See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Identify the ozonolysis products. The final step (ozonolysis of alkene C with O3 then Zn/H2O) gives two carbonyl products: pentan-3-one (3-pentanone, CH3CH2COCH2CH3) and acetaldehyde (CH3CHO). Ozonolysis cleaves a C=C double bond, so alkene C must be: CH3CH2CO-CH2-CH=CH-CH3 ... wait, let me think more carefully. Step 2: Retrosynthetic analysis of alkene C. Ozonolysis cleaves the double bond, giving pentan-3-one and acetaldehyde. The two carbons of the double bond become the carbonyl carbons. So C must be: CH3CH2CO-CH2-CH=CH-CH3, i.e., the double bond is between a CH and a CH3-bearing carbon, giving acetaldehyde (CH3CHO) from one side and 3-pentanone from the other side. Thus C is: CH3CH2C(=O)-CH2-CH=CH-CH3... No. Let me reconsider: one fragment is CH3CHO (acetaldehyde), meaning one carbon of the double bond carries only H and CH3. The other fragment is pentan-3-one: CH3CH2-CO-CH2-, meaning the other carbon of the double bond is CH2 attached to the ketone carbon. So C is: CH3CH2-CO-CH2-CH=CH-CH3... that gives a 7-carbon compound. But wait, let me recount: ozonolysis of CH3CH2-CO-CH2-CH=CHCH3 gives CH3CH2-CO-CH2-CHO (not pentan-3-one) + CH3CHO. That doesn't work. Step 3: Re-examine. Pentan-3-one is CH3CH2-CO-CH2CH3. For this to come from ozonolysis, the double bond carbon that becomes the ketone carbonyl in 3-pentanone must have been: =C(CH2CH3)(CH2CH3) — a trisubstituted alkene carbon. So C = (CH3CH2)2C=CHCH3, i.e., 2-ethyl-1-butene... actually (CH3CH2)2C=CHCH3. Ozonolysis gives: (CH3CH2)2C=O = pentan-3-one, and CH3CHO = acetaldehyde. Yes! This works perfectly. Step 4: Alkene C is (CH3CH2)2C=CHCH3, i.e., 3-ethylpent-3-ene or more precisely 3-methylpent-3-ene... actually (CH3CH2)2C=CHCH3 is 3-ethyl-2-pentene or 3-methyl-3-heptene... Let me name it: the structure is CH3CH2-C(=CH-CH3)-CH2CH3, which is 3-ethylpent-3-ene... No: (CH3CH2)2C=CHCH3: carbon skeleton: C2H5-C(=CHCH3)-C2H5. This is 3-ethylpent-2-ene. Step 5: Alkene C forms from alcohol B by dehydration (H2SO4, heat, -H2O). So B is the alcohol corresponding to C: (CH3CH2)2C(OH)-CH(CH3)... wait. Dehydration of B gives C = (CH3CH2)2C=CHCH3. Alcohol B would be (CH3CH2)2C(OH)-CH2CH3 (tertiary alcohol) with dehydration giving the more substituted alkene, but that would give (CH3CH2)2C=CHCH3 only if there's a CH(CH3) group. Actually B = (CH3CH2)2C(OH)-CHCH3... Let me reconsider: B is a tertiary alcohol that dehydrates to give (CH3CH2)2C=CHCH3. B = CH3CH(OH)C(C2H5)2... i.e., 3-ethylpentan-3-ol with adjacent CH(CH3). No: if C = (C2H5)2C=CHCH3, then B obtained by adding OH across the double bond = (C2H5)2C(OH)-CH2CH3 or (C2H5)2CH-CH(OH)CH3. The Grignard addition gives a tertiary alcohol, so B = (C2H5)2C(OH)-CH2CH3 = 3-ethylpentan-3-ol. Dehydration gives (C2H5)2C=CHCH3 (Zaitsev product). Step 6: Grignard reaction. B = (C2H5)2C(OH)CH2CH3 = 3-ethylpentan-3-ol is formed from ketone A + C2H5MgBr. The Grignard reagent adds an ethyl group to the ketone carbonyl. So A + C2H5MgBr → B. B has three ethyl groups on the carbinol carbon: (C2H5)3C-OH? Wait: (C2H5)2C(OH)CH2CH3 means the carbon has OH, two ethyl groups, and one ethyl group = three ethyl groups on one carbon: (C2H5)3COH = triethylcarbinol. So B = (CH3CH2)3COH. Step 7: Ketone A is formed when C2H5MgBr adds to A to give (C2H5)3COH. The ketone must be (C2H5)2C=O = pentan-3-one = diethyl ketone = 3-pentanone = CH3CH2-CO-CH2CH3. This corresponds to option (b). Step 8: Verify option (b): A = diethyl ketone (pentan-3-one). A + C2H5MgBr → (after H2O) B = (C2H5)3COH (triethylcarbinol). B + H2SO4/heat → C = (C2H5)2C=CHCH3 (dehydration, Zaitsev). C + O3 then Zn/H2O → (C2H5)2C=O (pentan-3-one) + CH3CHO (acetaldehyde). This matches the given products exactly. Why other options fail: (a) butan-2-one would give a different tertiary alcohol and different ozonolysis products; (c) hexan-2-one would not give the correct ozonolysis fragments; (d) acetone would give a secondary alcohol with different dehydration products. Therefore, the correct answer is B.