Question:
The product P of the following sequence of reactions is:
The product P of the following sequence of reactions is:
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In ozonolysis, always break the C=C bond and convert each double-bond carbon into a carbonyl group.
Updated On: Jun 19, 2026
- Option 1
- Option 2
- Option 3
- Option 4
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the first reaction (E2 elimination).
The substrate is a bromoalkane. On treatment with alcoholic KOH and heat (\(\Delta\)), it undergoes \(\beta\)-elimination (E2 reaction), leading to formation of an alkene by removal of HBr.
Step 2: Formation of alkene intermediate.
The base abstracts a \(\beta\)-hydrogen anti to the leaving group (Br), resulting in formation of the most substituted alkene according to Zaitsev’s rule. This gives a stable internal alkene intermediate.
Step 3: Ozonolysis reaction.
The alkene is then subjected to ozonolysis: \[ (i)\ O_3 \quad (ii)\ Zn/H_2O \] Ozonolysis cleaves the double bond and converts each alkene carbon into carbonyl compounds (aldehydes or ketones depending on substitution).
Step 4: Identifying cleavage products.
Since the alkene formed is asymmetrical, cleavage produces a ketone on the more substituted carbon and an aldehyde on the terminal carbon. The carbon skeleton rearrangement matches the structure shown in option (4).
Step 5: Final confirmation.
Among all given products, only option (4) satisfies both: - Correct elimination product - Correct ozonolysis cleavage pattern
Final Answer: \[ \boxed{\text{Option (4)}} \]
The substrate is a bromoalkane. On treatment with alcoholic KOH and heat (\(\Delta\)), it undergoes \(\beta\)-elimination (E2 reaction), leading to formation of an alkene by removal of HBr.
Step 2: Formation of alkene intermediate.
The base abstracts a \(\beta\)-hydrogen anti to the leaving group (Br), resulting in formation of the most substituted alkene according to Zaitsev’s rule. This gives a stable internal alkene intermediate.
Step 3: Ozonolysis reaction.
The alkene is then subjected to ozonolysis: \[ (i)\ O_3 \quad (ii)\ Zn/H_2O \] Ozonolysis cleaves the double bond and converts each alkene carbon into carbonyl compounds (aldehydes or ketones depending on substitution).
Step 4: Identifying cleavage products.
Since the alkene formed is asymmetrical, cleavage produces a ketone on the more substituted carbon and an aldehyde on the terminal carbon. The carbon skeleton rearrangement matches the structure shown in option (4).
Step 5: Final confirmation.
Among all given products, only option (4) satisfies both: - Correct elimination product - Correct ozonolysis cleavage pattern
Final Answer: \[ \boxed{\text{Option (4)}} \]
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