Explain leaving-group effects on substitution rate
Compare R–Cl, R–Br and R–I using the same alkyl group, nucleophile concentration and temperature.
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The core idea
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Learning objectives
- Explain leaving-group effects on substitution rate
Control the carbon skeleton
Compare R–Cl, R–Br and R–I using the same alkyl group, nucleophile concentration and temperature.
A faster substitution can then be attributed to leaving-group behaviour rather than steric or nucleophile changes.
Identify a viable leaving group
The leaving group accepts the C–X bond electron pair as it departs; the curved arrow therefore ends on X.
A more stable, less basic anion and a weaker C–X bond generally support easier departure in the matched halide series.
Connect bond cleavage to both mechanisms
In SN2, C–X cleavage occurs in the concerted transition state as the nucleophile attacks.
In SN1, C–X ionisation forms the carbocation and is normally the slow step; leaving-group ability strongly affects that barrier.
A good leaving group lowers the cost of breaking the C–X bond because the departing species can accommodate the electron pair. Compare leaving groups on the same carbon skeleton so that bond strength and anion stability are the variables actually being tested.
Primary halide rate series
With 0.100 mol dm⁻³ CN⁻, equimolar 1-chlorobutane, 1-bromobutane and 1-iodobutane give relative substitution rates 1.0 : 32 : 78.
The matched data support I > Br > Cl as the leaving-group rate order for this series.
State two chemical reasons for 1-iodobutane reacting faster than 1-chlorobutane.
Check your answer
A strong answer should include weaker C–I bonding and greater stability/lower basicity of I⁻ relative to Cl⁻; do not change the carbon skeleton or nucleophile.
Common mistake: hydroxide departure
Supplied species: protonated tert-butanol, (CH₃)₃C–OH₂⁺.
Draw the C–O cleavage arrow, name what departs, and compare its basicity with OH⁻ leaving from an unprotonated alcohol.
Draw the cleavage arrow for protonated tert-butanol and name the departing species.
Check your answer
The arrow runs from the C–O bond onto oxygen, and neutral H₂O departs, leaving (CH₃)₃C⁺. H₂O is a weak base and a good leaving group; OH⁻ is a strong base and a poor one, which is why the alcohol is protonated first.
Use initial-rate evidence
Supplied data: matched tert-butyl halides undergo unimolecular substitution at equal [RX] and fixed [nucleophile]; the iodide run is 45 times faster than the chloride run.
Decide which step the leaving group affects and which quantity in the rate law changes.
A tert-butyl iodide run is 45 times faster than the chloride run at equal [RX]. Explain without writing rate = k[RX][I⁻].
Check your answer
Iodide departs more easily than chloride, so the ionisation barrier is lower and the substrate-specific k is larger. The rate law remains rate = k[tert-butyl halide]; the leaving group is not a separate concentration term.
Separate base strength from leaving ability
The strongest base is not the best leaving group; stable weak bases such as I⁻ are better leaving groups than strongly basic OH⁻.
Leaving-group order must be inferred under matched conditions and cannot be read from atomic mass alone.
Correct: ‘OH⁻ leaves faster than Br⁻ because oxygen is more electronegative.’
Check your answer
State that unprotonated OH⁻ is strongly basic and poor, whereas Br⁻ is a more stable leaving anion; protonation can make water depart.
Assess the departure step
Answer all the check questions leaving-group checks before trying the unseen sulfonate comparison.
Later, try the protonated-alcohol case; compare substituent effects on substitution rate.
A full comparison names the bond being broken, explains the stability of the departing ion and connects both to the measured barrier or rate. Avoid a bare memorised order with no chemical reason.
Attempt the stored final practice and mark the controlled variables, bond-breaking arrow and rate explanation.
Check your answer
Do not infer from product yield alone; next rank backside crowding and carbocation ionisation barriers in matched substrates.
Explain leaving-group effects on substitution rate scientific representation
The text alternative reads every row aloud and states the controlled substrate, nucleophile and temperature before describing the leaving-group trend.
About 5 minutes
| Series | Substrate | Leaving group | Relative bond strength | Relative CN⁻ rate | Electron-pair destination |
|---|---|---|---|---|---|
| halide | CH₃CH₂CH₂Cl | Cl⁻ | strong | 1 | C–Cl pair → Cl |
| halide | CH₃CH₂CH₂Br | Br⁻ | medium | 32 | C–Br pair → Br |
| halide | CH₃CH₂CH₂I | I⁻ | weak | 78 | C–I pair → I |
| oxygen | ROH | OH⁻ | C–O | poor | C–O pair → O |
| oxygen | ROH₂⁺ | H₂O | C–O | improved | C–O pair → O |
Text alternative: The text alternative reads every row aloud and states the controlled substrate, nucleophile and temperature before describing the leaving-group trend.