Problem 177 - A Complete Resource Book in Chemistry for JEE Main

Question

Identify the products formed \((\mathrm{X})\) and \((\mathrm{Y})\) in the following reaction \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{HCHO}_{\text {conc. } \overline{\mathrm{OH}}}(\mathrm{X})+(\mathrm{Y})\) (a) \((\mathrm{X})\) is \(\mathrm{HCOO}^{9}\) (b) (Y) is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\) (c) \((\mathrm{Y})\) is \(\mathrm{CH}_{3} \mathrm{OH}\) (d) \((\mathrm{X})\) is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{\ominus}\)

Step-by-Step Solution

Answer

(X) is \( HCOO^- \) and (Y) is \( C_6H_5CH_2OH \), matching options (a) and (b).

Step 1: Understanding the Reaction Setup

The given chemical reaction involves benzaldehyde \( (C_6H_5CHO) \) reacting with formaldehyde \( (HCHO) \) in the presence of concentrated \( \overline{OH} \). This typically indicates a Cannizzaro reaction, where an aldehyde undergoes disproportionation in the presence of a strong base.

Step 2: Identifying Cannizzaro Reaction Products

In a Cannizzaro reaction, one molecule of the aldehyde is oxidized to form a carboxylate ion, while another molecule is reduced to form an alcohol. In this case, two types of reactants are involved, so one molecule of formaldehyde will likely form formate \( (HCOO^-) \) and another benzaldehyde will form benzyl alcohol \( (C_6H_5CH_2OH) \).

Step 3: Determining (X) and (Y)

Since formaldehyde is converted into formate ion \( (HCOO^-) \), this means \( (X) \) is \( HCOO^- \). The benzaldehyde is reduced to benzyl alcohol \( C_6H_5CH_2OH \), hence \( (Y) \) is \( C_6H_5CH_2OH \).

Step 4: Matching with Given Options

Check the given options against our identified products: - Option (a) states \( (X) \) is \( HCOO^- \), which is correct.- Option (b) states \( (Y) \) is \( C_6H_5CH_2OH \), which is also correct.- Options (c) and (d) are incorrect as per our reaction analysis.

Key Concepts

Disproportionation in Cannizzaro Reaction

Disproportionation is a key concept in chemistry where a single substance is simultaneously oxidized and reduced, leading to the formation of two different compounds. In the Cannizzaro reaction, this process specifically occurs with aldehydes when no alpha hydrogen is present. This results in an interesting chemical behavior where some molecules of the aldehyde are converted into an acid (or its corresponding carboxylate ion), while others are reduced to an alcohol.

The disproportionation process relies heavily on a strong base to facilitate these transformations. It provides the necessary conditions for the aldehyde to embark on dual pathways, producing distinct products. This fundamental understanding of disproportionation helps clarify how the Cannizzaro reaction can produce two different end products from the same starting material, depending on which pathway the molecules take.

The Role of Benzaldehyde

Benzaldehyde, with the chemical formula \( C_6H_5CHO \), plays a significant role in the Cannizzaro reaction. Unlike many aldehydes, benzaldehyde lacks an alpha hydrogen, making it frequently encountered in Cannizzaro reactions.

During the reaction, benzaldehyde undergoes reduction. Instead of adding hydrogen to the entire molecule, specific interactions lead to part of it transforming into benzyl alcohol \( C_6H_5CH_2OH \). This outcome is possible thanks to the action of a strong base that pushes the reduction process forward.

• Starts as an aldehyde.
• Lacks alpha hydrogen.
• Transforms into benzyl alcohol through reduction.

Understanding its pathway in the Cannizzaro reaction gives insights into how certain molecular structures can predict specific chemical behaviors.

Formaldehyde and its Reactivity

Formaldehyde \( HCHO \) is another crucial player in the Cannizzaro reaction, notable for its high reactivity. Unlike benzaldehyde, formaldehyde is often chosen for its ability to undergo oxidation quickly.

In this specific reaction setting, formaldehyde converts into formate ion \( HCOO^- \). This represents the oxidized product in the Cannizzaro reaction sequence.

• Highly reactive.
• Oxidized to form formate ion.
• Complements the reduction reaction of benzaldehyde.

Formaldehyde's behavior in the Cannizzaro reaction highlights its versatility in participating in various chemical transformations.

Importance of a Strong Base

The presence of a strong base, like concentrated \( OH^- \), is essential in a Cannizzaro reaction. It creates a highly alkaline environment that encourages the disproportionation of aldehydes.

Here's why the strong base is important:

• Facilitates the breaking and rearranging of chemical bonds.
• Acts as a catalyst that drives the transformation of aldehydes into both alcohols and acids.
• Ensures that the reaction occurs smoothly and efficiently by stabilizing intermediate forms and ions.

Thus, the role of the strong base is more than just a mere presence; it is the driving force that enables the unique pathways seen in the Cannizzaro reaction. Without it, the dual transformation of oxidation and reduction would not be possible.