See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Work backward from known products. The final product is 1,1-diphenylcyclopropane (a cyclopropane ring bearing two phenyl groups on C1). The last step uses CH2I2/Zn/Cu (Simmons-Smith reaction), which converts an alkene to a cyclopropane by adding a CH2 group across the double bond. So (H) must be 1,1-diphenylethylene (Ph2C=CH2). Step 2: (G) --H+/Delta--> (H) = Ph2C=CH2. Dehydration of an alcohol under acid/heat gives the alkene. So (G) is Ph2C(OH)-CH3, i.e., 1,1-diphenylethanol... but wait: Simmons-Smith on Ph2C=CH2 gives 1,1-diphenylcyclopropane. So (H) = Ph2C=CH2 (1,1-diphenylethylene). Then (G) must be the alcohol Ph2C(OH)-CH3 (2-methyl-1,1-diphenyl-1-ol, i.e., Ph2C(OH)CH3) whose acid-catalyzed dehydration gives Ph2C=CH2. Step 3: (F) --(i) PhMgBr excess / (ii) H+--> (G) = Ph2C(OH)CH3. With excess PhMgBr (2 equivalents) reacting with an acid chloride: an acid chloride R-COCl reacts with 2 PhMgBr to give a tertiary alcohol R-C(Ph)2-OH after workup. So (G) = CH3-C(Ph)2-OH means (F) = CH3COCl (acetyl chloride). Step 4: (D) --SOCl2--> (F) = CH3COCl. SOCl2 converts a carboxylic acid to its acid chloride. So (D) = CH3COOH (acetic acid). Step 5: From the first sequence: (A) --NH2OH/Delta--> (B) --H2SO4--> (C) --H3O+--> (D) + (E). (D) = CH3COOH (acetic acid). Also, (E) --CHCl3/KOH,Delta--> CH3-N≡C (methyl isocyanide). CHCl3/KOH with a primary amine gives isocyanide (carbylamine reaction). So (E) = CH3NH2 (methylamine). This means (C) undergoes hydrolysis (H3O+) to give acetic acid (D) and methylamine (E). (C) is therefore N-methylacetamide (CH3CONHCH3). The Beckmann rearrangement of an oxime using H2SO4 gives an amide. So (B) is an oxime, and (C) = CH3CONHCH3 comes from Beckmann rearrangement of (B). The oxime (B) that rearranges to CH3CONHCH3 must be the oxime of methyl ethyl ketone? No - Beckmann rearrangement of CH3-C(=NOH)-CH3 (acetone oxime) gives CH3-CO-NH-CH3 (N-methylacetamide) because the group anti to OH migrates. Acetone oxime rearranges to N-methylacetamide. So (B) = acetone oxime, meaning (A) = acetone (CH3COCH3). Step 6: Molecular weight of acetone (CH3COCH3) = 12+3+1+3 + 12+16 + 12+3+1+3 = 58 g/mol. CH3COCH3: C3H6O = 3(12)+6(1)+16 = 36+6+16 = 58. Step 7: Verify other options fail: 120, 60, 182 do not correspond to acetone. MW = 58 matches option (a). Therefore, the correct answer is A.