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(A) The reaction of an alkene with $HBr$ proceeds via an electrophilic addition mechanism involving a carbocation intermediate.Step $1$: Protonation o

Vedclass · Accepted Answer

$CH_3CH_2CH_2C(Br)(CH_3)_2$

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(A) The reaction of an alkene with $HBr$ proceeds via an electrophilic addition mechanism involving a carbocation intermediate.Step $1$: Protonation of the double bond can form two possible secondary carbocations:$(I)$ $CH_3-CH_2-CH^{+}-CH(CH_3)_2$$(II)$ $CH_3-CH^{+}-CH_2-CH(CH_3)_2$Step $2$: Carbocation $(I)$ undergoes a $1,2-$hydride shift from the adjacent tertiary carbon to form a much more stable tertiary carbocation:$CH_3-CH_2-CH^{+}-CH(CH_3)_2 \xrightarrow{1,2-H 	ext{ shift}} CH_3-CH_2-CH_2-C^{+}(CH_3)_2$Step $3$: The bromide ion $(Br^{-})$ attacks this highly stable tertiary carbocation to form the major product:$CH_3-CH_2-CH_2-C^{+}(CH_3)_2 + Br^{-} \longrightarrow CH_3-CH_2-CH_2-C(Br)(CH_3)_2$Thus,the major product is $2-$bromo$-2-$methylpentane.

The major product formed in the following reaction is :
$CH_3CH=CHCH(CH_3)_2 \xrightarrow{HBr}$

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The major product formed in the following reaction is :$CH_3CH=CHCH(CH_3)_2 \xrightarrow{HBr}$