See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: Osazone formation involves reaction of a sugar with excess phenylhydrazine (3 equivalents). The reaction affects C1 and C2 of the aldose, converting C1 (aldehyde) to CH=N-NH-Ph and C2 (CHOH or C=O) to C=N-NH-Ph. The configuration at C3, C4, C5, and C6 is preserved unchanged. Step 1: Read the osazone structure. The osazone has: - C1: CH=N-NH-Ph - C2: C=N-NH-Ph - C3: HO-C-H (OH on left in Fischer projection) - C4: H-C-OH (OH on right) - C5: H-C-OH (OH on right) - C6: CH2OH Step 2: Identify which sugars give the same osazone. Two aldohexoses give the same osazone if they differ only at C1 and C2 (i.e., they are epimers at C2). Because osazone formation destroys the stereochemical information at C1 and C2, any aldose with identical configuration at C3, C4, C5 will give the same osazone. Step 3: Determine the parent sugar from the osazone. The configuration at C3-C5 in the osazone (which mirrors the original sugar) is: - C3: OH on left (same as HO-C-H) - C4: OH on right - C5: OH on right Step 4: D-glucose has the Fischer projection: C1: CHO, C2: H-C-OH (OH right), C3: HO-C-H (OH left), C4: H-C-OH (OH right), C5: H-C-OH (OH right), C6: CH2OH. This matches C3-C5 of the osazone. Step 5: D-mannose is the C2 epimer of D-glucose: C1: CHO, C2: HO-C-H (OH left), C3: HO-C-H (OH left), C4: H-C-OH (OH right), C5: H-C-OH (OH right), C6: CH2OH. This also matches C3-C5 of the osazone. Step 6: Since D-glucose and D-mannose differ only at C2, they are C2 epimers and yield the same osazone. D-Idose has a different configuration at C3 (OH on right), so it gives a different osazone. Step 7: Therefore, both D-glucose and D-mannose give the osazone shown. Why option (c) fails: D-Idose has configuration C3: OH on right, C4: OH on left, C5: OH on right, which does not match the osazone shown. Therefore, the correct answer is D.