Major product 'B' of the following reaction s | vedclass

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(B) The reaction of $2$-methylbut-$2$-ene with $Br_2$ in the presence of $CH_3OH$ (a polar protic solvent) proceeds via the formation of a bromonium i

Vedclass · Accepted Answer

$CH_3-C(I)(CH_3)-CH(Br)-CH_3$

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(B) The reaction of $2$-methylbut-$2$-ene with $Br_2$ in the presence of $CH_3OH$ (a polar protic solvent) proceeds via the formation of a bromonium ion intermediate. $CH_3OH$ acts as a nucleophile and attacks the more substituted carbon atom of the bromonium ion to form an ether,$A$ ($3$-bromo-$2$-methoxy-$2$-methylbutane). Reaction: $CH_3-C(CH_3)=CH-CH_3 + Br_2 + CH_3OH \rightarrow CH_3-C(OCH_3)(CH_3)-CH(Br)-CH_3$ $(A)$. When $A$ reacts with $HI$,the methoxy group $(-OCH_3)$ is protonated by $H^+$ to form an oxonium ion,which is a good leaving group. $I^-$ then attacks the tertiary carbon atom,displacing $CH_3OH$ to form the final product $B$ ($3$-bromo-$2$-iodo-$2$-methylbutane). Therefore,the major product $B$ is $CH_3-C(I)(CH_3)-CH(Br)-CH_3$.

Major product '$B$' of the following reaction sequence is.

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