Question

Glucose does not give Schiff’s test because of the formation of cyclic ____________.

Options

• β-L-(–)-Glucose

• forms of α-D-(+)-Glucose and β-D-(+)-Glucose

• α-D-(–)-Glucose

• β-D-(–)-Glucose

Answer 1

Glucose does not give Schiff’s test because of the formation of cyclic forms of α-D-(+)-Glucose and β-D-(+)-Glucose.

Explanation:

Schiff's test is given by aldehydes. Despite having the aldehyde group, glucose does not give Schiff’s test as the aldehydic group is not free here. It was proposed that one of the —OH groups may add to the —CHO group and form a cyclic hemiacetal structure. It was found that glucose forms a six-membered ring in which —OH at C-5 is involved in a ring formation. This explains the absence of —CHO group and also the existence of glucose in two cyclic hemiacetal forms α-D-(+)-Glucose and β-D-(+)-Glucose.