CH_3-CH(CH_3)-CH=CH_2 + HBr to X (major produ | vedclass

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(B) The reaction proceeds via electrophilic addition involving carbocation formation and rearrangement. $1.$ Protonation of $CH_3-CH(CH_3)-CH=CH_2$ oc

Vedclass · Accepted Answer

$CH_3-C(Br)(CH_3)-CH_2-CH_3$

Vedclass · Answer

(B) The reaction proceeds via electrophilic addition involving carbocation formation and rearrangement. $1.$ Protonation of $CH_3-CH(CH_3)-CH=CH_2$ occurs at the terminal carbon to form a secondary $(2^\circ)$ carbocation: $CH_3-CH(CH_3)-CH^+-CH_3$. $2.$ $A$ $1,2$-hydride shift occurs from the adjacent tertiary carbon to form a more stable tertiary $(3^\circ)$ carbocation: $CH_3-C^+(CH_3)-CH_2-CH_3$. $3.$ The nucleophile $Br^-$ then attacks the $3^\circ$ carbocation to yield the major product: $CH_3-C(Br)(CH_3)-CH_2-CH_3$ ($2$-bromo-$2$-methylbutane).

$CH_3-CH(CH_3)-CH=CH_2 + HBr \to X$ (major product). $X$ is:

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