Selective Reduction Of Nitro Compounds
Reduction of m-dinitrobenzene with ammonium sulphide selectively converts only one nitro group; what is the principal organic product formed in this selective reduction?
Select the correct option:
Solution
m-nitroaniline
Aromatic nitro compounds can be reduced fully or selectively depending on the reagent. Strong reducing systems such as tin or iron with hydrochloric acid reduce every nitro group to an amino group. However, a mild and selective reagent like ammonium sulphide (or H2S with a base) reduces only one of two nitro groups in a dinitro compound, leaving the other intact. Applied to m-dinitrobenzene, ammonium sulphide therefore converts one –NO2 to –NH2 while sparing the second, giving m-nitroaniline. The option m-phenylenediamine would require reduction of both nitro groups, which only a strong reductant achieves, so it is wrong here. Aniline is incorrect because there is no loss of a substituent; both substituents stay on the ring. Nitrobenzene is impossible because the carbon framework already bears two groups and none is removed. This selective partial reduction is a frequently tested application of ammonium sulphide noted in NCERT discussions of nitro chemistry, and the same idea underlies the controlled synthesis of unsymmetrical disubstituted benzenes. As a plausibility check, the product retains one electron-withdrawing nitro group and one electron-donating amino group, consistent with reduction of exactly one of the two equivalent positions and with the milder reducing power of the sulphide reagent.
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About This Question
- Subject
- chemistry
- Chapter
- organic compounds containing nitrogen
- Topic
- selective reduction of nitro compounds
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
m-nitroaniline
Aromatic nitro compounds can be reduced fully or selectively depending on the reagent. Strong reducing systems such as tin or iron with hydrochloric acid reduce every nitro group to an amino group. However, a mild and selective reagent like ammonium sulphide (or H2S with a base) reduces only one of two nitro groups in a dinitro compound, leaving the other intact. Applied to m-dinitrobenzene, ammonium sulphide therefore converts one –NO2 to –NH2 while sparing the second, giving m-nitroaniline. The option m-phenylenediamine would require reduction of both nitro groups, which only a strong reductant achieves, so it is wrong here. Aniline is incorrect because there is no loss of a substituent; both substituents stay on the ring. Nitrobenzene is impossible because the carbon framework already bears two groups and none is removed. This selective partial reduction is a frequently tested application of ammonium sulphide noted in NCERT discussions of nitro chemistry, and the same idea underlies the controlled synthesis of unsymmetrical disubstituted benzenes. As a plausibility check, the product retains one electron-withdrawing nitro group and one electron-donating amino group, consistent with reduction of exactly one of the two equivalent positions and with the milder reducing power of the sulphide reagent.
This medium difficulty chemistry question is from the chapter organic compounds containing nitrogen, covering the topic of selective reduction of nitro compounds. It appeared in the 2025 exam.
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