Determine the structure of the alditol formed when b-D-xylofuranose is treated with NaBH4 and then water. Explain how NaBH4 can reduce the hemiacetal group of the furanose.

• A.
• B.
• C.
• D.

Answer: (B)

Verified step by step guidance

1. Identify the structure of β-D-xylofuranose. It is a five-membered ring with an oxygen atom, and the hydroxyl groups are positioned at C-2, C-3, and C-4.
2. Understand the role of NaBH4. Sodium borohydride (NaBH4) is a reducing agent that can reduce aldehydes and ketones to alcohols. In this case, it will reduce the hemiacetal group of the furanose to an alditol.
3. Recognize that the reduction of the hemiacetal group involves the conversion of the anomeric carbon (C-1) from a hemiacetal to an alcohol. This means the ring will open, and the C-1 will now have a hydroxyl group.
4. Draw the open-chain form of the sugar after reduction. The structure will have a linear chain with hydroxyl groups on each carbon, and the terminal carbon (C-1) will have a CH2OH group.
5. Compare the possible alditol structures provided in the images. The correct alditol will have the same stereochemistry at C-2, C-3, and C-4 as the original β-D-xylofuranose, with an additional CH2OH group at C-1.