The $\beta$ anomer of a cyclic monosaccharide has the -OH group drawn down, below the ring.

Question

Quizplus · Accepted Answer

The statement is incorrect because the anomer designation (alpha or beta) refers to the position of the -OH group relative to the anomeric carbon, not its position above or below the ring. In the beta anomer, the -OH group is on the same side as the CH2OH group in the Fischer projection, which can be either above or below the ring in the Haworth projection depending on the orientation of the ring.

The β\betaβ Anomer of a Cyclic Monosaccharide Has the -OH Group Drawn

The $\beta$ Anomer of a Cyclic Monosaccharide Has the -OH Group Drawn