Q.Which amide does produce ethanamine by Hofmann bromamide degradation reaction ?
Concept understanding — Hofmann Bromamide Reaction
Hofmann Bromamide Degradation
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.
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).
The general reaction is:
R−CONHX2+BrX2+4NaOHR−NHX2+2NaBr+NaX2COX3+2HX2O
Or, in a more compact form:
R−CONHX2BrX2,NaOHR−NHX2+COX2
Step-by-Step Mechanism
- 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
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).
- Mechanism: Involves an isocyanate intermediate. This is a classic example of a rearrangement reaction.
A common mistake: thinking the amine has the same number of carbons as the amide. Count carefully — the carbonyl carbon is lost as COX2. For example, acetamide (CHX3CONHX2, 2 carbons) gives methylamine (CHX3NHX2, 1 carbon).
Example
Convert benzamide to aniline:
CX6HX5−CONHX2BrX2,NaOHCX6HX5−NHX2+COX2
Benzamide (7 carbons) gives aniline (6 carbons) — the carbonyl carbon is gone.
Why This Reaction Matters
The Hofmann bromamide degradation is a powerful method to:
- Prepare primary amines that are difficult to make by other methods.
- Shorten a carbon chain by one carbon — useful in synthetic organic chemistry.
- Convert an aromatic amide to an aromatic amine (like benzamide to aniline).
R−CONHX2BrX2,4NaOHR−NHX2+2NaBr+NaX2COX3+2HX2O
Remember: the reaction works because the migrating alkyl group stabilises the electron-deficient nitrogen during the rearrangement. The better the migrating group can stabilise a positive charge, the faster the reaction.
The Hofmann bromamide degradation reaction closes out the key preparation methods in the NCERT Class 12 Amines chapter, searched as "Hofmann bromamide reaction mechanism class 12 chemistry" and frequently tested in both CBSE boards and JEE Main.
Hofmann bromamide degradation (section 13.3.6) removes the amide's carbonyl carbon, giving a primary amine with one FEWER carbon -- so the amide needed is propanamide (3 carbons), not acetamide.
Propanamide (propionamide), CH3-CH2-CO-NH2.
Step 1. Recall Hofmann bromamide degradation's carbon-count rule (section 13.3.6). R-CO-NH2 + Br2 + 4 KOH(aq), heat, gives R-NH2 + 2 KBr + K2CO3 + 2 H2O -- the product amine has ONE CARBON FEWER than the starting amide, since the whole carbonyl-carbon unit is removed.
Step 2. Work backward from the target, ethanamine. Ethanamine (CH3-CH2-NH2) has two carbons; the amide needed must therefore have ONE MORE carbon, i.e. three carbons total.
Step 3. Identify the amide. A three-carbon amide is propanamide (propionamide), CH3-CH2-CO-NH2. Hofmann degradation of propanamide, per the general reaction, gives CH3-CH2-NH2 (ethanamine) + 2 KBr + K2CO3 + 2 H2O.
Propanamide (propionamide), CH3-CH2-CO-NH2.
Applying section 13.3.6's one-carbon-fewer rule for Hofmann bromamide degradation in reverse, from the two-carbon product amine back to its three-carbon parent amide
- Picking acetamide (which the chapter's own worked example converts to METHYLamine, one carbon, not ethanamine) instead of correctly adding, not subtracting, one carbon when working backward from product to starting amide.
- Confusing Hofmann degradation's carbon-count-DEcreasing behaviour with the carbon-count-preserving amide reduction of section 13.3.4, which instead needs the SAME number of carbons in amide and product amine.
Showing the 12 most recent of 17 on this concept.
- CBSE 2026Set A1 markMCQQ.By which of the following processes, is methyl amine prepared ?(a) Wurtz reaction(b) Hoffmann Bromamide reaction(c) Friedel-Crafts reaction(d) Kolbe's reaction
›Reveal solutionSolution
The Hoffmann bromamide degradation converts an amide to a primary amine with one fewer carbon; acetamide -> methylamine.
In the Hoffmann bromamide (degradation) reaction, an amide is treated with bromine and alkali (KOH), losing the carbonyl carbon and forming a primary amine with one carbon atom fewer:
CH3-CONH2 + Br2 + 4 KOH --> CH3-NH2 + K2CO3 + 2 KBr + 2 H2O
Starting from acetamide (2 carbons) this gives methylamine (1 carbon). The other named reactions do not produce amines.
✓Final answer(b) Hoffmann Bromamide reaction.
- CBSE 2025Set ANNUAL1 markMCQQ.C6H5CONH2 + Br2 + 4NaOH -----> Product, Product is(a) C6H5COOH(b) C6H5NC(c) C6H5NH2(d) C6H6
›Reveal solutionSolution
Treating a primary amide with Br2 and excess NaOH is the Hofmann bromamide degradation, which shortens the carbon chain by one and gives a primary amine.
C6H5CONH2+Br2+4NaOH→C6H5NH2+2NaBr+Na2CO3+2H2O
Benzamide (C6H5CONH2) loses its carbonyl carbon (lost as carbonate) during this degradation, converting the amide directly into aniline (C6H5NH2), an amine with one carbon less than the starting amide.
✓Final answer(c) C6H5NH2.
- CBSE 2025Set ANNUAL1 markMCQQ.Hofmann bromamide degradation reaction is shown by:(a) CH3CH2NH2(b) CH3CH2NO2(c) CH3CONH2(d) CH3CH2CN
›Reveal solutionSolution
The Hofmann bromamide degradation converts a primary AMIDE (RCONH2) into a primary amine with one carbon less, using Br2/KOH — only a compound with the amide (-CONH2) functional group can undergo it.
RCONH2 + Br2 + 4KOH → RNH2 + K2CO3 + 2KBr + 2H2O.
Of the four options, only CH3CONH2 (acetamide) has the required -CONH2 amide group: CH3CONH2 --Br2/KOH--> CH3NH2 (methylamine, one carbon fewer than the starting amide) + K2CO3 + 2KBr + 2H2O.
The others do not have an amide group: CH3CH2NH2 is already an amine, CH3CH2NO2 is a nitro compound, and CH3CH2CN is a nitrile.
✓Final answer(c) CH3CONH2 (acetamide).
- CBSE 2025Set ANNUAL1 markQ.Write the structure and IUPAC name of the amine produced by the Hofmann degradation of benzamide.
›Reveal solutionSolution
Hofmann degradation of benzamide (Br2/KOH) removes the carbonyl carbon as CO2 (via K2CO3) and converts the amide nitrogen directly into the ring's amino group, giving aniline.
The Hofmann bromamide degradation converts an amide, R−CONH2, into a primary amine, R−NH2, with the loss of one carbon atom (as carbonate), when treated with bromine in aqueous/alcoholic potassium hydroxide. Mechanistically it proceeds through an N-bromoamide, then a nitrene/isocyanate intermediate, and finally hydrolysis and decarboxylation of the resulting carbamic acid.
For benzamide, R=C6H5−:
C6H5CONH2+Br2+4KOH→C6H5NH2+K2CO3+2KBr+2H2O
The product, having lost the carbonyl carbon, is simply the benzene ring bearing the amino group directly:
Structure: a benzene ring with one –NH2 substituent, i.e. C6H5−NH2.
✓Final answerAniline (C6H5NH2), IUPAC name benzenamine (also called phenylamine).
- CBSE 2024Set B1 markQ.Write True or False: Primary amines is prepared by Hoffmann bromamide reaction.
›Reveal solutionSolution
Hofmann bromamide degradation is indeed a standard method for preparing primary amines from amides.
In the Hofmann bromamide degradation reaction, an amide is treated with bromine and concentrated aqueous/ethanolic sodium hydroxide (or KOH). This converts the amide (RCONH2) into a primary amine (RNH2) with the loss of one carbon atom (as CO2, via an isocyanate intermediate):
RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O
This is a well-known method for stepping down from an amide to the corresponding primary amine, so the statement is correct.
✓Final answerTrue.
- CBSE 2024Set ANNUAL1 markMCQQ.In Hofmann-bromamide reaction an amide is converted to :(a) Primary amine(b) Secondary amine(c) Tertiary amine(d) Aldehyde
›Reveal solutionSolution
The Hofmann bromamide reaction converts an amide into a primary amine with one carbon atom less than the starting amide.
When an amide (RCONH2) is treated with bromine in aqueous or ethanolic sodium hydroxide, it is converted to a primary amine (RNH2) containing one carbon atom less than the amide. This is called the Hofmann bromamide degradation reaction.
RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O
Mechanism outline: Br2 reacts with the amide N-H in presence of NaOH to give an N-bromoamide, which loses HBr (as its conjugate base) to form a nitrene-like intermediate (isocyanate via 1,2-shift of R from carbon to nitrogen), giving an isocyanate R-N=C=O; this is hydrolysed by NaOH to a carbamate salt, which loses CO2 (as Na2CO3) on decarboxylation to give the primary amine.
This reaction is important because it is the only common method that reduces the carbon chain by one carbon while converting a carbonyl-containing group directly to an amine.
✓Final answer(a) Primary amine.
- CBSE 2023Set F1 markMCQQ.Which of the following is formed when acetamide reacts with Br2/KOH?(a) Acetone(b) Methyl amine(c) Acetaldehyde(d) Ammonia
›Reveal solutionSolution
Hofmann bromamide degradation converts acetamide (CH3CONH2) to methylamine (CH3NH2), a primary amine with one fewer carbon.
An amide treated with bromine and aqueous KOH (or NaOH) undergoes the Hofmann bromamide (degradation) reaction, giving a primary amine that has one carbon fewer than the amide:
CH3CONH2 + Br2 + 4KOH → CH3NH2 + K2CO3 + 2KBr + 2H2O
Acetamide (CH3CONH2, 2 carbons) therefore gives methylamine (CH3NH2, 1 carbon). The -CO- carbon is lost as carbonate.
✓Final answer(B) Methyl amine (CH3NH2).
- CBSE 2023Set A1 markQ.Match the following. Column A item: 'Hoffmann bromide'. Choose its correct match from Column B:(a) Ether(b) Primary amine(c) Lactose(d) C12H22O11(e) Glucose(f) Negative ions(g) C6H5SO2Cl(h) +7
›Reveal solutionSolution
The Hofmann bromamide (Hoffmann bromide) degradation reaction converts an amide into a primary amine with one less carbon atom.
RCONH2+Br2+4NaOH→RNH2+Na2CO3+2NaBr+2H2O
The product of this reaction is always a primary amine (1° amine), so 'Hoffmann bromide (reaction)' matches with 'Primary amine'.
✓Final answerHoffmann bromide → (b) Primary amine.
- CBSE 2023Set ANNUAL1 markMCQQ.When acetamide is heated with an aqueous solution of sodium hydroxide and bromine, the product formed is(a) Methylamine(b) Ethylamine(c) Propylamine(d) None of these
›Reveal solutionSolution
The Hofmann bromamide degradation converts an amide R-CONH2 into the primary amine R-NH2, removing the carbonyl carbon entirely.
CH3CONH2 (acetamide) + Br2 + 4NaOH --> CH3NH2 (methylamine) + 2NaBr + Na2CO3 + 2H2O
The reaction proceeds via an N-bromoamide intermediate, then a nitrene/isocyanate rearrangement, ultimately extruding the carbonyl carbon as carbonate and leaving a primary amine with one carbon fewer than the starting amide. Since acetamide has 2 carbons, the product methylamine has 1 carbon.
✓Final answer(a) Methylamine.
- CBSE 2023Set ANNUAL1 markQ.What is Hoffmann's bromamide reaction? Give an example with a suitable chemical reaction. (½+½=1)
›Reveal solutionSolution
Hofmann's bromamide degradation converts an amide RCONH2 into a primary amine RNH2 having one fewer carbon atom, by treating it with bromine in concentrated aqueous/alcoholic sodium hydroxide.
Reaction (21 mark): When an amide is treated with bromine in the presence of concentrated NaOH solution, it degrades to give a primary amine containing one carbon atom less than the amide, along with the loss of the carbonyl carbon as carbonate:
R−CO−NH2+Br2+4NaOH→R−NH2+2NaBr+Na2CO3+2H2O
(Mechanistically, bromine first brominates the amide N-H to give an N-bromoamide, which loses HBr under base to form a nitrene/isocyanate intermediate that rearranges and hydrolyses, expelling CO2 (trapped as carbonate) to give the amine — this is why the product amine has one carbon less than the starting amide.)
Example (21 mark): Acetamide is converted to methylamine:
CH3−CO−NH2+Br2+4NaOH→CH3−NH2+2NaBr+Na2CO3+2H2O
This reaction is of great synthetic importance because it is used to step down a carbon chain by one carbon while converting an amide functional group directly into a primary amine.
✓Final answerHofmann's bromamide reaction: RCONH2Br2/NaOHRNH2 (one carbon less); e.g. CH3CONH2→CH3NH2 (methylamine).
- CBSE 2022Set ANNUAL1 markQ.Benzamide, C6H5CONH2 on reduction with LiAlH4 followed by hydrolysis gives an amine. It also undergoes Hofmann's bromamide reaction (with Br2 / KOH) and forms amine. What will be the difference between the two amines ?
›Reveal solutionSolution
The two routes give different amines: LiAlH₄ reduction keeps the carbon skeleton (benzylamine), while the Hofmann degradation strips out one carbon as CO₂ (aniline).
Route 1 — LiAlH4 reduction: Reduction of the amide carbonyl converts −CONH2 directly into −CH2NH2, without loss of any carbon:
C6H5CONH2i) LiAlH4ii) H2OC6H5CH2NH2 (benzylamine)
Route 2 — Hofmann bromamide degradation: Treating the amide with Br2/KOH converts it to an isocyanate intermediate which hydrolyses with loss of the carbonyl carbon as CO2, giving an amine with one carbon less than the starting amide, and with the −NH2 attached directly to the ring:
C6H5CONH2Br2/KOHC6H5NH2 (aniline)+CO2
So the key difference is that benzylamine has the amine on a −CH2− carbon (retaining the original carbon skeleton), while aniline has the amine bonded directly to the benzene ring, having lost the original carbonyl carbon.
✓Final answerLiAlH4 reduction gives benzylamine (C6H5CH2NH2); the Hofmann bromamide reaction gives aniline (C6H5NH2), one carbon shorter, because Hofmann degradation removes the carbonyl carbon as CO2.
- CBSE 2022Set ANNUAL1 markQ.Write the reagent required (denoted '?') for the following reaction: CH3CONH2 --?--> CH3NH2.
›Reveal solutionSolution
Amides are converted to amines with one fewer carbon by the Hofmann bromamide degradation, using bromine in aqueous alkali.
Identifying the reaction: CH3CONH2 (acetamide, 2 carbons) converting to CH3NH2 (methylamine, 1 carbon) is a degradation that removes the carbonyl carbon — the Hofmann bromamide (Hofmann rearrangement) reaction.
Reagent and conditions: Bromine (Br2) in aqueous sodium hydroxide (NaOH).
Mechanism outline: Br2/NaOH first N-brominates the amide to give an N-bromoamide; base removes the remaining N-H to form an anion which rearranges (migration of the alkyl/aryl group from carbon to nitrogen with loss of Br-) to an isocyanate; the isocyanate is then hydrolysed by the aqueous alkali to a primary amine, releasing CO2 as carbonate.
CH3CONH2Br2NaOHCH3NH2+Na2CO3+NaBr+H2O
✓Final answerReagent: Br2/NaOH(aq) (Hofmann bromamide degradation) — converts acetamide to methylamine, with loss of one carbon as carbonate.
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