The correct statement about C and D is:CH_3-C | vedclass

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(B) $1$. In the first reaction,$CH_3-C \equiv C-CH_3$ undergoes syn-addition of $H_2$ with $Pd-BaSO_4$ (Lindlar's catalyst) to form cis-but$-2-$ene $(

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$C$ is racemic mixture & $D$ is meso

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(B) $1$. In the first reaction,$CH_3-C \equiv C-CH_3$ undergoes syn-addition of $H_2$ with $Pd-BaSO_4$ (Lindlar's catalyst) to form cis-but$-2-$ene $(A)$.$2$. The subsequent anti-addition of $Br_2$ to cis-but$-2-$ene yields a racemic mixture of $2,3-$dibromobutane $(C)$.$3$. In the second reaction,$CH_3-C \equiv C-CH_3$ undergoes anti-addition of $H_2$ with $Na/Liq. NH_3$ (Birch reduction) to form trans-but$-2-$ene $(B)$.$4$. The subsequent anti-addition of $Br_2/H_2O$ (halohydrin formation) to trans-but$-2-$ene yields a meso compound $(D)$.

The correct statement about $C$ and $D$ is:
$CH_3-C \equiv C-CH_3$ $\xrightarrow{H_2/Pd-BaSO_4} A$ $\xrightarrow{Br_2/CCl_4} C$
$CH_3-C \equiv C-CH_3$ $\xrightarrow{Na/Liq. NH_3} B$ $\xrightarrow{Br_2/H_2O} D$

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The correct statement about $C$ and $D$ is:$CH_3-C \equiv C-CH_3$ $\xrightarrow{H_2/Pd-BaSO_4} A$ $\xrightarrow{Br_2/CCl_4} C$$CH_3-C \equiv C-CH_3$ $\xrightarrow{Na/Liq. NH_3} B$ $\xrightarrow{Br_2/H_2O} D$