Substitution versus Elimination: MDCAT Chemistry notes
Substitution versus Elimination for MDCAT: SN1, SN2, E1 and E2 mechanisms, carbocations, molecularity and order, aqueous vs alcoholic KOH and products.
Two competing reactions
A nucleophile such as $\mathrm{OH^-}$ or $\mathrm{C_2H_5O^-}$ is also a base. When it meets an alkyl halide it can either attack the $\alpha$-carbon and replace the halogen (substitution) or remove a proton from the $\beta$-carbon so that HX is lost and a C=C forms (elimination, dehydrohalogenation). Both reactions need the halide to leave, so a better leaving group ($\mathrm{I^- > Br^- > Cl^-}$) speeds up both in the same way. SN2 competes with E2, and SN1 competes with E1.
SN2 mechanism
- One step, concerted: the nucleophile attacks from the back side while the C–X bond breaks, through a single transition state (no intermediate).
- Rate = $k[\mathrm{RX}][\mathrm{Nu^-}]$: second order, bimolecular (molecularity 2).
- Gives inversion of configuration.
- Favoured by primary halides, which are least crowded. Primary halides react almost always by SN2.
SN1 mechanism
- Two steps: (1) the C–X bond breaks slowly to give a carbocation intermediate; (2) the nucleophile attacks the carbocation quickly.
- Rate = $k[\mathrm{RX}]$: first order, unimolecular (molecularity 1).
- The carbocation is sp$^2$, planar, with an empty p-orbital. The nucleophile can attack either face, so a chiral substrate gives a racemic mixture.
- Reactivity tertiary > secondary > primary, because the tertiary carbocation is most stable.
E2 and E1
E2 is a one-step, bimolecular elimination in which a strong base removes a $\beta$-hydrogen as the halide leaves. E1 is two-step and goes through the same carbocation as SN1, which then loses a proton. In both, the product is an alkene.
| Feature | SN1 | SN2 | E1 | E2 |
|---|---|---|---|---|
| Steps | 2 | 1 | 2 | 1 |
| Order / molecularity | 1 / 1 | 2 / 2 | 1 / 1 | 2 / 2 |
| Intermediate | Carbocation | None (transition state) | Carbocation | None |
| Best substrate | 3$^\circ$ | 1$^\circ$ | 3$^\circ$ | 3$^\circ$, 2$^\circ$ with strong base |
| Product | Substitution | Substitution | Alkene | Alkene |
What decides the winner
| Condition | Favours |
|---|---|
| Aqueous KOH/NaOH, lower temperature | Substitution → alcohol |
| Alcoholic KOH, heat | Elimination → alkene |
| Strong, bulky base (e.g. ethoxide with a 2$^\circ$/3$^\circ$ halide) | Elimination |
| Higher temperature | Elimination |
| Less polar solvent (alcohol rather than water) | Elimination |
Examples:
- 1-chlorobutane + aqueous NaOH → 1-butanol.
- 2-bromobutane + alcoholic KOH → 2-butene (major, the more substituted alkene).
- 2-bromopropane (or 1-bromopropane) with alcoholic KOH or sodium ethoxide → propene as the major product.
Common MDCAT traps
- "Bond breaking and bond forming at the same time" is SN2, not SN1 or E1.
- SN1 is two-step, first order, molecularity 1; SN2 is one-step, second order, molecularity 2.
- Strong bases and high temperature favour elimination, not substitution.
- Aqueous KOH gives alcohol; alcoholic KOH gives alkene. Zn dust and conc. $\mathrm{H_2SO_4}$ are not used for dehydrohalogenation.
- The SN1 intermediate is a carbocation, not a carbanion or free radical.
Quick revision
- SN1 order: 3$^\circ$ > 2$^\circ$ > 1$^\circ$; SN2 order: 1$^\circ$ > 2$^\circ$ > 3$^\circ$.
- Racemization in SN1 comes from the planar carbocation's empty p-orbital.
- Leaving group plays the same role in substitution and elimination.
- Elimination of HX from an alkyl halide gives an alkene.