MHT-CET Chemistry · Halogen Derivatives of Alkanes

Nucleophilic Substitution: SN1 and SN2

A nucleophile replaces the halide either in two steps through a planar carbocation (SN1 — tertiary fastest, racemisation at a chiral carbon) or in one concerted backside attack (SN2 — methyl fastest, inversion); the nucleophile's attacking atom decides the product, KCN giving a nitrile and AgCN an isocyanide.

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Why this matters

36 PYQs, none HARD, and now the largest page: 2026 alone added 24. Of the twelve set from 2021 to 2025, half ask a mechanism fact — which is NOT a feature of SN2 (a carbocation), which halide is fastest by SN1 or SN2, which substrate racemises; half ask a product — 2-bromobutane with aqueous NaOH, ethyl bromide with silver acetate or silver propanoate, the reagent for an alkyl nitrite or a nitrile. Two cards.

Concept 1 of 2: SN1 Against SN2: Mechanism, Rate Order and Stereochemistry

SN1: the C–X bond breaks first, a planar carbocation forms, the nucleophile attacks either face — so the rate depends only on the halide (tertiary fastest, stabilised cation) and a chiral centre is racemised. SN2: the nucleophile pushes in from the back while X leaves in ONE step — no intermediate, the rate depends on both reactants, crowding slows it (methyl fastest), and the configuration inverts.

Definition

SN2: single step, bimolecular, backside attack, simultaneous bond making and breaking, inversion (Walden). NO carbocation. Rate: methyl > 1° > 2° > 3°. Fastest among 1-bromobutane, 1-chlorobutane, 2-chlorobutane, 1-iodobutane: the primary IODIDE (best leaving group, unhindered).
SN1: two steps, unimolecular, planar carbocation, racemisation at a stereocentre. Rate: 3° > 2° > 1° — tert-butyl iodide fastest.
Leaving group: I > Br > Cl > F for both.
Racemisation needs a chiral C–X carbon: 2-chlorobutane racemises on SN1 hydrolysis; 2-chloropropane, 3-chloropentane and neopentyl chloride cannot.
tert-Butyl bromide + AgF (SN1) keeps its skeleton: 2-fluoro-2-methylpropane. Tertiary alcohols react fastest with HBr for the same reason.

Rate orders

\text{SN1: } 3^\circ > 2^\circ > 1^\circ > \text{CH}_3\text{X};\qquad \text{SN2: } \text{CH}_3\text{X} > 1^\circ > 2^\circ > 3^\circ

Worked example

(R)-2-bromooctane is hydrolysed once with aqueous NaOH (SN2) and once with water in acetone (SN1). Describe the product's stereochemistry in each case.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 2 · Q63Moderate

Example 1 · Halogen Derivatives of Alkanes · Nucleophilic Substitution Reactions (SN1 and SN2)

Which among the following is NOT a feature of

S_N2

mechanism?

Expecting the 3°-fastest order for SN2

Crowding blocks the backside. SN2 is fastest on the LEAST substituted carbon; tertiary halides go SN1 (or eliminate). The two orders are exact opposites.

Drill 6 more on sn1 against sn2: mechanism, rate order and stereochemistry

Concept 2 of 2: Which Nucleophile Gives Which Product

The product is the nucleophile bonded through its attacking atom. Aqueous OH⁻ gives the alcohol; alkoxide gives an ether; a silver carboxylate gives an ester; KCN attacks through carbon to give a nitrile while AgCN attacks through nitrogen to give an isocyanide; KNO₂ gives an alkyl nitrite while AgNO₂ gives a nitroalkane. The ionic potassium salts free the whole ion; the silver salts are largely covalent, so the atom bonded to silver is tied up and the other end attacks: nitrogen in Ag–C≡N, and nitrogen again in Ag–O–N=O.

Definition

$\text{R-X} + \text{aq. NaOH} \to \text{R-OH}$ : 2-bromobutane → butan-2-ol (alcoholic KOH would give but-2-ene).
$\text{R-X} + \text{R'ONa} \to \text{R-O-R'}$ (Williamson ether synthesis).
$\text{R-X} + \text{R'COOAg} \to \text{R'COOR}$ : ethyl bromide + silver acetate → ethyl acetate $\text{CH}_3\text{COOC}_2\text{H}_5$ ; + silver propanoate → ethyl propanoate.
$\text{R-X} + \text{KCN (alc.)} \to \text{R-CN}$ (nitrile, C-attack); $+ \text{AgCN} \to \text{R-NC}$ (isocyanide, N-attack).
$\text{R-X} + \text{KNO}_2 \to \text{R-O-N=O}$ (alkyl nitrite, O-attack); $+ \text{AgNO}_2 \to \text{R-NO}_2$ (nitroalkane, N-attack).
$\text{R-X} + \text{NH}_3 \to$ amines (Hofmann); $+ \text{NaSH} \to \text{R-SH}$ ; $+ \text{NaI} \to \text{R-I}$ (Finkelstein).

Ambident nucleophiles

\text{KCN} \to \text{R-C≡N};\quad \text{AgCN} \to \text{R-N≡C};\qquad \text{KNO}_2 \to \text{R-O-N=O};\quad \text{AgNO}_2 \to \text{R-NO}_2

Worked example

Give the products of 1-bromopropane with (i) alcoholic KCN, (ii) AgNO₂, (iii) silver acetate.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q76Moderate

Example 2 · Halogen Derivatives of Alkanes · Nucleophilic Substitution Reactions (SN1 and SN2)

Identify the product obtained in the following reaction.

\text{CH}_{3}\text{CH}_{2}\text{Br} + \text{CH}_{3}\text{COOAg} \xrightarrow{\Delta} x + \text{AgBr}

Writing the ester the wrong way round

The acyl part comes from the silver salt, the alkyl from the halide: silver ACETATE + ETHYL bromide is CH₃COO–C₂H₅, ethyl acetate — not ethyl-COO-methyl.

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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)

SN1 Against SN2: Mechanism, Rate Order and Stereochemistry
Rate orders
$\text{SN1: } 3^\circ > 2^\circ > 1^\circ > \text{CH}_3\text{X};\qquad \text{SN2: } \text{CH}_3\text{X} > 1^\circ > 2^\circ > 3^\circ$
Which Nucleophile Gives Which Product
Ambident nucleophiles
$\text{KCN} \to \text{R-C≡N};\quad \text{AgCN} \to \text{R-N≡C};\qquad \text{KNO}_2 \to \text{R-O-N=O};\quad \text{AgNO}_2 \to \text{R-NO}_2$

Watch out for (2)

Expecting the 3°-fastest order for SN2
→ SN1 Against SN2: Mechanism, Rate Order and Stereochemistry
Writing the ester the wrong way round
→ Which Nucleophile Gives Which Product

Test yourself on Halogen Derivatives of Alkanes

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Related notes

Basic Principles — carbocation stability behind the SN1 order →