Match List I with List II
| List II | List II |
| A. Gatterman-Koch reaction | I. Diphenyl |
| B. Stephen's reaction | II. Ethyl amine |
| C. Hoffmann 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.
| List II | List II |
| A. Gatterman-Koch reaction | III. Benzaldehyde |
| B. Stephen's reaction | IV. Acetaldehyde |
| C. Hoffmann Bromamide degradation | II. Ethyl amine |
| D. Fittig reaction | I. Diphenyl |
A. Gatterman-Koch reaction → III. Benzaldehyde
In this reaction, Benzene is treated with Carbon monoxide $(CO)$ and Hydrogen chloride $(HCl)$ in the presence of anhydrous $AlCl_{3}$ or $CuCl$. This introduces a formyl group $(-CHO)$ directly onto the ring.
B. Stephen's reaction → IV. Acetaldehyde
This is a reduction of nitriles. Methyl cyanide $(CH_{3}CN)$ is reduced using stannous chloride $(SnCl_{2})$ and $HCl$, followed by hydrolysis, to yield Acetaldehyde.
C. Hoffmann Bromamide degradation → II. Ethyl amine
This reaction converts an amide into a primary amine with one less carbon atom. If you start with Propanamide $(CH_{3}CH_{2}CONH_{2})$ and treat it with Bromine $(Br_{2})$ and $KOH$, the carbonyl group is removed, leaving you with Ethyl amine $(CH_{3}CH_{2}NH_{2})$.
D. Fittig reaction → I. Diphenyl
The Fittig reaction involves the treatment of haloarenes (like Chlorobenzene) with metallic Sodium in dry ether. Two aryl groups join together to form a diaryl compound, in this case, Diphenyl (also known as Biphenyl).