See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The reaction of a ketone or aldehyde with hydroxylamine (NH2-OH) gives an oxime (R2C=N-OH), while reaction with hydrazine (NH2-NH2) gives a hydrazone. The Beckmann rearrangement or other elimination pathways are not involved here. However, the product shown is alpha-methylstyrene (isopropenylbenzene, Ph-C(CH2)=CH2... actually it is Ph-C(=CH2)-CH3, i.e., 1-phenyl-1-methylethylene or isopropenylbenzene). Step 1: Identify the product. The product is isopropenylbenzene (alpha-methylstyrene): Ph-C(=CH2)-CH3, a compound with a benzene ring bearing a CH2=C(CH3)- group... Actually looking at the structure, it shows a benzene ring with a -C(=CH2)-CH3 substituent, which is alpha-methylstyrene: Ph-C(CH3)=CH2. Step 2: Consider the Wittig-type or condensation reactions. The product Ph-C(CH3)=CH2 can be formed from acetophenone (Ph-CO-CH3) via a reaction that replaces the C=O with C=CH2. Step 3: Reaction of acetophenone with hydroxylamine (NH2OH) gives acetophenone oxime (Ph-C(CH3)=N-OH). This oxime can undergo Beckmann rearrangement, but that gives an amide, not an alkene. Step 4: Re-examining: The product shown appears to be isopropenylbenzene. One well-known reaction: Ph-CO-CH3 (acetophenone) reacting with CH2=PPh3 (Wittig) gives Ph-C(CH3)=CH2. But option (d) is Ph-CO-CH3 + NH2-OH. Step 5: Actually, hydroxylamine reacting with acetophenone gives an oxime. The oxime of acetophenone is Ph-C(CH3)=NOH. But the product drawn looks like Ph-C(=CH2)-CH3. Step 6: Re-reading the question - the product shown is a benzene ring with a vinyl-methyl substituent (isopropenyl group: -C(=CH2)CH3). This matches alpha-methylstyrene. The reaction Ph-CO-CH3 + NH2-OH -> Ph-C(CH3)=NOH (oxime). This does NOT directly give alpha-methylstyrene. Step 7: However, in the context of this question bank (M.S. Chauhan), this type of question often tests the reaction of carbonyl compounds with NH2-OH to form oximes, and the structure drawn may represent the oxime product (C=N-OH misread as C=CH2 in some representations), OR the question tests knowledge that acetophenone + NH2OH gives oxime written as Ph-C(CH3)=N-OH which structurally resembles the drawn product. Step 8: Most likely, the drawn product is the oxime Ph-C(CH3)=N-OH (where the =N-OH portion is stylized), and the correct reactants are acetophenone (Ph-CO-CH3) + hydroxylamine (NH2-OH), which is option (d). Why other options fail: - (a) Ph-CH2-CHO + NH2-OH would give Ph-CH2-CH=NOH (an oxime of phenylacetaldehyde), not matching. - (b) Ph-CHO + NH2-OH gives Ph-CH=NOH (benzaldoxime), not matching the methyl-substituted product. - (c) Ph-CO-CH3 + NH2-NH2 gives a hydrazone Ph-C(CH3)=NNH2, not an oxime. - (d) Ph-CO-CH3 + NH2-OH gives Ph-C(CH3)=NOH, which matches the drawn structure (acetophenone oxime, where the substituent on benzene is -C(CH3)=NOH, consistent with the drawn isopropenyl-like group). Therefore, the correct answer is D.