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  4. Ethanamide on treatment with textB textr text2 text/NaOH, gives:

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Q. Ethanamide on treatment with $ \text{B}{{\text{r}}_{\text{2}}}\text{/NaOH,} $ gives:

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Solution:

Amides, on reaction with $ \text{B}{{\text{r}}_{\text{2}}}\text{/NaOH,} $ gives one carbon less amines. This is Hofmann-bromamide reaction. $ C{{H}_{3}}CON{{H}_{2}}\xrightarrow{B{{r}_{2}}/NaOH}C{{H}_{3}}N{{H}_{2}} $