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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$\left[\right.$ Given $\left.: \mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.86\, \mathrm{~K}\, \mathrm{~kg}\, \mathrm{~mol}^{-1}\right]$

Compound $(C)$ can show geometrical isomerism. Product $(E)$ of the reaction will be