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 positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution 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 Cu₁⁺). 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 (CH₃COCH₂).