E1 vs E2 elimination — Zaitsev's rule and Hofmann elimination comparison

Question

2-Bromo-2-methylbutane is treated with (a) $\text{C}_2\text{H}_5\text{O}^-$ in ethanol, and (b) dilute $\text{C}_2\text{H}_5\text{OH}$ alone. Predict the major product in each case. Which follows Zaitsev’s rule and which doesn’t? Identify the mechanism (E1 or E2) in each case.

(JEE Advanced 2021, similar pattern)

Solution — Step by Step

2-Bromo-2-methylbutane is a tertiary halide. Tertiary substrates can undergo both E1 and E2 depending on the base.

The structure: $\text{CH}_3\text{CH}_2\text{C}(\text{CH}_3)_2\text{Br}$ . There are two possible $\beta$ -hydrogens — on $\text{CH}_2$ (gives the more substituted alkene) and on $\text{CH}_3$ (gives the less substituted alkene).

$\text{C}_2\text{H}_5\text{O}^-$ is a strong, non-bulky base. With a tertiary substrate and a strong base, the mechanism is E2 (one-step, concerted).

E2 with a non-bulky base follows Zaitsev’s rule — the more substituted alkene is the major product.

Major product: 2-methylbut-2-ene (trisubstituted alkene).

Ethanol alone is a weak base and a polar protic solvent. Tertiary substrates in polar protic solvents without a strong base favour E1 (two-step, via carbocation intermediate).

E1 also follows Zaitsev’s rule because the more stable (more substituted) alkene forms preferentially from the carbocation. The carbocation loses the $\beta$ -hydrogen that leads to the most stable product.

Major product: again 2-methylbut-2-ene.

Hofmann elimination gives the less substituted alkene as the major product. This happens when:

The base is bulky (like $\text{(CH}_3)_3\text{CO}^-$ , tert-butoxide)
Or the leaving group is a bulky ammonium salt (quaternary ammonium, Hofmann exhaustive methylation)

Neither case (a) nor (b) here uses a bulky base, so both follow Zaitsev.

Why This Works

E2 is a one-step mechanism — the base pulls off the $\beta$ -hydrogen while the leaving group departs simultaneously. The transition state resembles the product alkene, so the more stable (more substituted) alkene is kinetically and thermodynamically favoured. That’s Zaitsev’s rule.

E1 proceeds through a carbocation. Once the carbocation forms, the proton that leaves is the one that produces the most stable alkene. Since the carbocation intermediate is flat, the base approaches the most accessible hydrogen — which again leads to the Zaitsev product.

Hofmann’s rule is the exception: steric bulk prevents the base from reaching the internal $\beta$ -hydrogen, so it grabs the more accessible terminal hydrogen instead — giving the less substituted alkene.

Alternative Method — Quick Decision Flowchart

For JEE problems, use this mental flowchart:

Strong base present? → E2. If base is bulky → Hofmann. If base is small → Zaitsev.
No strong base, tertiary substrate, polar protic solvent? → E1 → Zaitsev product.
Primary substrate + strong base? → E2 (but watch for competing $\text{S}_\text{N}2$ ).

JEE Advanced frequently tests the competition between E2 and $\text{S}_\text{N}2$ . With strong nucleophiles that are also strong bases ( $\text{OH}^-$ , $\text{C}_2\text{H}_5\text{O}^-$ ), tertiary substrates almost always give elimination. Primary substrates give substitution. Secondary substrates — that’s where the real decision-making happens.

Common Mistake

Students often assume E1 always gives a mixture with no selectivity. Wrong — E1 does follow Zaitsev’s rule. The carbocation preferentially loses the proton that forms the more substituted alkene. The only time Zaitsev fails is with bulky bases (Hofmann) or in certain cyclic systems where ring strain matters.

Another trap: confusing E1cb with E1. E1cb (elimination unimolecular conjugate base) occurs when the $\beta$ -hydrogen is acidic (e.g., next to a carbonyl). It’s rare in JEE Main but has appeared in JEE Advanced — the carbanion forms first, then the leaving group departs.

Question

Solution — Step by Step

Why This Works

Alternative Method — Quick Decision Flowchart

Common Mistake

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