Question

(a) Convert each cyclic monosaccharide into a Fischer projection of its acyclic form. (b) Name each monosaccharide. (c) Label the anomer as α or β.

Short answer

Answer

Compound A

Compound B

Step-by-step solution

Ball and Stick conformation

Ball and stick conformation represents the most staggard conformation. To convert cyclic monosaccharide into Fisher projection the below steps are followed:

• The groups that are below the plane in the cyclic form will be on the left side of the Fischer projection formula.
• The groups above the plane in the cyclic form will be on the right side of the Fischer projection formula.

The structure A

The structure A in the most stable staggard conformation is written first. Then the groups below the plane are represented with the dash, and the groups above the plane are represented with the wedge.

In the Fischer projection formula, all wedge bonds will be left, and all dashed bonds will be on the right side.

Compound A

The structure B

The structure B in the Haroth formula is written first. In the Fischer projection formula, all bonds below the plane will be on the left side, and all bonds above the plane will be on the right side.

The Fischer projection formula of compound B is as shown:

Compound B

The configuration at the anomer

The compound A is a $\mathrm{\alpha }$-anomer. At C-1, the –OH group is below the plane, and at C-4, the$-{\mathrm{CH}}_2\mathrm{OH}$ group is above the plane. So both are trans to each other.

The compound B is a $\mathrm{\alpha }$-anomer. At C-1, the –OH group is below the plane, and at C-5, the $-{\mathrm{CH}}_2\mathrm{OH}$group is above the plane. So both are trans to each other.