JEE Challenger
More from Biomolecules
ChemEasyAdv 2007 • Paper 2

Fehling's Reaction with Glucose and Product Composition

This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

STATEMENT-1 :\textbf{STATEMENT-1 :} Glucose gives a reddish-brown precipitate with Fehling's solution.

because\textbf{because}

STATEMENT-2 :\textbf{STATEMENT-2 :} Reaction of glucose with Fehling's solution gives CuO\text{CuO} and gluconic acid.

Options

A

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

B

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

C

Statement-1 is True, Statement-2 is False

Correct
D

Statement-1 is False, Statement-2 is True

Topics & Concepts

BiomoleculesCarbohydratesGlucoseFehlings TestReducing Sugars

Step-by-Step Solution

  1. Statement-1 is True: Glucose is an aldohexose containing a free aldehyde group in open-chain form. It acts as a reducing sugar and reduces Fehling's solution (alkaline cupric tartrate complex) to produce a characteristic reddish-brown precipitate.

  2. Statement-2 is False: In the reaction, Cu2+\text{Cu}^{2+} is reduced to Cu+\text{Cu}^+, forming a precipitate of cuprous oxide (Cu2O\text{Cu}_2\text{O}), not cupric oxide (CuO\text{CuO}):
    R-CHO+2Cu2++5OH−→R-COO−+Cu2O↓  (red-brown)+3H2O\text{R-CHO} + 2\text{Cu}^{2+} + 5\text{OH}^- \rightarrow \text{R-COO}^- + \text{Cu}_2\text{O}\downarrow\; (\text{red-brown}) + 3\text{H}_2\text{O}
    CuO\text{CuO} is black, whereas the reddish-brown precipitate is Cu2O\text{Cu}_2\text{O}. Hence, Statement-2 is False.

Correct Answer: C

PreviousNext