Chemistry - Carboxylic acid and their Derivatives Question with Solution | TestHub

Propanal -> propanoic acid ($\mathrm{Cr}{\mathrm{O}}_3$) -> propanoyl chloride ($\mathrm{SOC}{\mathrm{l}}_2$) -> propanamide ($\mathrm{N}{\mathrm{H}}_3$) -> ethylamine ($\mathrm{B}{\mathrm{r}}_2$/$\mathrm{KOH}$, Hofmann degradation). One carbon is lost at the final step. B.