Consider the given reaction: CH_3-CH=CH-CH_2- | vedclass

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(B) The reaction proceeds via an $S_N1$ mechanism.First,the hydroxyl group is protonated by $HBr$ and leaves as water to form a resonance-stabilized a

Vedclass · Accepted Answer

$CH_3-CH(Br)-CH=CH_2$

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(B) The reaction proceeds via an $S_N1$ mechanism.First,the hydroxyl group is protonated by $HBr$ and leaves as water to form a resonance-stabilized allylic carbocation: $[CH_3-CH=CH-CH_2^+ \leftrightarrow CH_3-CH^+-CH=CH_2]$.The nucleophile $Br^-$ then attacks the more stable secondary carbocation site,leading to the formation of $CH_3-CH(Br)-CH=CH_2$ as the major product $[P]$.

Consider the given reaction: $CH_3-CH=CH-CH_2-OH \xrightarrow[S_N1]{HBr} [P]$. In the given reaction,the product $[P]$ is:

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