This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONLY ONE is correct. Which of the following compounds on reduction will give a mixture of two diastereomers? (a) Glucose (b) Mannose (c) Fructose (d) Galactose

The compounds are glucose, mannose, fructose, and galactose. All four are monosaccharides with glucose, mannose, and galactose being aldohexoses while fructose is a ketohexose. When one of these compounds undergoes reduction, it typically leads to the formation of sugar alcohols.

Diastereomers are stereoisomers that are not mirror images of each other. For the given compounds to form two diastereomers upon reduction, there must be two chiral centers with opposite configurations in the resulting sugar alcohol.

To consider each option, we will examine the structures of the sugar alcohols formed and look for a compound that yields two diastereomers upon reduction. (a) Glucose reduction forms glucitol (sorbitol), which has the same stereochemistry at all chiral centers. (b) Mannose reduces to mannitol, which also maintains the same stereochemistry at all chiral centers. (c) Fructose reduces to a mixture of two sugar alcohols: D-arabinitol and D-mannitol. These two products have the same molecular formula, but different chiral centers: D-arabinitol has the opposite stereochemistry at the C2 and C4 carbons compared to D-mannitol. (d) Galactose reduction forms galactitol, which retains the same stereochemistry at all chiral centers.

Based on the analysis, only the reduction of fructose leads to the formation of two diastereomers (D-arabinitol and D-mannitol). Thus, the correct answer is: (c) Fructose.