Identify the correct set of stereochemical relationships amongst the following monosaccharides I-IV A. I and II are anomers; III and IV are epimers B. I and III are epimers; II and IV are anomers C. I and II are epimers; III and IV are anomers D. I and III are anomers; I and II are epimers

The correct answer is A. I and II are anomers; III and IV are epimers.

Anomers are stereoisomers that differ in the configuration at the anomeric carbon, which is the carbon atom that is bonded to the carbonyl group and to the hydroxyl group. Epimers are stereoisomers that differ in the configuration at only one chiral center.

In the given monosaccharides, I and II are anomers because they differ in the configuration at the anomeric carbon. III and IV are epimers because they differ in the configuration at the chiral center that is not the anomeric carbon.

Here is a diagram of the four monosaccharides, with the anomeric carbon and the chiral center that is not the anomeric carbon labeled:

The anomeric carbon is the carbon atom that is bonded to the carbonyl group and to the hydroxyl group. The chiral center that is not the anomeric carbon is the carbon atom that is bonded to four different groups.

In monosaccharides, the anomeric carbon is always the carbon atom that is bonded to the carbonyl group. The chiral center that is not the anomeric carbon can be any of the other carbon atoms in the monosaccharide.

In the given monosaccharides, I and II are anomers because they differ in the configuration at the anomeric carbon. I has the alpha configuration at the anomeric carbon, while II has the beta configuration.

III and IV are epimers because they differ in the configuration at the chiral center that is not the anomeric carbon. III has the alpha configuration at this chiral center, while IV has the beta configuration.

Therefore, the correct answer is A. I and II are anomers; III and IV are epimers.