Q.Suggest a route by which the following conversion can be accomplished: cyclohexanecarboxamide (a cyclohexane ring bearing a –C(=O)NH2 group) is to be converted into N-methylcyclohexanamine (a cyclohexane ring bearing a –NH–CH3 group).
Imagine you have an amide — a molecule with a carbonyl group (−CO−) attached to a nitrogen. You want to turn it into a primary amine, but you also want to chop off one carbon from the chain. That is exactly what the Hofmann bromamide reaction does: it shortens the carbon skeleton by one carbon and gives you an amine.
The Intuition
The reaction uses bromine (BrX2) in the presence of a strong alkali (like NaOH or KOH). The alkali first deprotonates the amide nitrogen, making it a strong nucleophile. This nucleophile attacks bromine, forming an N-bromoamide. Under the strongly basic conditions, this intermediate loses a bromide ion and undergoes a rearrangement — the alkyl group attached to the carbonyl carbon migrates from carbon to nitrogen. The result is an isocyanate intermediate (R−N=C=O). Finally, the isocyanate is hydrolysed by the aqueous alkali to give a primary amine and carbon dioxide.
The net effect: the carbonyl carbon is lost as COX2, and the alkyl group ends up attached to the nitrogen.
Important
The product amine has one fewer carbon than the starting amide. The lost carbon is the carbonyl carbon.
The Precise Statement
Hofmann bromamide degradation (also called Hofmann rearrangement) is the conversion of a primary amide to a primary amine with one fewer carbon atom, using bromine and an aqueous alkali (usually NaOH or KOH).
Deprotonation: The amide nitrogen is deprotonated by the strong base, forming an amide anion.
R−CONHX2+OHX−R−CONHX−+HX2O
Bromination: The amide anion attacks bromine, forming an N-bromoamide.
R−CONHX−+BrX2R−CONHBr+BrX−
Second deprotonation: The N-bromoamide is deprotonated again by the base.
R−CONHBr+OHX−R−CONBrX−+HX2O
Rearrangement: The alkyl group migrates from the carbonyl carbon to the nitrogen, with simultaneous loss of bromide ion. This forms an isocyanate.
R−CONBrX−R−N=C=O+BrX−
Hydrolysis: The isocyanate reacts with water to form a carbamic acid, which spontaneously decarboxylates (loses COX2) to give the primary amine.
R−N=C=O+HX2OR−NH−COOHR−NHX2+COX2
Tip
The rearrangement step (step 4) is the key. The alkyl group migrates with its bonding electrons — it is a 1,2-shift from carbon to the electron-deficient nitrogen. This is why the carbon skeleton shortens by one carbon.
Key Points for Exams
Starting material: Primary amide (R−CONHX2) only. Secondary or tertiary amides do not undergo this reaction.
Reagents: BrX2 and NaOH (or KOH). Sometimes ClX2 can be used instead of BrX2, but bromine is more common.
Product: Primary amine with one fewer carbon.
By-products: NaBr, NaX2COX3, HX2O (or COX2 if written in the simplified form). …
First shorten the amide by one carbon to a primary amine using Hofmann bromamide degradation, then introduce a methyl group on nitrogen (cleanly via formylation followed by LiAlH4 reduction). …
The amide is first degraded to cyclohexanamine (losing the carbonyl carbon) by Hofmann bromamide reaction, and the nitrogen is then methylated. Formylation followed by LiAlH4 reduction cleanly installs exactly one N-methyl group, giving N-methylcyclohexanamine.
Step 1 – Hofmann bromamide degradation
Treat cyclohexanecarboxamide (C6H11–CONH2) with Br2 and hot aqueous KOH. The amide loses its carbonyl carbon (as carbonate) and the –NH2 migrates onto the ring carbon:
An amide can be shortened by exactly one carbon to the corresponding primary amine via the Hofmann bromamide degradation; to then add exactly one methyl group onto that amine's nitrogen without over-alkylating, formylate the amine and reduce the resulting formamide with LiAlH4 rather than using CH3I directly.
Steps
Recognise an amide-to-one-carbon-shorter-amine conversion as a Hofmann bromamide degradation: RCONH2 + Br2 + 4KOH gives RNH2 + K2CO3 + 2KBr + 2H2O (the carbonyl carbon is lost as carbonate).
Apply this to the given amide to obtain the primary amine.
To install exactly one methyl group on that amine's nitrogen, avoid direct alkylation with CH3I (which over-alkylates to a mixture of secondary/tertiary amine and quaternary ammonium salt).
Instead, formylate the amine with HCOOH (or ethyl formate) to form the N-substituted formamide (-NH-CHO). …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
COMEDK 2025Set 2025-E1 markMCQ
Q.Two statements, one Assertion and the other Reason are given. Identify the correct option Assertion : Primary and secondary amides on treatment with Br2 and alcoholic NaOH yield primary and secondary amines respectively and the reaction involves stepping down the series. Reason : The reaction occurs due to the migration of alkyl group from Carbonyl carbon atom to the Nitrogen atom with elimination of carbonyl group as the carbonate salt.
(A) Assertion is correct but Reason is incorrect.
(B) Both Assertion and Reason are incorrect.
(C) Both Assertion and Reason are correct.
(D) Assertion is incorrect but Reason is correct.
›Reveal solutionSolution
Hofmann bromamide degradation works only on primary amides (giving primary amines, one C shorter) — secondary amides do NOT give secondary amines, so the Assertion is wrong; but the Reason's account of the mechanism (alkyl migration from carbonyl C to N, loss of the carbonyl as carbonate) is correct. Hence Assertion incorrect, Reason correct.
Assertion. In the Hofmann bromamide reaction, R-CONH2Br2/alc. NaOHR-NH2. It requires an unsubstituted primary amide (−CONH2, which has the N–H bonds the mechanism needs). N-substituted (secondary) amides cannot form the intermediate N-bromoamide/nitrene and so do NOT give secondary amines. The "stepping-down" (product has one fewer carbon) is true for the primary case, but the claim that secondary amides give secondary amines makes the Assertion incorrect. …
Q.An organic compound [X] (molecular formula- C5H11NO when reacted with Br2 / aq. NaOH yielded [Y] which reacts with CHCl3 and Ethanolic KOH to produce a foul smelling compound. Compound [Y] also reacts with HONO to produce Butan-1-ol with liberation of N2(g). Identify [X].
(A) CH3−(CH2)3−CONH2
(B) CH3−(CH2)2−CO−CH2−NH2
(C) CH3−CH2−CO−(CH2)2−NH2
(D) CH3−(CH2)2−CH(NH2)−CHO
›Reveal solutionSolution
The key is that [Y] must be a primary amine (C₄H₉NH₂) that gives butan-1-ol with HONO and a foul-smelling isocyanide with CHCl₃/KOH. Tracing back, [X] is a primary amide that undergoes Hofmann degradation to yield that amine. The correct option is (A).
Concept & Intuition
We are given a sequence of reactions starting from an organic compound X (C₅H₁₁NO). The reactions are classic name reactions:
Hofmann bromamide degradation: A primary amide (RCONH₂) reacts with Br₂/aq. NaOH to give a primary amine with one fewer carbon atom.
Carbylamine reaction: A primary amine reacts with CHCl₃ and alcoholic KOH to give a foul-smelling isocyanide (carbylamine).
Reaction with nitrous acid (HONO): A primary aliphatic amine gives an alcohol with loss of N₂ gas.
Since [Y] reacts with HONO to give butan-1-ol (C₄H₉OH) and N₂, [Y] must be butan-1-amine (n-butylamine, C₄H₉NH₂). That means [Y] is a primary amine with four carbons.
Now, [Y] comes from [X] via Hofmann degradation. In that reaction, the amide loses its carbonyl carbon as CO₂, so the amine has one fewer carbon than the amide. Therefore, [X] must be a primary amide with five carbons (C₅H₁₁NO). The only option that is a primary amide is (A).
Let’s verify step by step.
Step-by-step reasoning
Identify [Y] from its reactions
[Y] reacts with CHCl₃/ethanolic KOH → foul-smelling compound. This is the carbylamine reaction, characteristic of primary amines (R–NH₂).
[Y] reacts with HONO → butan-1-ol + N₂ gas. For primary aliphatic amines, HONO gives nitrogen gas and an alcohol (via diazonium intermediate). The product is butan-1-ol, so [Y] must be butan-1-amine:
CH3CH2CH2CH2NH2
(Molecular formula: C₄H₁₁N)
2. Work backwards to [X] via Hofmann degradation
Hofmann degradation:
RCONH2+Br2+4NaOH→RNH2+2NaBr+Na2CO3+2H2O
The amide loses its carbonyl carbon, so the amine has one fewer carbon.
Since [Y] is C₄H₉NH₂ (4 carbons), [X] must be a C₅ amide: RCONH₂ where R = C₄H₉.
The molecular formula of [X] is given as C₅H₁₁NO, which matches a saturated primary amide (C₅H₁₁NO).
Check the options
(A) CH₃–(CH₂)₃–CONH₂: This is pentanamide (valeramide). Hofmann degradation gives butan-1-amine. ✓ …