Match Reaction with Reagent Required
| List-I | List-II |
| Reaction | Reagent Required |
| (A) Glucose to n-Hexane | (I) NH2OH |
| (B) Glucose to Oxime | (II) HNO3 |
| (C) Glucose to Gluconic Acid | (III) HI, Δ |
| (D) Glucose to Saccharic Acid | (IV) Br2 , H2O |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
Glucose is an aldohexose: a straight chain of six carbon atoms with an aldehyde group (–CHO) at C1, four secondary –OH groups and one primary –CH2OH group at C6. Each reagent in the list attacks a different part of this chain, and these reactions are the evidence for the open-chain structure:
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \)
(A) Glucose to n-Hexane — This matches (III) HI, Δ. On prolonged heating with HI, all the oxygen atoms are removed and glucose is fully reduced to n-hexane. This shows that the six carbon atoms are joined in a straight chain.
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{HI},\ \Delta} \ce{CH3-CH2-CH2-CH2-CH2-CH3} \)
(B) Glucose to Oxime — This matches (I) NH2OH. Hydroxylamine adds to the carbonyl group and loses water, giving the oxime (–CH=N–OH). This shows that glucose has a C=O group.
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH + NH2OH -> CH(=N-OH)-(CHOH)4-CH2OH + H2O} \)
(C) Glucose to Gluconic Acid — This matches (IV) Br2 , H2O. Bromine water is a mild oxidising agent. It oxidises only the –CHO group to –COOH, giving the six-carbon acid, gluconic acid. This shows that the carbonyl group is an aldehyde group.
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{Br2}\text{ water}} \ce{COOH-(CHOH)4-CH2OH} \)
(D) Glucose to Saccharic Acid — This matches (II) HNO3. Nitric acid is a stronger oxidising agent. It oxidises both ends of the chain, the –CHO group and the primary –CH2OH group, giving the dicarboxylic acid, saccharic acid. This shows that glucose has a primary alcoholic group.
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{HNO3}} \ce{COOH-(CHOH)4-COOH} \)
NCERT (2026-27 edition):
Structure of 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. …
6. On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (–OH) group in glucose.”