Substitution & Elimination — SN1, SN2, E1 and E2

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Give an alkyl halide a reagent and four things can happen: SN1, SN2, E1 or E2. Exams test whether you can choose. The choice depends on only four questions, and once you ask them in order the answer is rarely in doubt.

What you will be able to do

• Write the mechanism, rate law and stereochemical result of SN1, SN2, E1 and E2
• Rank substrates, leaving groups and nucleophiles for each pathway, and explain the role of solvent
• Predict carbocation rearrangements (hydride and methyl shifts) and their products
• Choose the major alkene by Saytzeff’s or Hofmann’s rule, including anti-periplanar E2
• Predict products of ambident nucleophiles (KCN/AgCN, KNO₂/AgNO₂) and explain why aryl halides resist substitution

The big picture

Four questions, asked in this order, decide the path.

In one line: The substrate (methyl, 1°, 2°, 3°), the reagent (strong or weak, small or bulky, nucleophile or base), the solvent (protic or aprotic) and the temperature together decide between SN1, SN2, E1 and E2.

1. What is the substrate? Methyl and 1° favour SN2; 3° cannot do SN2.
2. What is the reagent? Strong nucleophile → SN2; strong base → E2; weak nucleophile/base → SN1 or E1.
3. What is the solvent? Polar aprotic speeds SN2; polar protic stabilises ions and favours SN1/E1.
4. Is it heated? Heat favours elimination over substitution.

Think of it like this: SN2 is a hold-up at a narrow doorway: the nucleophile pushes in from the back while the leaving group walks out the front, all in one move — crowding blocks it. SN1 is a two-step queue: the leaving group leaves first, the carbocation waits, and anyone can come in from either side.

	SN2	SN1	E2	E1
Steps	One (concerted)	Two (via carbocation)	One (concerted)	Two (via carbocation)
Rate law	k[RX][Nu⁻]	k[RX]	k[RX][B⁻]	k[RX]
Best substrate	CH₃ > 1° > 2°	3° > 2°; benzylic, allylic	3° > 2° > 1°	3° > 2°
Reagent	Strong nucleophile	Weak nucleophile	Strong base	Weak base
Solvent	Polar aprotic	Polar protic	—	Polar protic
Stereo / product	Inversion	Racemisation (mostly)	Anti-periplanar; Saytzeff	Saytzeff
Rearrangement	Never	Possible	Never	Possible

SN2 — bimolecular substitution

One step, backside attack, inversion.

The nucleophile attacks the carbon from the side opposite the leaving group. In the transition state the carbon is partly bonded to both (five groups around it, trigonal bipyramidal). As the leaving group departs the other three groups flip over, like an umbrella in the wind.

Rate = k [RX] [Nu⁻]

Second order — doubling either concentration doubles the rate

Key idea: SN2 always gives inversion of configuration (Walden inversion). (R)-2-bromobutane with OH⁻ by SN2 gives (S)-butan-2-ol. If the nucleophile changes the priority order, the label may not flip even though the geometry does — check priorities each time.

• Reactivity: CH₃X > 1° > 2° ≫ 3° (3° does not react by SN2). Bulky groups block the back side.
• Neopentyl bromide (CH₃)₃CCH₂Br is a 1° halide yet reacts extremely slowly by SN2 — the tert-butyl group next door blocks the attack.
• Polar aprotic solvents (DMSO, DMF, acetone) speed SN2: they solvate the cation but leave the anion “naked” and reactive.
• Finkelstein reaction: R–Cl + NaI in dry acetone → R–I + NaCl↓. NaCl is insoluble in acetone, which pulls the equilibrium forward.

Solved example 01 — Stereochemistry of SN2

(R)-2-bromobutane reacts with NaOH in acetone. Name the product with its configuration.

1. A 2° halide, strong nucleophile, polar aprotic solvent → SN2.
2. OH⁻ attacks from the back side; the configuration inverts.
3. OH takes the place of Br in the priority order (both are highest), so the label flips too.

Answer: (S)-butan-2-ol.

SN1 — unimolecular substitution

Two steps, a flat carbocation, a mixture of products.

1. Slow step: R–X → R⁺ + X⁻ (ionisation; this step fixes the rate).
2. Fast step: R⁺ + Nu⁻ → R–Nu.

Rate = k [RX]

First order — the nucleophile is not in the rate law; adding more of it does not speed the reaction

• Reactivity follows carbocation stability: 3° > 2° > 1° > CH₃. Benzylic and allylic cations (resonance-stabilised) also react fast.
• Polar protic solvents (water, alcohols, formic acid) stabilise both the carbocation and the leaving anion. Reactions in water or alcohol as solvent are called solvolysis.
• Stereochemistry: the carbocation is planar (sp²) and can be attacked from either face → racemisation. In practice there is slightly more inversion, because the departing X⁻ briefly shields its own face.
• Rearrangement: a less stable carbocation shifts H⁻ or CH₃⁻ from the next carbon to become more stable (2° → 3°).

Solved example 02 — Hydride shift

Predict the main product when 3-methylbutan-2-ol reacts with HBr.

1. Protonation and loss of water gives a 2° carbocation at C2.
2. A hydride shifts from C3 to C2, giving a more stable 3° carbocation at C3.
3. Br⁻ attacks the 3° carbocation.

Answer: 2-Bromo-2-methylbutane (not 2-bromo-3-methylbutane).

Solved example 03 — Methyl shift

Neopentyl bromide is heated in water. What is the main substitution product?

1. SN2 is blocked by the tert-butyl group; slow ionisation gives a very unstable 1° carbocation.
2. A methyl group shifts to form the 3° 2-methylbut-2-yl cation.
3. Water attacks the 3° carbocation; loss of H⁺ gives the alcohol.

Answer: 2-Methylbutan-2-ol (with some alkene by E1).

Check yourself

• C1. Which reacts faster by SN1: benzyl chloride or chlorobenzene?
• C2. What happens to the SN1 rate if [Nu⁻] is doubled?

Answers at the end.

Leaving groups and nucleophiles

Who leaves easily, and who attacks well.

Leaving group ability

A good leaving group is a weak base — stable once it has left with the electron pair. I⁻ > Br⁻ > Cl⁻ > F⁻. So reactivity is R–I > R–Br > R–Cl > R–F for both SN1 and SN2. OH⁻ is a poor leaving group, which is why alcohols need acid first (to make H₂O the leaving group).

Nucleophilicity

Rule	Order	Reason
Same attacking atom	RO⁻ > HO⁻ > ROH; NH₂⁻ > NH₃	Negative charge increases nucleophilicity
Across a period	CH₃⁻ > NH₂⁻ > OH⁻ > F⁻	Less electronegative atom holds electrons less tightly
Down a group, polar protic solvent	I⁻ > Br⁻ > Cl⁻ > F⁻	Small F⁻ is heavily H-bonded and caged by the solvent
Down a group, polar aprotic solvent	F⁻ > Cl⁻ > Br⁻ > I⁻	Ions are free; basicity decides
Bulk	CH₃O⁻ > (CH₃)₃CO⁻	tert-Butoxide is a strong base but a poor nucleophile

Ambident nucleophiles

Reagent	Bond type	Attacking atom	Product with R–X
KCN	Mainly ionic	C (the more nucleophilic end)	Alkyl cyanide R–C≡N
AgCN	Mainly covalent (Ag–C)	N (C is tied up)	Alkyl isocyanide R–N≡C
KNO₂	Ionic	O	Alkyl nitrite R–O–N=O
AgNO₂	Covalent	N	Nitroalkane R–NO₂

Key idea: K gives the “normal” atom, Ag forces the other one. With KCN and KNO₂ the free ion attacks through C and O respectively; the silver salts are covalent, so the bound atom can’t attack and the other end does.

E2 — bimolecular elimination

One step, a strong base, and a strict geometry rule.

The base removes a β-hydrogen while the C–X bond breaks and the C=C forms, all at once. Strong bases favour it: alcoholic KOH, NaOEt in ethanol, t-BuOK.

Rate = k [RX] [B⁻]

Second order

Key idea: Anti-periplanar requirement: the β-H and the leaving group must be on opposite sides, in one plane (dihedral angle 180°). In a cyclohexane ring both must be axial — if no axial β-H sits anti to an axial X, E2 cannot happen from that conformation.

Which alkene? Saytzeff versus Hofmann

• Saytzeff (Zaitsev) rule: with a small base, the more substituted (more stable) alkene is major. 2-Bromobutane + alc. KOH → but-2-ene (mainly trans) > but-1-ene.
• Hofmann product: a bulky base such as t-BuOK removes the more accessible H on the less substituted carbon → the less substituted alkene is major.
• Hofmann elimination of quaternary ammonium hydroxides also gives the less substituted alkene.

Solved example 04 — Base size changes the product

Give the major alkene from 2-bromo-2-methylbutane with (a) NaOEt in ethanol, (b) t-BuOK.

1. Two kinds of β-H: on CH₃ (gives a terminal alkene) and on CH₂ (gives an internal alkene).
2. (a) Small base: Saytzeff → the more substituted alkene.
3. (b) Bulky base: attacks the less hindered CH₃ hydrogens → Hofmann.

Answer: (a) 2-Methylbut-2-ene. (b) 2-Methylbut-1-ene.

E1 — unimolecular elimination

The carbocation loses a proton instead of gaining a nucleophile.

• Same first step as SN1: R–X → R⁺ + X⁻ (slow). Then a weak base (often the solvent) removes a β-H.
• Rate = k[RX]. Favoured by 3° substrates, weak bases, polar protic solvents and heat.
• Always competes with SN1; higher temperature shifts the balance toward elimination.
• Gives the Saytzeff product and can rearrange, just like SN1.
• Acid-catalysed dehydration of alcohols (conc. H₂SO₄, heat) is an E1 reaction for 2° and 3° alcohols: ease 3° > 2° > 1°.

Check yourself

• C3. Why does t-BuOK favour elimination over substitution?
• C4. Which alkene is major when 2-bromobutane is heated with alcoholic KOH?

Answers at the end.

Choosing the pathway

The decision table most questions reduce to.

Substrate	Strong nucleophile, weak base (I⁻, CN⁻, RS⁻, N₃⁻)	Strong base (OH⁻, RO⁻)	Bulky strong base (t-BuO⁻)	Weak nucleophile/base (H₂O, ROH)
CH₃X	SN2	SN2	SN2	No reaction (slow)
1°	SN2	SN2 (E2 if heated)	E2	Very slow
2°	SN2	E2 mainly	E2	SN1 / E1 (slow)
3°	SN1 / E1	E2	E2	SN1 (cold) / E1 (heat)

• Aqueous KOH → substitution (alcohol). Alcoholic KOH → elimination (alkene). The ethoxide formed in ethanol is a stronger base, and alcohol solvates OH⁻ less.
• Higher temperature always increases the proportion of elimination.

Solved example 05 — Aqueous vs alcoholic

2-Bromopropane is treated with (a) aqueous KOH, (b) alcoholic KOH and heat. Give the main products.

1. (a) Water as solvent, OH⁻ as nucleophile, no heat → substitution.
2. (b) Alcoholic medium, heat → E2 elimination.

Answer: (a) Propan-2-ol. (b) Propene.

Special substrates

Where the rules above need a second look.

• Allylic and benzylic halides are very reactive by both SN1 (resonance-stabilised cation) and SN2 (the π system stabilises the transition state).
• Vinyl and aryl halides (C–X on an sp² carbon) resist nucleophilic substitution: the C–X bond has partial double-bond character from resonance, the sp² carbon holds the bond more tightly, the phenyl cation is very unstable, and the ring’s π cloud repels the nucleophile.
• Aryl halides react under harsh conditions (Dow process: NaOH at 623 K, high pressure) or when electron-withdrawing groups sit ortho or para to X: 2,4,6-trinitrochlorobenzene is hydrolysed by warm water.
• A meta nitro group has little effect: it cannot delocalise the negative charge of the intermediate.

Key idea: Order of reactivity toward aqueous hydrolysis: 2,4,6-trinitrochlorobenzene > 2,4-dinitrochlorobenzene > 4-nitrochlorobenzene > chlorobenzene. Each o/p nitro group stabilises the negative intermediate by resonance.

Exam trap gallery

Where marks are lost in this chapter.

#	What students write	What is actually correct	Why
1	SN1 gives complete inversion.	SN2 gives inversion; SN1 gives mostly racemisation (slight excess inversion).	A planar carbocation can be attacked from both faces.
2	Increasing [Nu⁻] speeds up SN1.	SN1 rate = k[RX] only.	The nucleophile enters after the slow step.
3	3° halides react fastest by SN2.	3° halides do not react by SN2.	Three alkyl groups block backside attack.
4	Neopentyl bromide reacts fast by SN2 because it is 1°.	It is extremely slow by SN2.	The neighbouring tert-butyl group blocks the back side.
5	KCN and AgCN both give alkyl cyanides.	KCN → R–CN; AgCN → R–NC (isocyanide).	Ag–C bond in AgCN leaves N to attack.
6	Aqueous and alcoholic KOH give the same product.	Aqueous → alcohol; alcoholic → alkene.	Ethoxide in ethanol is a strong base; heat favours E2.
7	t-BuOK gives the Saytzeff alkene.	t-BuOK gives the Hofmann (less substituted) alkene.	A bulky base takes the most accessible hydrogen.
8	3-Methylbutan-2-ol + HBr gives 2-bromo-3-methylbutane.	Main product: 2-bromo-2-methylbutane.	2° cation rearranges to 3° by a hydride shift.
9	In polar protic solvents F⁻ is the best nucleophile among the halides.	In protic solvents I⁻ is best; F⁻ is best only in aprotic solvents.	H bonding cages the small F⁻ ion.

Memory hook: “SN2: two in the rate, turns it inside-out” — second order, inversion. · “SN1: one in the rate, flat in between” — first order, planar carbocation, racemisation. · “Small base Saytzeff, big base Hofmann.” · “Water substitutes, alcohol eliminates” — aqueous vs alcoholic KOH.

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. The rate of an SN2 reaction depends on: (Easy)

• (a) [RX] only
• (b) [Nu⁻] only
• (c) both [RX] and [Nu⁻]
• (d) neither

Q02. Which reacts fastest by SN2? (Easy)

• (a) CH₃Br
• (b) (CH₃)₂CHBr
• (c) (CH₃)₃CBr
• (d) (CH₃)₃CCH₂Br

Q03. Which reacts fastest by SN1? (Easy)

• (a) CH₃Cl
• (b) CH₃CH₂Cl
• (c) (CH₃)₂CHCl
• (d) (CH₃)₃CCl

Q04. The best leaving group is: (Easy)

• (a) F⁻
• (b) Cl⁻
• (c) Br⁻
• (d) I⁻

Q05. Ethyl bromide with alcoholic KOH mainly gives: (Easy)

• (a) ethanol
• (b) ethene
• (c) diethyl ether
• (d) ethane

Q06. An alkyl halide reacts with AgCN to give mainly: (Moderate)

• (a) alkyl cyanide
• (b) alkyl isocyanide
• (c) carboxylic acid
• (d) amine

Q07. A nitroalkane is the main product when R–X reacts with: (Moderate)

• (a) KNO₂
• (b) AgNO₂
• (c) NaNO₃
• (d) HNO₂

Q08. SN1 reaction of an optically active 2° halide gives: (Moderate)

• (a) complete retention
• (b) complete inversion
• (c) racemisation (with some excess inversion)
• (d) no reaction

Q09. Which solvent best speeds an SN2 reaction of NaCN with CH₃CH₂Br? (Moderate)

• (a) water
• (b) ethanol
• (c) DMSO
• (d) acetic acid

Q10. 2-Bromobutane with alcoholic KOH gives as major product: (Moderate)

• (a) but-1-ene
• (b) but-2-ene
• (c) butan-2-ol
• (d) butane

Q11. Chlorobenzene is less reactive than chloroethane toward nucleophiles because: (Moderate)

• (a) C–Cl bond has partial double-bond character
• (b) Cl is more electronegative in chlorobenzene
• (c) benzene is a stronger acid
• (d) chlorobenzene is non-polar

Q12. Which is hydrolysed most easily? (Moderate)

• (a) chlorobenzene
• (b) 4-nitrochlorobenzene
• (c) 3-nitrochlorobenzene
• (d) 2,4-dinitrochlorobenzene

Q13. 2-Bromo-2-methylbutane with t-BuOK gives mainly: (Tough)

• (a) 2-methylbut-2-ene
• (b) 2-methylbut-1-ene
• (c) 2-methylbutan-2-ol
• (d) 3-methylbut-1-ene

Q14. 3-Methylbutan-2-ol with HBr gives mainly: (Tough)

• (a) 2-bromo-3-methylbutane
• (b) 1-bromo-3-methylbutane
• (c) 2-bromo-2-methylbutane
• (d) 3-methylbut-1-ene

Q15. For E2 in a cyclohexyl halide, the H and the leaving group must be: (Tough)

• (a) both equatorial
• (b) both axial and anti
• (c) cis and equatorial
• (d) one axial, one equatorial on the same side

Answer key with explanations

1. (c) both [RX] and [Nu⁻] — Bimolecular, one step.
2. (a) CH₃Br — Least hindered.
3. (d) (CH₃)₃CCl — Most stable 3° carbocation.
4. (d) I⁻ — Weakest base.
5. (b) ethene — Alcoholic KOH → elimination.
6. (b) alkyl isocyanide — Attack through N.
7. (b) AgNO₂ — Covalent AgNO₂ attacks through N.
8. (c) racemisation (with some excess inversion) — Planar carbocation.
9. (c) DMSO — Polar aprotic.
10. (b) but-2-ene — Saytzeff.
11. (a) C–Cl bond has partial double-bond character — Resonance with the ring.
12. (d) 2,4-dinitrochlorobenzene — Two o/p nitro groups.
13. (b) 2-methylbut-1-ene — Bulky base → Hofmann.
14. (c) 2-bromo-2-methylbutane — Hydride shift to 3° cation.
15. (b) both axial and anti — Anti-periplanar = trans-diaxial.

Advanced problems

Each needs two ideas from the chapter.

Solved example 06 — Rate change

The reaction of (CH₃)₃CBr with water is run with [(CH₃)₃CBr] doubled and an added nucleophile (OH⁻) at double concentration. By what factor does the rate change?

1. 3° substrate, water as solvent → SN1.
2. Rate = k[RX]: doubling [RX] doubles the rate.
3. [Nu⁻] does not appear in the rate law.

Answer: The rate doubles.

Solved example 07 — Which side wins?

Ethyl bromide is treated with (a) NaOEt in ethanol at room temperature, (b) t-BuOK. Predict the main products.

1. (a) 1° substrate, small strong nucleophile/base, no heat → SN2.
2. (b) Bulky base cannot reach the back side easily → E2.

Answer: (a) Diethyl ether (Williamson synthesis). (b) Ethene.

Check-yourself answers

• C1. Benzyl chloride: its benzylic carbocation is resonance-stabilised; chlorobenzene does not ionise.
• C2. No change — [Nu⁻] is not in the SN1 rate law.
• C3. It is a strong base but too bulky to reach the carbon for backside attack, so it removes a β-H instead.
• C4. But-2-ene (mainly trans), by Saytzeff’s rule.

One-page revision

Read this page the night before the exam.

SN2

• Rate = k[RX][Nu⁻]; one step
• Backside attack → inversion
• CH₃ > 1° > 2° ≫ 3°; neopentyl very slow
• Polar aprotic (DMSO, DMF, acetone)

SN1

• Rate = k[RX]; carbocation
• 3° > 2°; benzylic, allylic fast
• Racemisation (slight excess inversion)
• Rearranges: H or CH₃ shift to 3°

Leaving groups and nucleophiles

• I⁻ > Br⁻ > Cl⁻ > F⁻ leaving
• Protic: I⁻ best Nu; aprotic: F⁻ best
• KCN → RCN; AgCN → RNC
• KNO₂ → R–ONO; AgNO₂ → R–NO₂

E2

• Rate = k[RX][B⁻]; strong base
• Anti-periplanar (trans-diaxial in rings)
• Small base: Saytzeff
• Bulky t-BuOK: Hofmann

E1 and choice

• Carbocation, weak base, heat
• Competes with SN1; heat → E1
• Aq. KOH → alcohol; alc. KOH → alkene
• 3° + strong base → E2

Special substrates

• Allylic, benzylic: fast by SN1 and SN2
• Aryl, vinyl: unreactive (partial C=X)
• o/p NO₂ activates aryl halides
• Dow process: 623 K, high pressure