Questions: Zaitsev and Hofmann Selectivity in Elimination Reactions

Question 1

A student predicts that treating 2-methylbutyltrimethylammonium hydroxide with base will produce 2-methylbut-2-ene (the more substituted alkene) as the major product, reasoning that it is thermodynamically more stable. What is the flaw in this prediction?

• A. The substrate cannot undergo elimination because quaternary ammonium ions are too stable to react
• B. The reaction will proceed via E1, not E2, so Zaitsev's rule does not apply here
• C. The trimethylammonium leaving group is bulky, which prevents the base from accessing the more hindered beta-hydrogen; the Hofmann product (less substituted alkene) is the major product
• D. The more substituted alkene is actually less stable in this case, so Zaitsev's rule predicts the less substituted product anyway

Answer: (C)

Quaternary ammonium leaving groups are large and bulky. In the E2 transition state, the base, the departing hydrogen, and the leaving group must be arranged anti-periplanar. The steric bulk of the –N(CH₃)₃⁺ group raises the activation energy for removing the more hindered beta-hydrogen (which would give the Zaitsev product), redirecting the reaction toward the less hindered hydrogen and the Hofmann (less substituted) alkene. The student correctly applies Zaitsev's rule but ignores the steric override.

Question 2

Potassium tert-butoxide reacts with 2-bromobutane under E2 conditions. Which outcome is expected?

• A. But-2-ene (Zaitsev product) forms predominantly because tert-butoxide is a strong, hindered base that favors E2 over SN2
• B. But-1-ene (Hofmann product) forms predominantly because tert-butoxide is bulky and preferentially abstracts the less hindered terminal hydrogen
• C. SN2 substitution predominates because tert-butoxide is an excellent nucleophile
• D. Both but-1-ene and but-2-ene form in equal amounts because tert-butoxide has no preference

Answer: (B)

Tert-butoxide is bulky due to its three methyl groups. It has difficulty reaching the more hindered internal beta-hydrogen (C-3) that would produce but-2-ene, but can easily access the less hindered terminal hydrogen (C-1). This steric preference overrides the thermodynamic bias toward the more substituted alkene, giving the Hofmann product. Bulky bases consistently direct E2 reactions toward less substituted alkenes.

Question 3

Hofmann elimination is a mechanistically distinct reaction type from E2 — it involves a different transition state geometry than standard E2 elimination.

• T. True
• F. False

Answer: False

Hofmann elimination is not a separate mechanism — it follows standard E2 mechanics with the same anti-periplanar transition state requirement. The only difference is that steric bulk (from a bulky base or bulky leaving group) raises the activation energy for the pathway leading to the more substituted (Zaitsev) alkene, redirecting the reaction toward the less substituted product. The mechanism itself is identical to any E2 reaction.

Question 4

A bulky leaving group alone — not just a bulky base — can direct an elimination reaction toward the Hofmann (less substituted) product.

• T. True
• F. False

Answer: True

Steric bulk near the reacting carbon can originate from either the leaving group or the base. Quaternary ammonium salts (–N(CH₃)₃⁺) are the classic example: the bulky leaving group creates steric congestion around the more substituted beta-carbon, making it harder for the base to achieve the required anti-periplanar geometry there. The result is identical to using a bulky base — the less hindered hydrogen is abstracted and the less substituted alkene predominates.

Question 5

Explain why Hofmann elimination produces the less substituted alkene, in terms of the E2 transition state geometry rather than just 'steric hindrance blocks access.'

Model answer: E2 elimination requires a specific anti-periplanar arrangement: the base, the beta-hydrogen being abstracted, the alpha-carbon, and the leaving group must all lie in the same plane with a 180° dihedral angle. When the leaving group or base is bulky, achieving this geometry at the more substituted beta-carbon is sterically demanding — multiple groups crowd the transition state. The activation energy for that pathway rises, making it slower than the less substituted pathway where the transition state is less congested. The reaction favors whichever anti-periplanar arrangement has lower steric strain in the transition state.

The key is that it is not simply about 'the base can't get close' — it is about the geometry of the entire transition state, which requires five atoms in near-coplanar alignment. Steric bulk disrupts this specific geometry at the more substituted carbon without necessarily blocking access to the alpha-carbon. Understanding the anti-periplanar requirement explains why Hofmann products result even when the base is not especially bulky, as long as the leaving group creates enough congestion at the more substituted position.