X→AlCl3HClY→KOHaqC2H5OH.In the above reaction is

$X \to_{{AlCl}_{3}}^{HCl} Y \overset{KOH (aq)}{\to} C_{2} H_{5} OH .$ In the above reaction is

A
$C_{2} H_{5} Cl$
B
$C_{2} H_{2}$
C
$C_{2} H_{4}$
D
$C_{2} H_{5} Br$

Solution:

${CH}_{2} = C H_{2} \overset{HCl / {AlCl}_{3}}{\to} {CH}_{3} {CH}_{2} Cl \to_{C_{2} H_{5} OH}^{{KOH}_{aq}} {CH}_{3} {CH}_{2} OH$

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