MCQ
Reaction $C{H_3}CON{H_2}\xrightarrow{{NaOBr}}$ gives
- A$C{H_3}Br$
- B$C{H_4}$
- C$C{H_3}COBr$
- ✓$C{H_3}N{H_2}$
✓
Answer
Correct option: D.
$C{H_3}N{H_2}$
d
(d).$C{H_3}CON{H_2}\xrightarrow{{NaOBr}}C{H_3}N{H_2}$
(d).$C{H_3}CON{H_2}\xrightarrow{{NaOBr}}C{H_3}N{H_2}$
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