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, \((\mathrm{CH_3CO})_2\mathrm{O}\) and nitric acid.

Despite having the \(-\mathrm{CHO}\) 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 \(-\mathrm{OH}\) groups add to the \(-\mathrm{CHO}\) group and form a cyclic hemiacetal structure.

How do you explain the following ?

(i) Presence of a carbonyl group in glucose

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

Hint:

Use the reactions in the passage: oxime and bromine water show the -CHO group, and acetylation shows five -OH groups on different carbons.

Answer 1

Step 1: Recall the open-chain structure.
Glucose is an aldohexose with six carbon atoms. Its open-chain formula is \(\mathrm{CHO-(CHOH)_4-CH_2OH}\).
The passage names the reagents that were used to prove this structure. We use these reactions one by one.

Step 2: Evidence for the carbonyl group (part i).
Glucose reacts with hydroxylamine to form an oxime. This is a typical reaction of an aldehyde or a ketone.
\[ \mathrm{C_5H_{11}O_5-CHO + NH_2OH \rightarrow C_5H_{11}O_5-CH=NOH + H_2O} \]
Bromine water is a mild oxidising agent. It oxidises the \(-\mathrm{CHO}\) group to \(-\mathrm{COOH}\) and gives gluconic acid.
\[ \mathrm{CHO-(CHOH)_4-CH_2OH \xrightarrow{Br_2/H_2O} COOH-(CHOH)_4-CH_2OH} \]
Bromine water attacks only an aldehyde, not a ketone. So the carbonyl group is an aldehyde group at the end of the chain.

Step 3: Evidence for five -OH groups (part ii).
Glucose reacts with acetic anhydride and gives a pentaacetate. Five acetyl groups are added, so five \(-\mathrm{OH}\) groups are present.
\[ \mathrm{C_6H_{12}O_6 + 5(CH_3CO)_2O \rightarrow C_6H_7O(OCOCH_3)_5 + 5CH_3COOH} \]
On prolonged heating with HI, glucose gives n-hexane. So all six carbon atoms lie in one straight chain.
A carbon that carries two -OH groups loses water at once and becomes a carbonyl carbon. Glucose is stable, so no two -OH groups sit on one carbon.
Hence the five -OH groups are attached to five different carbon atoms.
Nitric acid oxidises both ends of the chain to give saccharic acid. This shows that the last carbon carries a primary -OH group.

Final Answer:
(i) Oxime formation and oxidation to gluconic acid by bromine water show a -CHO group. (ii) Pentaacetate formation shows five -OH groups, and n-hexane from HI shows they sit on different carbons of an unbranched chain.