Number of stereo centers present in linear and cyclic structures of glucose are respectively:

We start by recalling the definition of a stereocentre (also called a chiral centre). A stereocentre is a carbon atom that is attached to four different groups, so that interchanging any two of those groups produces a non-superimposable mirror image.

Now we draw in mind the Fischer projection of D-glucose in its open-chain (linear) form:

$$$\mathrm{HO{-}CH_2{-}C(=O)-HC(OH)-HC(OH)-HC(OH)-HC(OH)-CH_2OH}$$$

Written carbon by carbon, the skeleton is

$$\begin{aligned} \text{C}_1 &: \; \mathrm{CHO} \\ \text{C}_2 &: \; \mathrm{H{-}C(OH)-} \\ \text{C}_3 &: \; \mathrm{H{-}C(OH)-} \\ \text{C}_4 &: \; \mathrm{H{-}C(OH)-} \\ \text{C}_5 &: \; \mathrm{H{-}C(OH)-} \\ \text{C}_6 &: \; \mathrm{CH_2OH} \end{aligned}$$

We examine every carbon one by one:

• $$\text{C}_1$$ is part of the aldehyde group $$\mathrm{-CHO}$$ and is double-bonded to oxygen; therefore it is not bonded to four different substituents. It is achiral.

• $$\text{C}_2$$ has the four different groups $$\mathrm{-CHO,\; OH,\; H,\; CH(OH)-}$$. This satisfies the chiral condition, so $$\text{C}_2$$ is a stereocentre.

• $$\text{C}_3$$ is attached to $$\mathrm{-C(=O)H,\; OH,\; H,\; CH(OH)-}$$ (all four groups different). Thus $$\text{C}_3$$ is also a stereocentre.

• $$\text{C}_4$$ in the same way has four different groups, so $$\text{C}_4$$ is a stereocentre.

• $$\text{C}_5$$ is bonded to $$\mathrm{-C(OH)-,\; OH,\; H,\; CH_2OH}$$, again four distinct groups; hence $$\text{C}_5$$ is a stereocentre.

• $$\text{C}_6$$ has two identical hydrogens (it is $$\mathrm{CH_2OH}$$), so it is not chiral.

Counting, the linear form possesses

$$4 \text{ stereocentres: } C_2,\, C_3,\, C_4,\, C_5.$$

Next we convert glucose into its more stable cyclic (pyranose) form. The aldehyde carbon $$\text{C}_1$$ reacts intramolecularly with the hydroxyl on $$\text{C}_5$$, producing a hemiacetal. In the Haworth representation this looks like:

$$$\text{(Fructose open-chain structure: HO-CH}_2\text{-C(=O)-HC(OH)-HC(OH)-HC(OH)-CH}_2\text{OH)}$$$

The crucial observation is that during ring formation $$\text{C}_1$$—formerly achiral—now becomes bonded to four different groups: $$\mathrm{O\,(ring)},\; H,\; OH\,(anomeric),\; \text{and } C_2$$. Therefore $$\text{C}_1$$ turns into a new stereocentre, called the anomeric carbon.

The pre-existing centres $$\text{C}_2,\, C_3,\, C_4,\, C_5$$ remain chiral as before. Hence, in the cyclic form we have

$$5 \text{ stereocentres: } C_1,\, C_2,\, C_3,\, C_4,\, C_5.$$

Summarising the count:

$$\text{Linear (open)} = 4,\quad \text{Cyclic (pyranose)} = 5.$$

Comparing with the given options, the pair (4 and 5) corresponds to Option C.

Hence, the correct answer is Option C.