Question 1

What is the E2 elimination mechanism?

Accepted Answer

E2 is a concerted bimolecular elimination. A strong base removes a β-hydrogen while the leaving group departs and the C=C bond forms in the same elementary step, so the rate is k[RX][base] and there is no carbocation intermediate.

Question 2

Why does E2 normally need anti-periplanar geometry?

Accepted Answer

The β-C–H bond and C–X bond normally align anti-periplanar, with a dihedral angle close to 180°. This anti-aligned zig-zag gives the developing π bond a low-strain orbital pathway; syn-periplanar elimination exists but is usually less favourable.

Question 3

Which product does 2-bromobutane give with alcoholic KOH?

Accepted Answer

Alcoholic KOH generally favours elimination, and 2-bromobutane gives but-2-ene as the major alkene under typical small-base conditions. But-1-ene is the minor product; aqueous KOH generally favours substitution to butan-2-ol instead.

Question 4

Why is E2 stereospecific?

Accepted Answer

The anti conformer fixes which β-hydrogen is removed and how substituents lie across the new double bond. Different anti conformers of the C3 route can therefore give E- and Z-but-2-ene; the E/trans alkene is generally more stable.

Question 5

What is Saytzeff versus Hofmann elimination?

Accepted Answer

A small base such as ethoxide usually gives the more substituted Saytzeff alkene as the major product. A bulky base such as t-BuOK preferentially abstracts the more accessible β-hydrogen, so the less substituted Hofmann alkene can become major.

Question 6

Which alkyl halides favour E2 over SN2?

Accepted Answer

Under comparable E2 conditions, secondary and tertiary halides favour E2 over SN2 because backside SN2 attack is sterically hindered. Primary halides usually need a bulky strong base for E2 to compete effectively, provided a β-hydrogen is present.

E2 Elimination: Anti-Periplanar Runway & Alkene Products in 3D

Practice

Learned it? Now prove it under the clock.