Question

The $\alpha -$ and $\beta -$ forms of glucose are:
A. isomers of D(+) glucose and L(-) glucose respectively
B. diastereomers of glucose
C. anomers of glucose
D. isomers which differ in the configuration of $C-2$

Answer 1

Hint: Think about the meanings of the terms dextro- and laevo-rotatory compounds, diastereomers, epimers, anomers and consider the difference in the positioning of the hydroxyl group that is seen in the $\alpha -$ and $\beta -$ forms.

Complete step by step answer:
The $\alpha -$ and $\beta -$ forms of glucose differ in the location of the $-OH$ group that is present on the first carbon of the molecule.
Let us consider all the given options one by one, look at their meanings and determine whether they are true or not.

- D-(+)- glucose and L-(-)-glucose
These D- and L- forms are the stereoisomers of glucose. They are non-superimposable mirror images of each other. In the Fischer projection form, if the $-OH$ group on the fifth carbon is found towards the right hand side, then that molecule is said to be D-(+)- glucose and if the $-OH$ group is towards the left side, then the molecule is considered to be L-(-)-glucose. In these forms, the stereochemistry differs at all centres and not just one, so this is not the correct answer.

- Diastereomers
Diastereomers are isomers of a molecule that differ in configuration at more than one stereocenters but not at all. Diastereomers are not mirror images of each other. Since the $\alpha -$ and $\beta -$ forms of glucose differ at only one chiral center, this is not the correct answer. Some examples of the diastereomers of glucose are:

- Anomers
Before looking at anomers, let us first check what epimers mean. Epimers are the isomers that differ at one and only once chiral center. There are 4 chiral centers in glucose. An epimer of glucose at the hemiacetal carbon i.e. the first carbon, will be considered an anomer. The $\alpha -$ and $\beta -$ forms of glucose differ at this position, so this is the correct answer.

Notice the position of the $-OH$ group on the hemiacetal carbon. This carbon is also known as the anomeric carbon. The representation of the anomers in the Fischer projection formula is:

- Configuration of $C-2$
We have discussed that the anomeric carbon is the first carbon and it is on $C-1$ that the isomerism is seen for the $\alpha -$ and $\beta -$ forms of glucose, so this option is incorrect.
So, the correct answer is “Option C”.

Note: Remember that option A cannot be true since Both the isomers D-glucose and L-glucose have their own $\alpha -$ and $\beta -$ forms of glucose, so the $\alpha -$ form cannot be associated with D-glucose and the $\beta -$ form with L-glucose or vice versa.