Problem 40 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In the Cannizzaro reaction given below, $$ 2 \mathrm{PhCHO} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{PhCH}_{2} \mathrm{OH}+\mathrm{PhCO}_{2}^{-}, $$ the slowest step is (a) the attack of \(-\mathrm{OH}\) at the carbonyl group, (b) the transfer of hydride to the carbonyl group, (c) the abstraction of proton from the carboxylic acid, (d) the deprotonation of \(\mathrm{PhCH}_{2} \mathrm{OH}\).

Step-by-Step Solution

Answer

(b) The transfer of hydride to the carbonyl group is the slowest step.

Step 1: Understand the Cannizzaro Reaction

The Cannizzaro reaction involves the disproportionation of an aldehyde without any hydrogen on the alpha carbon when treated with a strong base like \( \text{OH}^- \). This reaction typically yields one molecule of alcohol and one molecule of carboxylate salt from two molecules of aldehyde.

Step 2: Identify Reaction Mechanism Stages

In the Cannizzaro reaction, the mechanism involves several steps: 1) Attack by \( -\text{OH} \) at the carbonyl carbon, 2) Formation of a tetrahedral alkoxide intermediate, 3) Intramolecular hydride transfer producing one alcohol and one carboxylate anion, 4) Deprotonation steps for final stabilization.

Step 3: Rate-determining Step

The rate-determining step in a multi-step process is the slowest step. For the Cannizzaro reaction, the hydride transfer between two aldehyde molecules, which results in the conversion of one aldehyde to alcohol and the other to a carboxylate anion, is known as the slowest step. This is because breaking and forming of bonds directly impact the energy profile, making it slower relative to other steps which involve basic proton transfers.

Step 4: Answer the Question

Considering the mechanism of the Cannizzaro reaction and identifying the slow rate of hydride transfer, the correct answer is option (b), as it represents the identified slowest step, where one carbonyl group gets reduced and the other oxidized.

Key Concepts

Aldehyde Disproportionation

Aldehyde disproportionation is a key feature of the Cannizzaro reaction. It involves the transformation of two aldehyde molecules into two distinct products: an alcohol and a carboxylate ion. This transformation occurs without the need for any external reducing or oxidizing agents, instead using the aldehyde itself to facilitate the changes.

In this reaction, one aldehyde molecule is oxidized to form a carboxylate ion, while the other is reduced to form an alcohol. This dual change is why it is termed a disproportionation reaction. A fascinating aspect of this process is the "self-sufficiency" of the reaction, utilizing only the starting aldehydes and a strong base like hydroxide ions ( ext{OH}^{-}) to progress. This makes the Cannizzaro reaction a unique and simplifying feature in organic reactions where aldehydes without alpha hydrogens are involved.

The reaction demonstrates a beautiful example of internal manipulation of electrons between two identical molecules to achieve distinct and separate functional groups.

Reaction Mechanism

The Cannizzaro reaction follows a distinctive mechanism that unfolds through multiple steps. It starts with the attack of the hydroxide ion ( ext{OH}^{-}) on the carbonyl carbon of the aldehyde molecule. This initial attack is crucial for kicking off the reaction, leading to the formation of a tetrahedral intermediate known as an alkoxide ion.

Within this intermediate, the reaction facilitates an intramolecular transfer of hydride ion (H^{-}). This step is crucial, as it results in the redistribution of electrons, converting one aldehyde molecule into an alcohol and the other into a carboxylate ion.

Finally, deprotonation helps stabilize the resultant products, ensuring a complete and efficient transformation from reactants to products. This multi-step mechanism showcases intricate interactions within molecules, highlighting the elegance of the Cannizzaro reaction's approach to chemically restructuring aldehydes.

Hydride Transfer

Hydride transfer is a critical step in the Cannizzaro reaction, where a hydride ion (H^{-}) shifts from one aldehyde molecule to another. This transfer is propelled by the formation of a partially negative charge on the carbon where hydroxide initially attaches, creating a scenario where electron movement is both possible and favorable.

During this transfer, one aldehyde molecule donates a hydride ion, while the other accepts it. The donor essentially undergoes oxidation, transforming into a carboxylate ion, while the recipient is reduced, forming an alcohol.

• Oxidation: The donor aldehyde loses hydrogen, forming a carboxylate ion.
• Reduction: The recipient gains hydrogen, converting to alcohol.

This exchange epitomizes the term "disproportionation," as the same species is simultaneously oxidized and reduced in the presence of hydroxide, facilitating this elegant transformation.

Rate-Determining Step

In the context of the Cannizzaro reaction, the rate-determining step is the phase that limits the overall speed of the reaction. In this case, the hydride transfer step is identified as the slowest in the entire process. It acts as a bottleneck, dictating the pace at which the entire reaction progresses.

This sluggishness is due to the energy-intensive nature of breaking and forming bonds associated with the hydride ion movement between the aldehyde molecules. This step requires overcoming a significant energy barrier, which makes it notably slower than other parts of the reaction mechanism, such as the nucleophilic attack or deprotonation steps.

Understanding this concept is important, as the rate-determining step governs experimental conditions and influences how chemists might adjust parameters (like temperature and concentration) to optimize the reaction's efficiency.