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.

All Amines notes MHT-CET Chemistry strategy

Why this matters

32 PYQs, none HARD. Of the fifteen set from 2021 to 2025, 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

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

Three reductions

\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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q97Moderate

Example 1 · Amines · Preparation of Amines

Identify the product 'B' in the following sequence of reactions.

\text{CH}_3\text{Br}\xrightarrow{\text{KCN}}\text{A}\xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}}\text{B}

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.

Drill 9 more on reductions: nitrile, amide and nitro to amine

Concept 2 of 2: Hofmann Bromamide Degradation, Gabriel Synthesis and Ammonolysis

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: $\text{RCONH}_2 + \text{Br}_2 + 4\text{KOH} \to \text{RNH}_2 + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O}$ . Acetamide → methylamine. Loss in molar mass = CO = 28 g mol⁻¹.
Gabriel: phthalimide $\xrightarrow{\text{KOH}}$ potassium phthalimide $\xrightarrow{\text{R-X}}$ N-alkylphthalimide $\xrightarrow{\text{NaOH(aq)}}$ primary amine + sodium phthalate. NOT formed: phthalic acid. Aryl halides do not work.
Ammonolysis: benzyl chloride + NH₃ → benzylamine; then 2 CH₃I → $\text{C}_6\text{H}_5\text{CH}_2\text{N(CH}_3)_2$ (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

\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?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift II · Q67Moderate

Example 2 · Amines · Preparation of Amines

Which from following compounds is obtained when acetamide is warmed with bromine and excess conc.

KOH_{(aq)}

solution?

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.

Drill 4 more on hofmann bromamide degradation, gabriel synthesis and ammonolysis

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
$\text{RCN} \xrightarrow{4[\text{H}]} \text{RCH}_2\text{NH}_2;\quad \text{RCONH}_2 \xrightarrow{\text{LiAlH}_4} \text{RCH}_2\text{NH}_2;\quad \text{RNO}_2 \xrightarrow{6[\text{H}]} \text{RNH}_2$
Hofmann Bromamide Degradation, Gabriel Synthesis and Ammonolysis
Hofmann bromamide degradation
$\text{RCONH}_2 \xrightarrow{\text{Br}_2,\ \text{KOH (conc.)}} \text{RNH}_2 \quad (\text{one carbon fewer; } -28\ \text{g mol}^{-1})$

Watch out for (2)

Forgetting that cyanide adds a carbon
→ Reductions: Nitrile, Amide and Nitro to Amine
Losing CO₂ (44) instead of CO (28)
→ Hofmann Bromamide Degradation, Gabriel Synthesis and Ammonolysis

Test yourself on Amines

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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Related notes

Halogen Derivatives — KCN versus AgCN on the halide →