The E2 reaction (bimolecular elimination) is a single-step, concerted mechanism where a base removes a proton as the leaving group departs. Whether a reaction follows E2 depends on several key factors:
1. Strong Base
- A strong base is essential because it must remove a β-hydrogen at the same time the leaving group leaves.
- Examples: sodium ethoxide, potassium tert-butoxide, LDA (lithium diisopropylamide).
- Bulky strong bases often favor elimination over substitution, because they struggle to attack carbon directly (blocking SN2).
2. Substrate Structure
- Tertiary halides/alcohol derivatives: Strongly favor E2 (SN2 is blocked by steric hindrance).
- Secondary halides: Can go either way, but with a strong base and heat, E2 dominates.
- Primary halides: Usually prefer SN2, but bulky bases can force E2.
3. Leaving Group
- A good leaving group (like I⁻, Br⁻, Cl⁻, tosylate) makes elimination faster.
- Poor leaving groups slow down or prevent E2.
4. Solvent
- Polar aprotic solvents (like acetone, DMSO, DMF) favor E2 by keeping the base strong.
- Polar protic solvents (like alcohols, water) can weaken bases by solvation, slowing E2.
5. Heat
- Higher temperatures favor elimination over substitution, because elimination leads to higher entropy (more molecules or more degrees of freedom).
6. Geometry Requirement
- The β-hydrogen and the leaving group must be anti-periplanar (opposite each other in the same plane).
- This stereoelectronic requirement makes E2 stereospecific, influencing whether an E- or Z-alkene forms.
In summary:
E2 reactions are favored by a strong (often bulky) base, a suitable substrate (secondary or tertiary), a good leaving group, polar aprotic solvents, and higher temperatures. The geometry of the molecule also plays a role in determining the stereochemistry of the alkene product.