Match List I with List II
| List I | List II |
| A. Gatterman-Koch reaction | I. Diphenyl |
| B. Stephen's reaction | II. Ethyl amine |
| C. Hofmann Bromamide degradation | III. Benzaldehyde |
| D. Fittig reaction | IV. Acetaldehyde |
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
Three of these named reactions prepare carbonyl compounds or amines, and one joins aryl groups: Gatterman-Koch (benzaldehyde from benzene) and Stephen (aldehyde from a nitrile) give aldehydes, Hofmann bromamide degradation gives an amine with one carbon less, and the Fittig reaction couples two aryl halides.
(A) Gatterman-Koch reaction — This matches (III) Benzaldehyde.
\( \ce{C6H6 + CO + HCl} \xrightarrow{\text{anhyd. }\ce{AlCl3}\text{/CuCl}} \ce{C6H5CHO} \)
(B) Stephen’s reaction — This matches (IV) Acetaldehyde. The nitrile is reduced to an imine, which hydrolyses to the aldehyde:
\( \ce{CH3CN + SnCl2 + HCl -> CH3CH=NH} \xrightarrow{\ce{H3O+}} \ce{CH3CHO} \)
(C) Hofmann Bromamide degradation — This matches (II) Ethyl amine.
\( \ce{CH3CH2CONH2 + Br2 + 4NaOH -> CH3CH2NH2 + Na2CO3 + 2NaBr + 2H2O} \)
(D) Fittig reaction — This matches (I) Diphenyl.
\( \ce{2C6H5Cl + 2Na} \xrightarrow{\text{dry ether}} \ce{C6H5-C6H5 + 2NaCl} \)
Shortcut: (A)-(III) leaves Options (2) and (3); (B)-(IV) then leaves only Option (2).