Amines
Amines are important class of organic compounds which are basic in nature. They are generally derived by replacing one or more hydrogen atoms of ammonia molecule by alkyl / aryl group(s). In nature, they are generally found in proteins, vitamins, alkaloids and hormones. Two biologically active compounds namely adrenaline and ephedrine contain secondary amino group and are used to increase blood pressure.
Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore, sp3 hybridized and the geometry of amine is pyramidal. Due to the presence of unshared pair of electrons, the angle C-N-E (where E is C or H) is less than 109.5°.
Inductive effects, solvation effects, steric effects etc. affect the basic strength of amines. Amines are classified as primary (1°), secondary (2°), and tertiary (3°) depending upon the number of hydrogen atoms replaced by alkyl or aryl groups in ammonia molecule. Amines are engaged in intermolecular hydrogen bonding. Aliphatic and aromatic primary amines show positive carbylamine reaction. Amines can be prepared from various compounds like reduction of nitro, nitriles, amides etc. compounds. A colourless crystalline solid benzene diazonium chloride can be obtained from aniline by reacting with sodium nitrite and hydrochloric acid at 273-278 K.
The final structure of amine produced by Hofmann degradation of m-bromobenzamide is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3-bromoaniline (m-bromoaniline)
The passage mentions that amines can be prepared from amides. Amides give amines in two ways, which differ in the carbon count:
Hofmann bromamide degradation: an amide is treated with bromine and aqueous or ethanolic NaOH. The alkyl or aryl group moves from the carbonyl carbon to nitrogen, and the carbonyl carbon is lost as carbonate. The product is a primary amine with one carbon atom less than the amide.
\( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)
Reduction with LiAlH4: the C=O group becomes CH2, so the amine keeps all the carbon atoms of the amide.
\( \displaystyle \ce{R-CONH2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{R-CH2NH2} \)
In m-bromobenzamide the aryl group is 3-bromophenyl. In the Hofmann degradation this aryl group moves to nitrogen, so the –NH2 group ends up on the same ring carbon that carried –CONH2, still meta to Br:
\( \displaystyle \ce{3-BrC6H4CONH2 + Br2 + 4NaOH -> 3-BrC6H4NH2 + Na2CO3 + 2NaBr + 2H2O} \)
So the amine formed is 3-bromoaniline (structure 2), with –NH2 directly on the ring.
Why the other options are incorrect
Option (1) 3-bromobenzylamine — Incorrect. It keeps the carbonyl carbon as –CH2NH2. That is the product of LiAlH4 reduction of the amide, not of the Hofmann degradation, which removes that carbon.
Option (3) 3-bromo-N-methylaniline — Incorrect. The Hofmann degradation gives a primary amine. No methyl group is put on the nitrogen.
Option (4) benzene-1,3-diamine — Incorrect. Only the amide group reacts. The Br on the ring is not replaced, because aryl halides do not undergo nucleophilic substitution under these mild conditions.
NCERT (2026-27 edition):
Preparation of Amines – “6. Hoffmann bromamide degradation reaction
Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide.
\( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)”