Question

Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.

Answer 1

1. Aldehydes give 2, 4-DNP test, Schiff’s test, and react with NaHSO₃ to form the hydrogen sulphite addition product. However, glucose does not undergo these reactions.
2. The pentaacetate of glucose does not react with hydroxylamine. This indicates that a free −CHO group is absent from glucose.
3. Glucose exists in two crystalline forms – α and β. The α-form (m.p. = 419 K) crystallises from a concentrated solution of glucose at 303 K and the β-form (m.p. = 423 K) crystallises from a hot and saturated aqueous solution at 371 K. This behaviour cannot be explained by the open chain structure of glucose.

Answer 2

D-(+)-glucose does not undergo certain characteristic reactions of aldehydes, e.g., glucose does not form NaHSO₃ addition product.

Glucose reacts with NH₂OH to form an oxime but glucose pentaacetate does not. This implies that the aldehydic group is absent in glucose pentaacetate.

D-(+)-glucose exists in two stereoisomeric forms, i.e., α-glucose and β-glucose.

Both α-D-glucose and β-D-glucose undergo mutarotation in aqueous solution. Although the crystalline forms of α- and β-D-(+)-glucose are quite stable in aqueous solution, each form slowly changes into an equilibrium mixture of both.

D-(+)-glucose forms two isomeric methyl glucosides. Aldehydes normally react with two moles of methanol per mole of the aldehyde to form an acetal but D-(+)-glucose, when treated with methanol in the presence of dry HCl gas, reacts with only one mole of methanol per mole of glucose to form a mixture of two methyl D-glucosides i.e., methyl-α-D-glucoside (melting point 438 K, specific rotation +158°) and methyl-β-D-glucoside (melting point 308 K, specific rotation – 33°).