Question

Indicate which Haworth projection corresponds to the beta-pyranose form of the Fischer projection below. Compound A Compound B Compound C Compound D Compound E Compound A Compound B Compound C Compound D Compound E

Answer 1

Concepts and reason

The concept used to solve this problem is based on the Haworth projection of an organic compound.

The Haworth projection is basically used to represent sugars in the cyclic form. It tells which atom is above the plane and which is below the plane.

Fundamentals

The Haworth structure can be drawn from the Fischer projection. The characteristics of Haworth projection is as follow.

The carbon atom is implicit type of atom in the Haworth structure. The carbon present on the 1 number is known as anomeric carbon.

Hydrogen atoms are present on carbon atom.

A thicker line represents the atoms which are closer to observer.

The Fischer projection is as follows.

The compound C and E have five membered ring as follow.

Now in compound D, ${\rm{C}}{{\rm{H}}_2}{\rm{OH}}$ group present on lower position of the six-membered ring as follows.

In compound A, OH group present on the anomeric carbon is at lower position of the ring as follow.

Thus, the compounds A, C, D and E are not Haworth projection corresponds to $\beta$ -pyranose.

Now, from the Fischer projection draw the Haworth projection as follow.

The atoms on left side of Fischer projection will present on upper position in the six membered ring and the atoms on right side will present lower position in the ring.

Ans:

The Haworth projection of $\beta$ -pyranose is as follow.