Question

The order of reactivity of the compounds $C _{6} H _{5} CH _{2} Br , C _{6} H _{5} CH \left( C _{6} H _{5}\right) Br , C _{6} H _{5} CH \left( CH _{3}\right) Br$ and $C _{6} H _{5} C \left( CH _{3}\right)\left( C _{6} H _{5}\right) Br \text { in } S _{ N }^{2}$ reaction is

• A. $C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{CH_3}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C}H_3}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{H}}{\overset{\underset{|}{H}}{C}} - Br$
• B. $C_6H_5 - \underset{\overset{|}{H}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{CH_3}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{CH_3}}{C}} - Br$
• C. $C_6H_5 - \underset{\overset{|}{CH_3}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{H}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{CH_3}}{C}} - Br$
• D. $C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{H}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{CH_3}}{\overset{\underset{|}{H}}{C}} - Br < C_6H_5 - \underset{\overset{|}{C_6H_5}}{\overset{\underset{|}{CH_3}}{C}} - Br$

Answer 1

Answer: (A)

Lesser the steric hindrance, easier and higher the rate of reaction towards $SN ^{2}$