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Questions on amines love carbon counting. Give an amide with four carbons and ask for the amine, or give an amine and ask which amide made it. The Hofmann bromamide reaction answers both, because it always removes exactly one carbon. This post covers what NCERT says, the balanced equation, the mechanism for students who need it, worked examples different from NCERT's, and the usual traps.
Where is it in the syllabus?
The NCERT Class 12 chapter Amines (2026-27 reprint) lists it as the sixth method of preparing amines, spelt "Hoffmann bromamide degradation reaction". It describes treating an amide with bromine in aqueous or ethanolic sodium hydroxide, says an alkyl or aryl group migrates "from carbonyl carbon of the amide to the nitrogen atom", and notes that the amine "contains one carbon less than that present in the amide". The JEE (Advanced) 2026 brochure names "Hoffmann bromamide degradation" among amine reactions. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list "general methods of preparation" for organic compounds containing nitrogen, which in NCERT includes this reaction. Check your year's syllabus.
The chemist was August Wilhelm Hofmann, with one f, as his name appears on the original paper. He reported the reaction in 1881 in a paper on the action of bromine in alkaline solution on amides (Hofmann, Berichte, 1881). Our profile of August Wilhelm Hofmann covers his wider work.
What is the balanced equation?
RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O
Check the atoms: one carbon leaves the amide and ends up in Na₂CO₃; four sodium on each side; two bromine; five oxygen and six hydrogen on each side once you count R separately. Four moles of NaOH per mole of amide is a common one-mark question.
What happens in the mechanism?
NCERT gives only the migration step. OpenStax Organic Chemistry sets out the full sequence, which JEE Advanced students should know in outline:
Base removes an N–H proton and the amide is brominated on nitrogen, giving an N-bromoamide.
Base removes the second N–H proton.
The R group shifts from carbon to nitrogen as bromide leaves, giving an isocyanate, R–N=C=O.
Water adds to the isocyanate, and the resulting carbamic acid loses carbon dioxide to give the amine. In the strongly basic solution the CO₂ ends up as carbonate.
Because R moves from carbon straight onto nitrogen, the carbon skeleton of R is untouched. Only the carbonyl carbon is lost.
Worked examples
1. Ethanamide to methanamine. CH₃CONH₂ gives CH₃NH₂. If you start with 5.91 g of ethanamide (59.07 g/mol), that is 0.100 mol, so the theoretical yield is 0.100 × 31.06 ≈ 3.11 g of methanamine. (Sample quantity.)
2. 2-Phenylethanamide to benzylamine. C₆H₅CH₂CONH₂ gives C₆H₅CH₂NH₂ (phenylmethanamine). From 13.5 g of amide (135.17 g/mol), 0.0999 mol. Theoretical yield 0.0999 × 107.16 ≈ 10.7 g; at 80% yield (sample figure), about 8.6 g.
3. Backwards: which amide gives propan-2-amine? The amine is (CH₃)₂CH–NH₂. Put a C=O between the carbon skeleton and the nitrogen: (CH₃)₂CH–CONH₂, 2-methylpropanamide.
4. Backwards: which amide gives cyclohexylamine? Cyclohexanecarboxamide, C₆H₁₁CONH₂.
5. A conversion chain. Ethanoic acid → ethanamide (via the acid chloride and ammonia) → methanamine by Hofmann. This steps down the series by one carbon. NCERT's reduction of nitriles does the opposite, an ascent of the amine series by one carbon.
Which mistakes cost marks?
Keeping the carbon count. Butanamide gives propan-1-amine, not butan-1-amine.
Using a secondary amide. The reaction needs a primary amide, CONH₂. An N-substituted amide has no second N–H to lose.
Expecting secondary or tertiary amines. Hofmann bromamide gives primary amines only.
Confusing it with Gabriel phthalimide synthesis, which also gives primary amines but keeps the carbon count of the alkyl halide.
Writing 2NaOH in the balanced equation.
For the amine reactions that follow, such as diazonium salts and the Hinsberg test, see our guide to amines and diazonium salts. If a conversion question has several steps and you lose track of carbons, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning at each step.
A practice routine
Day 1: Write the balanced equation from memory and check the atom count. Then redo NCERT Example 9.3 without looking.
Day 2: Eight forward questions: amide to amine, including aromatic and branched amides.
Day 3: Five backwards questions from amine to amide.
Day 4: Three conversion chains that combine Hofmann with nitrile reduction, one stepping down and one stepping up.
Tonight, name the amine formed from pentanamide and the amide you would need to make ethanamine.
