The major product of the following reaction i | vedclass

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Question

(A) The reaction proceeds in two steps as $HBr$ is in excess:$1$. The $-OH$ group is substituted by $-Br$ to form allyl bromide: $CH_{2}=CH-CH_{2}-OH

Vedclass · Accepted Answer

$CH_{3}-CHBr-CH_{2}Br$

Vedclass · Answer

(A) The reaction proceeds in two steps as $HBr$ is in excess:$1$. The $-OH$ group is substituted by $-Br$ to form allyl bromide: $CH_{2}=CH-CH_{2}-OH + HBr ightarrow CH_{2}=CH-CH_{2}Br + H_{2}O$.$2$. The second equivalent of $HBr$ adds across the double bond following Markownikoff's rule to form $1,2-	ext{dibromopropane}$: $CH_{2}=CH-CH_{2}Br + HBr ightarrow CH_{3}-CH(Br)-CH_{2}Br$.

The major product of the following reaction is $CH_{2}=CH-CH_{2}-OH \xrightarrow[\text{Excess}]{HBr} \text{Product}$

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