E2 (Zaitsev product): practice and mechanism
Strong base + secondary/tertiary haloalkane + heat = E2. By Zaitsev's rule the more substituted, more stable alkene forms.
Example
A strong base with a secondary or tertiary haloalkane and heat leads to E2 elimination. Following Zaitsev's rule, the more substituted, more stable alkene is the main product.
Mechanism step by step
- 1
The base removes a hydrogen from the carbon next to the halogen-bearing carbon (β-H).
- 2
At the same time the bonding electron pair swings over to form the new C=C bond and the halide leaves (one step).
- 3
For this, H and halogen must be antiperiplanar (in one plane, on opposite sides).
- 4
With several β-hydrogens the more substituted alkene is favoured (Zaitsev).
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which reagent achieves this transformation?
Typical mistakes
- That is the substitution product (ether). With secondary and tertiary haloalkanes, a strong base and heat, elimination wins.
- An alcohol would come from substitution with water or hydroxide. Ethoxide acts as a strong base here and removes HBr: an alkene forms.
Often confused with: E2 with bulky base (Hofmann product) · SN1 (hydrolysis)
Frequently asked questions
Why is the more substituted alkene more stable?
Alkyl groups stabilise the double bond by hyperconjugation, seen in smaller heats of hydrogenation.
How do I tell whether E2 or SN2 wins?
Strong base + branched (secondary/tertiary) substrate + heat → E2. Good nucleophile + primary substrate → SN2.