Question

Carbohydrates are optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis. They have been broadly classified into three groups - monosaccharides, oligosaccharides and polysaccharides. The carbohydrates may also be classified as either reducing or non-reducing sugars. An important monosaccharide glucose is an aldohexose and is correctly named as D(+)-glucose. It was assigned the open structure on the basis of its reactions with HI, \(\mathrm{NH_2OH}\), \(\mathrm{Br_2}\) water, \(\mathrm{(CH_3CO)_2O}\) and nitric acid.

Despite having the - CHO group, glucose does not give Schiff's test and hydrogen sulphite addition product with \(\mathrm{NaHSO_3}\). It was found that glucose forms a six-membered ring in which one of the - OH groups add to the - CHO group and form a cyclic hemiacetal structure.

Answer the following questions :

How do you explain the following ?

(i) Presence of a carbonyl group in glucose

(ii) Presence of five - OH groups in glucose which are attached to different carbon atoms.

Hint:

Use the oxime or cyanohydrin test for the carbonyl group, and the pentaacetate with acetic anhydride for five OH groups.

Answer 1

Step 1: Understanding the Concept:
The open chain structure of glucose, \(\mathrm{HOCH_2-(CHOH)_4-CHO}\), was fixed from chemical tests. Each test shows one feature of the structure.

Step 2: Part (i), carbonyl group:
Glucose reacts with hydroxylamine and gives an oxime.
\[ \mathrm{HOCH_2(CHOH)_4CHO + NH_2OH \rightarrow HOCH_2(CHOH)_4CH{=}NOH + H_2O} \]
It also adds hydrogen cyanide and gives a cyanohydrin. Both are reactions of a carbonyl group.
An oxime and a cyanohydrin can form only if a C=O group is present.
Mild oxidation with bromine water gives gluconic acid, which shows that the carbonyl is an aldehyde group.

Step 3: Part (ii), five -OH groups on different carbons:
Glucose reacts with acetic anhydride and forms a pentaacetate. So there are five -OH groups.
\[ \mathrm{C_6H_{12}O_6 + 5\,(CH_3CO)_2O \rightarrow C_6H_7O(OCOCH_3)_5 + 5\,CH_3COOH} \]
Two -OH groups on the same carbon would form an unstable gem-diol that loses water.
Glucose is stable, so each of the five -OH groups must sit on a different carbon atom.

Final Answer:
(i) The oxime with \(\mathrm{NH_2OH}\) (and the cyanohydrin with HCN) shows a carbonyl group.
(ii) The pentaacetate with acetic anhydride shows five -OH groups, each on a different carbon because gem-diols are unstable.