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MHT-CET Chemistry · Amines

Preparation of Amines

Amines are made by reducing nitriles (Mendius, Na/ethanol, one carbon MORE than the halide), amides (LiAlH₄, same carbons) and nitro compounds (Sn/HCl), by the Hofmann bromamide degradation of an amide (one carbon FEWER) and by Gabriel phthalimide synthesis (primary only), or by ammonolysis and alkylation of halides.

Why this matters

15 PYQs, none HARD. Ten are reductions — CH₃Br → KCN → Na/EtOH → ethylamine (the chapter's most repeated sequence), which amide LiAlH₄ turns into ethanamine, how many H atoms reduce a nitrile (4) or a nitro compound (6), acetic acid → SOCl₂ → NH₃ → acetamide; five are Hofmann degradation (loss of CO, 28 g mol⁻¹; acetamide → methylamine), Gabriel (what is never formed), and ammonolysis followed by methylation. Two cards.

Concept 1 of 2

Reductions: Nitrile, Amide and Nitro to Amine

Intuition

Count carbons before and after. A nitrile made from R–X has one carbon more than R, and reducing it (Mendius, Na/ethanol or LiAlH₄, 4 H atoms) keeps them all: CH₃Br → CH₃CN → CH₃CH₂NH₂. An amide reduced by LiAlH₄ keeps its carbons: acetamide → ethanamine. A nitro compound needs 6 H atoms: nitroethane → ethylamine. Nothing is lost in any of these.

Definition

  • Mendius: RCN+4[H]→Na/C2H5OH or LiAlH4RCH2NH2\text{RCN} + 4[\text{H}] \xrightarrow{\text{Na/C}_2\text{H}_5\text{OH or LiAlH}_4} \text{RCH}_2\text{NH}_2. No carbon lost. Acetonitrile needs 4 H atoms per mole.
  • Sequence: CH3Br→KCNCH3CN→Na/EtOHCH3CH2NH2\text{CH}_3\text{Br} \xrightarrow{\text{KCN}} \text{CH}_3\text{CN} \xrightarrow{\text{Na/EtOH}} \text{CH}_3\text{CH}_2\text{NH}_2 (ethylamine). Butanenitrile comes from n-propyl chloride + alcoholic KCN.
  • Amide: CH3CONH2→LiAlH4CH3CH2NH2\text{CH}_3\text{CONH}_2 \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{NH}_2 — A for ethanamine is acetamide, not propanamide or C₂H₅CN (which gives propanamine).
  • Nitro: C2H5NO2+6[H]→Sn/HClC2H5NH2+2H2O\text{C}_2\text{H}_5\text{NO}_2 + 6[\text{H}] \xrightarrow{\text{Sn/HCl}} \text{C}_2\text{H}_5\text{NH}_2 + 2\text{H}_2\text{O} — 6 H atoms.
  • Making the amide: acetic acid →SOCl2\xrightarrow{\text{SOCl}_2} acetyl chloride →NH3\xrightarrow{\text{NH}_3} acetamide. A Grignard with NH₃ is just protonated: EtMgCl + NH₃ → ethane.

Three reductions

RCN→4[H]RCH2NH2;RCONH2→LiAlH4RCH2NH2;RNO2→6[H]RNH2\text{RCN} \xrightarrow{4[\text{H}]} \text{RCH}_2\text{NH}_2;\quad \text{RCONH}_2 \xrightarrow{\text{LiAlH}_4} \text{RCH}_2\text{NH}_2;\quad \text{RNO}_2 \xrightarrow{6[\text{H}]} \text{RNH}_2

Worked example

Starting from ethyl bromide, make propan-1-amine, and say how many H atoms the reduction step consumes.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1AminesMODERATE
Identify the product 'B' in the following sequence of reactions. CH3Br→KCNA→Na/C2H5OHB\text{CH}_3\text{Br}\xrightarrow{\text{KCN}}\text{A}\xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}}\text{B}

[Q97 · 14th May Shift 1 · 2024]

Forgetting that cyanide adds a carbon

Methyl bromide ends as ETHYLamine — the CN carbon becomes CH₂. 'Methylamine' is the offered wrong answer in every version of the sequence.

Concept 2 of 2

Hofmann Bromamide Degradation, Gabriel Synthesis and Ammonolysis

Intuition

Hofmann degradation is the one route that LOSES a carbon: an amide with Br₂ and concentrated KOH gives the amine with one carbon fewer, the CO leaving as carbonate — a drop of 28 g mol⁻¹. Gabriel synthesis makes primary amines only: potassium phthalimide + R–X → N-alkylphthalimide → hydrolysis gives the amine and phthalate; phthalic acid itself never appears under the basic conditions. Ammonolysis of R–X gives a mixture that alkylates on.

Definition

  • Hofmann degradation: RCONH2+Br2+4KOH→RNH2+K2CO3+2KBr+2H2O\text{RCONH}_2 + \text{Br}_2 + 4\text{KOH} \to \text{RNH}_2 + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O}. Acetamide → methylamine. Loss in molar mass = CO = 28 g mol⁻¹.
  • Gabriel: phthalimide →KOH\xrightarrow{\text{KOH}} potassium phthalimide →R-X\xrightarrow{\text{R-X}} N-alkylphthalimide →NaOH(aq)\xrightarrow{\text{NaOH(aq)}} primary amine + sodium phthalate. NOT formed: phthalic acid. Aryl halides do not work.
  • Ammonolysis: benzyl chloride + NH₃ → benzylamine; then 2 CH₃I → C6H5CH2N(CH3)2\text{C}_6\text{H}_5\text{CH}_2\text{N(CH}_3)_2 (tertiary). Excess R–X goes on to the quaternary salt.
  • Hofmann DEGRADATION (amide → amine, one C fewer) is not Hofmann ELIMINATION (quaternary salt → alkene).

Hofmann bromamide degradation

RCONH2→Br2, KOH (conc.)RNH2(one carbon fewer; −28 g mol−1)\text{RCONH}_2 \xrightarrow{\text{Br}_2,\ \text{KOH (conc.)}} \text{RNH}_2 \quad (\text{one carbon fewer; } -28\ \text{g mol}^{-1})

Worked example

Which amide gives propan-1-amine by Hofmann degradation, and which nitrile gives it by Mendius reduction?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2AminesMODERATE
Which from following compounds is obtained when acetamide is warmed with bromine and excess conc. KOH(aq)KOH_{(aq)} solution?

[Q67 · 21 April Shift II · 2025]

Losing CO₂ (44) instead of CO (28)

The amide's carbonyl leaves as carbonate, but the MOLECULE loses C=O: 12 + 16 = 28. Amine mass = amide mass − 28.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Reductions: Nitrile, Amide and Nitro to Amine

    Three reductions

    RCN→4[H]RCH2NH2;RCONH2→LiAlH4RCH2NH2;RNO2→6[H]RNH2\text{RCN} \xrightarrow{4[\text{H}]} \text{RCH}_2\text{NH}_2;\quad \text{RCONH}_2 \xrightarrow{\text{LiAlH}_4} \text{RCH}_2\text{NH}_2;\quad \text{RNO}_2 \xrightarrow{6[\text{H}]} \text{RNH}_2
  • Hofmann Bromamide Degradation, Gabriel Synthesis and Ammonolysis

    Hofmann bromamide degradation

    RCONH2→Br2, KOH (conc.)RNH2(one carbon fewer; −28 g mol−1)\text{RCONH}_2 \xrightarrow{\text{Br}_2,\ \text{KOH (conc.)}} \text{RNH}_2 \quad (\text{one carbon fewer; } -28\ \text{g mol}^{-1})

Watch out for (2)

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