- ✓Acetic acid
- BAcetaldehyde
- CMethylamine
- DFormic acid
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In this reaction, $\mathrm{RCONHBr}$ is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation reaction is an intramolecular reaction.
$1.$ How can the conversion of $(i)$ to $(ii)$ be brought about?
$(A)$ $\mathrm{KBr}$ $(B)$ $\mathrm{KBr}+\mathrm{CH}_3 \mathrm{ONa}$ $(C)$ $\mathrm{KBr}+\mathrm{KOH}$ $(D)$ $\mathrm{Br}_2+\mathrm{KOH}$
$2.$ Which is the rate determining step in Hofmann bromamide degradation?
$(A)$ Formation of $(i)$ $(B)$ Formation of $(ii)$ $(C)$ Formation of $(iii)$ $(D)$ Formation of $(iv)$
$3.$ What are the constituent amines formed when the mixture of $(i)$ and $(ii)$ undergoes Hofmann bromamide degradation?
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give the answer question $1$, $2$, and $3.$
| Column $I$ | Column $II$ | ||
| $(A)$ | ${\left( {C{H_3}} \right)_3}CH + 2{O_2}\xrightarrow[{\left[ O \right]}]{{KMn{O_4}}}$ | $(i)$ | $CH_3COOH+H_2O$ |
| $(B)$ | $2C{H_4} + {O_2}\xrightarrow[{100\,atm}]{{Cu/523\,\,K}}$ | $(ii)$ | $(CH_3)_3COH$ |
| $(C)$ | $C{H_4} + {O_2}\xrightarrow[\Delta ]{{M{o_2}{O_3}}}$ | $(iii)$ | $2CH_3OH$ |
| $(D)$ | $C{H_3} - C{H_3} + \frac{3}{2}{O_2}\xrightarrow{{{{\left( {C{H_3}COO} \right)}_2}Mn}}$ | $(iv)$ | $HCHO + H_2O$ |