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

Ozonolysis of Alkene C₆H₁₀

1. Ozonolysis of Alkene A: 

The molecular formula of the alkene is \( C_6H_{10} \). The ozonolysis of this alkene breaks it into two carbonyl compounds. After ozonolysis, the two products formed are acetaldehyde (B) and butan-2-one (C).

2. Identifying Alkene A:

The alkene with the formula \( C_6H_{10} \) and a structure that undergoes ozonolysis to give acetaldehyde and butan-2-one is likely 1,5-hexadiene. The structure of 1,5-hexadiene is: \[ \text{CH}_2\text{=CH-CH}_2\text{CH}_2\text{CH}_3 \]

3. Products of Ozonolysis:

Ozonolysis of 1,5-hexadiene results in the following products:

• Compound B - Acetaldehyde \( (\text{CH}_3\text{CHO}) \): Acetaldehyde is an aldehyde that gives a positive Fehling's test, where it reduces \( \text{Cu}^{2+} \) to \( \text{Cu}_2\text{O} \). It also reacts with iodine and NaOH to form iodoform (\( \text{CH}_3\text{COOH} \)).
• Compound C - Butan-2-one \( (\text{CH}_3\text{COCH}_2\text{CH}_3) \): Butan-2-one is a methyl ketone that reacts with iodine and NaOH to form iodoform (\( \text{CH}_3\text{COOH} \)). It does not give a Fehling's test because ketones are generally resistant to oxidation by Fehling's solution.

4. Conclusion:

- Alkene A is 1,5-hexadiene. - Compound B is acetaldehyde (\( \text{CH}_3\text{CHO} \)). - Compound C is butan-2-one (\( \text{CH}_3\text{COCH}_2\text{CH}_3 \)).