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.
(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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