Question

In the following reaction, A and B are:

$A\stackrel{HBr}{\to }{C}_2{H}_{5}Br\stackrel{B}{\to }A$

• A. $C_2{H}_4$ and alcoholic KOH/ $△$
• B. $C_2{H}_{5}Cl$ and aqueous KOH /$△$
• C. $C_2{H}_{5}OH$ and aq KOH/ $△$
• D. $C_2{H}_2$ and ${Br}_2$

Answer 1

Answer: (A)

$C_2{H}_4$ and alcoholic KOH/ $△$

Reaction of process:
$C_2{H}_4+HBr\to C_2{H}_{5}Br$

and

$C_2{H}_{5}Br\stackrel{alc.KOH/△}{\to }{C}_2{H}_4$

In first reaction A i.e, ethene undergoes hydrohalogenation to form ethyl bromide then in presence of alc. KOH it undergoes $\beta$ elimination reaction to form ethene again.