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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/06%3A_Modeling_Reaction_Kinetics/6.02%3A_Temperature_Dependence_of_Reaction_Rates/6.2.03%3A_The_Arrhenius_Law/6.2.3.06%3A_The_Arrhenius_Law_-_Pre-exponential_Factors |
The pre-exponential factor (\(A\)) is an important component of the Arrhenius equation, which was formulated by the Swedish chemist Svante Arrhenius in 1889. The pre-exponential factor is also known as the and represents the frequency of collisions between reactant molecules at a standard concentration. Although often described as temperature independent, it is actually dependent on temperature because it is related to molecular collision, which is a function of temperature. The units of the pre-exponential factor vary depending on the order of the reaction. In first order reactions, the units of the pre-exponential factor are reciprocal time (e.g., 1/s). Because the pre-exponential factor depends on frequency of collisions, it is related to and . \[ k = A e^{E_a/RT} \label{eq1} \] The Arrhenius equation introduces the relationships between rate and \(A\), \(E_a\), and \(T\), where \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, and \(T\) is the temperature. The pre-exponential factor, \(A\), is a constant that can be derived experimentally or numerically. It is also called the and describes how often two molecules collide. To first approximation, the pre-exponential factor is considered constant. When dealing with the collision theory, the pre-exponential factor is defined as \(Z\) and can be derived by considering the factors that affect the frequency of collision for a given molecule. Consider the most elementary bimolecular reaction: \[A + A \rightarrow Product\nonumber \] An underlying factor to the frequency of collisions is the space or volume in which this reaction is allowed to occur. Intuitively, it makes sense for the frequency of collisions between two molecules to be dependent upon the dimensions of their respective containers. By this logic, \(Z\) is defined the following way: \[Z = \dfrac{(\text{Volume of the cylinder}) (\text{Density of the particles})}{\text{time}}\nonumber \] Using this relationship, an equation for the collision frequency, \(Z\), of molecule \(A\) with \(A\) can be derived: \[Z_{AA} = 2N^2_Ad^2 \sqrt{\dfrac{\pi{k_{b}T}}{m_a}}\nonumber \] A similar reasoning is used for bimolecular reactions that involve the collisions of \(A\) and \(B\) \[A + B \rightarrow Product\nonumber \] for deriving the collision frequency, \(Z\) between \(A\) and \(B\). \(Z_{AB} = N_AN_Bd^2_{AB} \sqrt{\dfrac{8{k_{b}T}}{\mu}}\) Substituting the collision factor back into the original Arrhenius equation yields: \[\begin{align*} k &= Z_{AB}e^{\frac{-E_a}{RT}} \\[4pt] &= N_A\, N_B\, d^2_{AB} \sqrt{\dfrac{8{k_{b}T}}{\mu}}\,e^{\frac{-E_a}{RT}}\end{align*}\nonumber \] This equation produces a rate constant with the standard units of (M s ); however, on a molecular level, a rate constant with molecular units would be more useful. To obtain this constant, the rate is divided by \(N_A\,N_B\). This produces a rate constant with units (m molecule s ) and provides the following equation: \[k = Z_{AB} e^{\frac{-E_a}{RT}}\nonumber \] Divide both sides by \(N_AN_B\) \[\dfrac{k}{N_AN_B} = d^2_{AB} \sqrt{\dfrac{8 k_b T}{\mu}}e^{\frac{-E_a}{RT}}\nonumber \] \(Z_{AB}\) becomes \(z_{AB}\): \[\dfrac{Z_{AB}}{N{_A}N{_B }} = z_{AB}\nonumber \] Substituting back into the Arrhenius equation (Equation \ref{eq1}): \[k = z_{AB}e^{\frac{-E_a}{RT}}\nonumber \] The pre-exponential factor is now defined within the collision theory as the following: \[d^2_{AB} \sqrt{\dfrac{8{k_{b}T}}{\mu}}\nonumber \] \(A\) and \(Z\) are practically interchangeable terms for collision frequency. The derivations for \(Z\) often ignores the steric effect of molecules. For a reaction to occur, two molecules must collide in the correct orientation. Not every collision results in the proper orientation, and thus some do not yield a corresponding product. To account for this steric effect, the variable \(P\), which represents the probability of two atoms colliding with the proper orientation, is introduced. The Arrhenius equation is as follows: \[k = Pze^{\frac{-E_a}{RT}}\nonumber \] The probability factor, \(P\), is very difficult to assess and still leaves the Arrhenius equation imperfect. The collision theory deals with gases and neglects to account for structural complexities in atoms and molecules. Therefore, the collision theory estimation for probability is not accurate for species other than gases. The transition state theory attempts to resolve this discrepancy. It uses the foundations of thermodynamics to give a representation of the most accurate pre-exponential factor that yields the corresponding rate. The equation is derived through laws concerning Gibbs free energy, enthalpy and entropy: \[k = \dfrac{k_bT}{h} e^{\frac{\Delta S^o}{R}} e^{\frac{-\Delta H^o}{RT}}(M^{1-m})\nonumber \] \(d^2_{AB} \sqrt{\dfrac{8{k_{b}T}}{\mu}}\) The pre-exponential factor is a function of temperature. As indicated in Table 1, the factor for the collision theory and the transition state theory are both responsive to temperature changes. The collision theory factor is proportional to the square root of \(T\), whereas that of the transition state theory is proportional to \(T\). The empirical factor is also sensitive to temperature. As temperature increases, molecules move faster; as molecules move faster, they are more likely to collide and therefore affect the collision frequency, \(A\). | 5,381 | 0 |
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This page discusses the solubility of the hydroxides, sulfates and carbonates of the Group 2 elements—beryllium, magnesium, calcium, strontium and barium—in water Group II metal oxide basicity and hydroxide solubility in water increase as you go down the column. BeO and Be(OH)2 are amphoteric and react with acids and strong bases such as NaOH. MgO is basic and Mg(OH)2 is weakly basic and do not dissolve in NaOH solution. The oxides of calcium, strontium, and barium are basic and the hydroxides are strongly basic. The solubilities of the hydroxides in water follow the order: Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2. Group II metal oxides become more basic as you go down the column. This trend is easily seen if you compare the electronegativity of the group II metal to the electronegativity of oxygen. As you can see the electronegativities of the metals decrease down the column making the change in electronegativities increases down the group. The greater the difference in electronegativity the more ionic the metal-oxygen bond becomes. The more ionic the metal-oxygen bond the more basic the oxide is Group II metal hydroxides become more soluble in water as you go down the column. This trend can be explained by the decrease in the lattice energy of the hydroxide salt and by the increase in the coordination number of the metal ion as you go down the column. The larger the lattice energy the more energy it takes to break the lattice apart into metal and hydroxide ions. Since the atomic radii increase down the group it makes sense that the coordination numbers also increases because the larger the metal ion the more room there is for water molecules to coordinate to it. The following examples illustrate this trend: This simple trend is true provided hydrated beryllium sulfate is considered, but not anhydrous beryllium sulfate. The Nuffield Data Book quotes anyhydrous beryllium sulfate, BeSO , as insoluble, whereas the hydrated form, BeSO .4H O is soluble, with a solubility of about 39 g of BeSO per 100 g of water at room temperature. Solubility figures for magnesium sulfate and calcium sulfate also vary depending on whether the salt is hydrated or not, but the variations are less dramatic. Two common examples illustrate this trend: The carbonates become less soluble down the group. All the Group 2 carbonates are very sparingly soluble. Magnesium carbonate, for example, has a solubility of about 0.02 g per 100 g of water at room temperature. There is little data for beryllium carbonate, but as it reacts with water, the trend is obscured. The trend to lower solubility is, however, broken at the bottom of the group: barium carbonate is slightly more soluble than strontium sulfate. There are no simple examples of this trend. Jim Clark ( ) | 2,800 | 1 |
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Group 2 elements (beryllium, magnesium, calcium, strontium and barium) react oxygen. to generate metal oxides. This Module addressed why it is difficult to observe a tidy pattern of this reactivity. On the whole, the metals burn in oxygen to form a simple metal oxide. Beryllium is reluctant to burn unless it is in the form of dust or powder. Beryllium has a very strong (but very thin) layer of beryllium oxide on its surface, and this prevents any new oxygen getting at the underlying beryllium to react with it. \[ 2X_{(s)} + O_{2(g)} \rightarrow 2XO_{(s)}\] with \(X\) representing any group 2 metal. It is almost impossible to find any trend in the way the metals react with oxygen. It would be quite untrue to say that they burn more vigorously as you go down the Group. To be able to make any sensible comparison, you would have to have pieces of metal which were all equally free of oxide coating, with exactly the same surface area and shape, exactly the same flow of oxygen around them, and heated to exactly the same extent to get them started. What the metals look like when they burn is a bit problematical! Strontium and barium will also react with oxygen to form strontium or barium peroxide. Strontium forms this if it is heated in oxygen under high pressures, but barium forms barium peroxide just on normal heating in oxygen. Mixtures of barium oxide and barium peroxide will be produced. \[ Ba_{(s)} + O_{2(s)} \rightarrow BaO_{2(s)}\] The strontium equation would look just the same. The reactions of the Group 2 metals with air rather than oxygen is complicated by the fact that they all react with nitrogen to produce nitrides. In each case, you will get a mixture of the metal oxide and the metal nitride. The general equation for the Group is: \[ 3X_{(s)} + N_{2(g)} \rightarrow X_3N_{2(s)}\] For example, the familiar white ash you get when you burn magnesium ribbon in air is a mixture of magnesium oxide and magnesium nitride. \[ 2Mg_{(s)} + O_{2(g)} \rightarrow 2MgO_{(s)}\] \[ 3Mg_{(s)} + N_{2(g)} \rightarrow Mg_3N_{2(s)}\] There are no simple patterns in the way the metals burn. While it would be tempting to say that the reactions get more vigorous as you go down the Group, but it is not true. The overall amount of heat evolved when one mole of oxide is produced from the metal and oxygen also shows no simple pattern: If anything, there is a slight tendency for the amount of heat evolved to decrease as you go down the Group. But how reactive a metal seems to be depends on how fast the reaction happens (i.e., Kinetics) - not the overall amount of heat evolved (i.e., Thermodynamics). The speed is controlled by factors like the presence of surface coatings on the metal and the size of the activation energy. You could argue that the activation energy will fall as you go down the Group and that will make the reaction go faster. The activation energy will fall because the ionization energies of the metals fall. In this case, though, the effect of the fall in the activation energy is masked by other factors - for example, the presence of existing oxide layers on the metals, and the impossibility of controlling precisely how much heat you are supplying to the metal in order to get it to start burning. Beryllium, magnesium and calcium don't form peroxides when heated in oxygen, but strontium and barium do. There is an increase in the tendency to form the peroxide as you go down the Group. The peroxide ion, O looks like this: The covalent bond between the two oxygen atoms is relatively weak. Now imagine bringing a small 2+ ion close to the peroxide ion. Electrons in the peroxide ion will be strongly attracted towards the positive ion. This is then well on the way to forming a simple oxide ion if the right-hand oxygen atom (as drawn below) breaks off. We say that the positive ion the negative ion. This works best if the positive ion is small and highly charged - if it has a high charge density. A high charge density simply means that you have a lot of charge packed into a small volume. Ions of the metals at the top of the Group have such a high charge density (because they are so small) that any peroxide ion near them falls to pieces to give an oxide and oxygen. As you go down the Group and the positive ions get bigger, they don't have so much effect on the peroxide ion. For example, Barium peroxide can form because the barium ion is so large that it doesn't have such a devastating effect on the peroxide ions as the metals further up the Group. Nitrogen is often thought of as being fairly unreactive, and yet all these metals combine with it to produce nitrides, X N , containing X and N ions. Nitrogen is fairly unreactive because of the very large amount of energy is required to break the triple bond joining the two atoms in the nitrogen molecule, N . When something like magnesium nitride forms, you have to supply all the energy needed to form the magnesium ions as well as breaking the nitrogen-nitrogen bonds and then forming N ions. All of these processes absorb energy. This energy has to be recovered from somewhere to give an overall exothermic reaction - if the energy can't be recovered, the overall change will be endothermic and will not happen. Energy is evolved when the ions come together to produce the crystal lattice ( or enthalpy). The size of the lattice energy depends on the attractions between the ions. The lattice energy is greatest if the ions are small and highly charged - the ions will be close together with very strong attractions. In the whole of Group 2, the attractions between the 2+ metal ions and the 3- nitride ions are big enough to produce very high lattice energies. When the crystal lattices form, so much energy is released that it more than compensates for the energy needed to produce the various ions in the first place. The excess energy evolved makes the overall process exothermic. This is in contrast to what happens in of the Periodic Table (lithium, sodium, potassium, rubidium and cesium). Their ions only carry one positive charge, and so the lattice energies of their nitrides will be much less. Lithium is the only metal in Group 1 to form a nitride. Lithium has by far the smallest ion in the Group, and so lithium nitride has the largest lattice energy of any possible Group 1 nitride. Only in lithium's case is enough energy released to compensate for the energy needed to ionize the metal and the nitrogen - and so produce an exothermic reaction overall. In all the other Group 1 elements, the overall reaction would be endothermic. Those reactions don't happen, and the nitrides of sodium and the rest are not formed. Jim Clark ( ) | 6,687 | 2 |
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This page examines at the effect of heat on the carbonates and nitrates of the Group 2 elements (beryllium, magnesium, calcium, strontium and barium). It explains how the thermal stability of the compounds changes down the group. All the carbonates in this group undergo thermal decomposition to the metal oxide and carbon dioxide gas. The term "thermal decomposition" describes splitting up a compound by heating it. All the Group 2 carbonates and their resulting oxides exist as white solids. If "X" represents any one of the elements, the following describes this decomposition: \[XCO_3(s) \rightarrow XO(s) + CO_2(g)\] Down the group, the carbonates require more heating to decompose. The Group 2 nitrates undergo thermal decomposition to the metal oxide, nitrogen dioxide and oxygen gas. These compounds are white solids and brown nitrogen dioxide and oxygen gases are also given off when heated. Magnesium and calcium nitrates normally crystallize with water, and the solid may dissolve in its own water of crystallization to make a colorless solution before it starts to decompose. Again, if "X" represents any one of the elements: \[ 2X(NO_3)_2(s) \rightarrow 2XO(s) + 4NO_2(g) + O_2 (g)\] Down the group, the nitrates must also be heated more strongly before they will decompose. Group 2 nitrates also become more thermally stable down the group. Both carbonates and nitrates of Group 2 elements become more thermally stable down the group. The larger compounds further down require more heat than the lighter compounds in order to decompose. This page offers two different explanations for these properties: polarizability and energetics. Detailed explanations are given for the carbonates because the diagrams are easier to draw, and their equations are also easier. Exactly the same arguments apply to the nitrates. A small 2+ ion has a lot of charge packed into a small volume of space. In other words, it has a high charge density and has a marked distorting effect on any negative ions which happen to be near it. A bigger 2+ ion has the same charge spread over a larger volume of space, so its charge density is lower; it causes less distortion to nearby negative ions. A shorthand structure for the carbonate ion is given below: This structure two single carbon-oxygen bonds and one double bond, with two of the oxygen atoms each carrying a negative charge. In real carbonate ions all the bonds are identical, and the charges are distributed over the whole ion, with greater density concentrated on the oxygen atoms.In other words, the charges are delocalized. The next diagram shows the delocalized electrons. The shading is intended to show that there is a greater electron density around the oxygen atoms than near the carbon. If this ion is placed next to a cation, such as a Group 2 ion, the cation attracts the delocalized electrons in the carbonate ion, drawing electron density toward itself. The carbonate ion becomes polarized. If the carbonate is heated the carbon dioxide breaks free, leaving the metal oxide. The amount of heating required depends on the degree to which the ion is polarized. More polarization requires less heat. The smaller the positive ion is, the higher the charge density, and the greater effect it will have on the carbonate ion. As the positive ions get larger down the group, they affect on the carbonate ions near them less. More heat must be supplied for the carbon dioxide to leave the metal oxide. In other words, the carbonates become more thermally stable down the group. The argument is exactly the same for the Group 2 nitrates. The small cations at the top of the group polarize the nitrate ions more than the larger cations at the bottom do. This process is much more difficult to visualize due to interactions involving multiple nitrate ions. The enthalpy changes for the decomposition of the various carbonates indicate that the reactions are strongly endothermic, implying that the reactions likely require constant heating to proceed. Remember that the reaction in question is the following: \[XCO_{3(s)} \rightarrow XO_{(s)} + CO_{2(g)}\] The calculated enthalpy changes (in kJ mol ) are given in the table below (there is no available data for beryllium carbonate). The reactions are more endothermic down the group, as expected, because the carbonates become more thermally stable, as discussed above. Here's where things start to get difficult! If you aren't familiar with Hess's Law cycles (or with Born-Haber cycles) and with lattice enthalpies (lattice energies), you aren't going to understand the next bit. Don't waste your time looking at it. You can dig around to find the underlying causes of the increasingly endothermic changes as you go down the Group by drawing an enthalpy cycle involving the lattice enthalpies of the metal carbonates and the metal oxides. is the heat needed to split one mole of crystal in its standard state into its separate gaseous ions. For example, for magnesium oxide, it is the heat needed to carry out 1 mole of this change: \[ MgO_{(s)} \rightarrow Mg^{2+}_{(g)} + O^{2-}_{(g)}\] Lattice Energy (LE): + 3889 kJ/mol Here's where things start to get difficult! If you aren't familiar with Hess's Law cycles (or with Born-Haber cycles) and with lattice enthalpies (lattice energies), you aren't going to understand the next bit. Don't waste your time looking at it. Lattice enthalpy is more usually defined as the heat evolved when 1 mole of crystal is formed from its gaseous ions. In that case, the lattice enthalpy for magnesium oxide would be -3889 kJ mol . The term we are using here should more accurately be called the "lattice dissociation enthalpy". The cycle we are interested in looks like this: You can apply Hess's Law to this, and find two routes which will have an equal enthalpy change because they start and end in the same places. For reasons we will look at shortly, the lattice enthalpies of both the oxides and carbonates fall as you go down the Group. But they don't fall at the same rate. The oxide lattice enthalpy falls faster than the carbonate one. If you think carefully about what happens to the value of the overall enthalpy change of the decomposition reaction, you will see that it gradually becomes more positive as you go down the Group. The size of the lattice enthalpy is governed by several factors, one of which is the distance between the centres of the positive and negative ions in the lattice. Forces of attraction are greatest if the distances between the ions are small. If the attractions are large, then a lot of energy will have to be used to separate the ions - the lattice enthalpy will be large. The lattice enthalpies of both carbonates and oxides fall as you go down the Group because the positive ions are getting bigger. The inter-ionic distances are increasing and so the attractions become weaker. The lattice enthalpies fall at different rates because of the different sizes of the two negative ions - oxide and carbonate. The oxide ion is relatively small for a negative ion (0.140 nm), whereas the carbonate ion is large (no figure available). In the oxides, when you go from magnesium oxide to calcium oxide, for example, the inter-ionic distance increases from 0.205 nm (0.140 + 0.065) to 0.239 nm (0.140 + 0.099) - an increase of about 17%. In the carbonates, the inter-ionic distance is dominated by the much larger carbonate ion. Although the inter-ionic distance will increase by the same amount as you go from magnesium carbonate to calcium carbonate, as a percentage of the total distance the increase will be much less. Some made-up figures show this clearly. I can't find a value for the radius of a carbonate ion, and so can't use real figures. For the sake of argument, suppose that the carbonate ion radius was 0.3 nm. The inter-ionic distances in the two cases we are talking about would increase from 0.365 nm to 0.399 nm - an increase of only about 9%. The rates at which the two lattice energies fall as you go down the Group depends on the percentage change as you go from one compound to the next. On that basis, the oxide lattice enthalpies are bound to fall faster than those of the carbonates. The nitrate ion is bigger than an oxide ion, and so its radius tends to dominate the inter-ionic distance. The lattice enthalpy of the oxide will again fall faster than the nitrate. if you constructed a cycle like that further up the page, the same arguments would apply. Jim Clark ( ) | 8,485 | 3 |
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This page discusses the reactions of the Group 2 elements ( , , , and barium) with water, using these reactions to describe the trend in reactivity in . Beryllium reacts with steam at high temperatures (typically around 700°C or more) to give white beryllium oxide and hydrogen. \[ Be_{(s)} + H_2O_{(g)} \rightarrow BeO_{(s)} + H_{2(g)} \label{0}\] There is an additional reason for the lack of reactivity of beryllium compared with the rest of the Group. Beryllium has a strong resistant layer of oxide on its surface which lowers its reactivity at ordinary temperatures. However, the oxide layer breaks up above 750°C and exposes the beryllium metal surface below it, and so the protection then fails. Magnesium burns in steam to produce white magnesium oxide and hydrogen gas. \[ Mg_{(s)} + H_2O_{(g)} \rightarrow MgO_{(s)} + H_{2(g)} \label{1}\] Very clean magnesium ribbon has a mild reaction with cold water, given below. After several minutes, hydrogen gas bubbles form on its surface, and the coil of magnesium ribbon usually floats to the surface. However, the reaction is short-lived because the magnesium hydroxide formed is almost insoluble in water and forms a barrier on the magnesium preventing further reaction. \[ Mg_{(s)} + 2H_2O_{(l)} \rightarrow Mg(OH)_{2(s)} + H_{2(g)} \label{2}\] As a general rule, if a metal reacts with cold water, the metal hydroxide is produced. If it reacts with steam, the metal oxide is formed. This is because the metal hydroxides thermally decompose to the oxide and water. These metals react with cold water with increasing vigor to give the metal hydroxide and hydrogen. Strontium and barium have reactivities similar to that of lithium. Calcium, for example, reacts fairly vigorously and exothermically with cold water. Bubbles of hydrogen gas are given off, and a white precipitate (of calcium hydroxide) is formed, together with an alkaline solution (also of calcium hydroxide, which is slightly water-soluble). The equation for the reactions of any of these metals would is as follows: \[ X_{(s)} + 2H_2O_{(l)} + X(OH) \rightarrow X(OH)_{2 (aq\, or\, s)} + H_{2(g)} \label{3}\] The hydroxide solubilities increase down the group. Calcium hydroxide is mainly formed as a white precipitate (although some does dissolve). Less precipitate is formed down the group with increasing solubility. The enthalpy change of a reaction is a measure of the amount of heat absorbed or evolved when the reaction takes place. An enthalpy change is negative if heat is evolved, and positive if it is absorbed. Calculate the enthalpy change for the possible reactions between beryllium or magnesium and steam gives the following values: \[Be_{(s)} + H_2O_{(g)} \rightarrow BeO_{(s)} + H_{2(g)} \;\;\; \Delta H = -369\; kJ/mol\] \[Mg_{(s)} + H_2O_{(g)} \rightarrow MgO_{(s)} + H_{2(g)} \;\;\; \Delta H = -360\; kJ/mol \] Notice that both possible reactions are strongly exothermic, giving out almost identical amounts of heat. However, only the magnesium reaction actually happens. The explanation for the different reactivities must lie somewhere else. Similarly, calculating the enthalpy changes for the reactions between calcium, strontium or barium and cold water reveals that the amount of heat evolved in each case is almost exactly the same—about -430 kJ mol . The reason for the increase in reactivity must again lie elsewhere. The activation energy for a reaction is the minimum amount of energy which is needed in order for the reaction to take place. It does not matter how exothermic the reaction would be once it got started - if there is a high activation energy barrier, the reaction will take place very slowly, if at all. When Group 2 metals react to form oxides or hydroxides, metal ions are formed. The formation of the ions from the original metal involves various stages all of which require the input of energy - contributing to the activation energy of the reaction. These stages involve the input of: After this, there will be a number of steps which give out heat again - leading to the formation of the products, and overall exothermic reactions. The graph shows the effect of these important energy-absorbing stages as you go down Group 2. Notice that the ionization energies dominate this - particularly the second ionization energies. Ionization energies fall down the group. Because it gets easier to form the ions, the reactions will happen more quickly. The reactions of the Group 2 elements proceed more readily as the energy needed to form positive ions falls. This is mainly due to a decrease in ionization energy down the group. This leads to lower activation energies, and therefore faster reactions. Jim Clark ( ) | 4,702 | 4 |
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All molecules possess a certain minimum amount of energy. The energy can be in the form of kinetic energy or potential energy. When molecules collide, the kinetic energy of the molecules can be used to stretch, bend, and ultimately break bonds, leading to chemical reactions. If molecules move too slowly with little kinetic energy, or collide with improper orientation, they do not react and simply bounce off each other. However, if the molecules are moving fast enough with a proper collision orientation, such that the kinetic energy upon collision is greater than the minimum energy barrier, then a reaction occurs. The minimum energy requirement that must be met for a chemical reaction to occur is called the activation energy, \(E_a\). The reaction pathway is similar to what happens in Figure 1. To get to the other end of the road, an object must roll with enough speed to completely roll over the hill of a certain height. The faster the object moves, the more kinetic energy it has. If the object moves too slowly, it does not have enough kinetic energy necessary to overcome the barrier; as a result, it eventually rolls back down. In the same way, there is a minimum amount of energy needed in order for molecules to break existing bonds during a chemical reaction. If the kinetic energy of the molecules upon collision is greater than this minimum energy, then bond breaking and forming occur, forming a new product (provided that the molecules collide with the proper orientation). The activation energy (\(E_a\)), labeled \(\Delta{G^{\ddagger}}\) in Figure 2, is the energy difference between the reactants and the activated complex, also known as transition state. In a chemical reaction, the transition state is defined as the highest-energy state of the system. If the molecules in the reactants collide with enough kinetic energy and this energy is higher than the transition state energy, then the reaction occurs and products form. In other words, the higher the activation energy, the harder it is for a reaction to occur and vice versa. However, if a catalyst is added to the reaction, the activation energy is lowered because a lower-energy transition state is formed, as shown in Figure 3. Enzymes can be thought of as biological catalysts that lower activation energy. Enzymes are proteins or RNA molecules that provide alternate reaction pathways with lower activation energies than the original pathways. Enzymes affect the rate of the reaction in both the forward and reverse directions; the reaction proceeds faster because less energy is required for molecules to react when they collide. Thus, the (k) increases. As indicated by Figure 3 above, a catalyst helps lower the activation energy barrier, increasing the reaction rate. In the case of a biological reaction, when an enzyme (a form of catalyst) binds to a substrate, the activation energy necessary to overcome the barrier is lowered, increasing the rate of the reaction for both the forward and reverse reaction. See below for the effects of an enzyme on activation energy. Catalysts do not just reduce the energy barrier, but induced a completely different reaction pathways typically with multiple energy barriers that must be overcome. For example: In thermodynamics, the change in , ΔG, is defined as: where \( \Delta G^o \) is the change in Gibbs energy when the reaction happens at Standard State (1 atm, 298 K, pH 7). To calculate a reaction's change in Gibbs free energy that did not happen in standard state, the Gibbs free energy equation can be written as: \[ \Delta G = \Delta G^o + RT\ \ln K \label{2} \] where When the reaction is at , \( \Delta G = 0\). The equation above becomes: \[ 0 = \Delta G^o + RT\ln K \nonumber \] Solve for ΔG : \[ \Delta G^o = -RT \ln K \nonumber \] Similarly, in transition state theory, the Gibbs energy of activation, \( \Delta G ^{\ddagger} \), is defined by: \[ \Delta G ^{\ddagger} = -RT \ln K^{\ddagger} \label{3} \] and \[ \Delta G ^{\ddagger} = \Delta H^{\ddagger} - T\Delta S^{\ddagger}\label{4} \] where Combining equations 3 and 4 and then solve for \(\ln K^{\ddagger}\) we have the : \[ \ln K^{\ddagger} = -\dfrac{\Delta H^{\ddagger}}{RT} + \dfrac{\Delta S^{\ddagger}}{R} \nonumber \] As shown in the figure above, activation enthalpy, \(\Delta{H}^{\ddagger} \), represents the difference in energy between the ground state and the transition state in a chemical reaction. The higher the activation enthalpy, the more energy is required for the products to form. Note that this activation enthalpy quantity, \( \Delta{H}^{\ddagger} \), is analogous to the activation energy quantity, E , when comparing the Arrhenius equation (described below) with the Eyring equation: \[E_a = \Delta{H}^{\ddagger} + RT \nonumber \] In general, a reaction proceeds faster if E and \(\Delta{H}^{\ddagger} \) are small. Conversely, if E and \( \Delta{H}^{\ddagger} \) are large, the reaction rate is slower. As temperature increases, gas molecule velocity also increases (according to the ). This is also true for liquid and solid substances. The (translational) kinetic energy of a molecule is proportional to the velocity of the molecules (KE = 1/2 mv ). Therefore, when temperature increases, KE also increases; as temperature increases, more molecules have higher KE, and thus the fraction of molecules that have high enough KE to overcome the energy barrier also increases. The fraction of molecules with energy equal to or greater than E is given by the exponential term \(e^{\frac{-E_a}{RT}}\) in the Arrhenius equation: \[k = Ae^{\frac{-E_a}{RT}} \label{5} \] Taking the natural log of both sides of Equation \(\ref{5}\) yields the following: \[\ln k = \ln A - \frac{E_a}{RT} \label{6} \] Equation \(\ref{4}\) has the linear form y = mx + b. Graphing ln k vs 1/T yields a straight line with a slope of -E /R and a y-intercept of ln A., as shown in Figure 4. As indicated in Figure 5, the reaction with a higher E has a steeper slope; the reaction rate is thus very sensitive to temperature change. In contrast, the reaction with a lower E is less sensitive to a temperature change. Because radicals are extremely reactive, E for a radical reaction is 0; an arrhenius plot of a radical reaction has no slope and is independent of temperature. The activation energy can also be calculated directly given two known temperatures and a rate constant at each temperature. Using Equation (2), suppose that at two different temperatures T and T , reaction rate constants k and k : \[\ln\; k_1 = - \frac{E_a}{RT_1} + \ln A \label{7} \] and \[\ln\; k_2 = - \frac{E_a}{RT_2} + \ln A \label{8} \] Subtract \(ln\; k_2\) from \(ln\; k_1\): \[ \ln\; k_1 - \ln\; k_2 = \left (- \dfrac{E_a}{RT_1} + \ln A \right ) - \left(- \dfrac{E_a}{RT_2} + \ln A \right) \label{9} \] After rearrangement: \[ \ln \left (\dfrac{k_1}{k_2} \right ) = \left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right)\dfrac{E_a}{R} \label{10} \] Use the Arrhenius Equation: \(k = Ae^{-E_a/RT}\) Use the equation: \( \ln \left (\dfrac{k_1}{k_2} \right ) = \dfrac{-E_a}{R} \left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)\) Use the equation \(\Delta{G} = \Delta{H} - T \Delta{S}\) Use the equation \(\ln k = \ln A - \dfrac{E_a}{RT}\) to calculate the activation energy of the forward reaction No. Most enzymes denature at high temperatures. At some point, the rate of the reaction and rate constant will decrease significantly and eventually drop to zero. Once the enzyme is denatured, the alternate pathway is lost, and the original pathway will take more time to complete. | 7,603 | 5 |
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1. Beryllium is the least reactive and does not react with water even at red heat and does not react with N . Magnesium only reacts at reasonable rate with steam, calcium and strontium readily tarnish in moist air and barium tarnishes readily. 2. The ionic character of the compounds increases down the Group. Beryllium forms highly covalent compounds generally with tetrahedral geometries, [BF ] , [BeCl ] (infinite linear polymer). Magnesium forms more polar compounds with 6-coordination. Calcium, strontium, and barium form increasingly ionic compounds with higher coordination numbers (8 is particularly common). 3. The organometallic compounds of beryllium are covalent and rather unreactive. Magnesium forms two important series of organometallic compounds, RMgX and R Mg, which provide convenient sources of carbanions for organic synthesis. The organometallic compounds of calcium, strontium, and barium are generally more reactive and are insoluble in organic solvents. 4. The oxides become progressively more basic down the Group. BeO and Be(OH) are amphoteric and react with acids and strong bases such as NaOH. MgO is basic and Mg(OH) is weakly basic and do not dissolve in NaOH solution. The oxides of calcium, strontium, and barium are basic and the hydroxides are strongly basic. The solubilities of the hydroxides in water follow the order: Be(OH) < Mg(OH) < Ca(OH) < Sr(OH) < Ba(OH) 5. BeX (X = F, Cl, Br, or I) are covalent polymers, which are readily hydrolyzed and are Lewis acids forming adducts BeX L (L = Lewis base). Magnesium, calcium, strontium, and barium halides are essentially ionic and are soluble in water. 6. BeH is a covalent polymer, magnesium hydride is partially ionic and the hydrides of calcium, strontium, and barium are very ionic and hydridic in their properties. 7. Mg and Ca have the greatest tendency to form complexes especially with ligands which have oxygen donor atoms. For small highly charged anions the order of stability is generally: Mg > Ca > Sr > Ba but for the anions, NO , SO , and IO the stability order is: Mg < Ca < Sr < Ba The most important complexes of these metals are with EDTA . The order of stability for this and related polydentate ligands is: Mg < Ca > Sr > Ba The calcium complex, [Ca(EDTA)] , is particularly important because it is water soluble and allows EDTA to solubilize calcium carbonate. Polyphosphates, P O and P O are able to function similarly to solubilize hard water deposits of CaCO . The crown polyethers and cryptate ligands also form stable complexes with Ca and Mg . 8. Both Mg and Ca have important roles in biology due to their fast to moderate ligand exchange rates; slower than K and Na but faster than most transition metal cations. 9. The thermal stabilities of the nitrates, carbonates, and peroxides increase down the column. 10. The solubilities of the sulfates, nitrates, and chlorides increase down the group. 11. The solubilities of the halides in alcohols increase down the group. 1. The high enthalpy of atomization of beryllium causes it to be mechanically harder, higher melting, less dense, and less reactive than the heavier elements of the group. 2. The high charge/size ratio of Be leads to compounds that are more covalent and complexes that are more stable than those of the remaining Group II cations. Many of the compounds have anomalously low melting points, enthalpies of formation, and are more soluble in organic solvents. The compounds are stronger Lewis acids. The halides are hygroscopic and fume when exposed to moist air. 3. Unlike magnesium and the heavier metals of the group, beryllium oxide and hydroxide are amphoteric. 4. Beryllium salts are much less thermally stable because the high lattice energy of the oxide lowers the Gibbs energy change for the decomposition reaction. Similarly it odes not form a peroxide or superoxide. With ethyne, beryllium forms the carbide, Be C, rather than the ethnide presumably because the lattice energy of the carbide is very favorable. | 4,063 | 6 |
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In this article, formulae based mnemonics by classifying lone pair of electrons (localized or delocalized) have been highlighted in an innovative and time economic way to enhance interest of students’ on heterocyclic chemistry for determination of planarity by calculating Hybridization state of hetero atom and by prediction of Aromatic, Anti aromatic, non aromatic behavior of different heterocyclic compounds . Here, I have tried to hub three (03) time economic mnemonics by including two (02) formulae for the prediction of hybridization of hetero atom, aromatic and anti aromatic behavior of heterocyclic compounds. This article encourages students to solve multiple choice type questions (MCQs) on ‘Aromaticity of Heterocyclic compounds’ at different competitive examinations in a time economic way. The conventional methods1-7 for determination of hybridization state of hetero atom (planarity of molecule), prediction of aromatic and anti aromatic nature of heterocyclic compound is time consuming. Keeping this in mind, in this article, I have introduced three time economic innovative mnemonics by using two formulae for the prediction of hybridization state of hetero atom of heterocyclic compounds to determine its planarity and aromatic / anti aromatic / non aromatic nature of heterocyclic compounds containing one, two or more number of hetero atoms to make heterocyclic chemistry metabolic and interesting for students. This study also shows omission behavior of some heterocyclic compounds with respect to their aromatic/anti aromatic/non aromatic nature due to presence or absence of vacant d orbitals in DLP based hetero atoms and how lone pair electron discriminates prediction of hybridization state of hetero atom in heterocyclic compound with prediction of its Aromatic and Anti Aromatic nature. Time Economic Innovative Mnemonics in Heterocyclic Chemistry Lone Pair of electrons can be generally classified into two types as Delocalized lone pair of electron (DLP) and Localized lone pair of electron (LLP) as follows: i)Delocalized lone pair of electron (DLP): When lone pair of electron of hetero atom undergo delocalization through conjugation then it is to be treated as delocalized lone pair of electron (DLP). Hetero atom (atom containing lone pair of electron) which is directly attached with single bonds only from all ends is to be considered as DLP containing hetero atom and its lone pair is to be treated as (DLP). Eg. In Pyrrole lone pair of N atom is to be treated as DLP because it is directly attached with three single bonds only. ii)Localized lone pair of electron (LLP): When lone pair of electron of hetero atom does not undergo delocalization through conjugation then it is to be treated as Localized lone pair of electron (LLP). Hetero atom (atom containing lone pair of electron) which is directly attached with single and double bonds with the ring system is to be considered as LLP containing hetero atom and its lone pair is to be treated as localized lone pair of electron (LLP). Eg. In Pyridine lone pair of N atom is to be treated as LLP because it is directly attached with double and single bonds with the ring system. It was first devised by Hückel in 1931. Conventional method for prediction of Aromatic nature of organic compound: Conventional method for Anti Aromatic nature of organic Compound: Cyclic molecule, Planer molecule in which all bonded atoms lie in same plane (having sp2 hybridized) Conjugated molecule with conjugated π-electron system, 4nπ electrons, where, n is a positive integer (n = 0,1,2,3 etc.) Conventional method for identification of Non Aromatic Nature of organic Compound: If a compound violates any one of the above three conditions (1 or 2 or 3) then it is non aromatic in nature. Planarity of Heterocyclic Compounds with the prediction of Hybridization State Planarity of heterocyclic compounds depends on the nature of the hybridization state of carbon and hetero atoms present in it. When all atoms (carbon and hetero) in the heterocyclic compounds having sp2 hybridized then it is planar but when there is a mixing of sp2 and sp3 hybridization state then it is treated as non planar. Hybridization state theory Prof. Linus Pauling (1931) first developed the Hybridization state theory in order to explain the structure of molecules such as methane (CH4).This concept was developed for simple chemical systems but this one applied more widely later on and from today’s point of view it is considered an operative empirical for excusing the structures of organic and inorganic compounds along with their related problems. Conventional method for prediction of hybridization state: Hybridization state for a molecule can be calculated by the formula 0.5 (V+H−C+A), Where, V = Number of valance electrons in central atom, H = Number of surrounding monovalent atoms, C = Cationic charge, A = Anionic charge Formula 1: Prediction of hybridization state of hetero atom Power on the Hybridization state of the hetero atom = (Total no of σ bonds around each hetero atom - 1) This formula should be applicable up to 4 σ bonds. If the power of the hybridization state will be 03, 02 and 01 then the hybridization state will be sp3, sp2 and sp respectively. All single (-) bonds are σ bond, in double bond (=) there is one σ and one π. In addition to these each localized lone pair of electron (LLP) can be treated as one σ bond. Hybridization State of Hetero atom with the help of LLP to find out the planarity in Heterocyclic Compounds are shown in Table-1 below. Heterocyclic Compounds (Planar/non planar) Number of σ bonds around hetero atom (from single and double bonds) (A) LLP (localized Lone Pair of e-s) (B) Total Number of σ bonds around hetero atom (A+B) Power of the Hybridization state of the hetero atom (Corresponding Hybridization state) = (A+B)-1 Pyrrole (Planar) 03 0 (lone pair of electron undergo delocalization,DLP with the ring system) 03 02 (sp2 N) Furan (Planar) 02 01 (out of two lone pair of electrons, one undergo delocalization,DLP and other remain as LLP) 03 02 (sp2 O) Thiophene (Planar) 02 01 (out of two lone pair of electrons, one undergo delocalization, DLP and other remain as LLP) 03 02 (sp2 S) Pyridine (Planar) 02 01 03 02 (sp2 N) Indole (Planar) 03 0 03 02 (sp2 N) Quinoline (Planar) 02 01 03 02 (sp2 N) (Planar) 02 01 03 02 (sp2 N) (Planar) 03 (N1) 02 (N3) 0 (N1) 01 (N3) 03 03 02 (sp2 N1) 02 (sp2 N3) (Planar) 02 (N1) 02 (N3) 01 (N1) 01 (N3) 03 03 02 (sp2 N1) 02 (sp2 N3) (Planar) 02 (N1) 02 (N3) 02 (N7) 03 (N9) 01 (N1) 01 (N3) 01 (N7) 0 (N9) 03 03 03 03 02 (sp2 N1) 02 (sp2 N3) 02 (sp2 N7) 02 (sp2 N9) (Planar) 02 (N) 02 (S) 01 (N) 01 (S) (out of two lone pair of electrons on S, one undergo delocalization,DLP and other remain as LLP) 03 03 02 (sp2 N) 02 (sp2 S) (Planar) 02 (N) 02 (S) 01 (N) 01 (S) (out of two lone pair of electrons on S, one undergo delocalization,DLP and other remain as LLP) 03 03 02 (sp2 N) 02 (sp2 S) (Planar) 02 (N1) 02 (N1) 01 (N1) 01 (N1) 03 03 02 (sp2 N1) 02 (sp2 N4) (Planar) 02 (N1,N3 and N5) 01 (N1,N3 and N5) 03 02 (sp2 N1,N3,N5) (Planar) 03 (N) 02 (S) 0 (N) 01 (S) (out of two lone pair of electrons on S, one undergo delocalization, DLP and other is LLP) 03 03 02 (sp2 N) 02 (sp2 S) (Planar) 02 (both N) 01 (both N) 03 02 (sp2 both N) (Planar) 02 (N1,N2,N3,N4) 01 (N1,N2,N3,N4) 03 02 (sp2 All N) (Planar) 02 01 03 02 (sp2 N) (Planar) 02 01 03 02 (sp2 N) (Non Planar) 03 01 04 03 (sp3 N) (Non Planar) 02 02 04 03 (sp3 O) The present study will be an innovative mnemonic involving calculation of ‘A’ value by just manipulating the no of π bonds within the ring system and delocalized lone pair of electron (DLP) with one (01). The heterocyclic compound having cyclic, planar, conjugated (i.e. all the carbon atoms having same state of hybridization, sp2) with even number of ‘A’ value will be treated as aromatic in nature and with odd number of ‘A’ value will be treated as anti aromatic in nature. Formula 2: Evaluation of A Value to predict Aromatic and Anti Aromatic Nature A = πb+DLP+1(constant) = even no = Aromatic A = πb+DLP+1(constant) = odd no = Anti Aromatic where, πb = number of π bonds with in the ring system; DLP = Delocalized lone pair of electron. In case of a multi hetero atom based heterocyclic compound, containing both DLP and LLP hetero atoms, Aromatic and Anti Aromatic behaviour should be predicted with respect to DLP based hetero atom only. Benzothiazole (Figure 1), is a multi hetero atom based heterocyclic compound, containing both DLP and LLP hetero atoms. Here, for N, DLP = 0 , LLP = 1 and for S, DLP = 1, LLP =1, so, in this case ‘A’ value should be calculated with respect to S only not N. Here, A = 4 + 1 + 1 = 6 (even no) = Aromatic. But when heterocyclic compounds contain both LLP based hetero atoms then Aromaticity should be predicted with respect to that hetero atom which contains lowest possible position number as per IUPAC nomenclature or any one of the hetero atom. Imidazole (Figure 2) is a multi hetero atom based hetero cyclic compound in which, N1 is DLP based hetero atom and N3 is LLP based hetero atom. In this case Aromaticity should be predicted with respect to the DLP based hetero atom N1. For N1, A = πb+DLP+1(constant) = 2+1+1 = 4 (even No) - Aromatic Eg. Pyrimidine (Figure 3) is a multi hetero atom based hetero cyclic compound in which, both N1 & N3 are in same environment based hetero atoms (LLP based hetero atoms). In this case Aromaticity should be predicted with respect to N1 (lowest possible position number as per IUPAC nemenclature). For N1, A = πb+DLP+1(constant) = 3+0+1 = 4 (even no) - Aromatic Aromaticity of heterocyclic compounds have been illustrated in Table-2 Hetero Cyclic Compound (Cyclic, Planar, Conjugated) πb value [πb =number of π bonds with in the ring system] DLP A value [A = πb + DLP + 1(constant)] (even No /odd No) Remark on Nature of compound (Aromatic/Anti Aromatic) Pyrrole 2 1 2 + 1 + 1 = 4 (even No) Aromatic Furan 2 1 ( Here out of two lone pairs on O only one LP take part in delocalization) 2 + 1 + 1 = 4 (even No) Aromatic Thiophene 2 1 (Here out of two lone pairs on O only one LP take part in delocalization) 2 + 1 + 1 = 4 (even No) Aromatic Pyridine 3 0 3 + 0 + 1 = 4 (even No) Aromatic Indole 4 1 4 + 1 + 1 = 6 (even No) Aromatic Quinoline 5 0 5 + 0 + 1 = 6 (even No) Aromatic 05 0 5 + 0 + 1 = 6 (even No) Aromatic 02 01 (N1) 2 + 1 + 1 = 4 (even No) Aromatic (m-diazine) 03 0 (N1) 3 + 0 + 1 = 4 (even No) Aromatic 04 01 (N9) 4 + 1 + 1 = 6 (even No) Aromatic 02 01 (S) 2 + 1 + 1 = 4 (even No) Aromatic 04 01 (S) 4 + 1 + 1 = 6 (even No) Aromatic 03 0 3 + 0 + 1 = 4 (even No) Aromatic 03 0 3 + 0 + 1 = 4 (even No) Aromatic 07 0 7 + 0 + 1 = 8 (even No) Aromatic 03 0 3 + 0 + 1 = 4 (even No) Aromatic 04 0 4 + 0 + 1 = 5 (odd No) Anti aromatic 02 0 2 + 0 + 1 = 3 (odd No) Anti aromatic Hetero Cyclic Compound (Cyclic, non-planar) πb value [πb =number of π bonds with in the ring system] DLP A value [A = πb + DLP + 1(constant)] (even No/odd No) Remark on Nature of compound - - - Non Aromatic (non planar – sp3) - - - Non Aromatic (non planar – sp3) Omission behavior of some heterocyclic compounds with respect to their Aromatic / Anti Aromatic and Non Aromatic nature : Aromatic Behavior of some heterocyclic compounds containing different DLP based hetero atoms (one contains vacant d orbitals) : In Phenothiazine (Figure 4), there is two DLP based hetero atoms N and S. In between N and S, since S having vacant d orbitals, so, in this case ‘A’ value will be predicted with respect to DLP based S hetero atom which contains vacant d orbitals only. Here, A = πb + DLP + 1(constant) = 6 +1+1 = 8 (even no) = Aromatic. Non Aromatic Behavior of some heterocyclic compounds containing same DLP based heteroatom having no d orbitals: Omission behavior of some heterocyclic compounds will be observed (Figure 5 and 6),when there, is at least two hetero atoms (same or different) but both the hetero atoms do not have any d orbitals (such as O,N etc.) and they are in DLP based environment in the ring system. These molecules have been studied with advanced molecular orbital techniques known as ‘ab initio calculations’. ‘Ab initio quantum chemistry methods’ are computational chemistry methods based on quantum chemistry8. In the case of 1,2-dioxin, 1,4-dioxin and dibenzo-1,4-dioxin there is DLP based O atoms in all the molecules but still they will be non aromatic due to prevention of significant free electron delocalization (makes non conjugated). The π electrons from the carbon bonds and the lone pair electrons on the oxygen atoms do not overlap to a significant degree due to absence of vacant d orbitals in both O atoms in each case (pπ-dπ overlap is not possible here in conjugation). It makes these molecules non conjugated and thus allows the molecules to become non aromatic instead of aromatic (A value = even No). In the heterocyclic compounds, where, there is two DLP based N atoms instead of two DLP based O atoms or there is one DLP N atom along with one DLP O atom, the same phenomena of non aromatic behavior will be observed. Because, both N and O atoms do not have any vacant d orbitals, and hence pπ-dπ overlap is not possible here in conjugation. Anti Aromatic Behavior of some heterocyclic compounds containing same DLP based hetero atoms having vacant d orbitals: These compounds (Figure 7) are anti aromatic, here both S atoms, having vacant d orbitals, contain one DLP and one LLP and here both DLP of both S atoms participate in the delocalization. Hence, for the prediction of ‘A’ value, consider both DLP (DLP = 2). Here, A = πb + DLP + 1 (Constant) = 2 + 2 + 1 = 5 (odd No) = Anti Aromatic. It may be expected that these three time economic innovative mnemonics of heterocyclic chemistry will help the students of Undergraduate, Senior Undergraduate and Post-Graduate level to predict aromatic, anti aromatic and non aromatic character of heterocyclic compounds along with their omission behaviour. Experiment in vitro on 100 students showed that by using these two formulae students can save up to 5-10 minutes time in the examination hall to predict the aromatic, anti aromatic and non aromatic character of any heterocyclic compounds and their comparative study including omission behaviour with respect to the DLP and LLP based hetero atoms present on them. On the basis of this, I can strongly recommend to use these three time economic innovative mnemonics in the field of heterocyclic chemistry. External Links: | 14,532 | 7 |
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The elements in the group include beryllium (Be), magnesium (Mg), calcium (Ca), strontium (Sr), barium (Ba), and radium (Ra). Group 2 contains soft, silver metals that are less metallic in character than the Group 1 elements. Although many characteristics are common throughout the group, the heavier metals such as Ca, Sr, Ba, and Ra are almost as reactive as the . All the elements in Group 2 have two electrons in their valence shells, giving them an oxidation state of +2. This enables the metals to easily lose electrons, which increases their stability and allows them to form compounds via ionic bonds. The following diagram shows the location of these metals in the Periodic Table: The table below gives a detailed account of the descriptive chemistry of each of the individual elements. Notice an increase down the group in atomic number, mass, and atomic radius, and a decrease down the group for ionization energy. These common periodic trends are consistent across the whole periodic table. 1560 K 973 K The reactions of the alkaline earth metals differ from those of the Group 1 metals. Radium is radioactive and is not considered in this section. \[ Ca_{(s)} + H_{2\, (g)} \rightarrow CaH_{2\, (s)} \] \[ Sr_{(s)} + O_{2 (g)} \rightarrow SrO_{2\, (s)}\] \[ 3Mg_{(s)} + N_{(g)} \rightarrow Mg_3N_{2 (s)} \] \[ Mg_{(s)} + Cl_{2(g)} \rightarrow MgCl_{2(s)} \] \[ Ba_{(s)} + 2H_2O_{(l)} \rightarrow Ba(OH)_{2(aq)} + H_{2(g)} \] Beryllium was first identified in 1798 by Louis-Nicolas Vauquelin, while performing a chemical analysis on aluminum silicates. The element was originally named glucinum, and it was first isolated in 1828 by Antoine Bussy and Friedrich Wohler. In 1898, Paul Lebeau was able to produce the first pure samples of Beryllium by electrolyzing molten beryllium fluoride and sodium fluoride. It was later renamed beryllium ( from the Greek word meaning "to become pale") after a pale, beryllium-containing gemstone called a beryl. Beryllium is the very first element in Group 2, and has the highest melting point (1560 K) of any element in the group. It is very rare on Earth as well as in the universe and is not considered important for plant or animal life. In nature, it can only be found in compounds with other elements. In solutions, it remains in elemental form only for pH values below 5.5. Beryllium is extremely light with a high ionization energy, and it is used primarily to strengthen alloys. Beryllium has a strong affinity for oxygen at high temperatures, and thus it is extremely difficult to extract from ores. This beryllium is not commercially available as it cannot be economically mass produced. Since 1957, the majority of industrial beryllium is produced by reducing BeF with magnesium, making it more readily available. : Because beryllium is relatively light and has a wide temperature range, it has many mechanical uses. It can be used in aircraft production in nozzles of liquid-fueled spacecrafts, and mirrors in meteorological satellites. The famous Spitzer Space Telescope's optics are composed entirely of beryllium. One of the most important applications of beryllium is the production of radiation windows. As beryllium is almost transparent to x-rays, it can be used in windows for x-ray tubes. The minimal absorption by Beryllium greatly reduces heating effects due to intense radiation. : Beryllium is a monoisotopic element—it has only one stable isotope, Be. Another notable isotope is cosmogenic Be, which is produced by cosmic ray spallation of oxygen and nitrogen. This isotope has a relatively long half-life of 1.51 million years, and is useful in examining soil erosion and formation, as well as the age of ice cores. : Beryllium forms compounds with most non-metals. The most common compound is beryllium oxide (\(BeO\)) which does not react with water and dissolves in strongly basic solutions. Because of its high melting point, \(BeO\) is a good heat conductor in electrical insulators. It is also an amphoteric oxide, meaning it can react with both strong acids and bases. Cation: \(H_2O_{(l)} + BeO_{(s)} + 2H_3O^+_{(aq)} \rightarrow [Be(H_2O)_4]^{2+}_{(aq)}\) Anion: \(H_2O_{(l)} + BeO_{(s)} + 2OH^-_{(aq)} \rightarrow [Be(OH)_4]^{2-}_{(aq)}\) Magnesium was first discovered in 1808 by Sir Humphry Davy in England by the electrolysis of magnesia and mercury oxide. Antoine Bussy was the first to produce it in consistent form in 1831. It is the 8th most abundant element in the Earth's crust, constituting 2% by mass. It is also the 11th most common element in the human body: fifty percent of magnesium ions are found in bones, and it is a required catalyst for over three hundred different enzymes. Magnesium has a melting point of 923 K and reacts with water at room temperature, although extremely slowly. It is also highly flammable and extremely difficult to extinguish once ignited. As a precaution, when burning or ligh magnesium, UV-protected goggles should be worn, as the bright white light produced can permanently damage the retina. Magnesium can be found in over 100 different minerals, but most commercial magnesium is extracted from dolomite and olivine. The Mg ion is extremely common in seawater and can be filtered and then electrolyzed to produce pure magnesium. : In its elemental form, magnesium is used for structural purposes in car engines, pencil sharpeners, and many electronic devices such as laptops and cell phones. Due to it's bright white flame color, magnesium is also often used in fireworks. In a biological sense, magnesium is vital to the body's health: the Mg ion is a component of every cell type. Magnesium can be obtained by eating foods rich in magnesium, such as nuts and certain vegetables, or by eating supplementary diet pills. Chlorophyll, the pigment that absorbs light in plants, interacts heavily with magnesium and is necessary for photosynthesis. : The three stable isotopes of magnesium are Mg, Mg, and Mg. The lightest isotope, Mg, composes about 74% of Mg, whereas Mg is associated with meteorites in the solar system. Mg is the only known radioactive isotope of Magnesium and it has a half-life of around 21 hours. : Magnesium ions are essential for all life on Earth, and can be found mainly in seawater and the mineral carnallite. Some examples of magnesium compounds include: magnesium carbonate (MgCO ), a white powder used by athletes and gymnasts to dry their hands for a firm grip; and magnesium hydroxide (Mg(OH) ), or milk of magnesia, used as a common component of laxatives. Calcium was isolated in 1808 by Sir Humphry Davy by the electrolysis of lime and mercuric oxide. In nature, it is only found in combination with other elements. It is the 5th most abundant element in the Earth's crust, and is essential for living organisms. Calcium, in the presence of Vitamin D, is well known for its role in building stronger, denser bones early in the lives of humans and other animals. Calcium can be found in leafy green vegetables as well as in milk, cheese, and other dairy products. Calcium has a melting point of 1115 K and gives off a red flame when ignited. Calcium was not readily available until the early 20th Century. : Calcium is an important component in cement and mortars, and thus is necessary for construction. It is also used to aid cheese production. : The four stable isotopes of calcium are \(\ce{^{40}Ca}\), \(\ce{^{42}Ca}\), \(\ce{^{43}Ca}\), \(\ce{^{44}Ca}\). The most abundant isotope, Ca, composes about 97% of naturally occurring calcium. \(\ce{^{41}Ca}\) is the only radioactive isotope of calcium with a half life of 103,000 years. : The most common calcium compound is calcium carbonate (\(CaCO_3\)). Calcium carbonate is a component of shells in living organisms, and is used as a commercial antacid. It is also the main component of limestone. As shown below, three steps are required to to obtain pure \(CaCO_3\) from limestone: calcination, slaking, and carbonation. CaCO (s) → CaO(s) + CO (g) CaO(s) + H O(l) → Ca(OH) (s) Ca(OH) + CO (g) → CaCO (s) + H O(l) Another important calcium compound is calcium hydroxide (\(Ca(OH)_2\)). Often referred to as 'slack lime', it can be refined to form cement. The formation of calcium hydroxide is given below: \[CaO_{(s)}+H_2O_{(l)} \rightarrow Ca(OH)_2\] Strontium was first discovered in 1790 by Adair Crawford in Scotland and is named after the village it was discovered in, Strontian. In nature, it is only found in combination with other elements as it is extremely reactive. It is the 15th most abundant element on Earth and is commonly found in the form of the mineral celestite. Strontium metal is a slightly softer than calcium and has a melting point of K. : In it's pure form, Strontium is used in alloys. It can also be used in fireworks as it produces a scarlet flame color. strontium ranelate (\(C_{12}H_6N_2O_8SSr_2\)) is used to treat sufferers of osteoporosis and strontium chloride (\(SrCl_2\)) is used to make toothpaste for sensitive teeth. : Strontium has four stable isotopes: Sr, Sr, Sr, and Sr. About 82% of naturally occurring strontium comes in the form of Sr. : Some applications of strontium compounds include strontium carbonate (\(SrCO_3\)), strontium sulfate (\(SrSO_4\)), and strontium nitrate (\(Sr(NO_3)_2\)), which can be used as a red flame in fireworks. Radium was first discovered in 1898 by Marie Sklodowska-Curie, and her husband, Pierre Curie, in a pitchblende uranium ore in North Bohemia in the Czech Republic; however, it was not isolated as a pure metal until 1902. Radium is the heaviest and most radioactive of the alkaline earth metals and it reacts explosively with water. Radium appears pure white but when exposed to air it immediately oxidizes and turns black. Because radium is a decay product of uranium, it can be found in trace amounts in all uranium ores. The exposure or inhalation of radium can cause great harm in the form of cancer and other disorders. : There are 25 isotopes of radium are that known to exist, but only 4 are found in nature. However, none are stable. The isotope which has the longest half life is Ra, which is produced by the decay of Uranium. The four most stable isotopes are Ra, Ra, Ra, and Ra. The three most abundant: Ra, Ra and Ra decay by emitting alpha particles, whereas Ra decays emitting a beta particle. Most radium isotopes have relatively short half-lives. : Radium compounds are extremely rare in nature because of its short half-life and intense radioactivity. As such, radium compounds are found almost entirely in uranium and thorium ores. All known radium compounds have a crimson colored flame. The most important compound of radium is radium chloride (\(RaCl_2\)). Previously, it had only been found in a mixture with barium chloride, but as \(RaCl_2\) appeared to be less soluble than barium chloride, the mixture could continually be treated to form a precipitate. This procedure was repeated several times until the radioactivity of the precipitate no longer increased, as radium chloride could be electrolyzed using a mercury cathode to produce pure radium. Currently, \(RaCl_2\) is still used to separate radium from barium and it is also used to produce Radon gas, which can be used to treat cancer. \[Be_{(s)}+F_2 \rightarrow BeF\] | 11,369 | 9 |
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In 1889, Svante Arrhenius proposed the from his direct observations of the plots of rate constants vs. temperatures: \[k = Ae^{-\frac{E_a}{RT}} \label{eq1} \] The , E , is the minimum energy molecules must possess in order to react to form a product. The slope of the Arrhenius plot can be used to find the activation energy. The Arrhenius plot can also be used by extrapolating the line back to the y-intercept to obtain the pre-exponential factor, A. This factor is significant because A=p×Z, where p is a steric factor and Z is the collision frequency. The pre-exponential, or frequency, factor is related to the amount of times molecules will hit in the orientation necessary to cause a reaction. It is important to note that the Arrhenius equation is based on the . It states that particles must collide with proper orientation and with enough energy. Now that we have obtained the activation energy and pre-exponential factor from the Arrhenius plot, we can solve for the rate constant at any temperature using the Arrhenius equation. The Arrhenius plot is obtained by plotting the logarithm of the rate constant, k, versus the inverse temperature, 1/T. The resulting negatively-sloped line is useful in finding the missing components of the Arrhenius equation. Extrapolation of the line back to the y-intercept yields the value for ln A. The slope of the line is equal to the negative activation energy divided by the gas constant, R. As a rule of thumb in most biological and chemical reactions, the reaction rate doubles when the temperature increases every 10 degrees Celsius. Looking at the Arrhenius equation, the denominator of the exponential function contains the gas constant, R, and the temperature, T. This is only the case when dealing with moles of a substance, because R has the units of J/molK. When dealing with molecules of a substance, the gas constant in the dominator of the exponential function of the Arrhenius equation is replaced by the Boltzmann constant, k . The Boltzmann constant has the units J/K. At room temperature, k T, is the available energy for a molecule at 25 C or 273K, and is equal to approximately 200 wave numbers. It is important to note that the decision to use the gas constant or the Boltzmann constant in the Arrhenius equation depends primarily on the canceling of the units. To take the inverse log of a number, the number must be unitless. Therefore all the units in the exponential factor must cancel out. If the activation energy is in terms of joules per moles, then the gas constant should be used in the dominator. However, if the activation energy is in unit of joules per molecule, then the constant, K, should be used. The Arrhenius equation (Equation \ref{eq1}) can be rearranged to deal with specific situations. For example, taking the logarithm of both sides yields the equation above in the form y=-mx+b. \[\ln k = \dfrac{-E_a}{RT}+\ln A \label{eq2} \] Then, a plot of \(\ln k\) vs. \(1/T\) and all variables can be found. This form of the Arrhenius equation makes it easy to determine the slope and y-intercept from an Arrhenius plot. It is also convenient to note that the above equation shows the connection between temperature and rate constant. As the temperature increases, the rate constant decreases according to the plot. From this connection we can infer that the rate constant is inversely proportional to temperature. The integrated form of the Arrhenius equation is also useful (Equation \ref{eq3}). This variation of the Arrhenius equation involves the use of two Arrhenius plots constructed on the same graph to determine the activation energy. The above equation, shows temperature's effect on multiple rate constants. This allows easy inference of the rate constants' sensitivity to activation energy and temperature changes. If the activation energy is high for a given temperature range, then the rate constant is highly sensitive; changes in temperature have a significant effect on the rate constant. If the activation energy is low for a given temperature range, then the rate constant is less sensitive, and changes in temperature have little effect on the rate constant. This phenomenon is graphically illustrated in the example below: The graph above shows that the plot with the steeper slope has a higher activation energy and the plot with the flatter slope has a smaller activation energy. This means that over the same temperature range, a reaction with a higher activation energy changes more rapidly than a reaction with a lower activation energy. The Arrhenius plot may become non-linear if steps become rate-limiting at different temperatures. Such an example can be found with Fox and co-workers in 1972 with beta-glycoside transport in . . The differences in the transition temperatures are due to fatty acid composition in cell membranes. The transition state difference is a result of the sharp change of fluidity of the membrane. Another example includes a sudden drop at low 1/T (high temperatures), a result of protein denaturation. 1. T/F The E calculated from the Arrhenius equation gives an exact value. 2. Describe the relationship between temperature and E and give examples. 3. Using the following information: A= 1×10 sec E = 75×10 J/mol R= 8.314 J mol/K Calculate k at 27° C with proper units. 4. Using information from problem 3, calculate k at 37° C with proper units. 5. Using the integrated equation solve for E using: k =7.78×10 at T =273 K k =3.46×10 at T =298 K | 5,519 | 10 |
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To discuss the electronic states of atoms we need a system of notation for multi-electron wavefunctions. As we saw in Chapter 8, the assignment of electrons to orbitals is called the electron configuration of the atom. One creates an electronic configuration representing the electronic structure of a multi-electron atom or ion in its ground or lowest-energy state as follows. First, obey the , which requires that each electron in an atom or molecule must be described by a different spin-orbital. Second, assign the electrons to the lowest energy spin-orbitals, then to those at higher energy. This procedure is called the (which translates from German as build-up principle). The mathematical analog of this process is the construction of the approximate multi-electron wavefunction as a product of the single-electron atomic orbitals. For example, the configuration of the boron atom, shown schematically in the energy level diagram in Figure \(\Page {1}\), is written in shorthand form as 1s 2s 2p . As we saw in , the degeneracy of the 2s and 2p orbitals is broken by the electron-electron interactions in multi-electron systems. Rather than showing the individual spin-orbitals in the diagram or in the shorthand notation, we commonly say that up to two electrons can be described by each spatial orbital, one with spin function \(\alpha\) (electron denoted by an arrow pointing up) and the other with spin function \(\beta\) (arrow pointing down). This restriction is a manifestation of the Pauli Exclusion Principle mentioned above. An equivalent statement of the Pauli Exclusion Principle is that each electron in an atom has a unique set of quantum numbers (n,\(l , m_l , m_s\)). Since the two spin functions are degenerate in the absence of a magnetic field, the energy of the two electrons with different spin functions in a given spatial orbital is the same, and they are shown on the same line in the energy diagram. Write the electronic configuration of the carbon atom and draw the corresponding energy level diagram. Write the values for the quantum numbers (n, \(l , m_l , m_s\)) for each of the six electrons in carbon. We can deepen our understanding of the quantum mechanical description of multi-electron atoms by examining the concepts of electron indistinguishability and the Pauli Exclusion Principle in detail. We will use the following statement as a guide to keep our explorations focused on the development of a clear picture of the multi-electron atom: “When a multi-electron wavefunction is built as a product of single-electron wavefunctions, the corresponding concept is that exactly one electron’s worth of charge density is described by each atomic spin-orbital.” A subtle, but important part of the conceptual picture is that the electrons in a multi-electron system are not distinguishable from one another by any experimental means. Since the electrons are indistinguishable, the probability density we calculate by squaring the modulus of our multi-electron wavefunction also cannot change when the electrons are interchanged (permuted) between different orbitals. In general, if we interchange two identical particles, the world does not change. As we will see below, this requirement leads to the idea that the world can be divided into two types of particles based on their behavior with respect to permutation or interchange. For the probability density to remain unchanged when two particles are permuted, the wavefunction itself can change only by a factor of \(e^{i\varphi}\), which represents a complex number, when the particles described by that wavefunction are permuted. As we will show below, the \(e^{i\varphi}\) factor is possible because the probability density depends on the absolute square of the function and all expectation values involve \(\psi \psi ^*\). Consequently \(e^{i\varphi}\) disappears in any calculation that relates to the real world because \(e^{i\varphi} e^{-i\varphi} = 1\). We could symbolically write an approximate two-particle wavefunction as \(\psi (r_1, r_2)\). This could be, for example, a two-electron wavefunction for helium. To exchange the two particles, we simply substitute the coordinates of particle 1 (\(r_l\)) for the coordinates of particle 2 (\(r_2\)) and vice versa, to get the new wavefunction \(\psi (r_1, r_2)\). This new wavefunction must have the property that \[|\psi (r_1, r_2)|^2 = \psi (r_2, r_1)^*\psi (r_2, r_1) = \psi (r_1, r_2)^* \psi (r_1, r_2) \label {9-38}\] since the probability density of the electrons in the atom does not change upon permutation of the electrons. Permute the electrons in Equation \(\ref{9-13}\) (the product function for He wavefunction.) Equation \(\ref{9-38}\) will be true only if the wavefunctions before and after permutation are related by a factor of \(e^{i\varphi}\), \[\psi (r_1, r_2) = e^{i\varphi} \psi (r_1, r_2) \label {9-39}\] so that \[ \left ( e^{-i\varphi} \psi (r_1, r_2) ^*\right ) \left ( e^{i\varphi} \psi (r_1, r_2) ^*\right ) = \psi (r_1 , r_2 ) ^* \psi (r_1 , r_2) \label {9-40}\] If we exchange or permute two identical particles twice, we are (by definition) back to the original situation. If each permutation changes the wavefunction by \(e^{i \varphi}\), the double permutation must change the wavefunction by \(e^{i\varphi} e^{i\varphi}\). Since we then are back to the original state, the effect of the double permutation must equal 1; i.e., \[e^{i\varphi} e^{i\varphi} = e^{i 2\varphi} = 1 \label {9-41}\] which is true only if \(\varphi = 0 \) or an integer multiple of π. The requirement that a double permutation reproduce the original situation limits the acceptable values for \(e^{i\varphi}\) to either +1 (when \(\varphi = 0\)) or -1 (when \(\varphi = \pi\)). Both possibilities are found in nature. Use Euler’s Equality to show that \(e^{12\varphi} = 1\) when \(\varphi = 0\) or \(n \pi\) and consequently \(e^{i \varphi} = \pm 1\). Wavefunctions for which \(e^{i \varphi} = +1\) are defined as symmetric with respect to permutation, because the wavefunction is identical before and after a single permutation. Wavefunctions that are symmetric with respect to interchange of the particles obey the following mathematical relationship: \[e^{i\varphi} e^{i\varphi} = e^{i 2\varphi} = 1 \label {9-42}\] The behavior of some particles requires that the wavefunction be symmetric with respect to permutation. These particles are called bosons and have integer spin such as deuterium nuclei, photons, and gluons. The behavior of other particles requires that the wavefunction be antisymmetric with respect to permutation \((e^{i\varphi} = -1)\). A wavefunction that is antisymmetric with respect to electron interchange is one whose output changes sign when the electron coordinates are interchanged, as shown below: \[ \psi (r_2 , r_1) = e^{i\varphi} \psi (r_1, r_2) = - \psi (r_1, r_2) \label {9-43}\] These particles, called fermions, have half-integer spin and include electrons, protons, and neutrinos. Explain without any equations why there are only two kinds of particles in the world: bosons and fermions. In fact, an elegant statement of the Pauli Exclusion Principle is simply “electrons are fermions.” This statement means that any wavefunction used to describe multiple electrons must be antisymmetric with respect to permutation of the electrons, providing yet another statement of the Pauli Exclusion Principle. The requirement that the wavefunction be antisymmetric applies to all multi-electron functions \(\psi (r_1, r_2, \cdots r_i)\), including those written as products of single electron functions \(\varphi _1 (r_1) \varphi _2 (r_2) \cdots \varphi _i (r_i)\). Another way to simply restate the Pauli Exclusion Principle is that “electrons are fermions.” The first statement of the Pauli Exclusion Principle was that two electrons could not be described by the same spin orbital. To see the relationship between this statement and the requirement that the wavefunction be antisymmetric for electrons, try to construct an for two electrons that are described by the same spin-orbital. We can try to do so for helium. Write the He approximate two-electron wavefunction as a product of identical 1s spin-orbitals for each electron,\(\varphi _{1s_{\alpha}} (r_1) \) and \(\varphi _{1s_{\alpha}} (r_2) \): \[ \psi (r_1, r_2 ) = \varphi _{1s\alpha} (r_1) \varphi _{1s\alpha} (r_2) \label {9-44}\] To permute the electrons in this two-electron wavefunction, we simply substitute the coordinates of electron 1 (\(r_l\)) for the coordinates of electron 2 (\(r_2\)) and vice versa, to get \[ \psi (r_2, r_1 ) = \varphi _{1s\alpha} (r_2) \varphi _{1s\alpha} (r_1) \label {9-45}\] This is identical to the original function (Equatin \(\ref{9-44}\)) since the two single-electron component functions . The two-electron function has not changed sign, as it must for fermions. We can construct a wavefunction that is antisymmetric with respect to permutation symmetry only if each electron is described by a different function. What is meant by the term permutation symmetry? Explain why the product function \(\varphi (r_1) \varphi (r_2)\) could describe two bosons (deuterium nuclei) but can not describe two fermions (e.g. electrons). Let’s try to construct an antisymmetric function that describes the two electrons in the ground state of helium. Blindly following the first statement of the Pauli Exclusion Principle, that each electron in a multi-electron atom must be described by a different spin-orbital, we try constructing a simple product wavefunction for helium using two different spin-orbitals. Both have the 1s spatial component but one has spin function \(\alpha\) and the other has spin function \(\beta\) so the product wavefunction matches the form of the ground state electron configuration for He, \(1s^2\). \[ \psi (r_1, r_2 ) = \varphi _{1s\alpha} (r_1) \varphi _{1s\beta} (r_2) \label {9-46}\] After permutation of the electrons, this becomes \[ \psi (r_2, r_1 ) = \varphi _{1s\alpha} (r_2) \varphi _{1s\beta} (r_1) \label {9-47}\] which is different from the starting function (Equation \(\ref{9-46}\)) since \(\varphi _{1s\alpha}\) and \(\varphi _{1s\beta}\) are functions. However, an antisymmetric function must produce the same function multiplied by (–1) after permutation, and that is not the case here. We must try something else. To avoid getting a totally different function when we permute the electrons, we can make a linear combination of functions. A very simple way of taking a linear combination involves making a new function by simply adding or subtracting functions. The function that is created by subtracting the right-hand side of Equation \(\ref{9-47}\) from the right-hand side of Equation \(\ref{9-46}\) has the desired antisymmetric behavior. \[\psi (r_1, r_2) = \dfrac {1}{\sqrt {2}} [ \varphi _{1s\alpha}(r_1) \varphi _{1s\beta}(r_2) - \varphi _{1s\alpha}(r_2) \varphi _{1s\beta}(r_1)] \label {9-48}\] The constant on the right-hand side accounts for the fact that the total wavefunction must be normalized. Show that the linear combination in Equation \(\ref{9-48}\) is antisymmetric with respect to permutation of the two electrons. Replace the minus sign with a plus sign (i.e. take the positive linear combination of the same two functions) and show that the resultant linear combination is symmetric. Write a similar linear combination to describe the \(1s^12s^1\) excited configuration of helium. A linear combination that describes an appropriately antisymmetrized multi-electron wavefunction for any desired orbital configuration is easy to construct for a two-electron system. However, interesting chemical systems usually contain more than two electrons. For these multi-electron systems a relatively simple scheme for constructing an antisymmetric wavefunction from a product of one-electron functions is to write the wavefunction in the form of a determinant. John Slater introduced this idea so the determinant is called a . The Slater determinant for the two-electron wavefunction of helium is \[ \psi (r_1, r_2) = \frac {1}{\sqrt {2}} \begin {vmatrix} \varphi _{1s} (1) \alpha (1) & \varphi _{1s} (1) \beta (1) \\ \varphi _{1s} (2) \alpha (2) & \varphi _{1s} (2) \beta (2) \end {vmatrix} \label {9-49}\] and a shorthand notation for this determinant is \[ \psi (r_1 , r_2) = 2^{-\frac {1}{2}} Det | \varphi _{1s} (r_1) \varphi _{1s} (r_2) | \label {9-50}\] The determinant is written so the electron coordinate changes in going from one row to the next, and the spin orbital changes in going from one column to the next. The advantage of having this recipe is clear if you try to construct an antisymmetric wavefunction that describes the orbital configuration for uranium! Note that the normalization constant is \((N!)^{-\frac {1}{2}}\) for N electrons. Show that the determinant form is the same as the form for the helium wavefunction that is given in Equation \(\ref{9-48}\). Expand the Slater determinant in Equation \(\ref{9-49}\) for the He atom. Write and expand the Slater determinant for the electronic wavefunction of the Li atom. Write the Slater determinant for the carbon atom. If you expanded this determinant, how many terms would be in the linear combination of functions? Write the Slater determinant for the \(1s^12s^1\) excited state orbital configuration of the helium atom. Now that we have seen how acceptable multi-electron wavefunctions can be constructed, it is time to revisit the “guide” statement of conceptual understanding with which we began our deeper consideration of electron indistinguishability and the Pauli Exclusion Principle. What does a multi-electron wavefunction constructed by taking specific linear combinations of product wavefunctions mean for our physical picture of the electrons in multi-electron atoms? Overall, the antisymmetrized product function describes the configuration (the orbitals, regions of electron density) for the multi-electron atom. Because of the requirement that electrons be indistinguishable, we can’t visualize specific electrons assigned to specific spin-orbitals. Instead, we construct functions that allow each electron’s probability distribution to be dispersed across each spin-orbital. The total charge density described by any one spin-orbital cannot exceed one electron’s worth of charge, and each electron in the system is contributing a portion of that charge density. Critique the energy level diagram and shorthand electron configuration notation from the perspective of the indistinguishability criterion. Can you imagine a way to represent the wavefunction expressed as a Slater determinant in a schematic or shorthand notation that more accurately represents the electrons? (This is not a solved problem!) | 14,763 | 11 |
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This page discusses the reactions of the Group 2 elements (beryllium, magnesium, calcium, strontium and barium) with common acids. Each metal reacts with dilute hydrochloric acid, producing bubbles of hydrogen gas and a colorless solution of the metal chloride: \[ X + 2HCl \rightarrow XCl_2 + H_2\] These reactions become more vigorous down the group. These are more complicated, because of the formation of insoluble sulfates. These metals react with with dilute sulfuric acid just as they did with dilute hydrochloric acid; the reaction between magnesium and dilute sulfuric is familiar to many beginning chemists. Hydrogen gas is formed, along with colorless solutions of beryllium or magnesium sulfate. For example: \[ Mg + H_2SO_4 \rightarrow MgSO_4 + H_2\] Calcium sulfate is slightly soluble, and strontium and barium sulfates are essentially insoluble. When exposed to sulfuric acid, a layer of insoluble sulfate is formed on each of these metals, slowing or stopping the reaction entirely. In the calcium case, some hydrogen is produced, along with a white precipitate of calcium sulfate. These reactions are more complicated. When a metal reacts with an acid, the metal usually reduces hydrogen ions to hydrogen gas. The elemental metal is oxidized to metal cations in the process. However, nitrate ions are easily reduced to nitrogen monoxide and nitrogen dioxide. Metals reacting with nitric acid, therefore, tend to produce oxides of nitrogen rather than hydrogen gas. If the acid is relatively dilute, the reaction produces nitrogen monoxide, although this immediately reacts with atmospheric oxygen, forming nitrogen dioxide. If c Various sources disagree on whether beryllium reacts with nitric acid. Beryllium forms a strong oxide layer (similar to that of aluminum) which slows reactions down until it has been removed. Some sources say that beryllium does not react with nitric acid. However, procedures for making beryllium nitrate by reacting beryllium powder with nitric acid are readily available. One source uses semi-concentrated nitric acid, claiming that the gas evolved is nitrogen monoxide. This is a reasonable conclusion. The reactivity of beryllium seems to depend on its source, and how it was manufactured. It is possible that small amounts of impurities in the metal can affect its reactivity. The rest of the Group 2 metals produce hydrogen gas from very dilute nitric acid, but this gas is contaminated with nitrogen oxides. Colorless solutions of the metal nitrates are also formed. Taking magnesium as an example, if the solution is very dilute: \[ Mg + 2HNO_3 \rightarrow Mg(NO_3)_2 + H_2\] At moderate concentrations (even with very dilute acid, this occurs to some extent): \[ 3Mg + 8HNO_3 \rightarrow 3Mg(NO_3)_2 + 2NO + 4H_2O\] And with concentrated acid: \[ Mg + 4HNO_3 \rightarrow Mg(NO_3)_2 + 2NO_2 + 2H_2O\] Jim Clark ( ) | 2,882 | 12 |
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Addition of an unsymmetrical substance such as \(\ce{HX}\) to an unsymmetrical alkene theoretically can give two products: and Both products are seldom formed in equal amounts; in fact, one isomer usually is formed to the exclusion of the other. For example, the hydration of propene gives 2-propanol (not 1-propanol), and hydrogen chloride adds to 2-methylpropene to give -butyl chloride (not isobutyl chloride): To understand the reason for the pronounced selectivity in the orientation of addition of electrophiles, it will help to consider one example, hydrogen bromide addition to 2-methylpropene. Two different carbocation intermediates could be formed by attachment of a proton to one or the other of the double bond carbons: Subsequent reactions of the cations with bromide ion give -butyl bromide and isobutyl bromide. In the usual way of running these additions, the product is very pure -butyl bromide. How could be have predicted which product would be favored? The first step is to decide whether the prediction is to be based on (1) which of the two products is the , or (2) which of the two products if formed . If we make a decision on the basis of product stabilities, we take into account \(\Delta H^0\) values, entropy effects, and so on, to estimate the equilibrium constants \(K_\text{eq}\) for the reactants and each product. When the ratio of the products is determined by the ratio of their equilibrium constants, we say the overall reaction is subject to (or ) . Equilibrium control requires that the reaction be . When a reaction is carried out under conditions in which it is , the ratio of the products is determined by the relative rates of formation of the various products. Such reactions are said to be under . The products obtained in a reaction subject to kinetic control are not necessarily the same as those obtained under equilibrium control. Indeed, the equilibrium constant for interconversion of -butyl bromide and isobutyl bromide at \(25^\text{o}\) is 4.5, and if the addition of hydrogen bromide to 2-methylpropene were under equilibrium control, the products would be formed in this ratio: \[K_\text{eq} = \dfrac{\left[ \text{tert-butyl bromide} \right]}{\left[ \text{isobutyl bromide} \right]} = 4.5\] But the addition product is \(99+\%\) -butyl bromide so the reaction clearly is kinetically controlled, -butyl being formed considerably faster than isobutyl bromide. . So to account for the formation of -butyl bromide we have to consider why the -butyl cation is formed more rapidly than the isobutyl cation: As we have seen in Section 8-7B, alkyl groups are more electron donating than hydrogen. This means that the more alkyl groups there are on the positive carbon of the cation, the more stable and the more easily formed the cation will be. The reason is that electron-donating groups can partially compensate for the electron deficiency of the positive carbon. As a result, we can predict that the -butyl cation with three alkyl groups attached to the positive center will be formed more readily than the primary isobutyl cation with one alkyl group attached to the positive center. Thus the problem of predicting which of the two possible products will be favored in the addition of unsymmetrical reagents to alkenes under kinetic control reduces to predicting which of two possible carbocation intermediates will be formed most readily. With simple alkenes, we shall expect the preference of formation of the carbocations to be in the order: > > . The reaction scheme can be represented conveniently in the form of an energy diagram (Figure 10-10). The activation energy, \(\Delta H^1_\text{tert}\) for the formation of the -butyl cation is less than \(\Delta H^1_\text{prim}\) for the formation of the isobutyl cation because the tertiary ion is much more stable (relative to the reactants) than the primary ion, and therefore is formed at the faster rate. The second step, to form the product from the intermediate cation, is very rapid and requires little activation energy. Provided that the reaction is , it will take the lowest-energy path and form exclusively -butyl bromide. However, if the reaction mixture is allowed to stand for a long time, isobutyl bromide begins to form. Over a long period, the products equilibrate and, at equilibrium, the product distribution reflects the relative stabilities of the rather than the stability of the transition states for formation of the intermediates. A rather simple rule, formulated in 1870 and known as , correlates the direction of additions of \(\ce{HX}\) to unsymmetrical alkenes. This rule, an important early generalization of organic reactions, may be stated as follows: . It should be clear that Markownikoff's rule predicts that addition of hydrogen bromide to 2-methylpropene will give -butyl bromide. We can extend Markownikoff's rule to cover additions of substances of the general type \(\ce{X-Y}\) to unsymmetrically substituted alkenes when a clear-cut decision is possible as to whether \(\ce{X}\) or \(\ce{Y}\) is the more electrophilic atom of \(\ce{X-Y}\). If the polarization of the \(\ce{X-Y}\) bond is such that \(\ce{X}\) is positive, \(^{\delta \oplus} \ce{X-Y} ^{\delta \ominus}\), then \(\ce{X}\) will be expected to add as \(\ce{X}^\oplus\) to the alkene to form the more stable carbocation. This step will determine the direction of addition. For example, if we know that the \(\ce{O-Br}\) bond of \(\ce{HOBr}\) is polarized as \(\overset{\delta \ominus}{\ce{HO}} - \overset{\delta \oplus}{\ce{Br}}\), then we can predict that addition of \(\ce{HOBr}\) to 2-methylpropene will give 1-bromo-2-methyl-2-propanol: Pauling's value for the electronegativity of carbon makes it slightly more electron-attracting than hydrogen. However, we expect that the electron-attracting power of a carbon atom (or of other elements) will depend also on the electronegativities of the groups to which it is attached. In fact, many experimental observations indicate that carbon in methyl or other alkyl groups is significantly electron-attracting than hydrogen. Conversely, the \(\ce{CF_3}-\) group is, as expected, far electron-attracting than hydrogen. The direction of polarization of bonds between various elements may be predicted from Figure 10-11. For example, an \(\ce{O-Cl}\) bond should be polarized so the oxygen is negative; a \(\ce{C-N}\) bond should be polarized so the nitrogen is negative: \[\overset{\delta \ominus}{\ce{O}}---\overset{\delta \oplus}{\ce{Cl}} \: \: \: \: \: \overset{\delta \oplus}{\ce{C}}---\overset{\delta \ominus}{\ce{N}}\] We then can predict that, in the addition of \(\ce{HOCl}\) to an alkene, the chlorine will add preferentially to form the more stable of two possible carbon cations. Generally, this means that chlorine will bond to the carbon carrying the greater number of hydrogens: A number of reagents that are useful sources of electrophilic halogen are included in Table 10-2. Some of these reagents, notably those with \(\ce{O}-\)halogen or \(\ce{N}-\)halogen bonds, actually are sources of hypohalous acids, \(\ce{HOX}\), and function to introduce halogen and hydroxyl groups at carbon. There are very few good fluorinating agents whereby the fluorine is added as \(\ce{F}^\oplus\). For alkenes that have halogen or similar substituents at the doubly bonded carbons, the same principles apply as with the simple alkenes. That is, under kinetic control the preferred product will be the one derived from the more stable of the two possible intermediate carbon cations. Consider a compound of the type \(\ce{Y-CH=CH_2}\). If \(\ce{Y}\) is electron-attracting than hydrogen, then hydrogen halide should add in such a way as to put the proton of \(\ce{HX}\) on the \(\ce{YCH=}\) end and \(\ce{X}\) on the \(\ce{=CH_2}\) end. The reason is that the positive carbon is expected to be more favorably located if it is not attached directly to an electron-attracting substituent: The addition goes as predicted, . For example, \[\ce{CF_3-CH=CH_2} + \ce{HCl} \rightarrow \ce{CF_3-CH_2-H_2-Cl}\] Such substituents are relatively uncommon, and most of the reported \(\ce{H-X}\) additions have been carried out with \(\ce{Y}\) groups having unshared electron pairs on an atom connected directly to a carbon of the double bond: These substituents usually are strongly electronegative relative to hydrogen, and this often causes diminished reactivity of the double bond toward electrophiles. Nonetheless, : The electron-attracting power of the substituent is more than counterbalanced by stabilization of the intermediate cation by the ability of the substituents to delocalize their electrons to the adjacent positive carbon (see ). and (1977) | 8,768 | 13 |
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Complicated molecular structures increase the likelihood that the rate constant depends on the trajectories with which the reactants approach each other. This kind of is well-known to all students of organic chemistry. Consider the addition of a hydrogen halide such as HCl to the double bond of an alkene, converting it to a chloroalkane: Experiments have shown that the reaction only takes place when the HCl molecule approaches the alkene with its hydrogen-end, and in a direction that is approximately perpendicular to the double bond, as shown in below: The reason for this is apparent: HCl is highly polar owing to the high electronegativity of chlorine, so that the hydrogen end of the molecule is slightly positive. The double bond of ethene consists of two clouds of negative charge corresponding to the σ ( ) and π ( ) molecular orbitals. The latter, which extends above and below the plane of the C H molecule, interacts with and attracts the HCl molecule. If, instead, the HCl approaches with its chlorine end leading as in , electrostatic repulsion between the like charges causes the two molecules to repel each other before any reaction can take place. The same thing happens in : the electronegativity difference between carbon and hydrogen is too small to make the C–H bond sufficiently polar to attract the incoming chlorine atom. ) | 1,371 | 14 |
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It is common knowledge that chemical reactions occur more rapidly at higher temperatures. Milk turns sour much more rapidly if stored at room temperature rather than in a refrigerator; butter goes rancid more quickly in the summer than in the winter; and eggs hard-boil more quickly at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects tend to be more lethargic on cold days. The reason for this is not hard to understand. Thermal energy relates direction to motion at the molecular level. As the temperature rises, molecules move faster and collide more vigorously, greatly increasing the likelihood of bond cleavages and rearrangements. Whether it is through the , transition state theory, or just common sense, chemical reactions are typically expected to proceed faster at higher temperatures and slower at lower temperatures. By 1890 it was common knowledge that higher temperatures speed up reactions, often doubling the rate for a 10-degree rise, but the reasons for this were not clear. Finally, in 1899, the Swedish chemist Svante Arrhenius (1859-1927) combined the concepts of activation energy and the Boltzmann distribution law into one of the most important relationships in physical chemistry: Take a moment to focus on the meaning of this equation, neglecting the factor for the time being. First, note that this is another form of the exponential decay law discussed in the previous section of this series. What is "decaying" here is not the concentration of a reactant as a function of time, but the magnitude of the rate constant as a function of the exponent . And what is the significance of this quantity? Recalling that is the , it becomes apparent that the exponent is just the ratio of the activation energy to the average kinetic energy. The larger this ratio, the smaller the rate (hence the negative sign). This means that high temperature and low activation energy favor larger rate constants, and thus speed up the reaction. Because these terms occur in an exponent, their effects on the rate are quite substantial. The two plots below show the effects of the activation energy (denoted here by ) on the rate constant. Even a modest activation energy of 50 kJ/mol reduces the rate by a factor of 10 . Looking at the role of temperature, a similar effect is observed. (If the -axis were in "kilodegrees" the slopes would be more comparable in magnitude with those of the kilojoule plot at the above right.) The Arrhenius equation, \[k = A e^{-E_a/RT} \label{1} \] can be written in a non-exponential form that is often more convenient to use and to interpret graphically. Taking the logarithms of both sides and separating the exponential and pre-exponential terms yields \[\begin{align} \ln k &= \ln \left(Ae^{-E_a/RT} \right) \\[4pt] &= \ln A + \ln \left(e^{-E_a/RT}\right) \label{2} \\[4pt] &= \left(\dfrac{-E_a}{R}\right) \left(\dfrac{1}{T}\right) + \ln A \label{3} \end{align} \] Equation \ref{3} is in the form of \(y = mx + b\) - the equation of a straight line. \[ \ln k=\ln A - \dfrac{E_{a}}{RT} \nonumber \] So if one were given a data set of various values of \(k\), the rate constant of a certain chemical reaction at varying temperature \(T\), one could graph \(\ln (k)\) versus \(1/T\). From the graph, one can then determine the slope of the line and realize that this value is equal to \(-E_a/R\). One can then solve for the activation energy by multiplying through by -R, where R is the gas constant. This affords a simple way of determining the activation energy from values of observed at different temperatures, by plotting \(\ln k\) as a function of \(1/T\). For the isomerization of cyclopropane to propene, the following data were obtained (calculated values shaded in pink): From the calculated slope, we have \[\begin{align*} –\left(\dfrac{E_a}{R}\right) &= –3.27 \times 10^4 K \\ E_a &=– (8.314\, J\, mol^{–1} K^{–1}) (–3.27 \times 10^4\, K) \\[4pt] &= 273\, kJ\, mol^{–1} \end{align*} \] This activation energy is high, which is not surprising because a carbon-carbon bond must be broken in order to open the cyclopropane ring. (C–C bond energies are typically around 350 kJ/mol.) This is why the reaction must be carried out at high temperature. Because the ln -vs.-1/ plot yields a straight line, it is often convenient to estimate the activation energy from experiments at only two temperatures. To see how this is done, consider that \[\begin{align*} \ln k_2 -\ln k_1 &= \left(\ln A - \frac{E_a}{RT_2} \right)\left(\ln A - \frac{E_a}{RT_1} \right) \\[4pt] &= \color{red}{\boxed{\color{black}{ \frac{E_a}{R}\left( \frac{1}{T_1}-\frac{1}{T_2} \right) }}} \end{align*} \] The ln- term is eliminated by subtracting the expressions for the two ln- terms.) Solving the expression on the right for the activation energy yields \[ E_a = \dfrac{R \ln \dfrac{k_2}{k_1}}{\dfrac{1}{T_1}-\dfrac{1}{T_2}} \nonumber \] A widely used rule-of-thumb for the temperature dependence of a reaction rate is that a rise in the temperature approximately doubles the rate. This is not generally true, especially when a strong covalent bond must be broken. For a reaction that does show this behavior, what would the activation energy be? Center the ten degree interval at 300 K. Substituting into the above expression yields \[\begin{align*} E_a &= \dfrac{(8.314)(\ln 2/1)}{\dfrac{1}{295} – \dfrac{1}{305}} \\[4pt] &= \dfrac{(8.314\text{ J mol}^{-1}\text{ K}^{-1})(0.693)}{0.00339\,\text{K}^{-1} – 0.00328 \, \text{K}^{-1}} \\[4pt] &= \dfrac{5.76\, J\, mol^{–1} K^{–1}}{(0.00011\, K^{–1}} \\[4pt] &= 52,400\, J\, mol^{–1} = 52.4 \,kJ \,mol^{–1} \end{align*} \] It takes about 3.0 minutes to cook a hard-boiled egg in Los Angeles, but at the higher altitude of Denver, where water boils at 92°C, the cooking time is 4.5 minutes. Use this information to estimate the activation energy for the coagulation of egg albumin protein. The ratio of the rate constants at the elevations of Los Angeles and Denver is 4.5/3.0 = 1.5, and the respective temperatures are \(373 \; \rm{K }\) and \(365\; \rm{K}\). With the subscripts 2 and 1 referring to Los Angeles and Denver respectively: \[\begin{align*} E_a &= \dfrac{(8.314)(\ln 1.5)}{\dfrac{1}{365\; \rm{K}} – \dfrac{1}{373 \; \rm{K}}} \\[4pt] &= \dfrac{(8.314)(0.405)}{0.00274 \; \rm{K^{-1}} – 0.00268 \; \rm{K^{-1}}} \\ &= \dfrac{(3.37\; \rm{J\; mol^{–1} K^{–1}})}{5.87 \times 10^{-5}\; \rm{K^{–1}}} \\[4pt] &= 57,400\; \rm{ J\; mol^{–1}} \\[4pt] &= 57.4 \; \rm{kJ \;mol^{–1}} \end{align*} \] : This low value seems reasonable because thermal denaturation of proteins primarily involves the disruption of relatively weak hydrogen bonds; no covalent bonds are broken (although can interfere with this interpretation). Up to this point, the pre-exponential term, \(A\) in the Arrhenius equation (Equation \ref{1}), has been ignored because it is not directly involved in relating temperature and activation energy, which is the main practical use of the equation. However, because \(A\) multiplies the exponential term, its value clearly contributes to the value of the rate constant and thus of the rate. Recall that the exponential part of the Arrhenius equation expresses the fraction of reactant molecules that possess enough kinetic energy to react, as governed by the . This fraction can run from zero to nearly unity, depending on the magnitudes of \(E_a\) and of the temperature. If this fraction were 0, the Arrhenius law would reduce to \[k = A \nonumber \] In other words, \(A\) is the fraction of molecules that would react if either the activation energy were zero, or if the kinetic energy of all molecules exceeded \(E_a\) — admittedly, an uncommon scenario (although barrierless reactions have been characterized). What would limit the rate constant if there were no activation energy requirements? The most obvious factor would be the rate at which reactant molecules come into contact. This can be calculated from kinetic molecular theory and is known as the or , \(Z\). In some reactions, the relative orientation of the molecules at the point of collision is important, so a geometrical or (commonly denoted by \(\rho\)) can be defined. In general, we can express \(A\) as the product of these two factors: \[A = Z\rho \nonumber \] Values of \(ρ\) are generally very difficult to assess; they are sometime estimated by comparing the observed rate constant with the one in which \(A\) is assumed to be the same as \(Z\). The exponential term in the Arrhenius equation implies that the rate constant of a reaction increases exponentially when the decreases. Because the rate of a reaction is directly proportional to the rate constant of a reaction, the rate increases exponentially as well. Because a reaction with a small activation energy does not require much energy to reach the transition state, it should proceed faster than a reaction with a larger activation energy. In addition, the Arrhenius equation implies that the rate of an uncatalyzed reaction is more affected by temperature than the rate of a catalyzed reaction. This is because the activation energy of an uncatalyzed reaction is greater than the activation energy of the corresponding catalyzed reaction. Since the exponential term includes the activation energy as the numerator and the temperature as the denominator, a smaller activation energy will have less of an impact on the rate constant compared to a larger activation energy. Hence, the rate of an uncatalyzed reaction is more affected by temperature changes than a catalyzed reaction. \[ \ln k_{1}=\ln A - \dfrac{E_{a}}{k_{B}T_1} \label{a1} \] at \(T_1\) and \[ \ln k_{2}=\ln A - \dfrac{E_{a}}{k_{B}T_2} \label{a2} \] at \(T_2\). By \[ \ln A = \ln k_{2} + \dfrac{E_{a}}{k_{B}T_2} \label{a3} \] and substitute for \(\ln A\) into Equation \ref{a1}: \[ \ln k_{1}= \ln k_{2} + \dfrac{E_{a}}{k_{B}T_2} - \dfrac{E_{a}}{k_{B}T_1} \label{a4} \] This simplifies to: \[\begin{align*} \ln k_{1} - \ln k_{2} &= -\dfrac{E_{a}}{k_{B}T_1} + \dfrac{E_{a}}{k_{B}T_2} \\[4pt] \ln \dfrac{k_{1}}{k_{2}} &= -\dfrac{E_{a}}{k_{B}} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right ) \end{align*} \] \[ k=Ae^{^{\frac{-E_{a}}{RT}}} \nonumber \] or \[\ln k=\ln A - \frac{E_{a}}{RT} \nonumber \] with the following terms: \(k\): \(A\): The or frequency factor \(E_a\): The is the threshold energy that the reactant(s) must acquire before reaching the transition state. \(R\): The gas constant. \(T\): The absolute temperature at which the reaction takes place. E is the factor the question asks to be solved. Therefore it is much simpler to use \(\large \ln k = -\frac{E_a}{RT} + \ln A\) To find E , subtract ln A from both sides and multiply by -RT. This will give us: \( E_a=\ln A -\ln k)RT\) Substitute the numbers into the equation: \(\ ln k = \frac{-(200 \times 1000\text{ J}) }{ (8.314\text{ J mol}^{-1}\text{K}^{-1})(289\text{ K})} + \ln 9\) Use the equation ln(7/k )=-[(900 X 1000)/8.314](1/370-1/310) Use the equation 12 = 15e Use the equatioin ln(15/7)=-[(600 X 1000)/8.314](1/T - 1/389) | 11,107 | 15 |
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An especially valuable group of intermediates can be prepared by addition of an compound to carbon-carbon double or triple bonds: The reaction is called and is a versatile synthesis of organoboron compounds. One example is the addition of diborane, \(\ce{B_2H_6}\), to ethene. Diborane behaves as though it is in equilibrium with \(\ce{BH_3}\) \(\left( \ce{B_2H_6} \rightleftharpoons 2 \ce{BH_3} \right)\), and addition proceeds in three stages: The monoalkylborane, \(\ce{RBH_2}\), and the dialkylborane, \(\ce{R_2BH}\), seldom are isolated because they rapidly add to the alkene. These additions amount to reduction of both carbons of the double bond: Organoboranes can be considered to be organometallic compounds. Elemental boron does not have the properties of a metal, and boron-carbon bonds are more covalent than ionic. However, boron is more electropositive than either carbon or hydrogen and when bonded to carbon behaves like most metals in the sense that bonds are polarized with \(\ce{R}\) negative and boron positive: Hydroboration and the many uses of organoboranes in synthesis were developed largely by H. C. Brown and co-workers. In our discussion, we shall give more detail on hydroboration itself, and then describe several useful transformations of organoboranes. The simplest borane, \(\ce{BH_3}\), exists as the dimer, \(\ce{B_2H_6}\), or in complexed form with certain ethers or sulfides: Any of these \(\ce{BH_3}\) compounds adds readily to most alkenes at room temperature or lower temperatures. The reactions usually are carried out in ether solvents, although hydrocarbon solvents can be used with the borane-dimethyl sulfide complex. When diborane is the reagent, it can be generated either or externally through the reaction of boron trifluoride with sodium borohydride: \[3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{H_4} + 4 \ce{BF_3} \rightarrow 2 \ce{B_2H_6} + 3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{F_4}\] Hydroborations have to be carried out with some care, because diborane and many alkylboranes are highly reactive and toxic substances; many are spontaneously flammable in air. With unsymmetrical alkenes, hydroboration occurs so that : These additions are : Furthermore, when there is a choice, addition occurs preferentially from the less crowded side of the double bond: If the alkene is a bulky molecule, borane may add only one or two alkene molecules to give either mono- or dialkylborane, \(\ce{RBH_2}\) or \(\ce{R_2BH}\), respectively, as the following reactions show: These bulky boranes still possess \(\ce{B-H}\) bonds and can add further to a multiple bond, but they are highly selective reagents and add only if the alkene or alkyne is unhindered. This selectivity can be useful, particularly to 1-alkynes, which are difficult to stop at the alkenylborane stage when using diborane: With a bulky dialkylborane, such as di-(1,2-dimethylpropyl)borane, further addition to the alkenylborane does not occur. An especially selective hydroborating reagent is prepared from 1,5-cyclooctadiene and borane. The product is a bicyclic compound of structure \(1\) (often abbreviated as 9-BBN), in which the residual \(\ce{B-H}\) bond adds to unhindered alkenes with much greater selectivity than is observed with other hydroborating reagents. It is also one of the few boranes that reacts sufficiently slowly with oxygen that it can be manipulated in air. An example of the difference in selectivity in the hydroboration of -4-methyl-2-pentene with \(\ce{B_2H_6}\) and \(1\) follows: According to the electronegativity chart (Figure 10-11), the boron-hydrogen bond is polarized in the sense \(\overset{\delta \oplus}{\ce{B}} --- \overset{\delta \ominus}{\ce{H}}\). Therefore the direction of addition of \(\ce{B_2H_6}\) to propene is that expected of a mechanism whereby the electrophilic boron atom becomes bonded to the less-substituted carbon of the double bond. However, there is no firm evidence to suggest that a carbocation intermediate is formed through a stepwise electrophilic addition reaction. For this reason, the reaction often is considered to be a . The stepwise formulation explains why boron becomes attached to the less-substituted carbon, but does not account for the fact that the reactions show no other characteristics of carbocation reactions. This could be because of an expected, extraordinarily fast rate of hydride-ion transfer to the carbocation. A more serious objection to the stepwise mechanism is that alkynes react more rapidly than alkenes, something which normally is not observed for stepwise electrophilic additions (cf. ). Some alkylboranes rearrange at elevated temperatures \(\left( 160^\text{o} \right)\) to form more stable isomers. For example, the alkylborane \(2\), produced by hydroboration of 3-ethyl-2-pentene, rearranges to \(3\) on heating: In general, the boron in alkylboranes prefers to be at the of a hydrocarbon chain so it is bonded to a carbon where steric crowding around boron is least severe. Thus rearrangement tends to proceed in the direction Rearrangement is associated with the fact that hydroboration is reversible at elevated temperatures. This makes possible a sequence of elimination-addition reactions in which boron becomes attached to different carbons and ultimately leads to the most stable product that has boron bonded to the carbon at the end of the chain: Rearrangement of alkylboranes can be used to transform alkenes with double bonds in the middle of the chain into 1-alkenes; for example, \(\ce{RCH=CHCH_3} \rightarrow \ce{RCH_2-CH=CH_2}\). The procedure involves hydroboration of the starting alkene in the usual manner; the borane then is isomerized by heating. An excess of 1-decene (bp \(170^\text{o}\)) then is added to the rearranged borane and the mixture is reheated. Heating causes the alkylborane to dissociate into 1-alkene and \(\ce{HBR_2}\); the 1-decene "scavenges" the \(\ce{HBR_2}\) as it forms, thereby allowing a more volatile 1-alkene (bp \(<170^\text{o}\)) to be removed by simple distillation. Thus, for the rearrangement of 3-ethyl-2-pentene to 3-ethyl-1-pentene, Alkylboranes formed in the hydroboration of alkenes and alkynes seldom are isolated; for the most part they are used as reactive intermediates for the synthesis of other substances. In the reactions of alkylboranes, the \(\ce{B-C}\) bond is cleaved in the sense \(\ce{B}^\oplus - \ce{C}^\ominus\) so that carbon is transferred to other atoms, such as \(\ce{H}\), \(\ce{O}\), \(\ce{N}\), and \(\ce{C}\), its bonding electron pair: In the first of these reactions (Equation 11-2), a hydrocarbon is produced by the cleavage of a borane, \(\ce{R_3B}\), with aqueous acid, or better, with anhydrous propanoic acid, \(\ce{CH_3CH_2CO_2H}\). The overall sequence of hydroboration-acid hydrolysis achieves the reduction of a carbon-carbon multiple bond without using hydrogen and a metal catalyst or diimide (Table 11-3): The second reaction (Equation 11-3) achieves the synthesis of a alcohol by the oxidation of the alkylborane with hydrogen peroxide in basic solution. Starting with a 1-alkene, one can prepare a primary alcohol in two steps: This sequence complements the direct hydration of 1-alkenes, which gives alcohols: Hydroboration of an alkene and subsequent reactions of the product trialkylborane, either with hydrogen peroxide or with acid, appear to be highly stereospecific. For example, 1-methylcyclopentene gives exclusively -2-methylcyclopentanol on hydroboration followed by reaction with alkaline hydrogen peroxide. This indicates that, overall, : Hydroboration of an alkyne followed by treatment of the alkenylborane with basic peroxide provides a method of synthesis of aldehydes and ketones. Thus hydroboration of 1-hexyne and oxidation of the 1-hexenylborane, \(4\), with hydrogen peroxide gives hexanal by way of the enol: If \(4\) is treated with deuteriopropanoic acid, replacement of \(\ce{-BR_2}\) by deuterium occurs with of configuration, forming -hexene-1-\(\ce{D_1}\): The stereospecific oxidation of alkylboranes occurs with hydrogen peroxide by an interesting and important general type of rearrangement which, for these reactions, involves migration of an organic group from boron to oxygen. The first step in the oxidation depends on the fact that tricoordinate boron has only six electrons in its valence shell and therefore behaves as if it were electron-deficient. The first step is bond formation at boron by the strongly nucleophilic peroxide anion (from \(\ce{H_2O_2} + \ce{OH}^\ominus \rightleftharpoons ^\ominus \ce{OOH} + \ce{H_2O}\)) to give a tetracovalent boron intermediate: In the second step, an alkyl group moves from boron to the neighboring oxygen and, in so doing, displaces hydroxide ion. Reaction is completed by hydrolysis of the \(\ce{B-O}\) bond: All three groups on boron are replaced in this manner. The rearrangement step (Equation 11-5) is an example of many related rearrangements in which a group, \(\ce{R}\), migrates with its bonding electrons from one atom to an adjacent atom. We already have encountered an example in the rearrangement of carbocations ( ): The difference between the carbocation rearrangement and the rearrangement of Equation 11-5 is that \(\ce{R}\) migrates from boron to oxygen as \(\ce{HO}^\ominus\) departs in what might be considered an internal \(S_\text{N}2\) reaction. We can generalize this kind of reaction of boron with a substance, \(\ce{X-Y}\), as in Equation 11-6: An example of the use of an \(\ce{X-Y}\) reagent is conversion of alkylboranes to primary amines with hydroxylaminesulfonic acid, \(\ce{H_2NOSO_3H}\) (Equation 11-4). The key steps are attack of the nucleophilic nitrogen at boron, followed by rearrangement, and hydrolysis, and (1977) | 9,840 | 17 |
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A most important property of enolate anions, at least as far as synthesis is concerned, is their excellent nucleophilicity, which enables them to to double bonds and to participate in nucleophilic substitution. When the addition is to a carbonyl double bond, it is called an (Equation 17-4). Additions of enolate anions to carbon-carbon double bonds usually are classified as (Equation 17-5), and these are discussed in and . The principles of \(S_\text{N}\) nucleophilic reactions of enolate anions (Equation 17-6) will be considered in , and their synthetic applications in detail in Chapter 18. The products of aldol addition are \(\beta\)-hydroxy aldehydes ( ) or \(\beta\)-hydroxyl ketones ( ). A typical example is the reaction of ethanal with base and, if the conditions are reasonably mild, the product is 3-hydroxybutanal: The overall reaction corresponds to a dimerization of ethanal, that is, an addition of one ethanal molecule to another with formation of a new carbon-carbon bond. The synthetic value of the reaction lies in the fact that it can be used to build large molecules from smaller molecules (see ). Formation of the enolate anion, \(7\), by removal of an \(\alpha\) hydrogen by base is the first step in the aldol addition: The anion then adds to the carbonyl group of a second molecule of ethanal in a manner analogous to the addition of other nucleophiles to carbonyl groups (e.g., cyanide ion, ). The adduct so formed, \(8\), rapidly adds a proton to the alkoxide oxygen to form the aldol, 3-hydroxybutanal. This last step regenerates the basic catalyst, \(\ce{OH}^\ominus\): The two possible valence-bond structures of the enolate anion, \(7a\) and \(7b\), show that the anion should act as an - a nucleophile with nucleophilic properties associated with carbon and oxygen. The addition step in the aldol reaction therefore may be expected to take place in either of two ways: The anion could attack as a nucleophile to form a carbon-carbon bond, \(8\), leading ultimately to the aldol, \(9\), or it might attack as an nucleophile to form a carbon-oxygen bond, thereby leading to the hemiacetal, \(10\). By this reasoning, we should obtain a mixture of products \(9\) and \(10\). However, the aldol \(9\) is the only one of these two possible products that can be isolated: Why is only one of these products formed? To understand this, you must recognize that aldol reactions are reversible and therefore are subject to rather than control ( ). Although the formation of \(10\) is mechanistically reasonable, it is not reasonable on thermodynamic grounds. Indeed, while the overall \(\Delta H^0\) (for the vapor) calculated from bond energies is \(-4 \: \text{kcal mol}^{-1}\) for the formation of the aldol, it is \(+20.4 \: \text{kcal mol}^{-1}\) for the formation of \(10\).\(^2\) Therefore, the reaction is overwhelmingly in favor of the aldol as the more stable of the two possible products. The equilibrium constant is favorable for the aldol addition of ethanal, as in fact it is for most aldehydes. For ketones, however, the reaction is much less favorable. With 2-propanone (acetone) only a few percent of the addition product "diacetone alcohol", \(11\) is present at equilibrium: The 2-propanone is boiled and the hot condensate from the reflux condenser flows back over solid barium hydroxide contained in the porous thimble and comes to equilibrium with the addition product \(11\). The barium hydroxide is retained by the porous thimble and the liquid phase returns to the boiler where the 2-propanone, which boils \(110^\text{o}\) below the temperature at which \(11\) boils, is selectively vaporized and returns to the reaction zone to furnish more adduct. The key step in aldol addition requires an electron-pair donor (nucleophile) and an electron-pair acceptor (electrophile). In the formation of 3-hydroxybutanal or \(11\), both roles are played by one kind of molecule, but there is no reason why this should be a necessary condition for reaction. Many kinds of mixed aldol additions are possible. Consider the combination of methanal and 2-propanone. Methanal cannot form an enolate anion because it has no \(\alpha\) hydrogens. However, it is expected to be a particularly good electron-pair acceptor because of freedom from steric hindrance and the fact that it has an unusually weak carbonyl bond (\(166 \: \text{kcal}\) compared to \(179 \: \text{kcal}\) for 2-propanone). In contrast, 2-propanone forms an enolate anion easily but is relatively poor as the electrophile. Consequently the addition of 2-propanone to methanal should and does occur readily: The problem is not to get addition, but rather to keep it from going too far. Indeed, all six \(\alpha\) hydrogens of 2-propanone can be replaced easily by \(\ce{-CH_2OH}\) groups: A commercially important mixed addition involves ethanal and an excess of methanal in the presence of calcium hydroxide. Addition occurs three times and the resulting trihydroxymethylethanal (which has no \(\alpha\) hydrogens) undergoes a "crossed Cannizzaro" reaction with more methanal to give a tetrahydroxy alcohol known as "pentaerythritol:: Pentaerythritol is used widely in the preparation of surface coatings and in the formation of its tetranitrate ester, pentaerythrityl tetranitrate [PETN, \(\ce{C(CH_2ONO_2)_4}\)], which is an important high explosive. An important property of aldol addition products is the ease with which they eliminate water in the presence of either acids or bases. For example, when 3-hydroxybutanal is heated in the basic solution in which it is formed (by aldol addition of ethanal), 2-butenol results: The ease of dehydration compared with simple alcohols is related to the fact that the product is a alkenone. The stabilization energy of the conjugated system makes the equilibrium constant for dehydration especially favorable. In many cases the aldol adduct is only an intermediate in aldol reactions because it dehydrates more rapidly than it can be isolated. Such is most often the case when the dehydration product is a polyunsaturated conjugated aldehyde or ketone. 2-Propanone and bezenecarbaldehyde (benzaldehyde), for instance, give the unsaturated ketone \(12\) in cold aqueous sodium hydroxide solution: Although the equilibrium for aldol addition may be unfavorable, when dehydration of the aldol product is rapid, \(\ce{C-C}\) bond formation may be pushed to completion by conversion of the aldol to the \(\alpha\),\(\beta\)-unsaturated ketone. The mechanism of base-catalyzed dehydration of aldols involves formation of an enolate anion by removal of a proton from the \(\ce{C2}\) or carbon and subsequent elimination of the hydroxyl group as hydroxide ion: This last step is one of the rare examples in which the leaving group is \(\ce{OH}^\ominus\). Generally, hydroxide is a poor leaving group in substitution (\(S_\text{N}1\) or \(S_\text{N}2\)) or elimination (\(E1\) or \(E2\)) reactions (see Section 8-7C). Dehydration of aldols to \(\alpha\),\(\beta\)-unsaturated carbonyl compounds usually is achieved best with acidic catalysts. An example is the dehydration of the aldol from 2-propanone to give 4-methyl-3-penten-2-one: If this reaction were attempted under basic conditions, extensive reversion of the aldol to 2-propanone would occur (see ). Under acidic conditions, however, the process is a straightforward proton transfer to oxygen followed by elimination of water and proton transfer from carbon: Aldol reactions provide a valuable synthetic method for forming carbon-carbon bonds. They can be adapted to extend the length of a carbon chain, to form cyclic compounds, and to provide intermediates that can be transformed into more useful materials. An important feature of these intermediates is that functional groups useful for later reactions are located close to or on the carbons of the newly formed \(\ce{C-C}\) bond. There is an almost bewildering number of variations on the aldol reaction and we shall not mention all of them. The main thing to recognize in all of these reactions is that the acceptor molecule always is a carbonyl compound, best an aldehyde, sometimes a ketone, even an ester (see ). The donor molecule is some type of carbanion; usually, but not always, an enolate anion. However, any substance that has a \(\ce{C-H}\) acidity in the p\(K_\text{a}\) range of 25 or less can be converted easily to a carbanion, which in principle may serve as the donor in aldol additions. Examples are listed in Table 17-1 and include not only aldehydes and ketones but esters, nitriles, and nitro compounds. The use of a nitroalkane in aldol addition is shown in the following sequence. The use of esters as the donor is discussed further in . Cyclic products can be formed by aldol additions provided the donor carbanion and acceptor carbonyl are part of the molecule. For example, consider how the synthesis of 3-methyl-2-cyclohexanone could be achieved from acyclic substances. The carbon-carbon bond formed in this process of aldol addition closes the ring and ultimately becomes the double bond in the conjugated system when the aldol product undergoes dehydration. Working backwards, we have the sequence and the starting material for the synthesis therefore is 2,6-heptanedione. Because \(\Delta G^0\) for the formation of aldol products is not very favorable, cyclizations involving aldol reactions usually will not proceed to give strained carbocyclic rings. The importance of aldol reactions is in the synthesis of alcohols, especially 1-butanol and 2-ethyl-1-hexanol: Notice that the combination of hydroformylation ( ), aldol addition, dehydration, and hydrogenation takes a simple alkene (propene) to an alcohol with more than twice as many carbons. One of the reactions in the metabolism of carbohydrates by the glycolic pathway is a type of aldol addition. In this reaction \(D\)-fructose (as the 1,6-diphosphate ester) is formed from \(D\)-glyceraldehyde and 1,3-dihydroxypropanone (both as monophosphate esters). The process is readily reversible and is catalyzed by an enzyme known as : It seems likely that this reaction could occur in quite the same way as in the laboratory aldol reactions discussed so far, because the enolate anion of the donor molecule (dihydroxypropanone) is not expected to be formed in significant amount at the pH of living cells. In fact, there is strong evidence that the enzyme behaves as an amino \(\left( \ce{ENH_2} \right)\) compound and reacts with the carbonyl group of dihydroxypropanone to form an imine, analogous to the reactions described in : This implies that the imine form of dihydroxypropanone is a key intermediate in the overall aldol-type addition. How can the imine behave as the carbon in addition to the aldehyde carbonyl of glyceraldehyde 3-phosphate? It is unlikely to do so directly, but it can rearrange to an enamine which, as we will explain in , can act as a carbon nucleophile: Attack of the nucleophilic carbon of the enamine at the aldehyde carbonyl of glyceraldehyde 3-phosphate forms the aldol of the imine which, on hydrolysis, gives the aldol and regenerates the enzyme: By using the neutral enamine as the carbon nucleophile rather than an enolate anion, the biological system avoids the need for strongly basic reaction conditions in aldol addition. \(^2\)This value probably is too large by \(3\) to \(4 \: \text{kcal}\), because resonance stabilization of alkoxyalkanes has been ignored in this calculation. and (1977) | 11,538 | 18 |
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\(K_c\) and \(K_p\) are the equilibrium constants of gaseous mixtures. However, the difference between the two constants is that \(K_c\) is defined by molar concentrations, whereas \(K_p\) is defined by the partial pressures of the gasses inside a closed system. The equilibrium constants do not include the concentrations of single components such as liquids and solid and they may have units depending on the nature of the reaction (although do not). Here are some easy steps on writing gas equilibrium constants (this is the same for finding K , K , K , Q and etc.):. The standard example of writing Gas Equilibrium Constants are: \[ \ aA + bB \; \rightleftharpoons \; cC + dD \nonumber \] \[ K_c = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \nonumber \] \[ K_p = \dfrac{(C)^{c}(D)^{d}} {(A)^{a}(B)^{b}} \nonumber \] Consider the thermal decomposition of \( NH_4SH_{(s)} \): \[ NH_4SH_{(s)} \rightleftharpoons NH_{3 (g)} + H_2S_{(g)} \nonumber \] This also is related to K \[ K_c = \dfrac{[NH_{3},H_{2}S]}{[NH_{4}SH]} \nonumber \] but since \(NH_4SH\) is a solid, we get: \[ K_c = \dfrac{[NH_{3},H_{2}S]}{[1]} \nonumber \] \[ K_c = [NH_{3},H_{2}S] \nonumber \] As for K , it is the same as K , but instead of brackets [ ], K uses parentheses ( ): \[ K_p = \dfrac{(NH_{3})(H_{2}S)}{(NH_{4}SH)} \nonumber \] \[ K_p = \dfrac{(NH_{3})(H_{2}S)}{(1)} \nonumber \] \[ K_p = (NH_{3})(H_{2}S) \nonumber \] Consider the double replacement reaction of hydrogen and iodine gas: \[ H_2 (g) + I_2 (g) \rightleftharpoons 2HI (g) \nonumber \] \[ K_c = \dfrac{[HI]^{2}}{[H_{2},I_{2}]} \nonumber \] \[ K_p = \dfrac{(HI)^{2}}{(H_{2})(I_{2})} \nonumber \] \(K_c\) is an equilibrium constant in terms of molar and is usually defined as: \[ K_c = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \nonumber \] in the general reaction, \[ aA + bB \rightleftharpoons cC + dD\nonumber \] If a large \(K_c\) is formed then there are more products formed. Inversely, a small \(K_c\) indicates that the reaction favors the reactants. \(K_p\) is an equilibrium constant in terms of . and is usually defined as: \[ K_p = \dfrac{(C)^{c}(D)^{d}}{(A)^{a}(B)^{b}} \nonumber \] for the general reaction \[ aA + bB \rightleftharpoons cC + dD \nonumber \] : Reactants/Products all in a single phase. For example: \[A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}\nonumber \] Reactants/Products in more than one phase. For example: \[A_{(s)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(s)}\nonumber \] The relationship between the two equilibrium constants are: \[ K_p = K_c (RT)^{ \Delta{n}} \nonumber \] or \[K_c = \dfrac{K_p}{(RT)^{ \Delta{n}}}\nonumber \] where, The value of K depends on whether the solution being calculated for is using concentrations or partial pressures. The gas equilibrium constants relate to the equilibrium (K) because they are both derived from (PV = nRT). \(K_c\) is the concentration of the reaction, it is usually shown as: \[ \dfrac{c[C]c[D]}{c[A]c[B]} \nonumber \] \(K_p\) is the amount of partial pressure in the reaction, usually shown as: \[ \dfrac{p(C)p(D)}{p(A)p(B)} \nonumber \] As we have seen above, K = K (RT) ,we can derive this formula from the Ideal Gas Law.
We know that K is in terms Molarity \(\left(\dfrac{Moles}{Liters}\right)\), and we can also arrange the Ideal Gas Law (PV = nRT) as: \(\left(\dfrac{n}{L}\right) = \left(\dfrac{P}{RT}\right)\). We know that Partial Pressure is directly proportional to Concentration: \(P = \left(\dfrac{n}{L}\right) * RT\) Pressure can be in units of: Pascal (Pa), Atmosphere (atm), or Torr. Therefore we can replace K with Molarity: the equation become, K = K (RT) (RT) = \(\dfrac{(RT)^{c}(RT)^{d}}{(RT)^{a}(RT)^{b}}\) Also: \(\left(\dfrac{n}{L}\right) = \left(\dfrac{P}{RT}\right)\), can be shown as K = K (RT) K is also written the same as K and K : \(\ aA + bB \rightleftharpoons cC + dD\). \( K = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\) \[ 2 NOBr_{(g)} \rightleftharpoons 2 NO_{(s)} + Br_{2 (g)} \nonumber \] Given: To set up K , it is \(\dfrac{Products}{Reactants}\) \[ K = \dfrac{[NO]^2 \; [Br_{2}]}{[NOBr]^2}\nonumber \] \[ K =\dfrac{[0.1]^2 \; [0.3]}{[0.46]^2}\nonumber \] K = 0.0142 M N O (l) is an important component of rocket fuel, At 25 °C N O is a colorless gas that partially dissociates into NO . The color of an equilibrium mixture of these 2 gasses depends on their relative proportions, which are dependent on temperature. Equilibrium is established in the reaction \( N_2O_4 (g) \rightleftharpoons 2NO_2 (g) \) at 25 °C. Given: What is the K for this reaction? Step 1: Convert grams to moles mol N O = 7.64 g * \( \dfrac{1 mol N_2 O_4}{92.01 g} \) = 8.303 * 10 mol mol NO = 1.56 g * \( \dfrac{1 mol NO_2}{46.01 g} \) = 3.391 * 10 mol Step 2: Convert moles to Molarity (moles/L) [N O ] M \(\dfrac{8.303 * 10^{-2} mol N_{2}O_{4}}{3.00 L}\) = 0.0277 M [NO ] M = \(\dfrac{3.391 * 10^{-2} mol NO_{2}}{3.00 L}\) = 0.0113 M Step 3: Write the Equilibrium constant for K : \[K_{c} = \dfrac{[NO_{2}]^{2}}{[N_{2}O_{4}]} = \dfrac{[0.0113]^{2}}{[0.0277]} = 4.61 \times 10^{-3}\nonumber \] Calculate K for the reaction \[N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}\nonumber \] Given: When equilibrium is established the mole percent of Nitrogen Oxide (NO) at 1.8% \[N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}\nonumber \] \(N_2\) X = 0.018 X = (0.79-x)+(.021-x)+(2x)= 1 X = \(\dfrac{2x}{X_{total}}\) 0.018 = \(\dfrac{2x}{1}\) x = 0.009 K = \(\dfrac{p(NO)^{2}}{p(N_{2})p(O_{2})}\) PV = NRT \(P = \dfrac{\dfrac{[n(NO)(RT)]^{2}}{V^{2}}}{\dfrac{n(N_{2})(RT)}{V}\dfrac{n(O_{2})(RT)}{V}}\)
[Volume Cancels out] \(P = \dfrac{n(NO)^{2}}{n(N_{2})n(O_{2})}\) \(K_{P} = \dfrac{(2x)^{2}}{(0.79-x)(0.21-x)}\) x = 0.009 K = 2.1x10 The process of finding the Reaction Quotient (Q ) is the same as finding K and K , where the products of the reaction is divided by the reactants of the reaction \(\left(\dfrac{Products}{Reactants}\right)\) at any time not necessarily at equilibrium. If a problem asks you to find which way the reaction will shift in order to achieve equilibrium, and K is given, you would have to calculate for Q and compare the two numbers.
When comparing K and Q: A trick to remember to which what the reaction will favor is:
Put:
K _ Q (in alphabetical order! - or it will not work)
K < Q : K \(\leftarrow\) Q
The reaction will favor the reactants because reactants are on the left of the equation.
K > Q : K \(\rightarrow\) Q
The reaction will favor the products because products are on the right of the equation.
K = Q : NO CHANGE (See Relationship for more information) \[ CO (g) + H_2O (g) \rightleftharpoons CO_2 (g) + H_2 (g) \nonumber \] Given: K = 1.00 at about 1100 K CO = 1.00 mol H O = 1.00 mol CO = 2.00 mol H = 2.00 mol
Compared with their initial amounts, which of the substances will be present in a greater amount and which is in a lesser amount when equilibrium is established? Step 1: Write out the expression for Q \[ Q_c = \dfrac{[CO_2,H_2]}{[CO,H_2O]} \nonumber \] Step 2: Plug in the number of Molarity, since volume is not given, assume it is 1 Liters \[ Q_c = \dfrac{[2.00,2.00]}{[1.00,1.00]} \nonumber \] \[ Q_c = 4.00 \nonumber \] Step 3: Compare K with Q K = 1.00 (unitless) Q = 4.00 (unitless) K < Q 1.00 < 4.00 Therefore, the reaction will shift to the LEFT towards the reactants. A mixture of hydrogen, iodine, and hydrogen iodide, each at 0.0020 M, was introduced into a container heated to 783 Kelvins. At this temperature \(K_c = 46\), Predict if more HI or less will be formed. Step 1: Write out the reaction \[ H_2 (g) + I_2 (g) \leftrightharpoons 2HI (g) \nonumber \] Step 2: Write out the expression for Q \[ Q_c = \dfrac{[HI]^2}{[I_2,H_2]} \nonumber \] Step 3: Plug in the Molarity given Molarity = 0.0020 M \[ Q_c = \dfrac{[0.0020]^{2}}{[0.0020,0.0020]} \nonumber \] \[ Q_{c} = 1.00 \nonumber \] Step 4: Compare K with Q K = 46 (unitless) Q = 1.00 (unitless) K > Q 46 > 1.00 Below are practice problems for K and K : 1. Gaseous Hydrogen Iodide is placed in a closed container at 425 C, Where it partially decomposes to Hydrogen and Iodine: 2HI (g) \(\rightleftharpoons\) H (g) + I (g) The following are given: [HI] = 3.53 * 10 M [H ] = 4.79 * 10 M [I ] = 4.79 * 10 M What is the value of K at this temperature? 2. Write the K for the reaction and state where the reaction is Homogeneous or Heterogeneous. a) N (g) + O (g) \(\rightleftharpoons\) 2NO (g) b) FeO (s) + H (g) \(\rightleftharpoons\) Fe (s) + H O (g) 3. Determine values of K from the K value given: (number 7 from p. 655 in the textbook) 2NO (g) + O (g) \(\rightleftharpoons\) 2NO (g); Kp = 1.48 * 10 at 184 C 1. Write the K for the reaction and state where the reaction is Homogeneous or Heterogeneous. a) 2C H (g) + 2H O (g) \(\rightleftharpoons\) 2C H (g) + O (g) b) Ti (s) + 2Cl (g) \(\rightleftharpoons\) TiCl (g) 2. Determine values of K from the K value given: (number 8 from p. 655 in the textbook) 2H S (g) + CH (g) \(\rightleftharpoons\) 4H (g) + CS (g); K = 5.27 * 10 at 973 K. 3. The two common chlorides of Phosphorus, PCl and PCl , both important in the production of other phosphorous compounds, coexist in equilibrium through: (number 17 from p. 655 in the textbook) PCl (g) + Cl (g) \(\rightleftharpoons\) PCl (g) At 250 C, an equilibrium mixture in a 2.50 L flask contains 0.105 g PCl , 0.220 g PCl , and 2.12 g Cl . What are the values of (a) Kc and (b) Kp for this reaction? 1. 2HI (g) \(\rightleftharpoons\) H (g) + I (g) [HI] = 3.53 * 10 M [H ] = 4.79 * 10 M [I ] = 4.79 * 10 M K = \(\dfrac{[H_{2},I_{2}]}{[HI]^{2}}\) K = \(\dfrac{[4.79 * 10^{-4} M,4.79 * 10^{-4} M]}{[3.53 * 10^{-3} M]^{2}}\) K = \(\dfrac{[2.29441 * 10^{-7}] M^{2}}{[1.24609 * 10^{-5}] M^{2}}\) 2. a) N (g) + O (g) \(\rightleftharpoons\) 2NO (g) K = \(\dfrac{[NO]^{2}}{[N_{2},O_{2}]}\) The reaction is a homogeneous reaction because the reactants/products all have the same phase. b) FeO (s) + H (g) \(\rightleftharpoons\) Fe (s) + H O (g) K = \(\dfrac{[H_{2}O]}{[H_{2}]}\), FeO and Fe are solids so they are no included in equilibrium constants. The reaction is a heterogeneous reaction because the reactants/products have different phases. 3. Converting to K from K (number 7 from p. 655 in the textbook) 2NO (g) + O (g) \(\rightleftharpoons\) 2NO (g) Kp = 1.48 * 10 at 184 C We know that K = K (RT) , we are given K but not K , you can rearrange the equation to: \(K_{c} = \dfrac{K_{p}}{(RT)^{-\Delta{n}}}\) Which can also be written as: K = K (RT) Now that we have your formula, we need to convert 184 C to Kelvin, K = 184 + 273 = 457K K = \(\dfrac{[NO_{2}]^{2}}{[NO]^{2}[O_{2}]}\) -\(\Delta{n}\) = (total number of moles of products) - (total number of moles in reactants)
-\(\Delta{n}\) = (2) - (3) = -1
\(\Delta{n}\) = -(-1) R = 0.08206 \(\dfrac{Liter \; Atm}{Mole \; Kelvin}\) Now, plug in all the numbers we found: K = K (RT) K = (1.48 * 10 )[(0.08206)(457K)] [it would be the same if you used this equation: K = K (RT)] 1. a) 2C H (g) + 2H O (g) \(\rightleftharpoons\) 2C H (g) + O (g) K = \(\dfrac{[C {2}H_{6}]^{2}[O_{2}]}{[C_{2}H_{4}]^{2}[H_{2}O]^{2}}\) The reaction is a homogeneous reaction because the reactants/products all have the same phase. b) Ti (s) + 2Cl (g) \(\rightleftharpoons\) TiCl (g) K = \(\dfrac{[TiCl_{4}]}{[Cl_{2}]^{2}}\), Ti is a solid so it is not included in equilibrium constants. The reaction is a heterogeneous reaction because the reactants/products have different phases. 2. Find K , when K is given: (number 8 from p. 655 in the textbook) 2H S (g) + CH (g) \(\rightleftharpoons\) 4H (g) + CS (g)
K = 5.27 * 10 at 973 K. K = K (RT) , since Temperature is already converted to Kelvin and R = 0.08206 \(\dfrac{Liter \; Atm}{Mole \; Kelvin}\) We need to find Delta n: K = \(\dfrac{[H {2}]^{4}[CS_{2}]}{[H_{2}S]^{2}[CH_{4}O]}\)
\(\Delta{n}\) = (total number of moles of products) - (total number of moles in reactants)
\(\Delta{n}\) = (5) - (3) = 2
We can plug in our numbers: K = (5.27 * 10 )[(0.08206)(973)]
3. The two common chlorides of Phosphorus, PCl and PCl , both important in the production of other phosphorous compounds, coexist in equilibrium through: (number 17 from p. 655 in the textbook) PCl (g) + Cl (g) \(\rightleftharpoons\) PCl (g) At 250 C, an equilibrium mixture in a 2.50 L flask contains 0.105 g PCl , 0.220 g PCl , and 2.12 g Cl . What are the values of: (a) K [We need to convert grams to Molarity (mol/L), so we multiply grams with the molar mass and divide by Liters.] PCl = \(\dfrac{0.105 g}{2.50 L}\) \(* \dfrac{1 mol}{137.3 g}\) (Molar Mass) = 2.0173 * 10 M PCl = \(\dfrac{0.220 g}{2.50 L}\) \(* \dfrac{1 mol}{137.3 g}\) (Molar Mass) = 6.4093 * 10 M Cl = \(\dfrac{2.12 g}{2.50 L}\) \(* \dfrac{1 mol}{70.9 g}\) (Molar Mass) = 0.0119605 \( K_{c} = \dfrac{[PCl_{5}]}{[Cl_{2},PCl_{3}]}\) \( K_{c} = \dfrac{[2.0173 * 10^{-4}]}{[0.0119605,6.4093 * 10^{-4}]}\) (b) K K = K (RT) We need to find: \(\Delta{n}\) and K, R = 0.08206 \(\dfrac{L atm}{mol K}\) K = 250 C + 273 = 523 K \(\Delta{n}\) = (1) - (2) = -1 K = (26.32) [(0.08206 )(523)] | 13,145 | 20 |
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A number of useful reactions for the preparation of aldehydes and ketones, such as ozonization of alkenes and hydration of alkynes, have been considered in previous chapters. These and other methods of preparation are summarized in Tables 16-7 and 16-8 at the end of the chapter. Only a few rather general methods that we have not discussed will be taken up here. Aldehydes and ketones often can be prepared by oxidation of alkenes to 1,2-diols (Sections 11-7C and 11-7D), followed by oxidative cleavage of the 1,2-diols with lead tetraethanoate or sodium periodate. For example, Cleavage of glycols with these reagents proceeds according to the following stoichiometry: In Chapter 15 primary alcohols, \(\ce{RCH_2OH}\), were shown to be readily oxidized to aldehydes, \(\ce{RCHO}\), and secondary alcohols, \(\ce{R_2CHOH}\), to ketones, \(\ce{R_2CO}\), by inorganic reagents such as \(\ce{CrO_3}\) and \(\ce{KMnO_4}\). However, it is a problem to avoid overoxidation with primary alcohols because of the ease with which aldehydes are oxidized to acids, \(\ce{RCHO} \rightarrow \ce{RCO_2H}\). A milder oxidant is methylsulfinylmethane [dimethyl sulfoxide, \(\ce{(CH_3)_2S=O}\)], and this reagent can be used to prepare aldehydes from alcohols by way of an intermediate such as the ester or halide in which the \(\ce{OH}\) group is converted to a better leaving group: Whichever method is employed, the key step is the formation of an alkoxysulfonium salt, \(7\), by a displacement reaction involving dimethyl sulfoxide as an oxygen nucleophile. (Notice that the \(\ce{S=O}\) bond, like the \(\ce{C=O}\) bonds, is strongly polarized as \(\overset{\oplus}{\ce{S}} \ce{-} \overset{\ominus}{\ce{O}}\).) In the examples listed in Equations 16-12 through 16-15, the \(\ce{X}\) group is \(\ce{Br}\), \(\ce{-OSO_2R'}\), \(\ce{-O_2CCF_3}\), and , respectively. In the nest step a sulfur ylide, \(8\), is formed from the reaction of a base with \(7\), but the ylide evidently is unstable and fragments by an internal \(E2\) reaction to form an aldehyde: Conversion of a carboxylic acid to an aldehyde by direct reduction is not easy to achieve, because acids generally are difficult to reduce, whereas aldehydes are easily reduced. Thus the problem is to keep the reaction from going too far. The most useful procedures involve conversion of the acid to a derivative that either is more easily reduced than an aldehyde, or else is reduced to a substance from which the aldehyde can be generated. The so-called involves the first of these schemes; in this procedure, the acid is converted to an acyl chloride, which is reduced with hydrogen over a palladium catalyst to the aldehyde in yields up to \(90\%\). The rate of reduction of the aldehyde to the corresponding alcohol is kept at a low level by poisoning the catalyst with suflur: Metal hydrides, such as lithium aluminum hydride, also can be used to reduce derivatives of carboxylic acids (such as amids and nitriles see Table 16-6) to aldehydes. An example follows: Many carbonyl compounds can be synthesized by acid-catalyzed rearrangements of 1,2-diols (a type of reaction often called the "pinacol-pinacolone" rearrangement). The general characteristics of the reaction are similar to those of carbocation rearrangements (Section 8-9B). The acid assists the reaction by protonating one of the \(\ce{-OH}\) groups to make it a better leaving group. The carbocation that results then can undergo rearrangement by shift of the neighboring \(\ce{R}\) group with its pair of bonding electrions to give a new, thermodynamically more stable species with a carbon-oxygen double bond (see Section 16-7). The prototype of this rearrangement is the conversion of pinacol to pinacolone as follows: An important method of preparing carbonyl (and hydroxy) compounds, especially on an industrial scale, is through rearrangements of alkyl hydroperoxides: The peroxides can be made in some cases by direct air oxidation of hydrocarbons, and in others by sulfuric acid-induced addition of hydrogen peroxide (as \(\ce{H-O_2H}\)) to double bonds: (Notice that hydrogen peroxide in methanoic acid behaves differently toward alkenes in producing addition of \(\ce{HO-OH}\), Section 11-7D.) The direct air oxidation of hydrocarbons is mechanistically similar to that of benzenecarbaldehyde (Section 16-7). The rearrangements of hydroperoxides are acid-catalyzed and are analogous to carbocation rearrangements except that positive oxygen (with only valence electrons) instead of positive carbon is involved in the intermediate stage: In principle, either phenyl or methyl could migrate to the positive oxygen, but only phenyl migration occurs in this case. The rearrangement reaction is closely related to the Baeyer-Villiger reaction (Section 16-7). This reaction is important for a number of reasons. It is an industrial synthesis of aldehydes from alkenes by the addition of carbon monoxide and hydrogen in the presence of a cobalt catalyst. A prime example is the synthesis of butanal from propene, in which 2-methylpropanal also is formed: As you can see, the reaction formally amounts to the addition of methanal as \(\ce{H-CHO}\) to the alkene double bond. Because one additional carbon atom is introduced as a "formyl" \(\ce{CHO}\) group, the reaction often is called , although the older name, , is widely used. Hydroformylation to produce aldehydes is the first step in an important industrial route to alcohols. The intermediate aldehydes are reduced to alcohols by catalytic hydrogenation. Large quantities of \(\ce{C_4}\)-\(\ce{C_8}\) alcohols are prepared by this sequence: The history of the oxo reaction is also noteworthy. It was developed originally in Germany in the years following World War I. At that time, the German chemical industry was faced with inadequate supplies of petroleum. Many German chemists therefore turned to research on ways by which hydrocarbons could be synthesized from smaller building blocks, particularly carbon monoxide and hydrogen derived from coal. The success achieved was remarkable and led to alkane and alkene syntheses known as the : This reaction in turn led to the discovery that aldehydes were formed by the further addition of carbon monoxide and hydrogen to alkenes, and was further developed as the oxo process for production of alcohols. The combination \(\ce{CO} + \ce{H_2}\) often is called "synthetic gas". It is prepared by the reduction of water under pressure and at elevated temperatures by carbon (usually coke), methane, or higher-molecular-weight hydrocarbons: The aldehyde synthesis by hydroformylation of alkenes described in the preceding section can be achieved indirectly using boron hydrides. An oversimplified expression of this reaction is The overall reaction is quite complex but involves a rearrangement similar to that described for the hydroboration-oxidation of alkenes (Section 11-6E). The first step is hydroboration of the alkene to a trialkylborane. When the trialkylborane is exposed to carbon monoxide, it reacts (carbonylates) to form a tetracovalent boron, \(9\): The complex \(9\) is unstable and rearranges by transfer of an alkyl group from boron to the electron-deficient carbonyl carbon to give \(10\): Now, if a metal-hydride reducing agent, such as \(\ce{LiAlH_4}\), is present, the carbonyl group of \(10\) is reduced and \(11\) is formed: The reduction product, \(11\), can be converted to an aldehyde by oxidation with aqueous hydrogen peroxide, provided the pH is carefully controlled. (Remember, aldehydes are unstable in strong base.) You may have noticed that only of the three alkyl groups of a trialkylborane is converted to an aldehyde by the carbonylation-reduction-oxidation sequence. To ensure that carbonylation takes the desired course without wasting the starting alkene, hydroboration is achieved conveniently with a hindered borane, such as "9-BBN", \(12\). With \(12\), only the least-hindered alkyl group rearranges in the carbonylation step: Carbonylation of alkylboranes also can produce ketones. The conditions are similar to those in the aldehyde synthesis except that the hydride reducing agent is omitted. By omitting the reducing agent, a second boron-to-carbon rearrangement can occur. Oxidation then produces a ketone: Rearrangement will continue a third time (ultimately to produce a tertiary alcohol) unless movement of the alkyl group remaining on boron in \(13\) is prevented by steric hindrance. and (1977) | 8,506 | 21 |
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The halogens are located on the left of the noble gases on the periodic table. These five toxic, non-metallic elements make up Group 17 of the periodic table and consist of: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At). Although astatine is radioactive and only has short-lived isotopes, it behaves similar to iodine and is often included in the halogen group. Because the halogen elements have seven valence electrons, they only require one additional electron to form a full octet. This characteristic makes them more reactive than other non-metal groups. Halogens form diatomic molecules (of the form X , where X denotes a halogen atom) in their elemental states. The bonds in these diatomic molecules are non-polar covalent single bonds. However, halogens readily combine with most elements and are never seen uncombined in nature. As a general rule, fluorine is the most reactive halogen and astatine is the least reactive. All halogens form Group 1 salts with similar properties. In these compounds, halogens are present as halide anions with charge of -1 (e.g. Cl , Br , etc.). Replacing the -ine ending with an -ide ending indicates the presence of halide anions; for example, Cl is named "chloride." In addition, halogens act as oxidizing agents—they exhibit the property to oxidize metals. Therefore, most of the chemical reactions that involve halogens are oxidation-reduction reactions in aqueous solution. The halogens often form single bonds, when in the -1 oxidation state, with carbon or nitrogen in organic compounds. When a halogen atom is substituted for a covalently-bonded hydrogen atom in an organic compound, the prefix can be used in a general sense, or the prefixes , , , or can be used for specific halogen substitutions. Halogen elements can cross-link to form diatomic molecules with polar covalent single bonds. Chlorine (Cl ) was the first halogen to be discovered in 1774, followed by iodine (I ), bromine (Br ), fluorine (F ), and astatine (At, discovered last in 1940). The name "halogen" is derived from the Greek roots hal- ("salt") and -gen ("to form"). Together these words combine to mean "salt former", referencing the fact that halogens form salts when they react with metals. is the mineral name for rock salt, a natural mineral consisting essentially of sodium chloride (NaCl). Lastly, the halogens are also relevant in daily life, whether it be the fluoride that goes in toothpaste, the chlorine that disinfects drinking water, or the iodine that facilitates the production of thyroid hormones in one's body. \[F < Cl < Br < I < At\] F < Cl < Br < I < At) At < I < Br < Cl < F). At < I < Br < Cl < F). At < I < Br < Cl) The reactivities of the halogens decrease down the group Cl electronegativity A halide is formed when a halogen reacts with another, less electronegative element to form a binary compound. Hydrogen, for example, reacts with halogens to form halides of the form Hydrofluoric acid can etch glass and certain inorganic fluorides over a long period of time. It may seem counterintuitive to say that HF is the weakest hydrohalic acid because fluorine has the highest electronegativity. However, the H-F bond is very strong; if The halogens' colors are results of the absorption of visible light by the molecules, which causes electronic excitation. Fluorine absorbs violet light, and therefore appears light yellow. Iodine, on the other hand, absorbs yellow light and appears violet (yellow and violet are complementary colors, which can be determined using a The colors of the halogens grow darker down the group: In closed containers, liquid bromine and solid iodine are in equilibrium with their vapors, which can often be seen as colored gases. One third exception to the rule is this: if a halogen exists in its elemental form (X ), its oxidation state is zero. : Although fluorine is very reactive, it serves many industrial purposes. For example, it is a key component of the plastic (called by the DuPont company) and certain other polymers, often referred to as fluoropolymers. Chlorofluorocarbons (CFCs) are organic chemicals that were used as refrigerants and propellants in aerosols before growing concerns about their possible environmental impact led to their discontinued use. Hydrochlorofluorocarbons (HFCs) are now used instead. Fluoride is also added to toothpaste and drinking water to help reduce tooth decay. Fluorine also exists in the clay used in some ceramics. Fluorine is associated with generating nuclear power as well. In addition, it is used to produce fluoroquinolones, which are antibiotics. Below is a list of some of fluorine's important inorganic compounds. : Chlorine has many industrial uses. It is used to disinfect drinking water and swimming pools. Sodium hypochlorite (NaClO) is the main component of bleach. Hydrochloric acid, sometimes called muriatic acid, is a commonly used acid in industry and laboratories. Chlorine is also present in polyvinyl chloride (PVC), and several other polymers. PVC is used in wire insulation, pipes, and electronics. In addition, chlorine is very useful in the pharmaceutical industry. Medicinal products containing chlorine are used to treat infections, allergies, and diabetes. The neutralized form of hydrochloride is a component of many medications. Chlorine is also used to sterilize hospital machinery and limit infection growth. In agriculture, chlorine is a component of many commercial pesticides: DDT (dichlorodiphenyltrichloroethane) was used as an agricultural insecticide, but its use was discontinued. : Bromine is used in flame retardants because of its fire-resistant properties. It also found in the pesticide methyl bromide, which facilitates the storage of crops and eliminates the spread of bacteria. However, the excessive use of methyl bromide has been discontinued due to its impact on the ozone layer. Bromine is involved in gasoline production as well. Other uses of bromine include the production of photography film, the content in fire extinguishers, and drugs treating pneumonia and Alzheimer's disease. : Iodine is important in the proper functioning of the thyroid gland of the body. If the body does not receive adequate iodine, a goiter (enlarged thyroid gland) will form. Table salt now contains iodine to help promote proper functioning of the thyroid hormones. Iodine is also used as an antiseptic. Solutions used to clean open wounds likely contain iodine, and it is commonly found in disinfectant sprays. In addition, silver iodide is important for photography development. : Because astatine is radioactive and rare, there are no proven uses for this halogen element. However, there is speculation that this element could aid iodine in regulating the thyroid hormones. Also, At has been used in mice to aid the study of cancer. | 6,871 | 22 |
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Aluminum is found in varying amounts in nature as aluminosilicates (contains aluminum, silicon, and oxygen) in various types of clay. As the minerals are weathered they gradually breakdown into various forms of hydrated aluminum oxide, Al O .xH O, known as bauxite. The bauxite is purified by the . First the ore is mixed with a hot concentrated solution of sodium hydroxide. The NaOH will dissolve the oxides of aluminum and silicon but not other impurities such as iron oxides, which remains insoluble. The insoluble materials are removed by filtration. The solution which now contains the oxides of aluminum and silicon are next treated by bubbling carbon dioxide gas through the solution. Carbon dioxide forms a weak acid solution of carbonic acid which neutralizes the sodium hydroxide from the first treatment. This neutralization selectively precipitates the aluminum oxide, but leaves the silicates in solution. Again filtration is used for the separation. After this stage the purified aluminum oxide is heated to evaporate the water. Aluminum in the metal form is very difficult to obtain by using some of the traditional chemical methods involving carbon or carbon monoxide as reducing agents to reduce the aluminum ions to aluminum metal. One of the earliest and costly methods in 1850 was to reduce aluminum chloride with sodium metal to obtain aluminum metal and sodium chloride. (Sodium metal is not easy to obtain either). As a result some of the earliest aluminum metal was made into jewelry. In 1886, Charles Hall, an American (23 yrs. old), and Paul Heroult, a Frenchmen (23 yrs old), simultaneously and independently developed the process still in use today to make aluminum metal. The purified aluminum oxide is mixed with cryolite, a mixture of sodium fluoride and aluminum fluoride, and heated to about 980 degrees Celsius to melt the solids. The mixture melts at a much lower temperature than aluminum oxide would by itself. The hot molten mixture is electrolyzed at a low voltage of 4-5 volts, but a high current of 50,000-150,000 amps. Aluminum ions are reduced to aluminum metal at the cathode (the sides and bottom of the electrolysis cell). At the anode, oxygen is produced from the oxide ions. The anode material is carbon in the form of graphite, which also is oxidized and must be replaced quite frequently. The electricity used to produce aluminum is relatively high. One pound of aluminum requires 6-8 kilowatt-hours of electrical energy. This amount of aluminum can be used to make 23 pop cans or one 300 watt light bulb burning for one hour is required to make one pop can. | 2,628 | 23 |
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Most alkenes react readily with ozone \(\left( \ce{O_3} \right)\), even at low temperatures, to yield cyclic peroxidic derivatives known as . For example, These substances, like most compounds with peroxide \(\left( \ce{O-O} \right)\) bonds, may explode violently and unpredictably. Therefore ozonizations must be carried out with appropriate caution. The general importance of these reactions derives not from the ozonides, which usually are not isolated, but from their subsequent products. The ozonides can be converted by hydrolysis with water and reduction, with hydrogen (palladium catalyst) or with zinc and acid, to carbonyl compounds that can be isolated and identified. For example, 2-butene gives ethanal on ozonization, provided the ozonide is destroyed with water and a reducing agent which is effective for hydrogen peroxide: An alternative procedure for decomposing ozonides from di- or trisubstituted alkenes is to treat them with methanol \(\left( \ce{CH_3OH} \right)\). The use of this reagent results in the formation of an aldehyde or ketone and a carboxylic acid: The overall ozonization reaction sequence provides an excellent means for locating the positions of double bonds in alkenes. The potentialities of the method may be illustrated by the difference in reaction products from the 1- and 2-butenes: Ozonization of alkenes has been studied extensively for many years, but there is still disagreement about the mechanism (or mechanisms) involved because some alkenes react with ozone to give oxidation products other than ozonides. It is clear that the ozonide is not formed directly, but by way of an unstable intermediate called a . the molozonide then either isomerizes to the "normal" ozonide or participates in other oxidation reactions. Although the structure of normal ozonides has been established beyond question, that of the molozonide, which is very unstable even at \(-100^\text{o}\), is much less certain. The simplest and most widely accepted mechanism involves formation of a molozonide by a direct of ozone to the double bond.\(^1\) Isomerization of the molozonide appears to occur by a fragmentation-recombination reaction, as shown in Equations 11-7 and 11-8: Several oxidizing reagents react with alkenes under mild conditions to give, as the overall result, addition of hydrogen peroxide as \(\ce{HO-OH}\). Of particular importance are alkaline permanganate \(\left( \ce{MnO_4^-} \right)\) and osmium tetroxide \(\left( \ce{OsO_4} \right)\), both of which react in an initial step by a suprafacial cycloaddition mechanism like that postulated for ozone. Each of these reagents produces -1,2-dihydroxy compounds (diols) with cycloalkenes: Osmium tetroxide is superior to permanganate in giving good yields of diol, but its use is restricted because it is a very costly and very toxic reagent. Alkenes can be oxidized with peroxycarboxylic acids, \(\ce{RCO_3H}\), to give oxacyclopropanes (oxiranes, epoxides), which are three-membered cyclic ethers: The reaction, known as , is valuable because the oxacyclopropane ring is cleaved easily, thereby providing a route to the introduction of many kinds of functional groups. In fact, oxidation of alkenes with peroxymethanoic acid (peroxyformic acid), prepared by mixing methanoic acid and hydrogen peroxide, usually does not stop at the oxacyclopropane stage, but leads to ring-opening and the subsequent formation of a diol: This is an alternative scheme for the hydroxylation of alkenes (see ). However, the overall stereochemistry is opposite to that in permanganate hydroxylation. For instance, cyclopentene gives -1,2-cylcopentanediol. First the oxirane forms by suprafacial addition and then undergoes ring opening to give the trans product: The ring opening is a type of \(S_\text{N}2\) reaction. Methanoic acid is sufficiently acidic to protonate the ring oxygen, which makes it a better leaving group, thus facilitating nucleophilic attack by water. The nucleophile always attacks from the side remote from the leaving group: The peroxyacids that are used in the formation of oxacyclopropanes include peroxyethanoic \(\left( \ce{CH_3CO_3H} \right)\), peroxybenzoic \( \left( \ce{C_6H_5CO_3H} \right)\), and trifluoroperoxyethanoic \(\left( \ce{CF_3CO_3H} \right)\) acids. A particularly useful peroxyacid is 3-chloroperoxybenzoic acid, because it is relatively stable and is handled easily as the crystalline solid. The most reactive reagent is trifluoroperoxyethanoic acid, which suggests that the peroxyacid behaves as an electrophile (the electronegativity of fluorine makes the \(\ce{CF_3}\) group strongly electron-attracting). The overall reaction can be viewed as a , in which the proton on oxygen is transferred to the neighboring carbonyl oxygen more or less simultaneously with formation of the three-membered ring: A reaction of immense industrial importance is the formation of oxacyclopropane itself (most often called ethylene oxide) by oxidation fo ethene with oxygen over a silver oxide catalyst at \(300^\text{o}\): Oxacyclopropane is used for many purposes, but probably the most important reaction is ring opening with water to give 1,2-ethanediol (ethylene glycol, bp \(197^\text{o}\)). This diol, mixed with water, is employed widely in automotive cooling systems to provide both a higher boiling and lower freezing coolant than water alone: Propene and higher alkenes are not efficiently epoxidized by oxygen and \(\ce{Ag_2O}\) in the same way as ethene because of competing attack at other than the double-bond carbons. \(^1\)The ozone structure shown here with single electrons having paired spins on the terminal oxygens accords both with the best available quantum mechanical calculations and the low dipole moment of ozone, which is not consonant with the conventional \(\ce{O=} \overset{\oplus}{\ce{O}} - \overset{\ominus}{\ce{O}}\) structure. See W. A. Goddard III, T. H. Dunning, Jr., W. J. Hunt, and P. J. Hay, , 368 (1973). and (1977) | 5,997 | 24 |
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This page discusses the trends in some atomic and physical properties of the Group 17 elements (the halogens): fluorine, chlorine, bromine and iodine. Sections below describe the trends in atomic radius, electronegativity, electron affinity, melting and boiling points, and solubility. There is also a section on the bond enthalpies (and strengths) of halogen-halogen bonds (for example, the Cl-Cl bond) and of hydrogen-halogen bonds (e.g. the H-Cl bond). You can see that the atomic radius increases as you go down the Group. The radius of an atom is governed by As shown in the figure above, electronegativity decreases from fluorine to iodine; the atoms become less effective at attracting bonding pairs of electrons as they grow larger. This can be visualized using dots-and-crosses diagrams for hydrogen fluoride and hydrogen chloride. The bonding electrons between the hydrogen and the halogen experience the same net charge of +7 from either the fluorine or the chlorine. However, in the chlorine case, the nucleus is further away from the bonding pair. Therefore, electrons are not as strongly attracted to the chlorine nucleus as they are to the fluorine nucleus. The stronger attraction to the closer fluorine nucleus makes fluorine is more electronegative. As the halogen atoms get larger, any bonding pair is farther and farther away from the halogen nucleus, and so is less strongly attracted towards it. Hence, the elements become less electronegative as you go down the Group,. Jim Clark ( ) | 1,519 | 25 |
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This page examines rate constant variation with temperature and activation energy, as shown by the Arrhenius equation. The rate equation for a reaction between two substances, A and B, is the following: The rate equation shows the effect of changing the reactant concentrations on the rate of the reaction. All other factors affecting the rate—temperature and catalyst presence, for example—are included in the rate constant, which is only constant if the only change is in the concentration of the reactants. If the temperature is changed or a catalyst is added, for example, the rate constant changes. This is shown mathematically in the Arrhenius equation: The various symbols represent the following: The Arrhenius equation often takes this alternate form, generated by taking the natural logarithm of the standard equation: \[ \large \ln k = \ln A - \dfrac{E_a}{RT} \nonumber \] The Arrhenius equation can be used to determine the effect of a change of temperature on the rate constant, and consequently on the rate of the reaction. If the rate constant doubles, for example, so does the rate of the reaction. What is the kinetic effect of increasing temperature from 20°C to 30°C (293 K to 303 K)? The frequency factor, A, is approximately constant for such a small temperature change. This problem concerns the quantity e , the fraction of molecules with energies equal to or in excess of the activation energy. Let's assume an activation energy of 50 kJ mol , or, equivalently, J mol . The value of the gas constant, R, is 8.31 J K mol . At 20 °C (293 K) the value of the fraction is: Raising the temperature (to 303 K) increases the quantity: You can see that the fraction of the molecules able to react has almost doubled by increasing the temperature by 10°C. The rate of reaction is nearly doubled. A catalyst provides a reaction route with a lower activation energy. Suppose that the catalyzed activation energy is to 25 kJ mol . The calculation is repeated at 293 K: Compared with the corresponding value for an activation energy of 50 kJ mol , there is a significant increase in the fraction of molecules able to react. There are almost 30,000 times more molecules which can react in the presence of the catalyst compared to having no catalyst (using our assumptions about the activation energies). These calculations can also be done in reverse: given the rate of reaction or rate constants at several temperatures, the activation energy can be calculated. Jim Clark ( ) | 2,499 | 26 |
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An especially valuable group of intermediates can be prepared by addition of an compound to carbon-carbon double or triple bonds: The reaction is called and is a versatile synthesis of organoboron compounds. One example is the addition of diborane, \(\ce{B_2H_6}\), to ethene. Diborane behaves as though it is in equilibrium with \(\ce{BH_3}\) \(\left( \ce{B_2H_6} \rightleftharpoons 2 \ce{BH_3} \right)\), and addition proceeds in three stages: The monoalkylborane, \(\ce{RBH_2}\), and the dialkylborane, \(\ce{R_2BH}\), seldom are isolated because they rapidly add to the alkene. These additions amount to reduction of both carbons of the double bond: Organoboranes can be considered to be organometallic compounds. Elemental boron does not have the properties of a metal, and boron-carbon bonds are more covalent than ionic. However, boron is more electropositive than either carbon or hydrogen and when bonded to carbon behaves like most metals in the sense that bonds are polarized with \(\ce{R}\) negative and boron positive: Hydroboration and the many uses of organoboranes in synthesis were developed largely by H. C. Brown and co-workers. In our discussion, we shall give more detail on hydroboration itself, and then describe several useful transformations of organoboranes. The simplest borane, \(\ce{BH_3}\), exists as the dimer, \(\ce{B_2H_6}\), or in complexed form with certain ethers or sulfides: Any of these \(\ce{BH_3}\) compounds adds readily to most alkenes at room temperature or lower temperatures. The reactions usually are carried out in ether solvents, although hydrocarbon solvents can be used with the borane-dimethyl sulfide complex. When diborane is the reagent, it can be generated either or externally through the reaction of boron trifluoride with sodium borohydride: \[3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{H_4} + 4 \ce{BF_3} \rightarrow 2 \ce{B_2H_6} + 3 \overset{\oplus}{\ce{Na}} \overset{\ominus}{\ce{B}} \ce{F_4}\] Hydroborations have to be carried out with some care, because diborane and many alkylboranes are highly reactive and toxic substances; many are spontaneously flammable in air. With unsymmetrical alkenes, hydroboration occurs so that : These additions are : Furthermore, when there is a choice, addition occurs preferentially from the less crowded side of the double bond: If the alkene is a bulky molecule, borane may add only one or two alkene molecules to give either mono- or dialkylborane, \(\ce{RBH_2}\) or \(\ce{R_2BH}\), respectively, as the following reactions show: These bulky boranes still possess \(\ce{B-H}\) bonds and can add further to a multiple bond, but they are highly selective reagents and add only if the alkene or alkyne is unhindered. This selectivity can be useful, particularly to 1-alkynes, which are difficult to stop at the alkenylborane stage when using diborane: With a bulky dialkylborane, such as di-(1,2-dimethylpropyl)borane, further addition to the alkenylborane does not occur. An especially selective hydroborating reagent is prepared from 1,5-cyclooctadiene and borane. The product is a bicyclic compound of structure \(1\) (often abbreviated as 9-BBN), in which the residual \(\ce{B-H}\) bond adds to unhindered alkenes with much greater selectivity than is observed with other hydroborating reagents. It is also one of the few boranes that reacts sufficiently slowly with oxygen that it can be manipulated in air. An example of the difference in selectivity in the hydroboration of -4-methyl-2-pentene with \(\ce{B_2H_6}\) and \(1\) follows: According to the electronegativity chart (Figure 10-11), the boron-hydrogen bond is polarized in the sense \(\overset{\delta \oplus}{\ce{B}} --- \overset{\delta \ominus}{\ce{H}}\). Therefore the direction of addition of \(\ce{B_2H_6}\) to propene is that expected of a mechanism whereby the electrophilic boron atom becomes bonded to the less-substituted carbon of the double bond. However, there is no firm evidence to suggest that a carbocation intermediate is formed through a stepwise electrophilic addition reaction. For this reason, the reaction often is considered to be a . The stepwise formulation explains why boron becomes attached to the less-substituted carbon, but does not account for the fact that the reactions show no other characteristics of carbocation reactions. This could be because of an expected, extraordinarily fast rate of hydride-ion transfer to the carbocation. A more serious objection to the stepwise mechanism is that alkynes react more rapidly than alkenes, something which normally is not observed for stepwise electrophilic additions (cf. ). Some alkylboranes rearrange at elevated temperatures \(\left( 160^\text{o} \right)\) to form more stable isomers. For example, the alkylborane \(2\), produced by hydroboration of 3-ethyl-2-pentene, rearranges to \(3\) on heating: In general, the boron in alkylboranes prefers to be at the of a hydrocarbon chain so it is bonded to a carbon where steric crowding around boron is least severe. Thus rearrangement tends to proceed in the direction Rearrangement is associated with the fact that hydroboration is reversible at elevated temperatures. This makes possible a sequence of elimination-addition reactions in which boron becomes attached to different carbons and ultimately leads to the most stable product that has boron bonded to the carbon at the end of the chain: Rearrangement of alkylboranes can be used to transform alkenes with double bonds in the middle of the chain into 1-alkenes; for example, \(\ce{RCH=CHCH_3} \rightarrow \ce{RCH_2-CH=CH_2}\). The procedure involves hydroboration of the starting alkene in the usual manner; the borane then is isomerized by heating. An excess of 1-decene (bp \(170^\text{o}\)) then is added to the rearranged borane and the mixture is reheated. Heating causes the alkylborane to dissociate into 1-alkene and \(\ce{HBR_2}\); the 1-decene "scavenges" the \(\ce{HBR_2}\) as it forms, thereby allowing a more volatile 1-alkene (bp \(<170^\text{o}\)) to be removed by simple distillation. Thus, for the rearrangement of 3-ethyl-2-pentene to 3-ethyl-1-pentene, Alkylboranes formed in the hydroboration of alkenes and alkynes seldom are isolated; for the most part they are used as reactive intermediates for the synthesis of other substances. In the reactions of alkylboranes, the \(\ce{B-C}\) bond is cleaved in the sense \(\ce{B}^\oplus - \ce{C}^\ominus\) so that carbon is transferred to other atoms, such as \(\ce{H}\), \(\ce{O}\), \(\ce{N}\), and \(\ce{C}\), its bonding electron pair: In the first of these reactions (Equation 11-2), a hydrocarbon is produced by the cleavage of a borane, \(\ce{R_3B}\), with aqueous acid, or better, with anhydrous propanoic acid, \(\ce{CH_3CH_2CO_2H}\). The overall sequence of hydroboration-acid hydrolysis achieves the reduction of a carbon-carbon multiple bond without using hydrogen and a metal catalyst or diimide (Table 11-3): The second reaction (Equation 11-3) achieves the synthesis of a alcohol by the oxidation of the alkylborane with hydrogen peroxide in basic solution. Starting with a 1-alkene, one can prepare a primary alcohol in two steps: This sequence complements the direct hydration of 1-alkenes, which gives alcohols: Hydroboration of an alkene and subsequent reactions of the product trialkylborane, either with hydrogen peroxide or with acid, appear to be highly stereospecific. For example, 1-methylcyclopentene gives exclusively -2-methylcyclopentanol on hydroboration followed by reaction with alkaline hydrogen peroxide. This indicates that, overall, : Hydroboration of an alkyne followed by treatment of the alkenylborane with basic peroxide provides a method of synthesis of aldehydes and ketones. Thus hydroboration of 1-hexyne and oxidation of the 1-hexenylborane, \(4\), with hydrogen peroxide gives hexanal by way of the enol: If \(4\) is treated with deuteriopropanoic acid, replacement of \(\ce{-BR_2}\) by deuterium occurs with of configuration, forming -hexene-1-\(\ce{D_1}\): The stereospecific oxidation of alkylboranes occurs with hydrogen peroxide by an interesting and important general type of rearrangement which, for these reactions, involves migration of an organic group from boron to oxygen. The first step in the oxidation depends on the fact that tricoordinate boron has only six electrons in its valence shell and therefore behaves as if it were electron-deficient. The first step is bond formation at boron by the strongly nucleophilic peroxide anion (from \(\ce{H_2O_2} + \ce{OH}^\ominus \rightleftharpoons ^\ominus \ce{OOH} + \ce{H_2O}\)) to give a tetracovalent boron intermediate: In the second step, an alkyl group moves from boron to the neighboring oxygen and, in so doing, displaces hydroxide ion. Reaction is completed by hydrolysis of the \(\ce{B-O}\) bond: All three groups on boron are replaced in this manner. The rearrangement step (Equation 11-5) is an example of many related rearrangements in which a group, \(\ce{R}\), migrates with its bonding electrons from one atom to an adjacent atom. We already have encountered an example in the rearrangement of carbocations ( ): The difference between the carbocation rearrangement and the rearrangement of Equation 11-5 is that \(\ce{R}\) migrates from boron to oxygen as \(\ce{HO}^\ominus\) departs in what might be considered an internal \(S_\text{N}2\) reaction. We can generalize this kind of reaction of boron with a substance, \(\ce{X-Y}\), as in Equation 11-6: An example of the use of an \(\ce{X-Y}\) reagent is conversion of alkylboranes to primary amines with hydroxylaminesulfonic acid, \(\ce{H_2NOSO_3H}\) (Equation 11-4). The key steps are attack of the nucleophilic nitrogen at boron, followed by rearrangement, and hydrolysis, and (1977) | 9,840 | 27 |
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