CH_3-C equiv C-CH_3 , xrightarrow[{(2) Br_2}] | vedclass

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(C) $1$. The reaction of $CH_3-C \equiv C-CH_3$ (but$-2-$yne) with $H_2/Pd/CaCO_3$ (Lindlar's catalyst) is a syn-addition of hydrogen to the triple bo

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$(d, l)-2,3-	ext{Dibromobutane}$

Vedclass · Answer

(C) $1$. The reaction of $CH_3-C \equiv C-CH_3$ (but$-2-$yne) with $H_2/Pd/CaCO_3$ (Lindlar's catalyst) is a syn-addition of hydrogen to the triple bond,which yields $cis-	ext{but-2-ene}$.$2$. The subsequent reaction with $Br_2$ is an anti-addition of bromine to the $cis-	ext{alkene}$.$3$. Anti-addition of $Br_2$ to a $cis-	ext{alkene}$ results in the formation of a racemic mixture of $(d, l)-2,3-	ext{dibromobutane}$.

$CH_3-C \equiv C-CH_3 \, \xrightarrow[{(2) Br_2}]{{(1) H_2 /Pd/CaCO_3}} \, (X) \, , \, (X)$ is

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