Match List I with List II.
| List I Reaction |
List II Product |
A.  |
I.  |
B.  |
II.  |
C.  |
III.  |
| D. \(C_6H_5\overset{\oplus}{N}{\equiv}N\overset{\ominus}{Cl} \xrightarrow{CH_3CH_2OH}\) | IV.  |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-IV, C-II, D-I
Each List-I item is a standard conversion of benzene derivatives: the Hofmann bromamide degradation of an amide, the conversion of an acid into its acid chloride, the Gattermann-Koch formylation of benzene, and the reduction of a diazonium salt by ethanol.
(A) Benzamide with \( \ce{Br2} \)/\( \ce{OH^{-}} \) — This matches (III) aniline. In the Hofmann bromamide degradation the amide loses its carbonyl carbon and gives a primary amine with one carbon less.
\( \ce{C6H5CONH2 + Br2 + 4NaOH -> C6H5NH2 + Na2CO3 + 2NaBr + 2H2O} \)
(B) Benzoic acid with \( \ce{PCl3} \) — This matches (IV) benzoyl chloride. \( \ce{PCl3} \) replaces the –OH of the carboxyl group by –Cl.
\( \ce{3C6H5COOH + PCl3 -> 3C6H5COCl + H3PO3} \)
(C) Benzene with CO/HCl, \( \ce{AlCl3} \) — This matches (II) benzaldehyde. In the Gattermann-Koch reaction, CO and HCl in the presence of anhydrous \( \ce{AlCl3} \) (or CuCl) introduce a –CHO group.
\( \ce{C6H6 + CO} \xrightarrow{\ce{HCl},\ \text{anhyd. }\ce{AlCl3/CuCl}} \ce{C6H5CHO} \)
(D) Benzenediazonium chloride with \( \ce{CH3CH2OH} \) — This matches (I) benzene. Ethanol is a mild reducing agent; it replaces the diazo group by hydrogen and is itself oxidised to ethanal.
\( \ce{C6H5N2^+Cl^- + CH3CH2OH -> C6H6 + N2 + CH3CHO + HCl} \)
Shortcut: (A)-(III) appears only in Option (2).
Note: It seems that there is a mistake in the original NTA paper. In item (D) of List I, the reagent over the arrow is printed partly over the structure of item (IV), so only “CH3CH2–” can be read. It is shown here as CH3CH2OH (ethanol), the reagent that reduces a diazonium salt to benzene.