JEE Advanced 2017 Paper 1 Chemistry: Paragraph 1

First check that each option is a genuine transformation, since each one is. (I) Acetophenone with excess $\mathrm{I_2/NaOH}$ undergoes the haloform reaction, giving benzoic acid on acidification. (II) Benzaldehyde has no $\alpha$-hydrogen, so conc. alkali brings about the Cannizzaro reaction, giving benzyl alcohol and benzoate. (III) Cyclohexanone oxime in conc. $\mathrm{H_2SO_4}$ undergoes Beckmann rearrangement to $\varepsilon$-caprolactam. (IV) Phenylacetylene is hydrated by $\mathrm{Hg^{2+}/H^+}$ (Markovnikov), giving the enol that tautomerises to acetophenone. Molar masses: benzoic acid $\mathrm{C_7H_6O_2}=122$; benzyl alcohol $\mathrm{C_7H_8O}=108$; caprolactam $\mathrm{C_6H_{11}NO}=72+11+14+16=113$; acetophenone $\mathrm{C_8H_8O}=120$. Only (III), (iii), (R) fits $113$, so option C is correct.

Disproportionation means that the same species is oxidised and reduced in one reaction. In the Cannizzaro reaction of benzaldehyde, one molecule transfers a hydride to a second molecule: $2\mathrm{C_6H_5CHO}+\mathrm{NaOH}\to\mathrm{C_6H_5CH_2OH}+\mathrm{C_6H_5COONa}$. The carbonyl carbon goes from $+1$ in the aldehyde to $+3$ in the benzoate (oxidation) and to $-1$ in the alcohol (reduction), so (II), (ii), (Q) is the correct combination. In (I) the haloform reaction oxidises the methyl ketone side chain while iodine is the species reduced, so acetophenone is not both oxidised and reduced. Beckmann rearrangement (III) is a non-redox rearrangement with migration of an alkyl group to nitrogen, and alkyne hydration (IV) is also a non-redox addition followed by tautomerism. Therefore only option B is correct.

Phenylacetylene ($\mathrm{C_8H_6}$) must first be hydrated; the Markovnikov hydration with dilute $\mathrm{H_2SO_4/HgSO_4}$ is reagent (iv) and it gives the enol $\mathrm{C_6H_5C(OH)=CH_2}$, which tautomerises to acetophenone, (S). Acetophenone, an eight-carbon methyl ketone, is then cleaved by hypoiodite (reagent (i)) to benzoic acid ($\mathrm{C_7}$, one carbon fewer) and iodoform, $\mathrm{CHI_3}$, after acidification. The first step is therefore (IV), (iv), (S). Option A starts from acetophenone (I), which is the second-stage compound and not phenylacetylene. Option B applies hypoiodite directly to the terminal alkyne, which has no $\mathrm{CH_3CO}$ unit and does not give benzoic acid that way. Option C pairs the correct substrate and reagent with the wrong product, as benzoic acid is the product of the second step. Only option D is consistent.