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.

see explanation

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.

b. acid is the correct answer @jayxboof all social media

acid : acids are those which accepts an electron-pairs.

base : bases are those which donates ab electron-pairs.

alkali : it is a base that dissolves in water. it is basically a base and the ph greater than 7.

hydroxide ion : it is a poly-atomic ion which is negatively charged.

your answer would be a. 12.5 kg