1. Identify the stereochemical features of an aldohexose

An aldohexose has the open-chain structure:

$\text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH}$

So, in the open-chain form, the stereogenic centers are at:

$C_2, C_3, C_4, C_5$

That means there are 4 chiral centers in the open-chain aldohexose.

2. What does D-configuration fix?

For sugars, D/L is determined by the configuration at the last chiral carbon (here $C_5$).

So if the aldohexose is specified to be in its D-configuration, then the configuration at $C_5$ is fixed.

Thus, among the 4 original chiral centers:

• $C_5$ is fixed,
• $C_2, C_3, C_4$ can vary.

Hence the number of possible open-chain D-aldohexoses is:

$2^3 = 8$

3. Conversion to pyranose form

In pyranose formation, the aldehyde group at $C_1$ reacts with the hydroxyl group at $C_5$ to form a cyclic hemiacetal.

This creates a new stereocenter at the anomeric carbon $C_1$.

For each open-chain D-aldohexose, there are therefore 2 anomers:

• $\alpha$
• $\beta$

So total stereoisomers in pyranose form:

$8 \times 2 = 16$

4. Final answer

Therefore, the total number of stereoisomers in the pyranose form of a D-aldohexose is:

$\boxed{16}$

5. Comparison with stored correct answer

Stored correct answer = $8$

My derived answer = $16$

The stored answer appears to have counted only the D-aldohexoses in open-chain form and did not include the additional anomeric stereocenter formed in the pyranose ring.

Hence, I disagree with the stored answer.