See image — Biomolecules Chemistry Question

💡 Solution & Explanation

Concept: Fischer glycosidation (acid-catalyzed reaction of a sugar hemiacetal/anomeric OH with methanol) selectively converts the free anomeric hydroxyl group into a methyl glycoside (OMe), while leaving all other hydroxyl groups (including the inter-ring glycosidic oxygen and the other free OH groups) unchanged. Step 1 – Identify the reactive site. In the starting disaccharide, the left ring is the non-reducing end (its anomeric carbon is involved in the inter-ring glycosidic bond, so it is NOT free). The right ring is the reducing end – it has a free anomeric OH (shown as OH at the top-right of the right pyranose ring). This is the only hemiacetal/hemiketal OH present. Step 2 – Apply reaction conditions. MeOH/H+ promotes acid-catalyzed Fischer glycosidation exclusively at the free anomeric position (reducing end). The anomeric OH on the right ring is protonated, water leaves to form an oxocarbenium ion, and methanol attacks to give a methyl glycoside. The product has OMe at the anomeric position of the right ring. Step 3 – Determine stereochemistry of substitution. The question asks to predict the product; under thermodynamic/kinetic control the major or depicted product replaces the anomeric OH with OMe. In option (a), the OMe appears at the top-right (same face as the original OH was drawn), and the other OH on the right ring remains. All other HO groups on the left ring remain as HO (not methylated, because simple OH groups do not react with MeOH/H+ under these mild glycosidation conditions). Step 4 – Eliminate wrong options. - Option (b): OMe is placed at the bottom-right of the right ring (wrong position/stereochemistry relative to the anomeric center as shown, or places OMe on a non-anomeric OH). - Option (c): OMe appears on the left ring's non-anomeric OH rather than on the right ring's anomeric position – incorrect; acid-catalyzed methanol does not selectively methylate non-anomeric hydroxyls. - Option (d): MeO is placed on the left ring's hydroxyl – same error as (c); non-anomeric OH groups are not converted to methyl ethers under Fischer glycosidation conditions. Option (a) correctly shows OMe installed at the anomeric position of the reducing-end (right) ring, with all other hydroxyls unchanged. Therefore, the correct answer is A.