See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: This is an aldol-type condensation (crossed aldol / Tollens reaction) where acetone reacts with formaldehyde (HCHO) under basic conditions (KOH) to introduce hydroxymethyl (-CH2OH) groups. Step 1: Identify the starting material and product. Starting material: Acetone (CH3-CO-CH3), which has 6 alpha-hydrogens (3 on each methyl group). Product: The structure shown is (HOCH2)3C-C(=O)-C(CH2OH)3, where each of the two methyl groups of acetone has been fully substituted - all three H atoms on each methyl replaced by -CH2OH groups. Step 2: Count hydroxymethyl groups added. Left quaternary carbon: 3 CH2OH groups (replacing 3 H's from one CH3) Right quaternary carbon: 3 CH2OH groups (replacing 3 H's from other CH3) Total CH2OH groups introduced = 3 + 3 = 6 Step 3: Each HCHO molecule introduces one -CH2OH group via the Tollens/aldol reaction: CH3-CO-CH3 + HCHO → HOCH2-CH2-CO-CH3 (one step) This process continues until all 6 alpha-H atoms are replaced. Step 4: Total moles of HCHO consumed = 6 (one for each alpha-hydrogen replaced, since each CH2OH group comes from one HCHO molecule). Step 5: Why x = 6 and not less: The product clearly shows both methyl groups fully converted to C(CH2OH)3 units, meaning all 6 alpha-hydrogens have been replaced, requiring exactly 6 moles of HCHO. Therefore, the correct answer is 6.