Question

Elimination of HBr from 2-bromobutane affords a mixture of but-1-ene and but-2-ene. With sodium ethoxide as base, but-2-ene constitutes 81% of the alkene products, but with potassium tert-butoxide, but-2-ene constitutes only 67% of the alkene products. Offer an explanation for this difference.

Answer 1

see explanation

Step-by-step explanation:

The elimination reaction involving 2-bromobutane leads to the formation of two products with the predominant product depending on the nature of the base used.

With a bulky base such as potassium tert-butoxide, the secondary carbon atom is inaccessible due to steric hindrance. Therefore, but-1-ene predominates (higher percentage).

However, when a less bulky base such as sodium ethoxide is used, the secondary carbon atom is more easily accessible, leading to the predominant formation of the thermodynamically favourable but-2-ene.