Santonin's transformations yield a polyhydroxy acid with six reactive hydrogens

Steps:

• Santonin features a lactone ring and two carbon-carbon double bonds.
• Dilute H2SO4 hydrolyzes the lactone, forming a hydroxy carboxylic acid with one -OH and one -COOH (two acidic H atoms).
• Cold dilute acidified KMnO4 dihydroxylates both isolated double bonds, adding four -OH groups (four more acidic H atoms).
• Product Q thus has six -OH equivalents; each H reacts with Na metal.

Why D is correct:

• Q has six -OH groups; sodium reacts with each acidic H via 2Na + 2ROH → 2RONa + H2, displacing six H atoms per molecule.

Why the others are wrong:

• A. 2: Counts only hydrolysis acidic H, omitting dihydroxylation.
• B. 4: Assumes single double bond dihydroxylation plus hydrolysis.
• C. 5: Yields odd number, inconsistent with paired -OH additions.

Final answer: D