Question

Predict the order of reactivity of the following compounds in S_N¹ andS_N² reactions.
i. The four isomeric bromobutanes.
ii. C₆H₅CH₂Br, C₆H₅CH(C₆H₅)Br, C₆H₅CH(CH₃)Br, C₆H₅C(CH₃)(C₆H₅)Br

Answer 1

i. CH₃CH₂CH₂CH₂Br < (CH₃)₂CHCH₂Br < CH₃CH₂CH(Br)CH₃ < (CH₃)₃CBr (S_N¹ ) CH₃CH₂CH₂CH₂Br > (CH₃)₂CHCH₂Br > CH₃CH₂CH(Br)CH₃ > (CH₃)₃CBr (S_N² )

stable than derived from CH₃CH₂CH₂CH²Br because of greater electron donating inductive effect of (CH₃)₂CH – group. Therefore, (CH₃)₂CHCH₂Br is more reactive than CH₃CH₂CH₂CH₂Br in S_N1 reactions. CH₃CH₂CH(Br)CH₃ is a secondary bromide and (CH₃)₃CBr is a tertiary bromide. Hence, the above order is followed in S_N1 . The reactivity in S_N2 reactions follows the reverse order as the steric hindrance around the electrophilic carbon increases in that order.

ii. C₆H₅C(CH₃)(C₆H₅)Br > C₆H₅CH(C₆H₅)Br > C₆H₅CH(CH₃)Br > C₆H₅CH₂Br (S_N1 ) C₆ H₅C(CH₃)(C₆H₅)Br < C₆H₅CH(C₆H₅)Br < C₆H₅CH(CH₃)Br < C₆H₅CH₂Br (SN₂ ) Of the two secondary bromides, the carbocation intermediate obtained from C₆H₅CH(C₆H₅)Br is more stable than obtained from C₆H₅CH(CH₃)Br because it is stabilised by two phenyl groups due to resonance. Therefore, the former bromide is more reactive than the latter in S_N1 reactions. A phenyl group is bulkier than a methyl group. Therefore, C₆H₅CH(C₆H₅)Br is less reactive than C₆H₅CH(CH₃)Br in S_N2 reaction.