Question

Question: Acid-catalyzed bromination of pentan-2-one (CH₃COCH₂CH₂CH₂) forms two products: BrCH₂COCH₂CH₂CH₃ (A) and CH₃COCH₂(Br)CH₂CH₃(B). Explain why the major product is B, with the Br atom on the more substituted side of the carbonyl group.

Short answer

Answer

The acid-catalyzed bromination of pentan-2-one (CH₃COCH₂CH₂CH₃ ) forms two products, as shown below.

Product B is more stable as a more substituted enolized intermediate is formed after the addition of an acid.

Step-by-step solution

Halogenation in acids

Halogenation is carried out by treating a carbonyl compound with alpha hydrogens. Then they form enolate and it is followed by an attack with a halogen in the presence of an acid.

 Step 2: Stabilzation of the enolate

After acid catalyzation, the enolate that is the most substituted will form the major product after alkylation. This is because with the increase in the alkyl group, the stability of the alkene increases due to the +I effect shown by the substituted alkyl groups.

Representation of the enolates

Exaplanation

Hence, considering the key steps, the acid-catalyzed bromination of pentan-2-one ( CH₃COCH₂CH₃) forms two products, as shown below.

Representation of the products A and B.

Product B is more stable as a more substituted enolized intermediate is formed after the addition of an acid.