Which of the following compounds will not give a red precipitate when heated with Fehling's solution?

Benzaldehyde

The correct answer is option 4. Benzaldehyde.

The given compounds in the options are :

 To understand why benzaldehyde does not give a red precipitate with Fehling's solution, let us break down the reaction mechanism and the nature of Fehling's test.

Fehling's Solution: Fehling’s solution is a chemical reagent used to differentiate between aldehydes and ketones. It contains two parts:

Fehling's A: Aqueous copper(II) sulfate (\(CuSO_4\)), which provides \(Cu^{2+}\) ions.

Fehling's B: A solution of sodium potassium tartrate (Rochelle salt) in a strongly alkaline medium (usually \(NaOH\)), which keeps the copper ions in solution as a complex.

In the presence of an aldehyde group that can be oxidized, Fehling’s solution changes from a blue solution of \(Cu^{2+}\) ions to a brick-red precipitate of copper(I) oxide \((Cu_2O)\).

The \(Cu^{2+}\) ion is reduced to \(Cu^+\), while the aldehyde is oxidized to a carboxylate ion in the basic medium.

Mechanism of Reaction:

Oxidation of Aldehyde: In an aldehyde, the \(CHO\) group is oxidized to a carboxylate ion \((COO^-)\) in the alkaline conditions provided by Fehling's solution. The aldehyde is the reducing agent, and it donates electrons to the \(Cu^{2+}\) ions.

Reduction of Copper:

\(Cu^{2+}\) ions in Fehling's solution are reduced to \(Cu^+\), which precipitates as \(Cu_2O\) (a red precipitate).

Why Some Aldehydes React and Others Don't:

Aliphatic Aldehydes: Aliphatic aldehydes, such as methanal (formaldehyde) and ethanal (acetaldehyde), have simple structures with no conjugation or resonance stabilization. These aldehydes can easily donate electrons and undergo oxidation to form a carboxylate ion. As a result, they reduce \(Cu^{2+}\) ions, leading to the formation of the red precipitate.

Aromatic Aldehydes: Aromatic aldehydes, like benzaldehyde, have the \(CHO\) group attached directly to a benzene ring. The presence of the aromatic ring makes the \(C-H\) bond in the aldehyde group much more stable due to conjugation and resonance with the benzene ring. This stabilizes the aldehyde and makes it much less reactive. The benzene ring's delocalized \(\pi \)-electrons reduce the electron-donating ability of the carbonyl carbon, so it doesn’t oxidize as easily as aliphatic aldehydes.

Thus, benzaldehyde cannot reduce \(Cu^{2+}\) ions in Fehling's solution because:

The aldehyde group in benzaldehyde is not easily oxidized due to the resonance stabilization provided by the aromatic ring. This prevents the formation of \(Cu_2O\) and the characteristic red precipitate.

Behavior of Each Compound:

1. Methanal (Formaldehyde):

It is the simplest aldehyde, and since it lacks the stabilization effects seen in aromatic aldehydes, it is highly reactive. Methanal will readily oxidize, reducing \(Cu^{2+}\) to \(Cu^+\), and will form a red precipitate of \(Cu_2O\).

2. Ethanal (Acetaldehyde):

As an aliphatic aldehyde, ethanal is also highly reactive and will give a red precipitate with Fehling’s solution.

3. Cyclohexane carbaldehyde:

This is an aliphatic aldehyde with the structure of a simple \(CHO\) group attached to a cyclohexane ring. It can still undergo oxidation to the carboxylate form and will reduce \(Cu^{2+}\) to \(Cu^+\), giving a red precipitate.

4. Benzaldehyde:

In contrast to the compounds above, benzaldehyde is an aromatic aldehyde. Its aldehyde group is part of a conjugated system with the benzene ring, making it more stable and less reactive.  It cannot easily reduce Fehling’s solution because the \(C-H\) bond in the aldehyde group is much harder to oxidize.

Therefore, benzaldehyde does not give a red precipitate with Fehling's solution.

Conclusion:

Among the given options, benzaldehyde (an aromatic aldehyde) is the only compound that will not give a red precipitate when heated with Fehling's solution. This is due to the resonance stabilization of the carbonyl group by the benzene ring, making the aldehyde group less reactive towards oxidation.