Problem 46 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following reagents converts \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCHO}\) to \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH} ?\) (a) aq. \(\mathrm{NaOH}\) (b) Acidic \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) (c) \(\mathrm{Na}_{2} \mathrm{CrO}_{4} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\mathrm{NaNO}_{2} / \mathrm{HCl}\)

Step-by-Step Solution

Answer

The correct reagent is (a) aq. \(\mathrm{NaOH}\).

Step 1: Identify the Reaction Type

The conversion of \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{COCHO}\) (benzoylformaldehyde) to \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\) is an example of a Cannizzaro reaction, which involves the disproportionation of non-enolizable aldehydes in the presence of a strong base.

Step 2: Determine the Correct Reagent

The Cannizzaro reaction requires a strong base, typically concentrated \(\mathrm{NaOH}\). Among the given options, aqueous \(\mathrm{NaOH}\) fits this requirement.

Step 3: Verify with Reaction Mechanism

In the Cannizzaro reaction, one molecule of the aldehyde is oxidized to an acid, and another is reduced to an alcohol. Here, the aldehyde group \(\mathrm{CHO}\) in \(\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{COCHO}\) converts into the alcohol group \(\mathrm{CH}(\mathrm{OH})\) and the carboxylic acid group \(\mathrm{COOH}\).

Key Concepts

Disproportionation

Disproportionation is an intriguing chemical process that plays a key role in the Cannizzaro reaction. In this context, disproportionation involves a single chemical species being simultaneously oxidized and reduced. This means that one part of the molecule gains electrons (reduction) while another part loses electrons (oxidation).
For example, in the Cannizzaro reaction, an aldehyde is both reduced and oxidized to yield an alcohol and a carboxylic acid.
The dual nature of these transformations makes disproportionation critical in converting certain non-enolizable aldehydes without needing external oxidizing or reducing agents. This self-contained transformation highlights the elegant simplicity and efficiency of natural chemical processes.

Strong Base

In the Cannizzaro reaction, a strong base is essential for facilitating the disproportionation process. The strong base, often concentrated sodium hydroxide ( NaOH ), provides the necessary environment for the aldehyde to undergo simultaneous oxidation and reduction. Here’s how it works:

• The strong base deprotonates the aldehyde's hydrogen atom, generating an alkoxide intermediate.
• This negatively charged intermediate makes it possible for one aldehyde molecule to attack another, leading to a transfer of a hydride ion.

Using a strong base is crucial because it both triggers the initial reaction and stabilizes intermediates, allowing the transformation to proceed smoothly.

Non-Enolizable Aldehyde

The term "non-enolizable" refers to aldehydes that cannot easily form enols, a type of compound featuring a hydroxyl group adjacent to a carbon-carbon double bond. In the Cannizzaro reaction, non-enolizable aldehydes are crucial participants.
Unlike their enolizable counterparts, non-enolizable aldehydes lack alpha-hydrogen atoms. This structural characteristic prevents the aldehyde from undergoing typical aldol condensations, instead channeling the reaction to the Cannizzaro pathway.

• Because they cannot enolize, these aldehydes are more stable under basic conditions.
• The absence of enol formation means the reaction is directed towards disproportionation resulting in alcohol and carboxylic acid production.

Thus, non-enolizable aldehydes provide a straightforward pathway to follow the Cannizzaro mechanism.

Reaction Mechanism

Understanding the detailed reaction mechanism of the Cannizzaro process is key to grasping its ability to transform non-enolizable aldehydes like benzoylformaldehyde. Here is how it typically happens:

• The strong base deprotonates the aldehyde, creating a negatively charged alkoxide ion.
• This alkoxide ion acts as a nucleophile and attacks another aldehyde molecule.
• The addition of the hydride ion facilitates the formation of alcohol on one molecule.
• The other molecule is simultaneously oxidized to form a carboxylic acid.

Such a mechanism elegantly illustrates both the oxidizing and reducing processes occurring together. This bifurcated pathway ensures that the desired products, in this case, an alcohol and a carboxylic acid, are obtained efficiently from a single type of aldehyde without requiring separate reagents.