Linear (open-chain) glucose

The structure of glucose in open-chain form is: $\text{CHO} - \text{CH(OH)} - \text{CH(OH)} - \text{CH(OH)} - \text{CH(OH)} - \text{CH}_2\text{OH}$

Now identify stereocenters:

• $C_1$ is an aldehyde carbonyl carbon, so it is not chiral.
• $C_2, C_3, C_4, C_5$ are each attached to four different groups, so all are stereocenters.
• $C_6$ is $\text{CH}_2\text{OH}$, so it is not chiral.

Therefore, number of stereocenters in linear glucose is: $4$

Cyclic glucose

In cyclic form, glucose forms a hemiacetal by reaction of the $C_1$ aldehyde group with the $C_5$ hydroxyl group.

This creates a new chiral center at anomeric carbon $C_1$.

In the cyclic form:

• $C_1$ becomes chiral
• $C_2, C_3, C_4, C_5$ remain chiral

Hence total stereocenters in cyclic glucose are: $5$

Compare with options

• A: $4 \& 4$ ❌
• B: $4 \& 5$ ✅
• C: $5 \& 4$ ❌
• D: $5 \& 5$ ❌

Final Answer

The number of stereocenters in linear and cyclic glucose are respectively: $4 \text{ and } 5$

So, the correct option is B.