Q.An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound 'B' which on heating with Br₂ and KOH form a compound 'C' of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.

Showing the 12 most recent of 17 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.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

   …

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2. CBSE 2025Set ANNUAL1 markMCQ
   Q.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.

   $\text{C}_6\text{H}_5\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{NaBr} + \text{Na}_2\text{CO}_3 + 2\text{H}_2\text{O}$

   …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.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.

   …

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4. CBSE 2025Set ANNUAL1 mark
   Q.Write the structure and IUPAC name of the amine produced by the Hofmann degradation of benzamide.

   ›Reveal solutionSolution

   Hofmann degradation of benzamide (Br$_2$/KOH) removes the carbonyl carbon as $CO_2$ (via $K_2CO_3$) and converts the amide nitrogen directly into the ring's amino group, giving aniline.

   The Hofmann bromamide degradation converts an amide, $R-CONH_2$, into a primary amine, $R-NH_2$, 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, $R = C_6H_5-$:

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5. CBSE 2024Set B1 mark
   Q.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):

   …

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6. CBSE 2024Set ANNUAL1 markMCQ
   Q.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

   …

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7. CBSE 2023Set F1 markMCQ
   Q.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

   …

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8. CBSE 2023Set A1 mark
   Q.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.

   $RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O$

   …

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9. CBSE 2023Set ANNUAL1 markMCQ
   Q.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 …

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10. CBSE 2023Set ANNUAL1 mark
    Q.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 $RCONH_2$ into a primary amine $RNH_2$ having one fewer carbon atom, by treating it with bromine in concentrated aqueous/alcoholic sodium hydroxide.

    Reaction ($\tfrac12$ 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:

    $R-CO-NH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + 2NaBr + Na_2CO_3 + 2H_2O$

    (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 $CO_2$ (trapped as carbonate) to give the amine — this is why the product amine has one carbon less than the starting amide.)

    Example ($\tfrac12$ mark): Acetamide is converted to methylamine:

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11. CBSE 2022Set ANNUAL1 mark
    Q.Benzamide, $C_6H_5CONH_2$ on reduction with $LiAlH_4$ followed by hydrolysis gives an amine. It also undergoes Hofmann's bromamide reaction (with $Br_2$ / 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 — $LiAlH_4$ reduction: Reduction of the amide carbonyl converts $-CONH_2$ directly into $-CH_2NH_2$, without loss of any carbon:

    $C_6H_5CONH_2 \xrightarrow[\text{ii) }H_2O]{\text{i) }LiAlH_4} C_6H_5CH_2NH_2 \ (\text{benzylamine})$

    Route 2 — Hofmann bromamide degradation: Treating the amide with $Br_2/KOH$ converts it to an isocyanate intermediate which hydrolyses with loss of the carbonyl carbon as $CO_2$, giving an amine with one carbon less than the starting amide, and with the $-NH_2$ attached directly to the ring:

    $C_6H_5CONH_2 \xrightarrow{Br_2/KOH} C_6H_5NH_2 \ (\text{aniline}) + CO_2$

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12. CBSE 2022Set ANNUAL1 mark
    Q.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).

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