MCQ
$C{H_3}CON{H_2}\xrightarrow{{NaN{O_2}/HCl}}X$
  • $C{H_3}COOH$
  • B
    $C{H_3}CO\mathop N\limits^ + {H_3}C{l^ - }$
  • C
    $C{H_3}N{H_2}$
  • D
    $C{H_3}CHO$

Answer

Correct option: A.
$C{H_3}COOH$
a
(a)Amide, on treating with $HN{O_2},$ give acids.

$C{H_3}CON{H_2}\mathop {\xrightarrow{{NaN{O_2}/HCl}}}\limits_{(HN{O_2})} \mathop {C{H_3}COOH}\limits_{\,}  + {N_2} + {H_2}O$

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