles_1))
search_structure(pattern_1)
print("\nSMILES 2a: {}\n# my expectation: [1, 2, 3]".format(smiles_2a))
search_structure(pattern_2a)
print("\nSMILES 2b: {} (Mol from SMILES)\n# my expectation: [1, 2, 3]".format(smiles_2b))
search_structure(pattern_2b)
print("\nSMILES 2c: {} (Mol from SMARTS)\n# my expectation: [1, 2, 3]".format(smiles_2c))
search_structure(pattern_2c)
print("\nSMARTS 3 : {}\n# my expectation: [1, 2, 3]".format(smarts_3))
search_structure(pattern_3)
```
**Result:**
``` lang-none
SMILES 1 : CC1=CC=CC=C1
# my expectation: [1, 2, 3, 4, 5, 6]
Structures found: [1, 2, 3, 4]
SMILES 2a: CC1=C([H])C([H])=C([H])C([H])=C1[H]
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3, 4]
SMILES 2b: Cc1ccccc1 (Mol from SMILES)
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3, 4]
SMILES 2c: Cc1ccccc1 (Mol from SMARTS)
# my expectation: [1, 2, 3]
Structures found: [1, 2, 4]
SMARTS 3 : [cR1]1[cR1][cR1][cR1][cR1][cR1]1-[c,C]
# my expectation: [1, 2, 3]
Structures found: [1, 2, 3]
```
[1]: https://i.stack.imgur.com/iUM91.png
|
How to search substructures with explicit hydrogen in RDKit? |
How to perform IR spectroscopy of silica particles while inflow at different RH? |
I want to ask a question about the fragmentation of the $\ce{M_17}$ peak of Salicyclamide to a phenolic fragment.
I was reading the following paper giving a profile about [salicyclamide][1].
I want to determine a mechanism that gives the following product from the fragmentation scheme:
[![Overall fragmentation scheme based from the mass spec analysis][2]][2]
I am however, unable to so far produce a mechanism that shows the scheme above.
The figure below shows the closest attempt I have made so far.
[![Attempt at a reasonable mechanism][3]][3]
I feel my attempt is incorrect since there is a net charge change.
What mistake am I making that can rectify my result and provide a reasonable suggestion for the mechanism involved?
[1]: https://www.sciencedirect.com/science/article/pii/S0099542808602014?via%3Dihub
[2]: https://i.stack.imgur.com/IMn84.png
[3]: https://i.stack.imgur.com/Ma40j.png |
In the preparation of $\ce{NH3}$ from $\ce{NH4Cl}$ and $\ce{Ba(OH)2}$, my teacher said that when heating the test tube, the test tube faces downward as opposed to upwards. Why is this so? Wouldn't the solid just fall out?
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/AvEVU.png |
Why does the test tube face downards in gas production? |
The following question comes from *General Chemistry: Principles and Modern Applications, 11th ed.* by Pettruci.
> **Exercise 15-33.** A mixture consisting of 0.150 mol H<sub>2</sub> and 0.150 mol I<sub>2</sub> is brought to equilibrium at 445 °C, in a 3.25 L flask. What are the equilibrium amounts of H<sub>2</sub>, I<sub>2</sub>, and HI?
$$\ce{H2(g) + I2(g) <=> 2HI(g)}\qquad K_c=50.2\text{ at }445\,\mathrm{^\circ C}$$
First, I found the concentrations of H<sub>2</sub> and I<sub>2</sub>:
$$
\begin{aligned}
\phantom{}[\ce{H2}]=[\ce{I2}]&=\frac{0.150\text{ mol}}{3.25\text{ L}} \\
&=0.0461538\text{ M}
\end{aligned}
$$
Setting up the ICE table:
$$
\begin{array}{|l|c c c c c|}
\hline
&\ce{H2}&+&\ce{I2}&\rightleftharpoons&\ce{2HI} \\
\hline
\text{Initial}&0.0461538&&0.0461538&&0 \\
\text{Change}&-x&&-x&&+2x \\
\text{Equilibrium}&0.0461538-x&&0.0461538-x&&2x \\
\hline
\end{array}
$$
Then,
$$
K_c=50.2=\frac{(2x)^2}{(0.0461538-x)(0.0461538-x)}.
$$
Most solutions I've seen for similar problems involve taking the square root of both sides of this equation. However, they only take the principal root, effectively ignoring a second valid solution. Here, I'll do the proper solution for finding $x$:
$$
\begin{aligned}
\\
50.2&=\left(\frac{2x}{0.0461538-x}\right)^2 \\
\sqrt{50.2}&=\pm\frac{2x}{0.0461538-x}
\end{aligned} \\
\begin{aligned}
7.0852&=\frac{2x}{0.0461538-x}&&\text{or}&7.0852&=-\frac{2x}{0.0461538-x} \\
0.327009-7.0852x&=2x&&&0.327009-7.0852x&=-2x \\
0.327009&=9.0852x&&&0.327009&=5.0852x \\
x&=0.0359936&&&x&=0.064306
\end{aligned}
$$
As you can see, there are two positive (and therefore, valid) solutions for $x$. However, only one of the solutions results in a positive $[\ce{H2}]_\mathrm{eq}$:
$$
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}=0.0461538-x
\end{aligned}
\\
\begin{aligned}
\phantom{}[\ce{H2}]_\mathrm{eq}&=0.0461538-0.0359936&&\text{or}&[\ce{H2}]_\mathrm{eq}&=0.0461538-0.064306 \\
&=0.0101602\text{ M}&&&&=-0.0181522\text{ M}
\end{aligned}
$$
And $x=0.0359936$ does indeed lead to the correct answer according to the textbook.
My question is: was it just a coincidence that I was able to eliminate one of the possible solutions for $x$? In other words, is there a scenario where I could end up with two possible $[\ce{H2}]_\mathrm{eq}$s? If not, what forbids this?
**Edit:** To reiterate, I am aware that one of the solutions is to be eliminated for being "non-physical". My question is whether it is possible for both solutions to be physical. |
If you mean the test tube containing the solids (reaction tube), then that tube faces downwards to promote the flow of gas into the receiving tube. The idea is to let the gas sink away from the reaction zone and towards the receiving tube as it cools and becomes more dense.
The receiving tube, of course, faces down so that the ammonia gas, lighter than air, can float to the top after it meets the air. So you have the ammonia flowing down with cooling through the reaction tube and then flowing up with buoyancy into the receiving tube.
Now for my bonus question: Had you been producing carbon dioxide by heating magnesium carbonate, would you change the reaction tube? How about the receiving tube? |
I want to ask a question about the fragmentation of the $\ce{M_17}$ peak of Salicyclamide to a phenolic fragment.
I was reading the following paper giving a profile about [salicyclamide][1].
I want to determine a mechanism that gives the following product from the fragmentation scheme:
[![Overall fragmentation scheme based from the mass spec analysis][2]][2]
I am however, unable to so far produce a mechanism that shows the scheme above.
The figure below shows the closest attempt I have made so far.
[![Attempt at a reasonable mechanism][3]][3]
I feel my attempt is incorrect since there is a net charge change.
What mistake am I making that can rectify my result and provide a reasonable suggestion for the mechanism involved?
**I don't understand how this question can be voted for closure as it is "not relevant to chemistry".** This is a clear phenomena given in a research paper and I wanted to discuss the mechanism(s) behind such findings that have been observed in the paper - as noted in the help centre under *Questions relating to observed chemical phenomena*.
[1]: https://www.researchgate.net/publication/231350583_33-_Analytical_Profile_of_Salicylamide
[2]: https://i.stack.imgur.com/IMn84.png
[3]: https://i.stack.imgur.com/Ma40j.png |
I’m required to make a lab with this information only (particle size vs dissolving rate of a solid (molecular covalent compound-sucrose)? so far what I’m thinking is using three different kinds of sugar like table sugar,sugar cubes,and one of them grinding it but the issues is how do you find how they dissolving rate? |
I need helping making a lab out of this (particle size vs dissolving rate of a solid (molecular covalent compound-sucrose)? |
As it says: are there any reactions that require heat *and* produce heat simultaneously? |
Are there any reactions that are endothermic and exothermic? |
Is it necessary that hydrolysis of ester will only take place in excess of water? |
I've recently been taught that the 4f orbitals in lanthanides are "core-like", supposedly meaning they have radius smaller than the 4d orbitals, therefore they are not available on the outside of the atom to form covalent bonds with ligands. The lecturer then went on to say that, therefore, the bonding of lanthanides to ligands is entirely ionic.
It is my understanding that a bond is deemed ionic when there is a large electronegativity difference between the two atoms, i.e. the bonding molecular orbital is near-identical in energy to the atomic orbital of the more electronegative atom, as shown below for sodium fluoride:
[!["MO diagram of NaF - the bonding MO is considerably closer in energy to the F AO than the Na AO"][1]][1]
However, if this is the case, how can lanthanides form any bonds if their valence orbitals are not available? Surely orbital overlap with the ligand is still required to form a molecular orbital, even if the resulting orbital is ionic in nature?
[1]: https://i.stack.imgur.com/C9FPz.png |
How can ionic bonding in lanthanides occur without valence orbitals available for overlap? |
An example is the accidentally formed problematic explosive nitrogen trichloride from the chlorination of a warmed ammonium salt solution, like $\ce{NH4Cl}$. To quote a source:
>Why is $\ce{NCl3}$ so unstable?
>The enthalpy of formation of $\ce{NCl3}$ is + 232 kJ mol-1, a fairly large endothermic value, a good sign that a compound is unstable. This refers to the process:
>$\ce{½ N2(g) + 3/2 Cl2(g) -> NCl3(l)}$
>This means that the enthalpy change for the reverse reaction, the decomposition of $\ce{NCl3}$, is pretty exothermic.
>$\ce{NCl3(l) -> ½ N2(g) + 3/2 Cl2(g)}$
>The positive entropy change associated with the formation of the gaseous chlorine and nitrogen would also favour the decomposition, as would the very high N-N bond energy (~940 kJ mol-1). |
An example is the accidentally formed problematic explosive nitrogen trichloride from the chlorination of a warmed ammonium salt solution, like $\ce{NH4Cl}$, where the reaction can rapidly transition from an endothermic to an exothermic process (albeit not precisely simultaneously). To quote a source:
>Why is $\ce{NCl3}$ so unstable?
>The enthalpy of formation of $\ce{NCl3}$ is + 232 kJ mol-1, a fairly large endothermic value, a good sign that a compound is unstable. This refers to the process:
>$\ce{½ N2(g) + 3/2 Cl2(g) -> NCl3(l)}$
>This means that the enthalpy change for the reverse reaction, the decomposition of $\ce{NCl3}$, is pretty exothermic.
>$\ce{NCl3(l) -> ½ N2(g) + 3/2 Cl2(g)}$
>The positive entropy change associated with the formation of the gaseous chlorine and nitrogen would also favour the decomposition, as would the very high N-N bond energy (~940 kJ mol-1). |
The product you are suggesting is very unstable as compared to the ones given as answers due to steric factors. Thus it would not be formed.
>Since −OH and −CH3 are o,p-directing groups, shouldn’t the major product be 3-methyl-2-nitrophenol? Why not?
The reasoning that hydroxyl and methyl are o,p directing groups is not at supportive as to why 3-methyl-2-nitrophenol should be formed in preference to the ones given as answers. The o,p directing effect just rules out the meta product and steric factor decides among the remaining possibilities here. |
A possible example (reflective of a class of compounds, energetics) is the accidentally formed problematic explosive nitrogen trichloride from the chlorination of a warmed ammonium salt solution, like $\ce{NH4Cl}$, where the reaction can rapidly transition from an endothermic to an exothermic process (albeit not precisely simultaneously). To quote a source:
>Why is $\ce{NCl3}$ so unstable?
>The enthalpy of formation of $\ce{NCl3}$ is + 232 kJ mol-1, a fairly large endothermic value, a good sign that a compound is unstable. This refers to the process:
>$\ce{½ N2(g) + 3/2 Cl2(g) -> NCl3(l)}$
>This means that the enthalpy change for the reverse reaction, the decomposition of $\ce{NCl3}$, is pretty exothermic.
>$\ce{NCl3(l) -> ½ N2(g) + 3/2 Cl2(g)}$
>The positive entropy change associated with the formation of the gaseous chlorine and nitrogen would also favour the decomposition, as would the very high N-N bond energy (~940 kJ mol-1). |
Is an endothermic reaction always endothermic at all conditions of pressure, temperature etc.? Similarly for exothermic... I mean is sign of ∆H always the same as sign of ∆H°?
This might seem to be a vague doubt but I just feel that there can be a reaction with trends of energies of formation of reactants and products such that endothermic reaction changes to exothermic or vice versa subject to change in concentrations of substances or external factors like temperature, pressure etc.
This is just a random thought of mine and even an example in this direction would be helpful.
Thanks! |
A possible example (reflective of a class of compounds, energetics) is the accidentally formed problematic explosive nitrogen trichloride from the chlorination of a warmed ammonium salt solution, like $\ce{NH4Cl}$, where the reaction can rapidly transition from an endothermic to an exothermic process (albeit not precisely simultaneously, around one-thousandth of a second or less). To quote a source:
>Why is $\ce{NCl3}$ so unstable?
>The enthalpy of formation of $\ce{NCl3}$ is + 232 kJ mol-1, a fairly large endothermic value, a good sign that a compound is unstable. This refers to the process:
>$\ce{½ N2(g) + 3/2 Cl2(g) -> NCl3(l)}$
>This means that the enthalpy change for the reverse reaction, the decomposition of $\ce{NCl3}$, is pretty exothermic.
>$\ce{NCl3(l) -> ½ N2(g) + 3/2 Cl2(g)}$
>The positive entropy change associated with the formation of the gaseous chlorine and nitrogen would also favour the decomposition, as would the very high N-N bond energy (~940 kJ mol-1). |
A possible example (reflective of a class of compounds, energetics) is the accidentally formed problematic explosive nitrogen trichloride from the chlorination of a warmed ammonium salt solution, like $\ce{NH4Cl}$, where the reaction can rapidly transition from an endothermic to an exothermic process (albeit not precisely simultaneously, around one-thousandth of a second or less). Or, one could argue that equivalently a highly exothermic energetic is endothermic on creation where the reaction moves in the reverse direction.
To quote a source:
>Why is $\ce{NCl3}$ so unstable?
>The enthalpy of formation of $\ce{NCl3}$ is + 232 kJ mol-1, a fairly large endothermic value, a good sign that a compound is unstable. This refers to the process:
>$\ce{½ N2(g) + 3/2 Cl2(g) -> NCl3(l)}$
>This means that the enthalpy change for the reverse reaction, the decomposition of $\ce{NCl3}$, is pretty exothermic.
>$\ce{NCl3(l) -> ½ N2(g) + 3/2 Cl2(g)}$
>The positive entropy change associated with the formation of the gaseous chlorine and nitrogen would also favour the decomposition, as would the very high N-N bond energy (~940 kJ mol-1). |
> In other words, is there a scenario where I could end up with two possible [solutions]?
No, there is always a single solution. The reaction quotient Q assumes values from zero (no products) to infinity (no reactants). It varies monotonically with the extent of reaction (the x in the ICE table). So there is always a solution (because Q covers the full range of possible equilibrium constant), and it is always unique (because for a given set of starting values, every value of Q is unique - follows from the monotonic variation).
The only exception is a process where none of the concentration vary with x (when all species are pure solids or liquids). In this case, Q takes a single value (except for running out of reactant or product). If that value is equal to K, x can take a range of values. If Q is different from K, there is no solution, and there won't be any equilibrium (i.e. the reaction would go to completion). |
>How would be the equation and the ICE table, and what is the pH of the mixture of these two solutions?
There are three equations:
$$\ce{HCOOH <=> H+ + HCOO-}$$
$$\ce{CH3COOH <=> H+ + CH3COO-}$$
$$\ce{H2O <=> H+ + OH-}$$
Hydroxide is a (very) minor species, so you can neglect it. As I explain below, at the pH of the mixture, the acetate ion is also a minor species, so you can first neglect the dissociation of acetic acid, and then correct for it later.
I would not use an ICE table in this situation, at least not one that describes all three reactions.
**My strategy to determine the pH of the mixture**
First, you can determine the equilibrium constants for each acid from the pH values of their aqueous solutions (before mixing). The corresponding pKa values are 3.74 and 4.78 for formic and acetic acid, respectively.
After mixing, the concentration of total formic acid and total acetic acid drop by a factor of 2 (mutual dilution). Formic acid is the stronger acid, and would have a pH of 3.06 at this concentration in the absence of acetic acid (less acidic than pH = 2.9 because of dilution).
At pH 3.06, acetic acid dissociates very little (the difference between pH and pKa is more than 1.5, vs. less than 0.7 for formic acid). As a first approximation, we can say that the pH is determined by the stronger acid.
To figure out how much the presence of acetic acid does influence the pH, we can calculate how much it would dissociate at pH 3.06, and correct the pH accordingly. Our new pH will be more acidic, so we assumed a too acidic pH in our correction, overshooting a bit. After a couple of iterations of this, the pH comes out as 3.04.
So the biggest effect of adding the acetic acid comes from diluting the formic acid (pH would go from 2.9 to 3.06 just by adding water). A secondary effect comes from acetic acid dissociating a tiny bit, giving a pH of 3.04.
Here are results from the spreadsheet used for the calculations:
[![enter image description here][1]][1]
And here are the formulas used (to dampen the overshooting effect, the pH value used in each step is the average of the updated pH and the previous pH, not shown)
[![enter image description here][2]][2]
>I tried to do $\mathrm{pH}=\frac{1}{2}\left(\mathrm{pH}_1+\mathrm{pH}_2\right)$, but it’s trivial and doesn't work.
To convince yourself once and for all that pH averaging does not work, consider these two extreme examples:
1. Mixing 5 mL of 1 M HCl with 6 mL of 1 M NaOH (starting pH values are 0 and 14, resulting pH is roughly 13).
2. Mixing 10 ml of 1 M HCl with 90 mL of water (starting pH values are 0 and 7, resulting pH is 1)
[1]: https://i.stack.imgur.com/XVMe3.jpg
[2]: https://i.stack.imgur.com/XKUoi.jpg |
A possible example (reflective of a class of compounds, energetics) is the accidentally formed problematic explosive nitrogen trichloride from the chlorination of a warmed ammonium salt solution, like $\ce{NH4Cl}$, where the reaction can rapidly transition from an endothermic to an exothermic process (albeit not precisely simultaneously, around one-thousandth of a second or less). Or, one could argue that equivalently a highly exothermic energetic is endothermic on creation where the reaction moves in the reverse direction.
To quote a source:
>Why is $\ce{NCl3}$ so unstable?
>The enthalpy of formation of $\ce{NCl3}$ is + 232 kJ mol-1, a fairly large endothermic value, a good sign that a compound is unstable. This refers to the process:
>$\ce{½ N2(g) + 3/2 Cl2(g) -> NCl3(l)}$
>This means that the enthalpy change for the reverse reaction, the decomposition of $\ce{NCl3}$, is pretty exothermic.
>$\ce{NCl3(l) -> ½ N2(g) + 3/2 Cl2(g)}$
>The positive entropy change associated with the formation of the gaseous chlorine and nitrogen would also favour the decomposition, as would the very high N-N bond energy (~940 kJ mol-1).
Now, $\ce{NCl3}$ in the presence of organics will explode, however, a dust particle likely does not detonate nitrogen trichloride as I have witnessed its transport in an open test tube while being cooled (please, do not verify given the inherent dangers of this compound). Clearly, the dust results in some exothermic activity, but is insufficient in mass to precipitate an explosion in cooled $\ce{NCl3}$, implying some equilibrium process. |
A possible example (which is real and reflective of a class of compounds, energetics) is the accidentally formed problematic explosive nitrogen trichloride from the chlorination of a warmed ammonium salt solution, like $\ce{NH4Cl}$, where the reaction can rapidly transition from an endothermic to an exothermic process (albeit not precisely simultaneously, around one-thousandth of a second or less). Or, one could argue that equivalently a highly exothermic energetic is endothermic on creation where the reaction moves in the reverse direction.
To quote a source:
>Why is $\ce{NCl3}$ so unstable?
>The enthalpy of formation of $\ce{NCl3}$ is + 232 kJ mol-1, a fairly large endothermic value, a good sign that a compound is unstable. This refers to the process:
>$\ce{½ N2(g) + 3/2 Cl2(g) -> NCl3(l)}$
>This means that the enthalpy change for the reverse reaction, the decomposition of $\ce{NCl3}$, is pretty exothermic.
>$\ce{NCl3(l) -> ½ N2(g) + 3/2 Cl2(g)}$
>The positive entropy change associated with the formation of the gaseous chlorine and nitrogen would also favour the decomposition, as would the very high N-N bond energy (~940 kJ mol-1).
Now, $\ce{NCl3}$ in the presence of organics will explode, however, a dust particle likely does not detonate nitrogen trichloride as I have witnessed its transport in an open test tube while being cooled (please, do not verify given the inherent dangers of this compound). Clearly, the dust results in some exothermic activity, but is insufficient in mass to precipitate an explosion in cooled $\ce{NCl3}$, implying some equilibrium process. |
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