Question

Sugar which does not give reddish brown precipitate with Fehling‟s reagent is:

• A. Lactose
• B. Maltose
• C. Sucrose
• D. Glucose

Answer 1

The Correct Option is C

To identify the sugar that does not yield a reddish-brown precipitate with Fehling's reagent, an understanding of Fehling's test chemistry is required.

Fehling's solution, an alkaline copper(II) sulfate solution, detects reducing sugars. A reducing sugar reduces copper(II) ions to copper(I) oxide, which precipitates as a reddish-brown solid. The reaction is:

\(2Cu^{2+} + R-CHO + 4OH^- \rightarrow Cu_2O \ (s) + 2H_2O + R-COOH\)

Consider each option:

1. Lactose - As a reducing sugar with a free anomeric carbon, lactose reduces Fehling's solution, producing a reddish-brown precipitate.
2. Maltose - Maltose, like lactose, possesses a free anomeric carbon, classifying it as a reducing sugar that yields a reddish-brown precipitate with Fehling's reagent.
3. Glucose - Glucose is a known reducing sugar that gives a positive result in Fehling's test.
4. Sucrose - Sucrose is a non-reducing sugar. Its glycosidic bond involves the anomeric carbons of both glucose and fructose, leaving no free anomeric carbon available for reduction. Consequently, sucrose does not reduce Fehling's solution and does not form a reddish-brown precipitate.

Therefore, the sugar that does not produce a reddish-brown precipitate with Fehling's reagent is: Sucrose