H3 Chemistry 9813 · Study focus: H3 Chemistry: Explain leaving-group effects on substitution rate
H3 Chemistry: Explain leaving-group effects on substitution rate
Start from the governing chemical model, test it against evidence, then transfer the reasoning to an unfamiliar case.
Your success criteria
- Explain leaving-group effects on substitution rate
- Use named chemical evidence.
- Transfer the governing reason to an unfamiliar case.
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.
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.
| 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.
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.
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.
Open the feedback checkpoint after attempting
- Credit weaker C–I bonding and greater stability/lower basicity of I− relative to Cl−; do not change the carbon skeleton or nucleophile.
Repair hydroxide departure
Unprotonated OH− is a poor leaving group because it is a strong base.
Protonating an alcohol converts the departing group into neutral H2O, which is much more viable.
- Draw the cleavage arrow for protonated tert-butanol and name the departing species.
Open the feedback checkpoint after attempting
- The arrow runs from the C–O bond to oxygen; neutral H2O departs, not OH−.
Use initial-rate evidence
Matched tert-butyl halides undergoing unimolecular substitution show that the initial rate changes when X changes even though [nucleophile] is fixed.
The interpretation concerns the ionisation barrier, not a claim that the leaving group appears in the SN1 rate law as a separate concentration.
- A tert-butyl iodide run is 45 times faster than the chloride run at equal [RX]. Explain without writing rate = k[RX][I−].
Open the feedback checkpoint after attempting
- Iodide departure gives a lower ionisation barrier and a larger substrate-specific k; the simple rate law remains k[tert-butyl halide].
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.’
Open the feedback checkpoint after attempting
- 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
Complete all twelve leaving-group checks before opening the unseen sulfonate comparison.
After the delayed protonated-alcohol case, compare substituent effects on substitution rate.
- Attempt the stored assessment and mark the controlled variables, bond-breaking arrow and rate explanation.
Open the feedback checkpoint after attempting
- Do not infer from product yield alone; next rank backside crowding and carbocation ionisation barriers in matched substrates.