The open chain structure of D-glucose OHC–(CHOH)4–CH2OH fails to explain the following reactions :
(i) Though it contains the aldehyde (–CHO) group, glucose does not give 2,4-DNP test, Schiff’s test and it does not form the hydrogen sulphiteadditionproduct with NaHSO3.
(ii) The pentaacetate of glucose does not react with hydroxylamine (NH2OH) to form the oxime indicating the absence of free –CHO group.
(iii) The formation of two anomeric methyl glycosides by glucose on reactionwithCH3OH and dry HCl can be explained in terms of the cyclic structure. The equilibrium mixture of a-and b-glucose react separately with methanol in the presence of dry HCl gas to form the corresponding methyl D-glucosides(iv) Like methyl glycosides, glucose pentaacetate also exists in two anomeric formes as explained below :These pentaacetates donot have a free –OH group at C1 and hence are not hydrolysed in aqueous solution to produce the open chain aldehyde form and hence do not react with NH2OH to formglucose oxime.
(v) The existence of glucose in two crystalline forms termed as α and β-D-glucose can again be explained on the basis of cyclic structure of glucose and not by its open chain structure. It was proposed that one of the –OH groups may add to – CHO group and form a cyclic hemiacetal structure. It was found that glucose forms a 6-membered ring in which –OH at C – 5 is involved in ring formation. This explains the absence of –CHO group and also existence of glucose in two forms as shown below. These two forms exist in equilibrium with open chain structure.