Question

Select the member of each pair that shows the greater rate of \(S_{N} 2\) reaction with \(\mathrm{KN}_{3}\) in acetone.

Short answer

Answer: In Pair 1, CH3CH2Br (a) has a greater rate of SN2 reaction with KN3 in acetone. In Pair 2, CH3CH2I (a) has a greater rate of SN2 reaction with KN3 in acetone.

Step-by-step solution

Identify the given pairs

The first thing to do is to identify the given pairs. However, the exercise didn't provide any specific pair. To provide an example, let's consider two pairs: Pair 1: (a) CH3CH2Br (b) (CH3)3CBr Pair 2: (a) CH3CH2I (b) CH3CH2F Now we will compare the SN2 reactivity of these pairs with KN3 in acetone.

Compare the leaving groups and steric hindrance

For Pair 1: (a) CH3CH2Br: has a primary carbon (less steric hindrance) and bromine as the leaving group. (b) (CH3)3CBr: has a tertiary carbon (more steric hindrance) and bromine as the leaving group. SN2 reactions are favored by less steric hindrance, so CH3CH2Br will have a greater rate of the SN2 reaction with KN3 in acetone than (CH3)3CBr. For Pair 2: (a) CH3CH2I: has a primary carbon (less steric hindrance) and iodine as the leaving group. (b) CH3CH2F: has a primary carbon (less steric hindrance) and fluorine as the leaving group. Both molecules have similar steric hindrance due to primary carbon. However, iodine is a better leaving group compared to fluorine (as it is a weaker base). Hence, CH3CH2I will have a greater rate of SN2 reaction than CH3CH2F with KN3 in acetone.

Conclude the rate of SN2 reaction

Based on the comparison of leaving groups and steric hindrance: Pair 1: CH3CH2Br (a) would have a greater rate of SN2 reaction with KN3 in acetone than (CH3)3CBr (b). Pair 2: CH3CH2I (a) would have a greater rate of SN2 reaction with KN3 in acetone than CH3CH2F (b).