Question

In the above chemical reaction sequence “A’’ and “B” respectively are :

• A. $\text{O}_3, \, \text{Zn}/\text{H}_2\text{O} \, \text{and} \, \text{NaOH}_{(\text{alc.})}/\text{I}_2$
• B. $\text{H}_2\text{O}, \, \text{H}^+ \, \text{and} \, \text{NaOH}_{(\text{alc.})}/\text{I}_2$
• C. $\text{H}_2\text{O}, \, \text{H}^+ \, \text{and} \, \text{KMnO}_4$
• D. $\text{O}_3, \, \text{Zn}/\text{H}_2\text{O} \, \text{and} \, \text{KMnO}_4$

Answer 1

Answer: (A)

$\text{O}_3, \, \text{Zn}/\text{H}_2\text{O} \, \text{and} \, \text{NaOH}_{(\text{alc.})}/\text{I}_2$

Answer 2

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."