The reaction sequence is given as:CH_3-CH=CH- | vedclass

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(B) $1$. The reaction of $CH_3-CH=CH-CH_3$ with $Br_2$ in $CCl_4$ results in the formation of a vicinal dihalide,$CH_3-CH(Br)-CH(Br)-CH_3$ ($2$,$3$-di

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$CH_3 - C \equiv C - CH_3$

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(B) $1$. The reaction of $CH_3-CH=CH-CH_3$ with $Br_2$ in $CCl_4$ results in the formation of a vicinal dihalide,$CH_3-CH(Br)-CH(Br)-CH_3$ ($2$,$3$-dibromobutane).$2$. Treatment with $alc. KOH$ causes the first dehydrohalogenation (elimination of $HBr$) to form a vinyl bromide,$CH_3-CH=C(Br)-CH_3$.$3$. Subsequent treatment with a stronger base,$NaNH_2$,causes the second dehydrohalogenation to form the alkyne,$CH_3-C \equiv C-CH_3$ (but$-2-$yne).$4$. Thus,the major product $(A)$ is $CH_3-C \equiv C-CH_3$.

The reaction sequence is given as:
$CH_3-CH=CH-CH_3$ $\xrightarrow{Br_2/CCl_4} \text{Intermediate}$ $\xrightarrow[(ii) NaNH_2]{(i) alc. KOH} (A)$
Product $(A)$ is:

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The reaction sequence is given as:$CH_3-CH=CH-CH_3$ $\xrightarrow{Br_2/CCl_4} \text{Intermediate}$ $\xrightarrow[(ii) NaNH_2]{(i) alc. KOH} (A)$Product $(A)$ is: