CH3COOH (acetic acid) cannot reduce Fehling solution while HCOOH, HCHO and CH3CHO reduce Fehling solution due to the presence of CHO group

Fehling solution is used to test for the presence of reducing sugars, such as glucose, fructose, and lactose. It is composed of two separate solutions: Fehling solution A, which contains copper(II) sulfate, and Fehling solution B, which contains sodium hydroxide and potassium sodium tartrate.

Among the given options, acetic acid (option B) cannot reduce Fehling solution. Here's why:

Explanation:
1. Fehling solution consists of Cu2+ ions in solution, which are reduced to Cu+ ions when a reducing sugar is present.
2. The reducing sugar acts as a reducing agent and donates electrons to the Cu2+ ions, causing them to be reduced.
3. Fehling solution A contains copper(II) sulfate, which provides the Cu2+ ions.
4. Fehling solution B contains sodium hydroxide and potassium sodium tartrate, which serve as stabilizing agents.
5. When a reducing sugar is present, it reacts with Fehling solution to produce a reddish-brown precipitate of Cu2O (copper(I) oxide).
6. Acetic acid (option B) is not a reducing sugar. It does not have an aldehyde or ketone functional group, which is necessary for a compound to act as a reducing sugar.
7. Acetic acid is a weak acid and does not have the ability to reduce the Cu2+ ions in Fehling solution.
8. Therefore, when acetic acid is added to Fehling solution, no reduction reaction occurs, and no precipitate is formed.

In summary, acetic acid cannot reduce Fehling solution because it is not a reducing sugar. Reducing sugars, such as glucose, fructose, and lactose, have the ability to donate electrons and reduce the Cu2+ ions in Fehling solution, resulting in the formation of a reddish-brown precipitate of Cu2O.