Question

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

Short answer

Answer: D-glucitol (D-sorbitol) and D-mannitol.

Step-by-step solution

Identify the structure of D-fructose

Draw the structure of D-fructose. It is a ketohexose, meaning it has a ketone functional group and six carbon atoms. The structure of D-fructose is: ``` HO HO | | HO-CH2-CHO-CH2OH | CH=O ```

Reaction with sodium borohydride (NaBH4)

Now, we need to understand the reaction of NaBH4 with the ketone group in D-fructose. Sodium borohydride is a selective reducing agent that will convert the ketone group to an alcohol. This occurs through a nucleophilic attack by hydride ion (H-) on the carbonyl carbon, followed by the addition of a proton source.

Identify the two alditols

When the ketone group of D-fructose reacts with sodium borohydride, the carbonyl (=O) is reduced to an alcohol (-OH) group. This will result in two alditols being formed. These two alditols are: ``` D-glucitol or D-sorbitol HO HO | | HO-CH2-CHO-CH2OH | CH2OH ``` and ``` D-mannitol HO HO | | HO-CH2-CHO-CH2OH | HO-CH2 ``` Both alditols are formed due to the equatorial and axial positions of the hydroxyl (-OH) group on the second carbon atom in D-fructose.

Summarize the resulting alditols

After the reduction of D-fructose with sodium borohydride (NaBH4), two alditols are formed: D-glucitol (D-sorbitol) and D-mannitol.