Questions: E2 Elimination Mechanism and Hoffmann's Rule
5 questions to test your understanding
Score: 0 / 5
Question 1 Multiple Choice
2-Bromobutane undergoes E2 elimination with potassium tert-butoxide (KOtBu). What is the major product, and why?
A2-Butene (more substituted), because Zaitsev's rule always predicts the thermodynamically stable product
B1-Butene (less substituted), because the bulky base cannot reach the more hindered internal β-hydrogen
C2-Butene (more substituted), because the anti-periplanar requirement favors the internal hydrogen
DA mixture of equal amounts of 1-butene and 2-butene, since base size doesn't affect regiochemistry
KOtBu is sterically bulky. The internal β-hydrogen (adjacent to the more substituted carbon) is blocked by surrounding methyl groups, preventing the base from abstracting it. The external β-hydrogen (on the less substituted carbon) is more accessible, so 1-butene (the Hoffmann product) is the major product. Option A describes Zaitsev selectivity with a small base — it's wrong here because Hoffmann's rule overrides Zaitsev when the base is bulky.
Question 2 Multiple Choice
The anti-periplanar requirement in E2 elimination means that...
AThe base must attack from the same face as the leaving group
BThe β-hydrogen and leaving group must be 180° apart when viewed along the breaking C–C bond
CThe reaction only proceeds when the substrate is a secondary alkyl halide
DThe reaction is reversible, allowing the molecule to adopt the required conformation
Anti-periplanar means the H being abstracted and the leaving group must be on opposite sides (180° dihedral) along the C–C bond axis — this is a strict geometric requirement for the single concerted step of E2. It applies regardless of base size and regardless of whether Zaitsev or Hoffmann selectivity operates. The anti-periplanar arrangement allows simultaneous bond breaking (C–H and C–LG) and π bond formation.
Question 3 True / False
Switching from sodium ethoxide (NaOEt) to potassium tert-butoxide (KOtBu) as the base for an E2 reaction can change the major alkene product without changing the substrate.
TTrue
FFalse
Answer: True
This is exactly the point of Hoffmann's rule. The substrate is the same; only the base changes. NaOEt is small and strong — it accesses the more hindered hydrogen and gives the Zaitsev (more substituted) product. KOtBu is bulky and strong — it is sterically blocked from the hindered hydrogen and abstracts the accessible one, giving the Hoffmann (less substituted) product. Same mechanism (E2, anti-periplanar), different regioselectivity driven purely by base steric bulk.
Question 4 True / False
Hoffmann's rule states that E2 eliminations generally produce the less substituted alkene as the major product.
TTrue
FFalse
Answer: False
Hoffmann's rule only applies when a sterically bulky base is used. With a small, strong base (like NaOEt or NaOH), E2 elimination follows Zaitsev's rule — the more substituted (more stable) alkene is the major product. Hoffmann selectivity is a consequence of steric inaccessibility of the more hindered β-hydrogen, not an absolute rule that always overrides Zaitsev.
Question 5 Short Answer
Explain why a bulky base like KOtBu produces the less substituted alkene in E2 elimination, even though the more substituted alkene is thermodynamically more stable.
Think about your answer, then reveal below.
Model answer: The bulky tert-butoxide base is physically blocked from abstracting the more hindered β-hydrogen (adjacent to the more substituted carbon) because surrounding alkyl groups create steric congestion. The base can only access the less hindered β-hydrogen (adjacent to the less substituted carbon), so it abstracts that one instead. The reaction is kinetically controlled — the product distribution reflects which β-hydrogen the base can reach, not which product is more stable.
E2 is concerted and irreversible under these conditions, so thermodynamic stability of the product is irrelevant once the transition state determines which hydrogen is abstracted. The Zaitsev product would be thermodynamically preferred, but the bulky base cannot reach the transition state that leads to it. This is a classic example of kinetic vs thermodynamic control: base size determines kinetic accessibility, not the product stability.