Question

Glucose can occur in three forms: two cyclic forms and one open-chain structure. In aqueous solution, only a tiny fraction of the glucose is in the open-chain form. Yet test for the presence of glucose depend on reaction with the aldehyde group, which is found only in the open chain form. Explain why these tests work

Short answer

The open-chain structure is in the equilibrium with the cyclic hemiacetal.

Step-by-step solution

To find which is found only in the open chain form

Aldoses are contained in an aldehyde group and several hydroxyl groups.

If the aldehyde group and the hydroxyl group are part of the same molecule, cyclic hemiacetal results. Cyclic hemiacetals are particularly stable if they result in five or six-membered rings. In fact, five- and six-membered cyclic hemiacetals are often more stable than their open-chain forms.

For most common aldohexoses, the equilibrium favors six-membered rings with a hemiacetal linkage between the aldehyde carbon and the hydroxyl group on C5.

Step 2:  To Explain why these tests work

The tests for the presence of glucose depend on reaction with the aldehyde group which is only available for the reaction in the open-chain structure. As we know, a small proportion of open-chain form is present at equilibrium. But as the aldehyde group reacts, the equilibrium between the cyclic forms of glucose and the open-chain structure will shift to produce more of the open-chain structure.

Hence The open-chain structure is in equilibrium with the cyclic hemiacetal.