In the following reaction,A and B respectivel | vedclass

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(A) The given reaction sequence is $A$ $\xrightarrow{HBr} C_2H_5Br$ $\xrightarrow{B} A$.Step $1$: $A$ reacts with $HBr$ to form $C_2H_5Br$ (ethyl brom

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$C_2 H_4$ and alcoholic $KOH / \Delta$

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(A) The given reaction sequence is $A$ $\xrightarrow{HBr} C_2H_5Br$ $\xrightarrow{B} A$.Step $1$: $A$ reacts with $HBr$ to form $C_2H_5Br$ (ethyl bromide). This indicates that $A$ is ethene $(C_2H_4)$.Reaction: $CH_2=CH_2   HBr ightarrow CH_3-CH_2Br$.Step $2$: $C_2H_5Br$ reacts with reagent $B$ to regenerate $A$ $(C_2H_4)$.This is a dehydrohalogenation reaction,which requires an alcoholic solution of $KOH$ and heat $(\Delta)$.Reaction: $CH_3-CH_2Br   	ext{alc. } KOH \xrightarrow{\Delta} CH_2=CH_2   KBr   H_2O$.Therefore,$A$ is $C_2H_4$ and $B$ is alcoholic $KOH / \Delta$.

In the following reaction,$A$ and $B$ respectively are
$A \xrightarrow{HBr} C_2H_5Br \xrightarrow{B} A$

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In the following reaction,$A$ and $B$ respectively are$A \xrightarrow{HBr} C_2H_5Br \xrightarrow{B} A$