If 2,4-dimethyl pentane is subjected to free radical chlorination reaction, how many different monochlorinated product would be formed?

When 2,4-dimethylpentane undergoes free radical chlorination, four different monochlorinated products are formed.

Explanation

• The possible positions for chlorination in 2,4-dimethylpentane are:
  • At the terminal methyl groups (two distinct types: one at C-1 and one at C-5),
  • At the central methylene (C-3),
  • At the secondary carbons (C-2 and C-4) create chiral centers, producing a racemic (±) mixture counted as a single product for structural (constitutional) isomers.

Thus, based on the structures and the image:

1. Chlorination at C-1 (terminal methyl)
2. Chlorination at C-2 (secondary, creates chiral center; racemic pair counted as one product)
3. Chlorination at C-3 (central methylene)
4. Chlorination at C-5 (terminal methyl)

Total = 4 structural isomers (including one racemic mixture as a single product).image.jpg