Substitution vs Elimination

Spec Mapping — OCR H432 Module 4.2.2(e) — Substitution vs elimination in haloalkanes, covering the competition between nucleophilic substitution and β-elimination when a haloalkane meets a base; the effect of solvent (aqueous vs ethanolic), temperature, base/nucleophile strength and the class of the haloalkane (primary, secondary, tertiary) on the product distribution; the SN1/SN2/E1/E2 four-way mechanism map at A-Level depth; Zaitsev selectivity for unsymmetric substrates; the synthetic implications for the four-corner interconversion of alcohols, alkenes and haloalkanes (refer to the official OCR H432 specification document for exact wording).

The chemistry of haloalkanes plus a base is the OCR A-Level student's first encounter with competing reaction pathways, in which one set of starting materials gives two very different products and the chemist's job is to choose conditions that steer the reaction towards one outcome rather than the other. With aqueous hydroxide and gentle heat, a haloalkane gives an alcohol by nucleophilic substitution. With ethanolic hydroxide and reflux temperature, the same haloalkane gives an alkene by β-elimination. Same reagent, different solvent — completely different product. Underneath the simple solvent rule sit four mechanisms (SN1, SN2, E1, E2) whose relative rates respond to (i) the class of the haloalkane (primary, secondary, tertiary), (ii) the strength and steric bulk of the base, (iii) the temperature, and (iv) the dielectric properties of the solvent. This lesson develops the synoptic comparison: side-by-side equations of substitution and elimination on the same substrate, the E2 mechanism with curly arrows (not required for OCR H432 marks but useful for interpretation), the four-factor table that controls the pathway choice, Zaitsev product selectivity for unsymmetric substrates, the synthetic "square" of interconversion between alcohols, alkenes and haloalkanes, and a moles-based worked calculation of the percentage yield down each branch.

Key Mechanism: E2 (Elimination Bimolecular) is the concerted base-promoted loss of HX from adjacent carbons of a haloalkane to give an alkene. Three curly arrows: (i) base lone pair takes the β-H, (ii) the σ(C-H) electrons fold into a new π(C=C), (iii) the C-X bond electrons leave with the halide. The rate is bimolecular: rate = k[R-X][base], hence the "2". The competing concerted substitution is SN2; the two pathways share the same kinetic order and the same starting materials but produce a π-bond instead of a C-Nu bond.

The two reactions side by side

Take 2-bromopropane, $(CH_3)_2CHBr$ , and treat it with sodium hydroxide. Under aqueous conditions you get the alcohol; under ethanolic conditions you get the alkene.

Substitution (aqueous NaOH, warm):

$CH_3\text{-}CHBr\text{-}CH_3 + OH^- \xrightarrow{\text{aq, warm}} CH_3\text{-}CHOH\text{-}CH_3 + Br^-$

Hydroxide attacks the δ+ carbon of the C-Br bond, the bromide leaves, and the product is propan-2-ol.

Elimination (ethanolic NaOH, hot/reflux):

$CH_3\text{-}CHBr\text{-}CH_3 + OH^- \xrightarrow{\text{ethanolic, hot}} CH_3\text{-}CH\text{=}CH_2 + Br^- + H_2O$

The hydroxide acts as a base rather than a nucleophile, removing a β-H from the methyl carbon. The σ(C-H) electrons become the new π(C=C), bromide leaves, and the product is propene (with water as co-product).

flowchart TD
  A[2-Bromopropane + NaOH] --> B{Conditions?}
  B -->|Aqueous NaOH<br/>warm 40-60 C| C["SUBSTITUTION<br/>Propan-2-ol + NaBr"]
  B -->|Ethanolic NaOH<br/>hot reflux 70+ C| D["ELIMINATION<br/>Propene + NaBr + H2O"]

Same starting material, same reagent, same overall stoichiometry of one molecule of base per molecule of haloalkane — but the solvent and temperature steer the outcome.

The four mechanisms

At A-Level OCR identifies four mechanistic options. Their kinetic order and rate-determining step are:

Mechanism	Full name	Rate law	RDS	Substrate preference
SN2	Substitution, Nucleophilic, Bimolecular	rate = k[R-X][Nu]	Concerted attack	Primary
SN1	Substitution, Nucleophilic, Unimolecular	rate = k[R-X]	C-X ionisation	Tertiary
E2	Elimination, Bimolecular	rate = k[R-X][base]	Concerted	Primary or secondary with strong base
E1	Elimination, Unimolecular	rate = k[R-X]	C-X ionisation	Tertiary

The bimolecular pathways (SN2 and E2) require both the substrate and the base/nucleophile to be present in the rate-determining transition state. The unimolecular pathways (SN1 and E1) go via a carbocation intermediate formed in a slow C-X ionisation step that does not involve the nucleophile or base.

The E2 transition state

flowchart LR
  A["Base: + H-C(beta)-C(alpha)-X"] -->|concerted 3 arrows<br/>antiperiplanar geometry| B["Base-H + C=C + X-"]

The three concerted curly arrows are:

From the lone pair on the base to the β-H.
From the C-H σ-bond electrons into the C-C σ-framework (folding to form a new π-bond).
From the C-X bond electrons onto the leaving halide.

For E2 to work, the β-H and the X must lie on opposite sides of the C-C axis (antiperiplanar geometry, ~180°). This geometric constraint matters at university level; at OCR A-Level it is enough to state that the H and X are removed from adjacent carbons.

The β-hydrogen requirement

Elimination needs a hydrogen on a carbon adjacent to the C-X carbon (the β-carbon). If there is no β-H, elimination is impossible and only substitution can occur. Examples:

Haloalkane	β-H present?	Possible reactions
Halomethane, CH₃X	No	Substitution only
Halomethyl-benzene, PhCH₂X	No (benzylic α-H is not β to X)	Substitution only
Neopentyl halide, (CH₃)₃CCH₂X	No (no H on the (CH₃)₃C carbon)	Substitution only — slow, SN2 sterically blocked
1-Bromopropane, CH₃CH₂CH₂Br	Yes (on C2)	Both, substitution dominant under aqueous conditions
2-Bromopropane, (CH₃)₂CHBr	Yes (on both methyls)	Both, controlled by solvent and temperature
2-Bromo-2-methylpropane, (CH₃)₃CBr	Yes (on each methyl)	Elimination dominant in nearly any base

The four-factor table — what controls the pathway

OCR examines a simple but reliable set of rules. Commit this table to memory:

Factor	Favours Substitution	Favours Elimination
Solvent	Water (aqueous)	Ethanol (ethanolic)
Temperature	Warm (~40-60 °C)	Hot, reflux (~70 °C and above)
Base/nucleophile size	Small, high charge density (OH⁻, CN⁻, NH₃)	Bulky, sterically demanding (e.g. tert-butoxide)
Class of haloalkane	Primary	Tertiary
Base concentration	Dilute	Concentrated

Solvent — why aqueous gives substitution and ethanolic gives elimination

In water, hydroxide ion is heavily solvated by hydrogen bonding (water is an excellent H-bond donor to the O⁻ lone pairs). The solvent cage tempers the basicity of OH⁻ (a hydrogen-bonded proton-acceptor cannot easily accept a new proton from a β-C). The exposed lone pair still points outward and remains a competent nucleophile for SN2 attack on the δ+ carbon. So aqueous OH⁻ substitutes.

In ethanol, OH⁻ is much less effectively solvated — ethanol is a poorer H-bond donor than water, so the lone pairs on O⁻ are less shielded. The exposed, less-hindered O⁻ behaves more as a strong base, abstracting the β-H to drive elimination. The same hydroxide species, identical chemical formula, but a different micro-environment.

Temperature

Elimination is favoured at higher temperature for two reasons:

Entropy. Substitution converts one molecule of R-X plus one of base into one molecule of R-Nu plus one halide (no net change in number of particles). Elimination converts one of R-X plus one of base into one alkene plus water plus one halide (one extra particle). The Gibbs equation $\Delta G = \Delta H - T\Delta S$ tells us the entropy-favoured pathway becomes more competitive as $T$ rises.
Activation energy. Elimination usually has a slightly higher $E_a$ than substitution for the same substrate. Higher temperature compresses the rate gap by populating both transition states more equally.

Base steric bulk

Small nucleophiles like OH⁻ and CN⁻ can attack the δ+ carbon for substitution. Bulky bases like tert-butoxide $(CH_3)_3CO^-$ cannot fit past the alkyl substituents of the haloalkane carbon, but can still reach the more exposed β-H. So bulky bases tip the balance towards elimination.

Substrate class

flowchart LR
  A[Primary R-CH2-X] -->|aq OH-, warm| B[Alcohol via SN2 - clean]
  A -->|ethanolic OH-, hot| C[Some alkene, alcohol still significant]
  D[Secondary R2CH-X] -->|aq OH-, warm| E[Mostly alcohol, some alkene]
  D -->|ethanolic OH-, hot| F[Mostly alkene]
  G[Tertiary R3C-X] -->|aq OH-, warm| H[Mixture, elimination already significant]
  G -->|ethanolic OH-, hot| I[Alkene almost exclusively via E1 or E2]

Primary substrates prefer substitution because their α-carbon is sterically open for SN2 backside attack.
Tertiary substrates prefer elimination because (i) the α-carbon is sterically blocked for SN2, and (ii) the carbocation formed by SN1/E1 ionisation can lose a β-H more easily than be captured by a nucleophile (entropic and electronic reasons).
Secondary substrates sit between and give mixtures unless conditions are tuned carefully.

Worked example 1 — predicting the product

Predict the main product and give full conditions for each pair of reagents.

(a) 2-bromobutane + KOH

2-Bromobutane is secondary.

For predominantly substitution to butan-2-ol: aqueous KOH, warm (~50 °C), reflux for 30 min.
For predominantly elimination to a mixture of but-1-ene and (E/Z)-but-2-ene with but-2-ene as the Zaitsev major: ethanolic KOH, hot reflux.

(b) 1-bromopropane + KOH

1-Bromopropane is primary.

For predominantly substitution to propan-1-ol: aqueous KOH, warm. Clean SN2 outcome.
For elimination to propene: ethanolic KOH at reflux works but the yield is modest; better synthetic routes to propene exist (cracking, dehydration of propan-1-ol).

This is tertiary. The α-carbon is sterically blocked by three methyls, so SN2 is essentially impossible. Even in aqueous conditions a tertiary haloalkane gives a substantial fraction of elimination (via E1 from the carbocation). With ethanolic KOH at reflux, the main product is 2-methylpropene $(CH_3)_2C=CH_2$ , with only traces of the substitution product 2-methylpropan-2-ol.

Worked example 2 — multiple alkene products from one substrate

Heat butan-2-yl chloride, CH₃-CHCl-CH₂-CH₃, with ethanolic KOH at reflux. Draw and name all elimination products. Predict which is the major product and justify.

Butan-2-yl chloride is secondary. Elimination from C2 can lose a β-H from either C1 (a methyl, three equivalent H) or C3 (a methylene, two equivalent H):

Loss of H from C1 → but-1-ene, $CH_2=CH\text{-}CH_2\text{-}CH_3$ — monosubstituted, no stereoisomerism (terminal C=C).
Loss of H from C3 → but-2-ene, $CH_3\text{-}CH=CH\text{-}CH_3$ — disubstituted, exists as both E-but-2-ene (trans, methyls on opposite sides) and Z-but-2-ene (cis, methyls on the same side).

So there are three distinct alkene products. By Zaitsev's rule, the more substituted alkene (but-2-ene) is the major product because the higher degree of substitution stabilises the C=C through hyperconjugation and inductive donation from the methyl groups. Among the two stereoisomers, E-but-2-ene is favoured over Z by ~4 kJ mol⁻¹ because the methyl groups are on opposite sides and steric strain is minimised. A typical product mixture might be E-but-2-ene 60 %, Z-but-2-ene 25 %, but-1-ene 15 % — though the exact ratio depends on conditions and is not formally examined.

Worked example 3 — the synthesis "square"

The substitution-vs-elimination machinery, combined with the dehydration and substitution of alcohols (Lesson 3) and the addition reactions of alkenes (Module 4.1), gives you a complete four-corner interconversion of alcohols, alkenes and haloalkanes:

flowchart TD
  A[Alkene] -->|H2O / dil H2SO4<br/>hydration, Markovnikov| B[Alcohol]
  B -->|conc H3PO4, 170 C<br/>dehydration E1| A
  C[Haloalkane] -->|aq NaOH, warm reflux<br/>substitution SN2/SN1| B
  B -->|NaX + conc acid<br/>reflux substitution| C
  C -->|ethanolic NaOH, hot reflux<br/>elimination E2/E1| A
  A -->|HX gas<br/>electrophilic addition| C

The four edges of the square — each with its own reagent and conditions — let you convert any one of the three functional groups into any other. Almost every A-Level synthesis question requires you to navigate at least one edge.

Conversion	Reagent	Solvent / conditions	Mechanism
Alcohol → Alkene	conc H₃PO₄ (or H₂SO₄)	170 °C, distil	E1 (acid-catalysed)
Alkene → Alcohol	dilute H₂SO₄ then H₂O	warm, reflux	Markovnikov addition
Alcohol → Haloalkane	NaX + conc H₂SO₄ (Cl, Br); NaI + conc H₃PO₄	reflux	Acid-catalysed SN
Haloalkane → Alcohol	aqueous NaOH	warm reflux	SN2 (1°) or SN1 (3°)
Haloalkane → Alkene	ethanolic NaOH	hot reflux	E2 (1°/2°) or E1 (3°)
Alkene → Haloalkane	HX gas (or HX(aq))	room temp	Electrophilic addition

Lesson 6: Substitution vs Elimination in Haloalkanes