Which of the following correctly represents the Haworth projection of the $\alpha$-pyranose form of $D$-glucose derived from its Fischer projection?

• A. The anomeric hydroxyl group is positioned down (axial) on carbon 1, and the ${\mathrm{CH}}_2{\mathrm{OH}}_{}$ group is above the ring.
• B. The anomeric hydroxyl group is positioned up (equatorial) on carbon 1, and the ${\mathrm{CH}}_2{\mathrm{OH}}_{}$ group is above the ring.
• C. The anomeric hydroxyl group is positioned up (equatorial) on carbon 1, and the ${\mathrm{CH}}_2{\mathrm{OH}}_{}$ group is below the ring.
• D. The anomeric hydroxyl group is positioned down (axial) on carbon 1, and the ${\mathrm{CH}}_2{\mathrm{OH}}_{}$ group is below the ring.

Answer: (A)

Verified step by step guidance

1. Start by identifying the Fischer projection of D-glucose, noting the orientation of hydroxyl groups on each carbon, especially at C-1 (the anomeric carbon) and C-5, which determines the ring closure.
2. Recall that the Haworth projection is a cyclic representation formed when the C-5 hydroxyl attacks the aldehyde at C-1, creating a hemiacetal and forming a six-membered pyranose ring.
3. Determine the position of the anomeric hydroxyl group (on C-1) in the Haworth projection: for the alpha (α) anomer, the OH group is on the opposite side of the ring relative to the CH2OH group at C-5.
4. Translate the Fischer projection substituents to the Haworth projection: groups on the right in the Fischer projection point down in the Haworth ring, and groups on the left point up.
5. Conclude that in the α-D-pyranose form of glucose, the anomeric hydroxyl group at C-1 is positioned down (axial), and the CH2OH group at C-5 is positioned above the ring.