MCQ
Hofmann degradation is given by:
- ✓Imide
- BAcid chloride
- CAcid anhydride
- DNone of these
$\mathrm{CH}_{3} \mathrm{CONH}_{2}+\mathrm{Br}_{2}+4 \mathrm{KOH} \rightarrow \mathrm{CH}_{3} \mathrm{NH}_{2}+\mathrm{K}_{2} \mathrm{CO}_{3}+2 \mathrm{KBr}+2 \mathrm{H}_{2} \mathrm{O}$
This is Hoffmann bromamide degradation reaction. The amine formed contains one carbon
atom less than amide.
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$1-$ Bromopropane is reacted with reagents in List $I$ to give product in List $II$
| List$I-$ Reagent | List$II-$Product |
| $A$ $KOH (\text { alc) }$ | $I$ Nitrile |
| $B$ $KCN \text { (alc) }$ | $II$ Ester |
| $C$ $AgNO _2$ | $III$ Alkene |
| $D$ $H _3 CCOOAg$ | $IV$ Nitroalkane |


