Question

Name the two alditols formed by \(\mathrm{NaBH}_{4}\) reduction of D-fructose.

Short answer

Answer: The two alditols formed from the NaBH4 reduction of D-fructose are D-Mannitol and D-Sorbitol.

Step-by-step solution

Draw the structure of D-fructose

D-fructose, a monosaccharide, has the following structure: \(CH_{2}OH\text{-}CH(OH)CO\text{-}CH(OH)CH(OH)CH_{2}OH\) We need to identify the carbonyl group in D-fructose, as that is the part that will be reduced.

Identify the carbonyl group

The carbonyl group in D-fructose is a ketone, specifically located on the second carbon atom (from the top). This is the site that will be reduced by NaBH4: \(CH_{2}OH\text{-}CH(OH)\underline{CO}\text{-}CH(OH)CH(OH)CH_{2}OH\)

Understand the NaBH4 reduction

Sodium borohydride (NaBH4) is a reducing agent that converts the carbonyl group (ketone/aldehyde) into an alcohol. In the case of D-fructose, the ketone group will be reduced to an alcohol.

Write the structures of the reduced compounds

During the NaBH4 reduction of D-fructose, the ketone group on the second carbon atom will be reduced to an alcohol, as shown below: D-Fructose: \(CH_{2}OH\text{-}CH(OH)CO\text{-}CH(OH)CH(OH)CH_{2}OH\) Alditol 1: \(CH_{2}OH\text{-}CH(OH)CH_{2}OH\text{-}CH(OH)CH(OH)CH_{2}OH\) However, the reduction can result in two stereoisomers due to the change in the configuration around the second carbon atom. Thus, we also need to draw the second alditol: Alditol 2: \(CH_{2}OH\text{-}CH(OH)CH(OH)\text{-}CH(OH)CH(OH)CH_{2}OH\)

Name the two alditols

Based on the structures obtained in Step 4, the two alditols formed by NaBH4 reduction of D-fructose are: 1. D-Mannitol (Alditol 1) 2. D-Sorbitol (Alditol 2)