See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: Phenyl osazone formation involves reaction of a sugar with excess phenylhydrazine. The osazone reaction affects only C-1 and C-2 of the sugar (the carbonyl carbon and the adjacent CHOH). Therefore, two sugars that differ ONLY in the configuration at C-1 and/or C-2 (i.e., they are epimers at C-1 or C-2, or both) will yield the same osazone, because those stereocenters are destroyed during osazone formation. Reasoning: D-Glucose and D-Mannose are C-2 epimers of each other — they have identical configurations at C-3, C-4, C-5, and C-6, and differ only at C-2. When osazone is formed, the configuration at C-2 is obliterated (it becomes C=N-NHPh), so both sugars give the same osazone, known as glucosazone (or D-glucosazone). Why other options fail: (a) D-Glucose and D-Allose differ at C-3 (and C-2), so they do NOT give the same osazone. (b) D-Glucose and D-Altrose (likely intended as D-Altrose, written as D-Alfrose in the question) differ at C-2 and C-3, so they do NOT give the same osazone. (d) D-Glucose and D-Talose differ at C-2, C-3, and C-4, so they do NOT give the same osazone. Only D-Glucose and D-Mannose are C-2 epimers (differing solely at C-2), ensuring identical osazone products. Therefore, the correct answer is C.