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 NaHSO₄ 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 − ∝ andβ. The ∝-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 but 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 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 43 8 K, specific rotation +158°) and methyl – β – D – glucoside (melting point 308 K, specific rotation – 33°).