A dibromide X (C_4H_8Br_2) on dehydrohalogena | vedclass

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(A) $1$. The dibromide $X$ is $2,3-	ext{dibromobutane}$ $(CH_3-CH(Br)-CH(Br)-CH_3)$.$2$. Dehydrohalogenation of $2,3-	ext{dibromobutane}$ removes tw

Vedclass · Accepted Answer

$X = 	ext{2,3-dibromobutane}, Z = 	ext{Na/NH}_3(l)$

Vedclass · Answer

(A) $1$. The dibromide $X$ is $2,3-	ext{dibromobutane}$ $(CH_3-CH(Br)-CH(Br)-CH_3)$.$2$. Dehydrohalogenation of $2,3-	ext{dibromobutane}$ removes two molecules of $HBr$ to form $Y$,which is $2-	ext{butyne}$ $(CH_3-C \equiv C-CH_3)$.$3$. Reduction of $2-	ext{butyne}$ with $Z = 	ext{Na/NH}_3(l)$ (Birch reduction) yields $trans-2-	ext{butene}$.$4$. $trans-2-	ext{butene}$ is a non-polar isomer of $C_4H_8$ because its dipole moment $\mu = 0$ due to symmetry.$5$. Therefore,$X$ is $2,3-	ext{dibromobutane}$ and $Z$ is $	ext{Na/NH}_3(l)$.

$A$ dibromide $X$ $(C_4H_8Br_2)$ on dehydrohalogenation gave $Y$,which on reduction with $Z$ gave a non-polar isomer of $C_4H_8$. What are $X$ and $Z$ respectively?

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