See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Identify compound P from its reaction. P reacts with KOH/heat to give an alpha,beta-unsaturated ketone shown as CH3-CO-CH=C(CH3) -- examining the structure drawn, it is an enone with a methylenated carbon: the product is (CH3)CO-CH=C(CH3)-- but looking at the drawn structure more carefully, the product of (P) with KOH/heat is an alpha,beta-unsaturated ketone that appears to be 4-methylpent-3-en-2-one [CH3COCH=C(CH3)2] or similar. Actually the drawn product shows a ketone (C=O with CH3) connected via C=C to two methyl groups, i.e., CH3-CO-CH=C(CH3) -- this is mesityl oxide-like but the structure shown is: acetyl group (CH3CO-) with the double bond going to =C< with two methyls: that would be (CH3)CO-CH=C(CH3)2 = 4-methylpent-3-en-2-one. But wait -- for ozonolysis of R to give P and Q, P must be an aldehyde or ketone fragment from ozonolysis. So P (the ozonolysis product) must be a carbonyl compound: CH3-CO-CH=... no. Ozonolysis cleaves a C=C double bond to give two carbonyl fragments. So R has a C=C bond that upon ozonolysis gives P and Q. Q (from KOH/heat giving Ph-CH2-OH + Ph-CO2-) is benzaldehyde (PhCHO) -- this is a Cannizzaro reaction (no alpha-H aldehyde disproportionates with KOH to give alcohol + carboxylate). So Q = PhCHO (benzaldehyde). Step 2: P reacts with KOH/heat (aldol condensation) to give the alpha,beta-unsaturated ketone shown. The product shown is CH3-CO-CH=C(CH3) with two methyls on the terminal carbon, i.e., the self-aldol of acetone would give mesityl oxide: (CH3)2C=CH-CO-CH3. That matches the drawn structure (4-methylpent-3-en-2-one). So P = acetone (CH3COCH3). Step 3: Ozonolysis of R gives P (acetone, CH3COCH3) and Q (benzaldehyde, PhCHO). Ozonolysis cleaves C=C: the two fragments are Ph-CH= and =C(CH3)2. So R must be Ph-CH=C(CH3)2. This is option (b): Ph-CH=C(CH3)2. Step 4: Verify option (b): Ph-CH=C(CH3)2 upon ozonolysis gives PhCHO (Q = benzaldehyde, which undergoes Cannizzaro with KOH to give PhCH2OH + PhCO2-) and (CH3)2C=O (acetone, P, which undergoes aldol condensation with KOH/heat to give mesityl oxide = 4-methylpent-3-en-2-one). This matches perfectly. Step 5: Why other options fail: (a) Ph-CH=CH-CH3 would give PhCHO + CH3CHO; acetaldehyde has alpha-H so it undergoes aldol not Cannizzaro, and aldol of acetaldehyde gives crotonaldehyde not the drawn ketone. (c) Ph-C(CH3)=CH-CH3 would give Ph-CO-CH3 (acetophenone) + CH3CHO. (d) CH3-C(CH3)=CH2 would give acetone + formaldehyde; formaldehyde Cannizzaro gives methanol + formate, not PhCH2OH + PhCO2-. Therefore, the correct answer is B.