What is the colour of copper compound formed in Fehling's test for aliphatic aldehydes?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Red brown
Fehling’s test is a classical chemical test used to detect the presence of reducing sugars (like glucose) or aliphatic aldehydes. It distinguishes between aldehydes and ketones based on their ability to reduce Cu²⁺ ions to Cu⁺ ions.
Composition of Fehling's Solution:
Fehling's solution is prepared by mixing two solutions:
Fehling's A: Aqueous solution of copper(II) sulfate (CuSO₄). This gives the solution a blue color due to the presence of Cu²⁺ ions.
Fehling's B: A solution of potassium sodium tartrate (Rochelle salt) in a strong alkali (NaOH). This helps to maintain the Cu²⁺ ions in solution by complexing them as [Cu(C₄H₄O₆)₂]²⁻.
Reaction with Aliphatic Aldehydes:
When an aliphatic aldehyde (R-CHO) is mixed with Fehling's solution and heated, the aldehyde is oxidized to a carboxylate ion, while the Cu²⁺ ions from Fehling’s solution are reduced to Cu⁺ ions.
Reaction Steps:
Oxidation of the Aldehyde: The aldehyde group (-CHO) is oxidized to a carboxylate anion (-COO⁻):
\(R-CHO \longrightarrow R-COOH\)
Reduction of Copper (Cu²⁺ to Cu⁺):
The Cu²⁺ ions are reduced to Cu⁺ ions, which form Cu₂O (copper(I) oxide), a red-brown precipitate:
\(2Cu^{2+} + R-CHO + 5OH^{-} \longrightarrow Cu_2O(s) + R-COO^- + 3H_2O\)
Appearance of the Red-Brown Precipitate:
The reduced copper species, Cu₂O (copper(I) oxide), is insoluble in water and precipitates out as a brick-red (or red-brown) solid.
The formation of this red-brown precipitate indicates the presence of a reducing sugar or an aliphatic aldehyde.
Color Change:
Initially: Fehling’s solution is blue due to the Cu²⁺ ions.
After Reaction: The blue color disappears as the Cu²⁺ ions are reduced, and the solution forms a red-brown precipitate of Cu₂O.
Why Red-Brown:
Cu₂O is copper(I) oxide, which has a characteristic brick-red (red-brown) color. This is the visual indicator of a positive Fehling’s test for aldehydes.
Fehling's Test Example:
For an aliphatic aldehyde like glucose: Glucose (an aldehyde) reduces Cu²⁺ ions, forming Cu₂O as a red-brown precipitate and producing gluconic acid as a by-product.
Summary:
Fehling’s test is used to detect reducing aldehydes or reducing sugars.
The blue Cu²⁺ ions in Fehling’s solution are reduced to Cu₂O, a red-brown (brick-red) precipitate.
The formation of this red-brown precipitate indicates a positive Fehling’s test for aldehydes or reducing sugars.
Thus, the color of the copper compound formed is red-brown.