E1 and E2 Elimination: Saytzeff, Hofmann and Which Alkene Forms

The two mechanisms

E2 is a single concerted step. A base removes a hydrogen from the carbon adjacent to the leaving group while the leaving group departs and the double bond forms. Rate depends on both substrate and base, so kinetics are second order.

E1 has two steps. The leaving group departs to give a carbocation, then a base removes a proton from an adjacent carbon. Only the first step is rate-determining, so kinetics are first order and independent of base concentration.

The anti-periplanar requirement

E2 needs the C–H bond being broken and the C–LG bond being broken to lie in the same plane and point opposite ways. The reason is orbital overlap: the electrons from the breaking C–H bond must feed directly into the antibonding orbital of the C–LG bond, and that alignment is only available anti-periplanar.

Saytzeff versus Hofmann orientation

Saytzeff is the default because more substituted alkenes are more stable, and the transition state has partial double bond character that reflects that stability.

Hofmann orientation appears in two situations. With a bulky base, the base cannot reach the more hindered internal hydrogen and removes an accessible terminal one instead. With a charged leaving group such as a quaternary ammonium, the transition state is more carbanion-like than alkene-like, so alkene stability stops controlling the outcome and the most acidic, least hindered hydrogen is removed.

Choosing between substitution and elimination

The same reagents can do both, and questions frequently ask which dominates.

• Strong bulky base → elimination, since attack at carbon is blocked but proton removal is not.
• Strong non-bulky nucleophile that is a weak base → substitution.
• High temperature → elimination, because it produces more particles and is entropically favoured.
• Tertiary substrate with any base → elimination dominates; SN2 is blocked and E2 is not.

E1cb, briefly

Where the beta hydrogen is unusually acidic — typically because a carbonyl sits adjacent — the base can remove it first to give a carbanion, which then expels the leaving group. This is E1cb, and it is recognised by an acidic beta hydrogen combined with a poor leaving group.

Frequently asked questions

Why must elimination be anti-periplanar rather than syn?

Because the orbital overlap required for the concerted process is only achieved in the anti arrangement. Syn elimination does occur where the anti geometry is impossible, but it is much slower and needs special systems.

Does E1 have a stereochemical requirement?

No. The carbocation is planar, so there is free rotation before the proton is removed. This is a clean discriminator between the mechanisms.

Why does a bulky base change the product?

It cannot reach the internal hydrogens surrounded by alkyl groups, so it takes a terminal one. The product is decided by access, not stability.

How do I predict the major alkene when several are possible?

Default to Saytzeff, then check for a bulky base or a charged leaving group. If either is present, switch to Hofmann.