Question

In the following reaction, $A$ and $B$, respectively are
$A \ce{->[{H B r}]}C_{2} H_{5} B r \ce{->[{B}]}A$

• A. $C _{2} H _{4}$, alc. $KOH/\Delta$
• B. $C _{2} H _{5} Cl$, aq. $KOH / \Delta$
• C. $CH _{3} OH$, aq. $KOH / \Delta$
• D. $C _{2} H _{2}, PBr _{3}$

Answer 1

Answer: (A)

$H _{2} C = CH _{2} \ce{->[{ HBr }]}CH _{3} CH _{2} Br\ce{->[\text {Alc. KOH (B)}]} H _{2} C = CH _{2}$
Hence, $A$ and $B$ are respectively $C _{2} H _{4}$ and alc. $KOH$.
Alc. $KOH$ causes the dehydrohalogenation (ie, removal of hydrogen and halogen) reactions.