MHT-CET Chemistry · Amines

Chemical Reactions and Basicity of Amines

Amines are bases — in water the secondary alkylamine is strongest and aniline weakest, so pKb runs the other way — and their reactions sort them by degree: acylation and the Hinsberg test need an N–H, the carbylamine test needs NH₂, exhaustive methylation and Hofmann elimination take any amine to an alkene.

All Amines notes MHT-CET Chemistry strategy

Why this matters

28 PYQs, 3 HARD — the largest page in the chapter. Fourteen are basicity in one wording or another (highest or lowest pKb, the aqueous methylamine order, the stability of R₃NH⁺); six are the tests (Hinsberg's reagent by name and formula, carbylamine's product and which amine gives it, which amine cannot be acylated); eight are exhaustive methylation and Hofmann elimination, including the two HARD rows on which alkene leaves a triethylpropylammonium salt. Three cards.

Concept 1 of 3

Basicity: the Aqueous Order and pKb

Intuition

Alkyl groups push electrons onto N (+I), so in the gas phase basicity rises 1° < 2° < 3°. In WATER the ammonium ion must also be solvated, and a tertiary ion has only one N–H to hydrogen-bond with — so the secondary amine wins and the tertiary drops to third: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃. Aniline is far weaker than all of them because its lone pair is delocalised into the ring. Low pKb = strong base.

Definition

Aqueous: $(\text{CH}_3)_2\text{NH} > \text{CH}_3\text{NH}_2 > (\text{CH}_3)_3\text{N} > \text{NH}_3$ ; for ethyl: $(\text{C}_2\text{H}_5)_2\text{NH} > (\text{C}_2\text{H}_5)_3\text{N} > \text{C}_2\text{H}_5\text{NH}_2 > \text{NH}_3$ . 'Expected' (+I only, as the 2025 paper keys): $\text{NH}_3 < \text{RNH}_2 < \text{R}_2\text{NH} < \text{R}_3\text{N}$ .
Strongest base among ammonia, ethylamine, diethylamine, triethylamine: diethylamine. Lowest pKb among N-methylethanamine, propan-2-amine, NH₃, aniline: N-methylethanamine.
Aromatic amines: aniline weakest (highest pKb); benzylamine $\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$ is aliphatic and much stronger. Order: p-toluidine > aniline > p-nitroaniline.
Conjugate-acid stability: $\text{R}_3\text{NH}^+ > \text{R}_2\text{NH}_2^+ > \text{RNH}_3^+ > \text{NH}_4^+$ (+I); NH₄⁺ least stable.
pKb decreasing: $\text{NH}_3 > \text{RNH}_2 > \text{R}_2\text{NH}$ .

Aqueous basicity

\text{R}_2\text{NH} > \text{RNH}_2 \gtrless \text{R}_3\text{N} > \text{NH}_3 \gg \text{C}_6\text{H}_5\text{NH}_2;\qquad \text{low } pK_b = \text{strong base}

Worked example

Arrange aniline, ammonia, methylamine and dimethylamine by increasing pKb, and say which of N-methylaniline and benzylamine is the stronger base.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1AminesMODERATE

Which from following is a correct increasing order of basic strength of compounds?

[Q97 · 23 April Shift I · 2025]

Reading pKb as basic strength

pKb is −log Kb: the STRONGER the base, the SMALLER the number. 'Highest pKb' asks for the weakest base — aniline — and the strongest base has the lowest.

Drill 13 more on basicity: the aqueous order and pkb

Concept 2 of 3

Acylation, the Carbylamine Test and Hinsberg's Reagent

Intuition

Each reaction needs a particular hydrogen on nitrogen. Acylation (acetyl chloride, acetic anhydride) replaces an N–H, so primary and secondary amines react and tertiary ones do not. The carbylamine reaction needs BOTH hydrogens: only a primary amine with chloroform and alcoholic KOH gives the foul-smelling isocyanide. Hinsberg's reagent, benzenesulphonyl chloride, gives a KOH-soluble sulphonamide with a primary amine, an insoluble one with a secondary amine, and nothing with a tertiary amine.

Definition

Acylation: $\text{RNH}_2$ or $\text{R}_2\text{NH}$ + $\text{CH}_3\text{COCl}$ → N-acyl amide. Does NOT react: $(\text{C}_2\text{H}_5)_3\text{N}$ , N,N-dimethylaniline, ethyldimethylamine. N-Methylaniline DOES.
Carbylamine: $\text{RNH}_2 + \text{CHCl}_3 + 3\text{KOH} \to \text{R-NC}$ (alkyl isocyanide) + 3KCl + 3H₂O. Foul smell; primary amines only — ethylamine yes, dimethylamine and trimethylamine no.
Hinsberg's reagent = benzenesulphonyl chloride $\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$ . 1°: sulphonamide soluble in alkali; 2°: insoluble; 3°: no reaction.
Aniline + Br₂ water → 2,4,6-tribromoaniline; aniline + H₂SO₄ → sulphanilic acid; nitration of aniline needs acetylation first.

Carbylamine reaction

\text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{R-N≡C} + 3\text{KCl} + 3\text{H}_2\text{O}

Worked example

Three liquids are ethylamine, diethylamine and triethylamine. Use Hinsberg's reagent and the carbylamine test to identify each.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2AminesEASY

Which one of the following compounds does not react with acetyl chloride?

[Q71 · 9th May Shift 2 · 2023]

Giving the carbylamine test to a secondary amine

It needs two N–H hydrogens to form the isocyanide; secondary and tertiary amines give no smell. Acylation, by contrast, needs only ONE N–H — so secondary amines acylate but do not give carbylamine.

Drill 5 more on acylation, the carbylamine test and hinsberg's reagent

Concept 3 of 3

Exhaustive Methylation and Hofmann Elimination

Intuition

An amine keeps attacking methyl iodide until nitrogen carries four groups — a quaternary ammonium iodide; from a primary amine that takes THREE CH₃I. Moist Ag₂O swaps the iodide for hydroxide, and heating eliminates: the hydroxide takes a β-hydrogen from the LEAST substituted alkyl group and that group leaves as an alkene (Hofmann, anti-Saytzeff), the rest staying as a tertiary amine. On a triethylpropylammonium salt the ethyl's β-H is the more accessible, so ETHENE leaves — n moles of it from n moles of salt.

Definition

$\text{CH}_3\text{NH}_2 \xrightarrow{3\,\text{CH}_3\text{I}} (\text{CH}_3)_4\text{N}^+\text{I}^-$ : three moles of iodomethane (two N–H replaced, then the lone pair).
Hofmann elimination: $\text{R}_4\text{N}^+\text{X}^- \xrightarrow{\text{moist Ag}_2\text{O}} \text{R}_4\text{N}^+\text{OH}^- \xrightarrow{\Delta} \text{alkene} + \text{R}_3\text{N} + \text{H}_2\text{O}$ . This is the ELIMINATION; $\text{RCONH}_2 \to \text{RNH}_2$ is the DEGRADATION; $\text{RNH}_2 \to \text{R}_4\text{NX}$ is exhaustive methylation.
Least substituted alkene leaves: $\text{C}_3\text{H}_7\text{N}^+(\text{C}_2\text{H}_5)_3\text{I}^-$ → ethene (n mol per n mol salt) + triethylamine-free amine $\text{C}_3\text{H}_7\text{N(C}_2\text{H}_5)_2$ ; not propene.
Diethyldimethylammonium hydroxide → $\text{CH}_2\text{=CH}_2$ + $\text{CH}_3\text{CH}_2\text{N(CH}_3)_2$ .

Hofmann elimination

\text{R}_3\text{N}^+\text{-CH}_2\text{CH}_3\ \text{OH}^- \xrightarrow{\Delta} \text{R}_3\text{N} + \text{CH}_2\text{=CH}_2 + \text{H}_2\text{O} \quad (\text{least substituted alkene})

Worked example

Ethylamine is treated with excess CH₃I, then moist Ag₂O, then heated. Give the alkene and the amine formed, and the number of CH₃I consumed.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3AminesHARD

Identify the alkene obtained as major product in the following Hofmann elimination reaction.

\text{C}_3\text{H}_7\overset{+}{N}(\text{C}_2\text{H}_5)_3\ \text{I}^-\xrightarrow{\text{moist Ag}_2\text{O}/\Delta} A\xrightarrow{\Delta,-\text{H}_2\text{O}}

Alkene + Amine

[Q78 · 3rd May 2nd Shift · 2023]

Eliminating the propyl group because it is 'bigger'

Hofmann elimination removes the β-hydrogen that is easiest to reach — on the LEAST substituted, least hindered alkyl. Ethyl beats propyl: ethene forms, propene does not.

Drill 7 more on exhaustive methylation and hofmann elimination

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 (3)

Basicity: the Aqueous Order and pKb
Aqueous basicity
$\text{R}_2\text{NH} > \text{RNH}_2 \gtrless \text{R}_3\text{N} > \text{NH}_3 \gg \text{C}_6\text{H}_5\text{NH}_2;\qquad \text{low } pK_b = \text{strong base}$
Acylation, the Carbylamine Test and Hinsberg's Reagent
Carbylamine reaction
$\text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{R-N≡C} + 3\text{KCl} + 3\text{H}_2\text{O}$
Exhaustive Methylation and Hofmann Elimination
Hofmann elimination
$\text{R}_3\text{N}^+\text{-CH}_2\text{CH}_3\ \text{OH}^- \xrightarrow{\Delta} \text{R}_3\text{N} + \text{CH}_2\text{=CH}_2 + \text{H}_2\text{O} \quad (\text{least substituted alkene})$

Watch out for (3)

Reading pKb as basic strength
→ Basicity: the Aqueous Order and pKb
Giving the carbylamine test to a secondary amine
→ Acylation, the Carbylamine Test and Hinsberg's Reagent
Eliminating the propyl group because it is 'bigger'
→ Exhaustive Methylation and Hofmann Elimination

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