Match List-I with List-II.
| List-I (Reaction of glucose) | List-II (What it shows about the structure of glucose) |
|---|---|
| (A) Prolonged heating with HI gives n-hexane | (I) Five –OH groups attached to different carbon atoms |
| (B) Reaction with hydroxylamine gives an oxime | (II) The carbonyl group is an aldehydic group |
| (C) Bromine water gives gluconic acid | (III) A carbonyl group is present |
| (D) Acetic anhydride gives a stable pentaacetate | (IV) Six carbon atoms linked in a straight chain |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-IV, B-III, C-II, D-I
The open chain structure of glucose was worked out from its reactions; each reaction tells one thing about the molecule.
(A) Prolonged heating with HI gives n-hexane — This matches (IV) Six carbon atoms linked in a straight chain. HI removes all the oxygen and leaves the carbon skeleton as n-hexane, an unbranched chain:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow[\Delta]{\ce{HI}} \ce{CH3-(CH2)4-CH3} \)
(B) Reaction with hydroxylamine gives an oxime — This matches (III) A carbonyl group is present. Hydroxylamine reacts with a >C=O group; HCN adds in the same way to give a cyanohydrin. These reactions show a carbonyl group, but not whether it is an aldehyde or a ketone:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH ->[NH2OH] CH(=N-OH)-(CHOH)4-CH2OH + H2O} \)
(C) Bromine water gives gluconic acid — This matches (II) The carbonyl group is an aldehydic group. A mild oxidising agent oxidises an aldehyde group to –COOH without breaking the chain, giving a six carbon acid:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{Br2}\text{ water}} \ce{COOH-(CHOH)4-CH2OH} \)
(D) Acetic anhydride gives a stable pentaacetate — This matches (I) Five –OH groups attached to different carbon atoms. Each –OH is acetylated, so five acetyl groups enter; since the pentaacetate is stable, no carbon carries two –OH groups:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \ce{+ 5(CH3CO)2O -> CHO-(CH-OCOCH3)4-CH2-OCOCH3 + 5CH3COOH} \)
NCERT (2026-27 edition):
10.1.2.1 Glucose – “2. On prolonged heating with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain.”
“3. Glucose reacts with hydroxylamine to form an oxime and adds a molecule of hydrogen cyanide to give cyanohydrin. These reactions confirm the presence of a carbonyl group (>C = O) in glucose.”
“4. Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water. This indicates that the carbonyl group is present as an aldehydic group.”
“5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms.”