Describe the two-step SN1 mechanism
Model carbocation formation followed by nucleophile attack and connect the slow first step to the rate law.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Use the named reagents, solvents, temperatures and pressure conditions; begin curly arrows at a bond or lone pair and distinguish typical mechanisms from absolute rules. Written practice is self-checked
Before you begin
- Use covalent formulas, nucleophiles, electrophiles, homolytic and heterolytic fission, and the alkene and alcohol pathways in Topics 13–14.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What forms in the first SN1 step?
What you will learn
- Show heterolytic C–X fission
- Show nucleophile attack on a carbocation
- Relate the slow step to unimolecular kinetics
The leaving group departs first
In SN1, the C–X bond breaks heterolytically first, with the bond pair moving to X. This forms X⁻ and a positively charged carbon intermediate.
This ionisation step is slow and determines the rate. Its rate depends on halogenoalkane concentration: rate ∝ [R–X]. The nucleophile concentration does not appear in this simplified rate law.
The nucleophile attacks in a later step
After the carbocation forms, a nucleophile donates a lone pair to the positive carbon and forms the new bond. The reaction is stepwise, unlike SN2.
Tertiary halogenoalkanes often favor SN1 because alkyl groups stabilize the carbocation. The carbocation is a short-lived intermediate, not an isolated stable final product.
Pause and explain
Which concentration appears in the simple SN1 rate law?
Worked example
Outline the SN1 mechanism for a tertiary bromoalkane reacting with water.
Show the worked solution
- The C–Br bond pair moves to Br, forming Br⁻ and a tertiary carbocation.
- Water then donates a lone pair to the carbocation.
- Proton loss can give the alcohol product.
Answer Step 1 (slow): R₃C–Br → R₃C⁺ + Br⁻. Step 2: H₂O attacks R₃C⁺; deprotonation gives R₃C–OH.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Write the two stages of SN1 hydrolysis of a tertiary halogenoalkane.
Give me a hint
Show the leaving group departure before water attack.
Compare my reasoning
- Move the C–X bond pair to X to form R₃C⁺ and X⁻.
- Water donates a lone pair to R₃C⁺.
- Remove a proton to form R₃C–OH.
R₃C–X → R₃C⁺ + X⁻ (slow); R₃C⁺ + H₂O → R₃C–OH₂⁺, then deprotonation gives R₃C–OH.
Look for these in your work
- Ionisation first
- Nucleophile second
- Alcohol product completed
If the halogenoalkane concentration doubles at constant temperature, how does the simplified SN1 rate change?
Give me a hint
Apply rate ∝ [R–X].
Compare my reasoning
- The rate is first order in halogenoalkane.
- The nucleophile concentration is not in the rate expression.
- Doubling [R–X] doubles the rate.
The rate doubles if other conditions are unchanged.
Look for these in your work
- Rate expression used
- Nucleophile excluded
- Rate change predicted
A learner draws a single concerted transition for a tertiary halogenoalkane and labels it SN1. What is missing?
Give me a hint
SN1 requires ionisation before nucleophile attack.
Compare my reasoning
- Show heterolytic C–X fission first.
- Add the carbocation intermediate.
- Show nucleophile attack as the following step.
Show two steps: slow carbocation formation, then nucleophile attack.
Look for these in your work
- Stepwise nature stated
- Intermediate included
- Attack order corrected
Common mistakes
- Putting nucleophile attack before carbocation formation in SN1
- Including [Nu] in the simplified SN1 rate law
- Treating the carbocation as a final product
What can you explain now?
What forms in the first SN1 step?
Compare with the explanation
A carbocation and a halide ion
Heterolytic C–X fission sends the bond pair to X.
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Make it stick
Which step controls the SN1 rate, and why does the nucleophile concentration not appear in the simple rate law?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.