This question involves determining the total number of stereoisomers for the pyranose form of a given D-aldohexose. Let's break down the concepts step by step.

Step 1: Understanding the Aldohexose Structure
An aldohexose is a six-carbon sugar with an aldehyde functional group. The given structure has four chiral centers (asymmetric carbons) in its open-chain form. In the D-configuration, the highest numbered chiral carbon (C5) has the hydroxyl group on the right in Fischer projection.

Step 2: Pyranose Form Formation
The pyranose form is a cyclic hemiacetal formed when the aldehyde group reacts with the hydroxyl group on C5, creating a six-membered ring. This cyclization generates a new chiral center at the anomeric carbon (C1).

Step 3: Counting Stereoisomers
In the open-chain form, there are 4 chiral centers, so the number of stereoisomers is $2^4=16$. However, in the pyranose form, the anomeric carbon (C1) can have two configurations: α and β. Since the ring formation does not eliminate any existing chiral centers (C2, C3, C4, C5 remain chiral), the total stereoisomers in pyranose form are due to the 5 chiral centers (C1, C2, C3, C4, C5). Thus, the number is $2^5=32$.

Final Answer: The total number of stereoisomers in the pyranose form is 32.

Related Topics

Carbohydrate Stereochemistry: Aldohexoses have multiple chiral centers leading to numerous stereoisomers. The D/L configuration specifies the orientation around the penultimate carbon. Cyclization to pyranose forms introduces an additional chiral center at the anomeric carbon, doubling the number of stereoisomers compared to the open-chain form.

Formulae

Number of stereoisomers = $2^n$, where n is the number of chiral centers.