JEE Advanced 2021 Paper 2 Chemistry: Paragraph 1

The compound $\mathrm{C_8H_8O}$ giving a 2,4-DNP precipitate and the iodoform test, but no Tollens' test, is the methyl ketone acetophenone, $\mathrm{C_6H_5COCH_3}$. The iodoform reaction is $\mathrm{C_6H_5COCH_3+3I_2+4NaOH\to C_6H_5COONa+CHI_3+3NaI+3H_2O}$; the sodium benzoate is acidified to benzoic acid $\mathrm{B}$. One mole of ketone gives exactly one mole of $\mathrm{CHI_3}$ (the yellow precipitate), molar mass $394\ \mathrm{g\,mol^{-1}}$. The number of moles of $\mathrm{A}$ is $6.0/120=0.050$, so the mass of iodoform is $0.050\times394=19.7$ g. Option (B) doubles the answer by assuming two moles of product, (C) halves it, and (D) triples it by assuming three moles of iodoform (confusing it with the three moles of $\mathrm{I_2}$ used).

The amide $\mathrm{C}$ is benzamide, $\mathrm{C_6H_5CONH_2}$. In the Hofmann degradation the amide is N-brominated, deprotonated, and rearranges through an isocyanate, with migration of the phenyl group to nitrogen. The isocyanate is then hydrolysed to the carbamate and loses carbonate to give aniline $\mathrm{D}$. The overall balanced equation is $\mathrm{RCONH_2+Br_2+4KOH\to RNH_2+K_2CO_3+2KBr+2H_2O}$. Counting charges and atoms: potassium $4=2+2$, oxygen $1+4=3+2$, hydrogen $2+4=2+4$. Thus four moles of KOH are consumed. Two moles are used to neutralise the HBr formed and the other two moles convert the liberated $\mathrm{CO_2}$ into carbonate, while $\mathrm{NaNO_2/HCl}$ and 2-naphthol then give the azo dye in later steps.