Match List-I with List-II : $\begin{array}{|l|l|} \hline \text{List-I } & \text{List-II } \\ \text{(Reaction) } & \text{(Product formed) } \\ \hline \text{(a) Gabriel synthesis} & \text{(i) Benzaldehyde} \\ \text{(b) Kolbe synthesis} & \text{(ii) Ethers} \\ \text{(c) Williamson synthesis} & \text{(iii) Primary amines} \\ \text{(d) Etard synthesis} & \text{(iv) Salicyclic acid} \\ \hline \end{array}$ Choose the correct answer from the options given below :

• A

  (a) - (iii), (b) - (i), (c) - (ii), (d) - (iv)
• B

  (a) - (ii), (b) - (iii), (c) - (i), (d) - (iv)
• C

  (a) - (iv), (b) - (iii), (c) - (i), (d) - (ii)
• Correct answer
  D

  (a) - (iii), (b) - (iv), (c) - (ii), (d) - (i)

Solution

(a) Gabriel synthesis $\longrightarrow$ (iii) Primary amines
Pthalimide is alkylated with primary or unbranched secondary alkyl halides or sulphonate. The product thus obtained is hydrolysed to give primary amines.



Treatment of toluene with chromyl chloride in an inert solvent like $\mathrm{CS}_2$ or $\mathrm{CCl}_4$ forms an insoluble complex $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}\left(\mathrm{OCrCl}_2 \mathrm{OH}\right)_2$ which further on hydrolysis produces benzaldehyde.

Hence, the correct answer is option: 'D'.