Elimination Reactions

E1 elimination begins with the spontaneous, heterolytic ionisation of the C–X bond, identical to the first step of S_N1. The halide departs with both electrons, generating a carbocation intermediate. This is the slow, rate-determining step; the rate depends only on the concentration of the substrate: rate = k[substrate]. In the fast second step, a base (often the solvent, e.g. ethanol or water) abstracts a β-hydrogen from a carbon adjacent to the carbocation, and the electrons shift to form the π bond. E1 is favoured by tertiary substrates (more stable carbocations), polar protic solvents, and weak bases/nucleophiles at elevated temperatures. Because a free carbocation forms, E1 can be accompanied by carbocation rearrangements (hydride or methyl shifts).

E2 is a concerted, single-step mechanism. A strong base attacks the β-hydrogen at the same time as the C–X bond breaks — there is no intermediate. The transition state involves simultaneous C–H bond breaking, π bond formation, and C–X bond breaking. The rate is bimolecular: rate = k[substrate][base]. Crucially, E2 requires an anti-periplanar geometry: the β-hydrogen and the leaving group must be 180° apart (anti arrangement) in the transition state. This strict stereoelectronic requirement means that only the β-hydrogen on the correct face of the molecule can be eliminated, giving stereospecific products. E2 is favoured by tertiary or secondary substrates, strong bulky bases (e.g. KOH, NaOEt, LDA), and high base concentrations.

When E1 or E2 elimination can yield two or more alkene regioisomers, Zaitsev's rule (also spelled Saytzeff) states that the thermodynamically more stable, more highly substituted alkene is the major product. For example, elimination of 2-bromobutane with KOH gives predominantly but-2-ene (an internal, disubstituted alkene) rather than but-1-ene (a monosubstituted terminal alkene). The increased stability arises from hyperconjugation: more alkyl groups donating electron density into the π* orbital lower the overall energy. Zaitsev's rule applies reliably when a small, unhindered base such as KOH or NaOH is used.

When a bulky base such as potassium tert-butoxide (KOC(CH₃)₃) or triethylamine is used in an E2 reaction, the base cannot easily access the more hindered β-hydrogens close to the central carbon. Instead, it preferentially abstracts the least hindered, most accessible β-hydrogen on the terminal carbon, giving the less substituted (Hofmann) alkene as the major product. This outcome is kinetically controlled: the product with the lowest activation energy (least steric strain in the transition state) dominates. Hofmann elimination is also used to describe elimination from quaternary ammonium salts with silver oxide/water, always yielding the least substituted alkene.

Elimination always competes with nucleophilic substitution. Increasing temperature strongly favours elimination over substitution because elimination has a larger activation entropy (two molecules become three in E2, or a more disordered carbocation forms in E1). Strong, bulky bases favour E2 over S_N2. Tertiary substrates undergo E1/S_N1 in polar protic solvents (no strong base). Primary substrates with a strong base give mainly E2 or S_N2. Polar aprotic solvents (DMSO, acetone) increase the reactivity of anionic nucleophiles/bases, often tipping the balance toward substitution unless the base is very bulky.

• \ce{CH3CH2CH(Br)CH3} — 2-Bromobutane (—): Secondary haloalkane; the substrate undergoing elimination. The bromine is the leaving group on C-2.
• \ce{KOH} — Potassium hydroxide (—): Strong, small base that abstracts the β-hydrogen; favours the Zaitsev (more substituted) alkene product.
• \ce{CH3CH=CHCH3} — But-2-ene (—): Disubstituted internal alkene; the Zaitsev (major) product, more thermodynamically stable due to greater hyperconjugation.
• \ce{CH3CH2CH=CH2} — But-1-ene (—): Monosubstituted terminal alkene; the Hofmann (minor) product under these conditions.
• \Delta — Heat (—): Elevated temperature favours elimination over substitution by increasing the contribution of the entropy term to the activation free energy.

Problem. 2-Bromo-2-methylbutane is treated separately with (a) aqueous KOH/ethanol at 80 °C and (b) potassium tert-butoxide in tert-butanol at 80 °C. Predict the major organic product in each case and state the mechanism (E1 or E2) and the applicable regiochemistry rule.

Answer. Condition (a) KOH: major product is 2-methylbut-2-ene (trisubstituted alkene, Zaitsev, E2). Condition (b) KOtBu: major product is 2-methylbut-1-ene (less substituted terminal alkene, Hofmann, E2).

• Nucleophilic Substitution: SN1 and SN2 — Substitution and elimination compete directly in haloalkane reactions; understanding SN1/SN2 is essential for predicting whether a nucleophile/base gives substitution or elimination products.
• Leaving Groups and Halide Reactivity — The identity and quality of the leaving group (F < Cl < Br < I) directly affects both E1 and E2 rates and the competition between elimination and substitution.
• Alkene Stability and Hyperconjugation — Zaitsev's rule is explained by the greater thermodynamic stability of more substituted alkenes, which arises from hyperconjugation — a concept central to understanding organic stability.
• Conformational Analysis and Stereochemistry — E2 elimination requires an exact anti-periplanar arrangement of the leaving group and β-hydrogen; conformational analysis explains why certain substrates (especially cyclohexane derivatives) give only specific stereoisomers.
• Carbocation Stability and Rearrangements — E1 elimination proceeds through a carbocation intermediate; understanding carbocation stability (primary < secondary < tertiary) and possible rearrangements (hydride/methyl shifts) is critical for predicting E1 products.