Question

An alkene A with molecular formula C\(_6\)H\(_{10}\) on ozonolysis gives a mixture of two compounds B and C. Compound B gives a positive Fehling's test and also reacts with iodine and NaOH solution. Compound C does not give Fehling's test but forms iodoform. Identify the compounds A, B, and C.

Hint:

Ozonolysis of alkenes cleaves the double bond, producing carbonyl compounds such as aldehydes and ketones.

Answer 1

The molecular formula of the alkene is C\(_6\)H\(_{10}\). The ozonolysis of this alkene would break it into two carbonyl compounds. Let's break down the reaction:

1. Ozonolysis of Alkene A:

Ozonolysis of C\(_6\)H\(_{10}\) gives two products, B and C. The alkene is likely 1,5-hexadiene, which splits into acetaldehyde (B) and butan-2-one (C) after ozonolysis.

• Compound B (Acetaldehyde): Acetaldehyde is an aldehyde, which gives a positive Fehling’s test (as it reduces Cu²⁺ to Cu₂O). It also reacts with iodine and NaOH, forming iodoform (CH₃COOH).
• Compound C (Butan-2-one): Butan-2-one is a methyl ketone, which reacts with iodine and NaOH to form iodoform (CH₃COOH) but does not give a Fehling’s test (as ketones are generally resistant to oxidation by Fehling's solution).

Thus,

• Alkene A is 1,5-hexadiene.
• Compound B is acetaldehyde (CH₃CHO).
• Compound C is butan-2-one (CH₃COCH₂CH₃).