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Question

(B) The reaction of $3	ext{-methylbutan-2-ol}$ with $HBr$ proceeds via an $S_N1$ mechanism.$1$. Protonation of the alcohol group followed by the loss

Vedclass · Accepted Answer

$CH_{3}-C(Br)(CH_{3})-CH_{2}-CH_{3}$

Vedclass · Answer

(B) The reaction of $3	ext{-methylbutan-2-ol}$ with $HBr$ proceeds via an $S_N1$ mechanism.$1$. Protonation of the alcohol group followed by the loss of a water molecule generates a secondary carbocation: $CH_{3}-CH(CH_{3})-CH^+-CH_{3}$.$2$. This secondary carbocation undergoes a $1,2	ext{-hydride shift}$ to form a more stable tertiary carbocation: $CH_{3}-C^+(CH_{3})-CH_{2}-CH_{3}$.$3$. Finally,the bromide ion $(Br^-)$ attacks the tertiary carbocation to form the major product: $2	ext{-bromo-2-methylbutane}$,which is $CH_{3}-C(Br)(CH_{3})-CH_{2}-CH_{3}$.

Consider the following reaction and identify the product $(P)$.

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