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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