- \( \text{O}_3, \, \text{Zn}/\text{H}_2\text{O} \, \text{and} \, \text{NaOH}_{(\text{alc.})}/\text{I}_2 \)
- \( \text{H}_2\text{O}, \, \text{H}^+ \, \text{and} \, \text{NaOH}_{(\text{alc.})}/\text{I}_2 \)
- \( \text{H}_2\text{O}, \, \text{H}^+ \, \text{and} \, \text{KMnO}_4 \)
- \( \text{O}_3, \, \text{Zn}/\text{H}_2\text{O} \, \text{and} \, \text{KMnO}_4 \)
The Correct Option is A
Solution and Explanation
Step 1: Ozonolysis (formation of compound A):} The double bond in the cycloalkene undergoes ozonolysis in the presence of ozone (\( \text{O}_3 \)) followed by reduction with Zn/\( \text{H}_2\text{O} \). This cleaves the double bond, producing two aldehyde groups on adjacent carbons. 2.
Step 2: Haloform reaction (formation of compound B): The aldehyde (or ketone) group in compound A reacts with \( \text{NaOH}_{(\text{alc})} \) and \( \text{I}_2 \) (haloform reaction). This cleaves the terminal methyl ketone or aldehyde group to produce sodium formate (\( \text{HCOONa} \)) and iodoform (\( \text{CHI}_3 \)), leaving a carboxylic acid group. The final product, compound B, contains a carboxylate ion (\( \text{COO}^- \)) and a secondary alcohol group. The complete reaction mechanism ensures the correct conversion of "A" to "B."
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