Santonin's oxidation to tetraol followed by lactone hydrolysis Steps:

• Santonin (C₁₅H₁₈O₃) features two C=C bonds and a γ-lactone ring.
• Dilute acidified KMnO₄ causes syn dihydroxylation of both double bonds, forming a tetraol-lactone (C₁₅H₂₂O₇) with four OH groups.
• This tetraol-lactone retains the lactone but now has four alcoholic OH groups.
• Cold H₂SO₄ hydrolyzes the lactone, opening it to add one alcoholic OH and one COOH group, yielding Q (C₁₅H₂₄O₈). Why D is correct:
• Q has five alcoholic OH and one COOH, providing six O-H bonds whose hydrogens react with Na via ROH + Na → RONa + ½H₂ and RCOOH + Na → RCOONa + ½H₂. Why the others are wrong:
• A: Ignores multiple oxidations and hydrolysis, yielding too few active H.
• B: Counts only one diol (two OH) plus lactone opening (two active H), totaling four.
• C: Overcounts or miscounts COOH H as two, but it's one per group. Final answer: D