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172 CHAPTER 6 
 
stable, tertiary allylic carbocation, which is resonance 
stabilized, as shown below. 
 
 
 
(f) This carbocation is secondary, and it can rearrange 
via a hydride shift (shown below) to give a more stable, 
tertiary carbocation: 
 
 
 
(g) This carbocation is tertiary and will not rearrange. 
 
6.42. 
(a) The C–Br bond is broken, indicating the loss of a 
leaving group (Brˉ), while the C–O bond is formed, 
indicating a nucleophilic attack. This is, in fact, a 
concerted process in which nucleophilic attack and loss 
of the leaving group occur in a simultaneous fashion. 
One curved arrow is required to show the nucleophilic 
attack, and another curved arrow is required to show loss 
of the leaving group: 
 
 
(b) We identify the bond broken (CH3CH2—Br), and the 
bond formed (CH3CH2—OH). Using the data in Table 
6.1, ΔH for this reaction is expected to be approximately 
(285 kJ/mol) – (381 kJ/mol). The sign of ΔH is therefore 
predicted to be negative, which means that the reaction 
should be exothermic. 
(c) Two chemical entities are converted into two 
chemical entities. Both the reactants and products are 
acyclic. Therefore, ΔS for this process is expected to be 
approximately zero. 
(d) ΔG has two components: (ΔH) and (–TΔS). Based 
on the answers to the previous questions, the first term 
has a negative value and the second term is insignificant. 
Therefore, ΔG is expected to have a negative value. This 
is confirmed by the energy diagram, which shows the 
products having lower free energy than the reactants. 
(e) The position of equilibrium is dependent on the sign 
and value of ΔG. As mentioned in part e, ΔG is 
comprised of two terms. The effect of temperature 
appears in the second term (–TΔS), which is insignificant 
because ΔS is approximately zero. Therefore, an 
increase (or decrease) in temperature is not expected to 
have a significant impact on the position of equilibrium. 
(f) This transition state corresponds with the peak of the 
curve, and has the following structure: 
 
 
 
(g) The transition state in this case is closer in energy to 
the reactants than the products, and therefore, it is closer 
in structure to the reactants than the products (the 
Hammond postulate). 
(h) If we inspect the rate equation, we see that the sum of 
the exponents is two, so this reaction is second order. 
(i) According to the rate equation, the rate is linearly 
dependent on the concentration of hydroxide. Therefore, 
the rate will be doubled if the concentration of hydroxide 
is doubled. 
(j) At higher temperature, more molecules will have the 
requisite energy of activation necessary for the reaction 
to occur, so the rate will increase with increasing 
temperature. 
 
6.43. 
(a) Keq does not affect the rate of the reaction. It only 
affects the equilibrium concentrations. 
(b) ΔG does not affect the rate of the reaction. It only 
affects the equilibrium concentrations. 
(c) Temperature affects the rate of the reaction, by 
increasing the number of collisions that result in a 
reaction. 
(d) ΔH does not affect the rate of the reaction. It only 
affects the equilibrium concentrations. 
(e) Ea greatly affects the rate of the reaction. Lowering 
the Ea will increase the rate of reaction. 
(f) ΔS does not affect the rate of the reaction. It only 
affects the equilibrium concentrations. 
 
6.44. In order to determine if reactants or products are 
favored at high temperature, we must consider the effect 
of temperature on the sign of ΔG. Recall that ΔG has 
two components: (ΔH) and (–TΔS). The reaction is 
exothermic, so the first term (ΔH) has a negative value, 
which contributes to a negative value of ΔG. This favors 
products. At low temperature, the second term will be 
insignificant and the first term will dominate. Therefore, 
the process will be thermodynamically favorable, and the 
reaction will favor the formation of products. However, 
at high temperature, the second term becomes more 
significant. In this case, two moles of reactants are 
converted into one mole of product. Therefore, ΔS for 
this process is negative, which means that (–TΔS) is 
positive. At high enough temperature, the second term 
(–TΔS) should dominate over the first term (ΔH), 
generating a positive value for ΔG. Therefore, the 
reaction will favor reactants at high temperature. 
 
 
 
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