Final product of the reactionCH_3-CH=CH_2 xri | vedclass

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(A) Step $1$: $CH_3-CH=CH_2 + Br_2 \xrightarrow{CCl_4} CH_3-CH(Br)-CH_2Br$ $(A)$Step $2$: $CH_3-CH(Br)-CH_2Br + 3NaNH_2 \rightarrow CH_3-C \equiv C^-N

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$CH_3-CH_2-CH=CH_2$

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(A) Step $1$: $CH_3-CH=CH_2 + Br_2 \xrightarrow{CCl_4} CH_3-CH(Br)-CH_2Br$ $(A)$Step $2$: $CH_3-CH(Br)-CH_2Br + 3NaNH_2 \rightarrow CH_3-C \equiv C^-Na^+ + 2NaBr + 2NH_3$. Upon workup,this gives $CH_3-C \equiv CH$ $(B)$Step $3$: $CH_3-C \equiv CH + NaNH_2 \rightarrow CH_3-C \equiv C^-Na^+$. Then $CH_3-C \equiv C^-Na^+ + CH_3-Br \rightarrow CH_3-C \equiv C-CH_3$ $(C)$Step $4$: $CH_3-C \equiv C-CH_3 + H_2 \xrightarrow{Pd-BaSO_4} cis-CH_3-CH=CH-CH_3$ $(D)$The final product $D$ is $cis-but-2-ene$.

Final product of the reaction
$CH_3-CH=CH_2$ $\xrightarrow[CCl_4]{Br_2} A$ $\xrightarrow[(3 \ mol)]{NaNH_2} B$ $\xrightarrow{CH_3-Br} C$ $\xrightarrow[Pd-BaSO_4]{H_2} D$

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Final product of the reaction$CH_3-CH=CH_2$ $\xrightarrow[CCl_4]{Br_2} A$ $\xrightarrow[(3 \ mol)]{NaNH_2} B$ $\xrightarrow{CH_3-Br} C$ $\xrightarrow[Pd-BaSO_4]{H_2} D$