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Question

Predict the order of reactivity of the four isomeric bromobutanes in $$ S_{N}1 $$ and $$ S_{N}2 $$ reactions.


Solution

$$ CH_{3}CH_{2}CH_{2}CH_{2}Br< (CH_{3})_{2}CHCH_{2}Br< CH_{3}CH_{2}CH(Br)CH_{3}< (CH_{3})_{3}CBr \ \ \ \  (S_{N}1) $$

$$ CH_{3}CH_{2}CH_{2}CH_{2}Br< (CH_{3})_{2}CHCH_{2}Br< CH_{3}CH_{2}CH(Br)CH_{3}> (CH_{3})_{3}CBr \ \ \ \  (S_{N}2) $$


Of the two primary bromides, the carbocation intermediate derived from $$ (CH_{3})_{2}CHCH_{2}Br $$ is more stable than that derived from $$ CH_{3}CH_{2}CH_{2}CH_{2}Br $$ because of greater electron donating inductive effect of $$ (CH_{3})_{2}CH $$ group. So, $$ (CH_{3})_{2}CHCH_{2}Br $$ is more reactive than $$ CH_{3}CH_{2}CH_{2}CH_{2}Br $$ in $$ S_{N}1 $$ reactions.$$ CH_{3}CH_{2}CH(Br)CH_{3} $$ is a secondary bromide and in $$ S_{N}2 $$ reactions follows the reverse order as the steric hindrance around the electrophilic carbon increases in that order.

Chemistry

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