See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the final product and work backwards. The final product after Wolff-Kishner-type conditions (NH2-NH2 / HO- / Delta) is ethylbenzene (C8H10: a benzene ring with an ethyl group). The Wolff-Kishner reduction converts a carbonyl (C=O) to a CH2. So intermediate B (C8H8O) must be a carbonyl compound that gives ethylbenzene upon Wolff-Kishner reduction. Step 2 - Identify intermediate B. Ethylbenzene is Ph-CH2-CH3. Wolff-Kishner of a ketone R-CO-R' gives R-CH2-R'. So B must be Ph-CO-CH3, i.e., acetophenone (C8H8O). This matches the molecular formula C8H8O. Step 3 - Ozonolysis of A (C16H16) gives B (C8H8O). Ozonolysis cleaves a C=C double bond. Since A is C16H16 and ozonolysis gives one product B of formula C8H8O (two equivalents, since 2 x C8H8O accounts for all carbons and oxygens added), each fragment from ozonolysis is acetophenone. This means both sides of the double bond in A give Ph-CO-CH3 upon ozonolysis. Step 4 - Determine the structure of A. If ozonolysis of A (an alkene) gives two molecules of acetophenone (Ph-CO-CH3), then A must be the alkene Ph(CH3)C=C(CH3)Ph, i.e., 1,2-diphenyl-1,2-dimethylethylene (stilbene analog with two methyl groups). This has the molecular formula: each carbon of the double bond carries one Ph and one CH3 group. Formula: C6H5-C(CH3)=C(CH3)-C6H5 = C16H16. Confirmed. Step 5 - Match to answer options. Option (b) shows Ph(CH3)C= =C(CH3)Ph as the (E)-isomer (trans configuration, two separate structures shown side by side indicating stereoisomers or the same connectivity drawn differently). Option (c) shows Ph(CH3)C= =C(Ph)(CH3), which is the same connectivity as (b) - just the (Z)-isomer or the same compound drawn from the other perspective. Both (b) and (c) represent the same constitutional structure Ph(CH3)C=C(CH3)Ph but as different geometric isomers (E and Z). Both would give acetophenone upon ozonolysis, and both have formula C16H16. Option (a) is Ph2C=C(CH3)2 (1,1-diphenyl-2-methylpropene), which upon ozonolysis would give Ph2C=O (benzophenone, C13H10O) and (CH3)2C=O (acetone), not acetophenone. So (a) is incorrect. Step 6 - Conclusion. Both geometric isomers (b) and (c) of 1,2-diphenyl-1,2-dimethylethylene satisfy all conditions, so the answer is (d) both (b) and (c). Therefore, the correct answer is D.