Appeared in: 2026 (5 sets)
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 $ ext{D}(+) ext{--glucose}$. It was assigned the open structure on the basis of its reactions with $ ext{HI}$, $ ext{NH}_2 ext{OH}$, $ ext{Br}_2 ext{ water}$, $( ext{CH}_3 ext{CO})_2 ext{O}$ and nitric acid.
Despite having the $- ext{CHO}$ group, glucose does not give Schiff's test and hydrogen sulphite addition product with $ ext{NaHSO}_3$. It was found that glucose forms a six-membered ring in which one of the $- ext{OH}$ groups add to the $- ext{CHO}$ group and form a cyclic hemiacetal structure.
Answer the following questions :
(a) How do you explain the following ?
(i) Presence of a carbonyl group in glucose
(ii) Presence of five $- ext{OH}$ groups in glucose which are attached to different carbon atoms.
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Official Evaluation Steps
(a) (i) Presence of a carbonyl group in glucose:
- Step 1 (Governing Reaction): The presence of a carbonyl group ($>C=O$) is confirmed by the reaction of glucose with hydroxylamine ($NH_2OH$) to form an oxime and with hydrogen cyanide ($HCN$) to form a cyanohydrin.
- Step 2 (Step-by-Step Working): $$CHO(CHOH)_4CH_2OH + NH_2OH \rightarrow CH=NOH(CHOH)_4CH_2OH + H_2O$$ $$CHO(CHOH)_4CH_2OH + HCN \rightarrow CH(OH)(CN)(CHOH)_4CH_2OH$$
- Step 3 (Final Answer): The formation of these addition products confirms the presence of a carbonyl group in the open-chain structure of glucose. (1 Mark)
- Step 1 (Governing Reaction): The presence of five hydroxyl ($-OH$) groups is confirmed by the acetylation of glucose with acetic anhydride $(CH_3CO)_2O$.
- Step 2 (Step-by-Step Working): Glucose reacts with acetic anhydride to form glucose pentaacetate. The reaction indicates that five acetyl groups are introduced into the molecule, replacing five hydrogen atoms of the hydroxyl groups. $$CHO(CHOH)_4CH_2OH + 5(CH_3CO)_2O \rightarrow CHO(CHOCOCH_3)_4CH_2OCOCH_3 + 5CH_3COOH$$
- Step 3 (Final Answer): Since glucose pentaacetate is a stable compound, it confirms that the five $-OH$ groups are attached to five different carbon atoms, as a single carbon atom holding more than one $-OH$ group would be unstable. (1 Mark)