Aniline is treated with (i) NaNO2 / dil. HCl at 273-278 K, then (ii) SnCl2/HCl to give X. X is then warmed with propanone (acetone) to give Y. Y is

propanone phenylhydrazone, C6H5NHN=C(CH3)2

propanone azine, (CH3)2C=N-N=C(CH3)2

N-isopropylaniline, C6H5NHCH(CH3)2

propanone phenylhydrazone, C6H5NHN=C(CH3)2

JEE Advanced levelSingle correct optionDifficulty 3 of 3nc071

Aniline is treated with (i) $NaNO X_{2}$ / dil. $HCl$ at 273-278 K, then (ii) $SnCl X_{2} / HCl$ to give X. X is then warmed with propanone (acetone) to give Y. Y is

SnCl X_{2} / HCl

reduces the diazonium salt to phenylhydrazine,

C X_{6} H X_{5} NHNH X_{2}

(X). Its terminal

NH X_{2}

condenses with the carbonyl group of propanone, losing water, to give the hydrazone

C X_{6} H X_{5} NHN = C (CH X_{3}) X_{2}

. The anil would need aniline itself, and the azine would need hydrazine.