Statement I: Glucose gives a reddish-brown precipitate with Fehling’s solution.
Statement II: In the reaction of glucose with Fehling’s solution, CuO and gluconic acid are formed.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Statement I is correct but Statement II is false
Fehling’s reagent contains \( \ce{Cu^2+} \) ions in alkaline solution. An aldehyde reduces them to copper(I) oxide, \( \ce{Cu2O} \), and is itself oxidised to the carboxylate anion.
Statement I: Glucose gives a reddish-brown precipitate with Fehling’s solution. — True. Glucose is an aldohexose: its aldehyde group reduces Fehling’s solution, so glucose is a reducing sugar and gives the reddish-brown precipitate.
Statement II: In the reaction of glucose with Fehling’s solution, CuO and gluconic acid are formed. — False. Glucose is oxidised to gluconic acid (present as its anion in the alkaline reagent), but the copper product is \( \ce{Cu2O} \), copper(I) oxide, the reddish-brown precipitate. Copper is reduced from +2 to +1, so CuO (copper in the +2 state) cannot be the product.
NCERT (2026-27 edition):
10.1.1 Classification of Carbohydrates – “… All those carbohydrates which reduce Fehling’s solution and Tollens’ reagent are referred to as reducing sugars. All monosaccharides whether aldose or ketose are reducing sugars.”
10.1.2.1 Glucose – “Glucose is an aldohexose and is also known as dextrose.”
(ii) Fehling’s test – “… On heating an aldehyde with Fehling’s reagent, a reddish brown precipitate is obtained. Aldehydes are oxidised to corresponding carboxylate anion.”
\( \displaystyle \ce{R-CHO + 2Cu^2+ + 5OH^- -> RCOO^-} + \underset{\text{Red-brown ppt}}{\ce{Cu2O}} + \ce{3H2O} \)
10.1.2.1 Glucose – “4. Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on reaction with a mild oxidising agent like bromine water.”