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.
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.
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Solution and Explanation
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 (\( \text{CH}_3\text{COCH}_2\text{CH}_3 \)).
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