3. Substitution and Elimination
Alkyl halides react two ways: substitution swaps the halogen for a nucleophile, and elimination removes the halogen plus a neighboring H to form an alkene. Each splits into a one-step and a two-step mechanism: $S_N1$, $S_N2$, $E1$, $E2$.
Mechanism Overview
- $S_N2$: one step — the nucleophile attacks from the back as the leaving group departs. Rate = $k[\text{RX}][\text{Nu}^-]$. Inverts stereochemistry.
- $S_N1$: two steps — the leaving group leaves first, forming a carbocation, which the nucleophile then captures. Rate = $k[\text{RX}]$. Racemizes stereochemistry.
- $E2$: one step — a base removes a β-hydrogen anti-periplanar to the leaving group as the double bond forms.
- $E1$: two steps — carbocation forms first, then a base removes a β-hydrogen.
Carbocation stability: tertiary > secondary > primary, mainly through hyperconjugation — neighboring C–H σ bonds donating electron density into the empty p orbital. This is why $S_N1$/$E1$ favor tertiary substrates.
Predicting the Mechanism
- Substrate: methyl/primary → $S_N2$ or $E2$; tertiary → $S_N1$ or $E1$.
- Nucleophile/base: strong, small nucleophiles (e.g., $\text{OH}^-$, $\text{I}^-$) → $S_N2$; strong bulky bases (e.g., $t$-BuOK) → $E2$; weak nucleophiles (e.g., $\text{H}_2\text{O}$, $\text{ROH}$) → $S_N1$/$E1$.
- Solvent: polar protic solvents (water, alcohols) stabilize carbocations → $S_N1$/$E1$; polar aprotic solvents (DMSO, acetone) → $S_N2$.
- Temperature: heat favors elimination (entropy: two molecules from one).
Zaitsev's Rule
Elimination favors the more substituted alkene — the double bond with the most carbon substituents.
Example: $(\text{CH}_3)_2\text{C}(\text{Br})\text{CH}_2\text{CH}_3$ with KOH and heat undergoes $E2$ to give mainly 2-methyl-2-butene (trisubstituted) over 2-methyl-1-butene (disubstituted).
Worked predictions:
$(\text{CH}_3)_3\text{CBr} + \text{H}_2\text{O}$ → $S_N1$ → tert-butanol (weak nucleophile, tertiary substrate).
$\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{NaOH (conc.)}$ → $S_N2$ → 1-propanol (strong nucleophile, primary substrate).
$(\text{CH}_3)_3\text{CBr} + \text{KOH}$, heat → $E2$ → 2-methylpropene.
$(\text{CH}_3)_3\text{CBr} + \text{H}_2\text{O}$ → $S_N1$ → tert-butanol (weak nucleophile, tertiary substrate).
$\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{NaOH (conc.)}$ → $S_N2$ → 1-propanol (strong nucleophile, primary substrate).
$(\text{CH}_3)_3\text{CBr} + \text{KOH}$, heat → $E2$ → 2-methylpropene.
Practice Problems
- Predict the mechanism and product: (CH3)2CHBr + NaI in acetone.
- Predict the mechanism and product: (CH3)3CCl + CH3OH.
- Which alkene is the major product of E2 on 2-bromobutane with KOH: 1-butene or 2-butene? Why?
- Why does t-BuOK (potassium tert-butoxide) favor elimination over substitution?
Answers: 1. SN2 → (CH3)2CHI (strong nucleophile, secondary substrate, polar aprotic solvent) 2. SN1 → (CH3)3COCH3 (weak nucleophile, tertiary substrate) 3. 2-butene — more substituted, per Zaitsev's rule 4. It is a strong but bulky base; steric hindrance blocks backside attack (SN2) while it can still abstract an exposed β-hydrogen (E2).