Compare E1 and E2 profiles and rate laws

E1 ionisation forms a carbocation before deprotonation; E2 couples proton removal, π-bond formation and leaving-group departure in one step.

GCE A-Level H3 Chemistry 9813-2027

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Learning objectives

Compare E1 and E2 profiles and rate laws

Match kinetics to elementary steps

E1 ionisation forms a carbocation before deprotonation; E2 couples proton removal, π-bond formation and leaving-group departure in one step.

Use concentration-rate data together with profile topology and curved arrows rather than identifying a mechanism from one label alone.

Rate laws and profile features

The simple E1 rate law is rate = k[RX], first order overall; its profile has two maxima separated by a carbocation minimum.

The E2 rate law is rate = k[RX][base], second order overall; its concerted profile has one maximum and no intermediate minimum.

Relate orders to mechanism

Changing base concentration does not alter the rate-determining unimolecular ionisation in E1, although base removes a proton in the later step.

In E2, both substrate and base occur in the sole elementary step, so doubling either concentration doubles rate when the other and temperature are fixed.

The base appears in the E2 rate law because it removes the proton in the only step. It is absent from the simple E1 rate law because C–X ionisation occurs first and controls the rate, even though a base is used later.

Extract an E2 rate constant

At 298 K, [2-bromobutane] = 0.040 mol dm⁻³, [ethoxide] = 0.050 mol dm⁻³ and rate = 1.20 × 10⁻⁴ mol dm⁻³ s⁻¹.

k = rate/([RX][base]) = 0.060 dm³ mol⁻¹ s⁻¹.

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Recalculate k and predict the rate if [ethoxide] doubles.

Check your answer

A strong answer gives 0.060 dm³ mol⁻¹ s⁻¹ and 2.40 × 10⁻⁴ mol dm⁻³ s⁻¹ at fixed substrate.

Read an E1 profile

Supplied reaction: tert-butyl bromide eliminating by E1, with ionisation followed by base removal of a β-H.

Draw one maximum for each elementary step and a minimum for each intermediate, then decide which step controls the rate law.

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Annotate reactant, two transition states, carbocation intermediate and product on the profile.

Check your answer

Reactant, then the first and higher maximum for C–Br ionisation, then the carbocation minimum, then a second, usually smaller maximum for β-H removal, then the alkene product. Rate = k[tert-butyl bromide], set by ionisation.

Use an initial-rate table

Supplied data: at fixed [RX], doubling [water] leaves the rate unchanged; doubling [RX] doubles the rate.

Take the order in each species from the effect of doubling it, then write the rate law and derive the units of k with rate in mol dm⁻³ s⁻¹.

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Infer the rate law from the stated factors and give k units.

Check your answer

Zero order in water and first order in RX, so rate = k[RX] and k has units of s⁻¹. This is consistent with rate-determining E1 ionisation, but the kinetics does not show the later steps.

Do not count arrows as steps

Three simultaneous curved arrows in E2 depict one elementary step and one transition state.

E1 uses more than one arrow-bearing step because the carbocation is a real intermediate at an energy minimum, not a transition state.

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Better reasoning: ‘E2 has three transition states because it has three curved arrows.’

Check your answer

Reject it; the arrows are simultaneous electron movements within one concerted transition state.

Triangulate mechanism evidence

Report reaction orders, calculate k with units, count profile maxima/minima and map each curved arrow to the proposed step.

Next compare how these distinct pathways choose among available β-sites and alkene regioisomers.

Use concentration factors to establish the rate law, calculate k with units, and match the law to a one-maximum E2 or two-maximum E1 profile. Agreement between all three is stronger than any one clue.

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State the minimum evidence for distinguishing E1 from E2.

Check your answer

Your answer should include base dependence, overall order, profile topology, presence/absence of carbocation and concerted/stepwise arrows.