Explain substituent effects on substitution rate

Alkyl substitution has opposite kinetic consequences for backside SN2 attack and carbocation-forming SN1 ionisation.

GCE A-Level H3 Chemistry 9813-2027

On this page

Learning objectives

Explain substituent effects on substitution rate

Specify the pathway before ranking substrates

Alkyl substitution has opposite kinetic consequences for backside SN2 attack and carbocation-forming SN1 ionisation.

A rate ranking is meaningless unless substrate, leaving group, nucleophile and the operative pathway are stated.

Separate steric and electronic effects

Crowding at the reacting carbon obstructs the SN2 approach trajectory.

Alkyl groups stabilise developing or complete carbocation charge, lowering the barrier for SN1 ionisation.

Build both matched series

For methyl, primary, secondary and tertiary halides with a fixed strong nucleophile, the SN2 tendency falls as substitution increases.

For matched ionisation conditions and the same leaving group, tertiary carbocation formation is more favourable than secondary, while primary and methyl carbocations are strongly disfavoured.

Always locate the branching relative to the reacting carbon. Branching at or next to that carbon can obstruct SN2 approach, while groups that stabilise developing positive charge can lower an SN1 ionisation barrier; these are different effects on different pathways.

Backside attack data

With 0.050 mol dm⁻³ CN⁻, equal methyl bromide, bromoethane, 2-bromopropane and tert-butyl bromide give relative substitution rates 100 : 38 : 3 : less than 0.01 for the SN2 channel.

The fall follows increasing steric obstruction at the carbon bearing Br.

Try this

Draw the backside approach for bromoethane and tert-butyl bromide and identify the steric collision in the tertiary case.

Check your answer

A strong answer should include an approach opposite C–Br and three methyl groups shielding the tertiary carbon; do not draw a tertiary SN2 intermediate.

Carbocation stability series

Supplied substrates for SN1 ionisation: tert-butyl bromide, 2-bromopropane and bromoethane.

Classify the carbocation each would form and count the alkyl groups that stabilise its positive charge, then link that to the ionisation barrier.

Try this

Rank tert-butyl bromide, 2-bromopropane and bromoethane for a stated SN1 ionisation and justify the extremes.

Check your answer

tert-Butyl bromide > 2-bromopropane >> bromoethane. Three alkyl groups stabilise the developing tertiary charge, giving the lowest ionisation barrier; bromoethane would form an unstable primary carbocation, giving the highest.

Detect branching beyond the first carbon

Supplied substrates: 1-bromobutane and neopentyl bromide, both primary, each with CN⁻.

Classification as ‘primary’ is a starting point, not a substitute for inspecting the whole local geometry: check what is attached to the carbon next to the reacting carbon.

Try this

Compare SN2 attack on 1-bromobutane and neopentyl bromide with CN⁻.

Check your answer

1-Bromobutane is faster. Neopentyl bromide's reacting carbon is primary, but the adjacent tert-butyl framework severely obstructs backside approach.

Common mistake: one universal rate order

Tertiary is not universally fastest: it favours SN1 but blocks SN2.

Primary is not universally fastest: it favours SN2 but gives an unstable carbocation for SN1.

Try this

Correct: ‘More alkyl groups always increase substitution rate.’

Check your answer

State opposite pathway effects—alkyl substitution stabilises SN1 charge development but hinders SN2 backside approach.

Assess matched substrate effects

Complete the substrate check questions before trying the unseen branching rate set.

Later, try a different cyclopropyl-free question; connect the substrate trends to complete SN1 and SN2 rate laws and profiles.

When comparing substrates, keep the nucleophile and leaving group fixed, label the reacting carbon and state whether steric access or charge stabilisation controls the rate difference.

Try this

Attempt the substituent-rate final practice question and mark pathway, steric/electronic cause and controlled variables.

Check your answer

Use only the matched evidence supplied, then test the proposed pathway against concentration changes and profile topology.