Elimination E2 Hofmann

Alkyl Halide Reactions: Alkene formation using Bulky Bases (E2 Hofmann Product)

Alkyl Halide Reactions: E2 Hofmann (Bulky Bases)

Sterically demanding bases such as tert-butoxide, LDA, DBU, or DBN remove the most accessible β-hydrogen while the leaving group departs, generating the Hofmann alkene. Despite the regioselective switch, the underlying principle is still anti-periplanar geometry: only a β-hydrogen aligned anti to the C–X bond can participate, and cyclic substrates must supply a trans-diaxial pairing. Use this guide to keep bulky-base eliminations predictable—in rings, in hindered chains, and in exam settings where Hofmann vs Zaitsev decisions matter.

Quick Summary

Class: E2 (bimolecular, concerted) with strong, bulky bases; rate = k[substrate][base].
Geometry: Anti-periplanar β-hydrogen required; rings must present a trans-diaxial pair.
Regiochemistry: Hofmann (less substituted) alkene dominates because bulky bases target the most accessible β-hydrogen; conjugation can still trump sterics.
Substrate order: Tertiary ≳ secondary > primary; primary requires very strong base and heat.
Solvent/temperature: Polar aprotic solvents or tert-alcohols support basicity; heat favors elimination over substitution.
Concerted event: No carbocation intermediates or rearrangements—everything happens in one transition state.

Mechanism (3 Steps)

Step 1 – Anti alignment – Rotate the substrate until a β-hydrogen lines up anti to the C–X bond; in cyclohexanes the leaving group and β-hydrogen must both be axial (trans-diaxial).
Step 2 – Concerted elimination – The bulky base removes that β-hydrogen while electrons shift to the C=C bond and the leaving group departs simultaneously.
Step 3 – Alkene outcome – Label the resulting double bond as Hofmann when the anti-accessible hydrogen resides on the less substituted carbon; note the E or Z configuration imposed by the anti trajectory.

E2 Hofmann Step 1: anti alignment with bulky base — Step 1 — Identify the β-hydrogen the bulky base can reach while maintaining anti alignment.

E2 Hofmann Step 2: concerted elimination with bulky base — Step 2 — Push the three arrows together: base → β-H, C–H → C=C, C–X → X⁻.

E2 Hofmann Step 3: Hofmann alkene product — Step 3 — Emphasize the Hofmann alkene and note any accompanying conjugation badge.

Worked Examples

Reactant panel showing a hindered secondary chloride — 2-chloro-3-methylpentane + t-BuOK (t-BuOH, reflux) — the bulky base reaches the terminal β-H first, so the Hofmann alkene dominates.

Reagent panel showing potassium tert-butoxide — 2-chloro-3-methylpentane + t-BuOK (t-BuOH, reflux) — the bulky base reaches the terminal β-H first, so the Hofmann alkene dominates.

Reactant panel showing tert-hexyl bromide — Tert-hexyl bromide + LDA (THF, warm) — only the least hindered β-hydrogen is accessible, so the terminal alkene forms.

Reagent panel showing LDA — Tert-hexyl bromide + LDA (THF, warm) — only the least hindered β-hydrogen is accessible, so the terminal alkene forms.

Reactant panel showing a bridgehead chloride — Bridgehead chloride + DBU — steric congestion in both base and substrate steers the reaction to the Hofmann alkene.

Reagent panel showing DBU — Bridgehead chloride + DBU — steric congestion in both base and substrate steers the reaction to the Hofmann alkene.

Reactant panel showing cyclohexylmethyl bromide — Cyclohexylmethyl bromide + DBU (aprotic solvent, heat) — the ring flip furnishes the anti β-H, giving the less substituted double bond.

Reagent panel showing DBU base — Cyclohexylmethyl bromide + DBU (aprotic solvent, heat) — the ring flip furnishes the anti β-H, giving the less substituted double bond.

Practical Tips

Heat plus bulky base is the recipe for Hofmann; mention this whenever differentiating from small-base Zaitsev conditions.
For cyclohexanes, draw the two chairs and highlight the one with an axial leaving group; mark the axial β-hydrogen to reinforce the trans-diaxial rule.
Remind students that elimination happens even when SN2 is impossible (e.g., neopentyl); this makes bulky-base E2 a powerful synthetic option.
Point out safety and handling considerations—DBU/DBN are pungent, and strong bases demand dry apparatus.

Exam-Style Summary

Bulky bases enforce Hofmann selectivity by abstracting the most accessible anti β-hydrogen. Show the conformer with anti alignment, push the three concerted arrows, and label the less substituted alkene (noting any conjugation that overrides sterics).

Keep these pitfalls in mind:

Elimination stalls without an accessible anti β-hydrogen—expect recovery or competing pathways.
Upgrade poor leaving groups (e.g., convert chlorides to sulfonates) to keep the reaction moving.
Conjugation can outweigh Hofmann control if it delivers a markedly more stable alkene.
Bulky bases are weak nucleophiles, so elimination can still be slow on extremely hindered systems—set expectations accordingly.

Interactive Toolbox

Mechanism Solver — explore bulky-base presets (t-BuOK, LDA, DBU, DBN) and observe how anti alignment drives Hofmann products.
Reaction Solver — test how base sterics, temperature, and substrate class influence Hofmann vs Zaitsev outcomes.
IUPAC Namer — practice naming terminal alkenes and noting any E/Z descriptors.

Study Tools