Question

Treatment of methyl \(\beta\)-D-glucopyranoside with benzaldehyde forms a six- membered cyclic acetal. Draw the most stable conformation of this acetal. Identify each new chiral center in the acetal.

Short answer

Answer: The most stable conformation of the cyclic acetal formed is the chair conformation, where bulky substituents like benzyl and methyl groups are placed in the equatorial positions to minimize steric interactions. In the product, the original chiral centers (C2, C3, C4, and C5) are retained, and a new chiral center, C1, is introduced due to the acetal formation.

Step-by-step solution

Reacting methyl β-D-glucopyranoside with benzaldehyde

First, we need to understand how the reaction between methyl β-D-glucopyranoside and benzaldehyde occurs. Methyl β-D-glucopyranoside has a hemiacetal functional group at the anomeric carbon (C1 carbon) and benzaldehyde has an aldehyde group. The hemiacetal functional group reacts with the aldehyde group to form a cyclic acetal. The product formed will retain all the existing chiral centers in methyl β-D-glucopyranoside and may introduce new chiral centers.

Drawing the most stable conformation

To draw the most stable conformation of the cyclic acetal formed, we need to consider the steric interactions between the substituents. The most stable conformation will have the least number of steric clashes. The acetal will be a substituted six-membered ring with an acetal oxygen bridge. The chair conformation will be the most stable conformation for a six-membered ring. In the chair conformation, the bulky substituents are placed in the equatorial position in order to minimize steric interactions. In this case, the benzyl and methyl groups will both be placed in the equatorial position. The other groups (hydroxyl and others) will occupy their respective axial and equatorial positions based on the stereochemistry of the starting material (methyl β-D-glucopyranoside).

Identifying the new chiral centers

In the product formed, the original chiral centers (C2, C3, C4, and C5) in methyl β-D-glucopyranoside are retained. We need to check if there are any new chiral centers created during the reaction. The acetal carbon (C1) is a new chiral center, as it was originally part of the hemiacetal group in the reactant, and now it is connected to the benzyl group in the product. There are no other new chiral centers formed in the reaction.