Monosaccharides are carbohydrates which cannot be hydrolysed into simpler molecules. Those containing an aldehyde group are called aldoses while those containing a keto group are called ketoses. The aldehyde group is present at C1 while the keto group is usually present at C2. All monosaccharides reduce Tollens’ reagent as well as Fehling’s solution and hence are called reducing sugars. Glucose has a six-membered cyclic structure called the pyranose structure, while fructose in sucrose has a five-membered cyclic structure called the furanose structure. During ring formation, C1 of glucose becomes chiral, and hence glucose exists in two cyclic forms called the α-anomer and the β-anomer. When two molecules of the same or different monosaccharides combine through a glycosidic linkage, a disaccharide is formed.
Glucose is treated separately with (i) bromine water and (ii) HI on prolonged heating. Which of the following are the main organic products of (i) and (ii), respectively?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Gluconic acid and n-hexane
(i) Bromine water is a mild oxidising agent. It oxidises only the –CHO group of glucose to –COOH and gives the six carbon carboxylic acid, gluconic acid. This shows that the carbonyl group of glucose is an aldehydic group.
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow{\ce{Br2}\text{ water}} \ce{COOH-(CHOH)4-CH2OH} \)
\( \displaystyle \ce{C6H12O6 + Br2 + H2O -> C6H12O7 + 2HBr} \)
(ii) HI, on prolonged heating, reduces glucose completely: every oxygen-containing group is removed and the hydrocarbon n-hexane is left. Since the product is a straight-chain hexane, all six carbon atoms of glucose are linked in a straight chain.
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow[\Delta]{\ce{HI}} \ce{CH3-CH2-CH2-CH2-CH2-CH3} \)
\( \displaystyle \ce{C6H12O6 + 14HI -> C6H14 + 6H2O + 7I2} \)
Why the other options are incorrect
Option (1) Saccharic acid and n-hexane — Incorrect. Saccharic acid, a dicarboxylic acid, is formed by the stronger oxidising agent nitric acid, which oxidises the –CH2OH group as well. Bromine water stops at gluconic acid.
Option (2) Gluconic acid and cyclohexane — Incorrect. HI gives an open chain hydrocarbon, n-hexane; this is the evidence that the six carbons of glucose form a straight chain, not a carbocyclic ring.
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.”
“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.”