Compare SN1 and SN2 profiles, rate laws and steady-state reasoning
SN2 is concerted: nucleophile attack and leaving-group departure occur through one transition state with no intermediate.
Continue where you stopped
The core idea
On this page
Learning objectives
- Compare SN1 and SN2 profiles, rate laws and steady-state reasoning
Separate one-step and two-step pathways
SN2 is concerted: nucleophile attack and leaving-group departure occur through one transition state with no intermediate.
SN1 is stepwise: slow ionisation forms a carbocation, followed by faster nucleophile capture.
Link molecularity and rate law
The simple SN2 law is rate = k[RX][Nu], first order in each reactant and second order overall.
The simple SN1 law is rate = k[RX], first order overall because the nucleophile is absent from the ionisation step.
Use the steady-state statement correctly
In an SN1 steady-state treatment, the carbocation remains at low concentration because its rate of formation is approximately balanced by its rates of consumption.
9813 requires this qualitative account; deriving a mathematical steady-state rate expression is outside the stated requirement.
Use the rate law as a clue to the slow event: SN1 rate depends only on haloalkane because ionisation is slow, whereas SN2 rate depends on both partners because attack and departure occur together.
Initial-rate discrimination
Runs for an alkyl bromide give: doubling [RX] doubles rate; doubling [CN⁻] also doubles rate.
The evidence supports rate = k[RX][CN⁻] and a bimolecular rate-determining event.
Calculate the predicted rate factor when [RX] is tripled and [CN⁻] is halved.
Check your answer
The rate changes by 3 × 1/2 = 1.5; credit explicit use of both first-order dependencies.
Read the energy profiles
Supplied mechanism: a two-step substitution, C–X ionisation followed by nucleophile capture.
Draw one maximum for each elementary step and a minimum for each intermediate, then decide which step usually has the higher barrier.
Label reactants, both transition states, carbocation and products on a two-step substitution profile.
Check your answer
Reactants, then the first maximum (C–X ionisation, commonly the higher), then the carbocation minimum, then the second maximum (nucleophile capture), then products. An SN2 profile would instead have one maximum and no intermediate.
Interpret a low intermediate concentration
Supplied observation: during an SN1 reaction, the carbocation concentration measured at any moment is very low.
Compare how fast the carbocation forms with how fast it is consumed, and distinguish an intermediate from a transition state.
Explain how a carbocation can be mechanistically essential yet remain difficult to detect during an SN1 reaction.
Check your answer
Its formation is slow (rate-determining ionisation) and its consumption is fast, so it stays at a low steady-state concentration while product keeps forming. Low concentration is not absence: it is a real intermediate at an energy minimum, not a transition state.
Common mistake: shared-rate-law reasoning
SN1 and SN2 form substitution products through different elementary sequences and therefore have different concentration dependencies.
A good leaving group can speed either mechanism without making their rate laws identical.
Correct: ‘Both mechanisms contain nucleophile and haloalkane, so both rates must be k[RX][Nu].’
Check your answer
State that SN2 includes nucleophile in its single rate-determining step, whereas SN1 ionisation precedes nucleophile capture and gives k[RX].
Assess kinetic convergence
Complete the rate/profile check questions before trying the unseen bromide data table.
Later, try the chloride profile; use stereochemical evidence and ion-pair effects to distinguish pathways.
For an unfamiliar data table, vary one concentration at a time, find each order, write the rate law with units for k and check that the proposed energy profile has the matching number of steps.
Attempt the kinetics final practice question and mark rate orders, k units, profile topology and steady-state statement.
Check your answer
No mathematical steady-state derivation is required; next compare inversion with two-face capture and qualify any product ratio.
Compare SN1 and SN2 profiles, rate laws and steady-state reasoning scientific representation
The text alternative lists every maximum and minimum in sequence and states each concentration-rate comparison without relying on line colour or shape alone.
About 5 minutes
SN2
- [Nu···C···X]‡; one maximum
- rate=k[RX][Nu]; k dm³ mol⁻¹ s⁻¹
SN1
- ionisation TS
- R⁺ intermediate; formation ≈ consumption
- capture TS
- rate=k[RX]; k s⁻¹
| Pathway | Independent change | Rate response | Rate law | k units |
|---|---|---|---|---|
| SN2 | double [RX] | rate doubles | k[RX][Nu] | dm³ mol⁻¹ s⁻¹ |
| SN2 | double [Nu] | rate doubles | k[RX][Nu] | dm³ mol⁻¹ s⁻¹ |
| SN1 | double [RX] | rate doubles | k[RX] | s⁻¹ |
| SN1 | double [Nu] | unchanged | k[RX] | s⁻¹ |
Text alternative: The text alternative lists every maximum and minimum in sequence and states each concentration-rate comparison without relying on line colour or shape alone.