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

The final product is isopropylbenzene, formed by the deamination of D ($(NaN{O}_2$/HCl), followed by reaction with $H_{3}P{O}_2$. This indicates D is an isopropylaniline.

Compound D is produced from C via Hofmann bromamide degradation ($B{r}_2$/NaOH), which converts an amide ($-CON{H}_2$) to an amine ($-N{H}_2$). Thus, C is p-isopropylbenzamide.

Working further back, C is formed from B ($N{H}_3$,Δ), meaning B is p-isopropylbenzoic acid.

B was formed by the $KMn{O}_4$ oxidation of A, where a second alkyl group (like a methyl group in p-cymene) was oxidized to a carboxyl group.

Comparing this to the options, structure (C) is the correct representation of the amide p-isopropylbenzamide.