Glucose and Galactose are having identical configuration in all the positions except position.

We first recall the definition of an epimer. Two monosaccharides that differ only in the configuration around a single chiral carbon atom (other than the anomeric carbon) are called epimers. Mathematically, if a molecule has several chiral centres $$C\_1, C\_2, C\_3, \ldots , C\_n$$ and two structures are identical at all centres except one $$C\_k$$, then they are epimers at $$C\_k$$.

Now, D-Glucose and D-Galactose are both aldohexoses, so each contains six carbon atoms arranged linearly in the Fischer projection as:

$$\text{CHO - C}_2\text{H(OH) - C}_3\text{H(OH) - C}_4\text{H(OH) - C}_5\text{H(OH) - CH}_2\text{OH}$$

In the Fischer projection for D-Glucose, the orientations of the OH groups (from $$C\_2$$ to $$C\_5$$) are:

$$C_2$$: Right, $$C_3$$: Left, $$C_4$$: Right, $$C_5$$: Right

For D-Galactose, the corresponding orientations are:

$$C_2$$: Right, $$C_3$$: Left, $$C_4$$: Left, $$C_5$$: Right

We carefully compare these two sets.

At $$C\_2$$ both molecules have the OH group on the right. At $$C\_3$$ both have the OH on the left. At $$C\_5$$ both have the OH on the right.

The only discrepancy appears at $$C\_4$$:

• In D-Glucose, the OH at $$C\_4$$ is on the right.
• In D-Galactose, the OH at $$C\_4$$ is on the left.

Since every other chiral centre matches, D-Glucose and D-Galactose differ only in the configuration at carbon number 4. Therefore they are C-4 epimers.

Looking back at the options, Option B states “C - 4”. This matches our conclusion.

Hence, the correct answer is Option B.