Question

Treatment of \(\beta\)-D-glucose with methanol in the presence of an acid catalyst converts it into a mixture of two compounds called methyl glucosides (Section 25.3A). In these representations, the six-membered rings are drawn as planar hexagons. (a) Propose a mechanism for this conversion and account for the fact that only the - \(\mathrm{OH}\) on carbon 1 is transformed into an \(-\mathrm{OCH}_{3}\) group. (b) Draw the more stable chair conformation for each product. (c) Which of the two products has the chair conformation of greater stability? Explain.

Short answer

Answer: The -OH group on carbon 1 is transformed into an -OCH3 group due to its higher reactivity as it is part of the hemiacetal functional group, which is more susceptible to nucleophilic attack. The product with the chair conformation of greater stability is the one where the -OCH3 group and other substituents (like other -OH groups) are in equatorial positions, as these positions minimize steric hindrance and 1,3-diaxial interactions.

Step-by-step solution

Propose a mechanism for the conversion of β-D-glucose into methyl glucosides.

The reaction starts with the protonation of the hydroxyl group on carbon 1 by the acidic catalyst. This makes the oxygen more electrophilic, and methanol can then act as a nucleophile. Methanol attacks the electrophilic oxygen, leading to the formation of an intermediate. Finally, the intermediate loses a proton, resulting in the formation of two different methyl glucosides.

Explain why only the -OH group on carbon 1 is transformed into an -OCH3 group.

The reason only the -OH group on carbon 1 undergoes the transformation into an -OCH3 group is due to its higher reactivity compared to other hydroxyl groups. The -OH group on carbon 1 is part of the hemiacetal functional group, which is more susceptible to nucleophilic attack.

Draw the more stable chair conformations for each product.

To draw the more stable chair conformations for each product consider the fact that equatorial positions are more stable than axial positions for substituents. For each product, draw a chair conformation in which the -OCH3 group and as many of the other substituents as possible is in an equatorial position.

Determine the chair conformation of greater stability and provide an explanation.

The product with the chair conformation of greater stability is the one where the -OCH3 group and other substituents (like other -OH groups) are in equatorial positions. The reason for the greater stability is that equatorial positions minimize steric hindrance and 1,3-diaxial interactions.