The order of reactivity of the compounds C_6H | vedclass

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(A) The reactivity of alkyl halides in $S_N2$ reactions is primarily governed by steric hindrance. The nucleophile attacks from the side opposite to t

Vedclass · Accepted Answer

$C_6H_5C(CH_3)(C_6H_5)Br < C_6H_5CH(C_6H_5)Br < C_6H_5CH(CH_3)Br < C_6H_5CH_2Br$

Vedclass · Answer

(A) The reactivity of alkyl halides in $S_N2$ reactions is primarily governed by steric hindrance. The nucleophile attacks from the side opposite to the leaving group. As the number of bulky groups attached to the electrophilic carbon increases,the steric hindrance increases,making the $S_N2$ reaction slower.The compounds are:$1$. $C_6H_5CH_2Br$ (Primary,least hindered)$2$. $C_6H_5CH(CH_3)Br$ (Secondary,more hindered than primary)$3$. $C_6H_5CH(C_6H_5)Br$ (Secondary,more hindered than $C_6H_5CH(CH_3)Br$ due to the bulky phenyl group)$4$. $C_6H_5C(CH_3)(C_6H_5)Br$ (Tertiary,most hindered)Thus,the order of reactivity in $S_N2$ is: $C_6H_5C(CH_3)(C_6H_5)Br < C_6H_5CH(C_6H_5)Br < C_6H_5CH(CH_3)Br < C_6H_5CH_2Br$.

The order of reactivity of the compounds $C_6H_5CH_2Br$,$C_6H_5CH(C_6H_5)Br$,$C_6H_5CH(CH_3)Br$ and $C_6H_5C(CH_3)(C_6H_5)Br$ in $S_N2$ reaction is

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