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

42 PYQs, 5 HARD — the largest page in the chapter. Of the twenty-seven set from 2021 to 2025, 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); seven 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

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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 23 April Shift I · Q97Moderate

Example 1 · Amines · Chemical Reactions and Basicity of Amines

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

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

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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 2 · Q80Easy

Example 2 · Amines · Chemical Reactions and Basicity of Amines

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

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

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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 1 · Q78Hard

Example 3 · Amines · Chemical Reactions and Basicity of Amines

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

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 6 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

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.

Take the Amines chapter test Drill all 44 questions