Question

Which of the following is/are correct about Cannizzaro reaction? (a) Step-1 involves attack of \(\mathrm{OH}^{\odot}\) on carbonyl carbon of aldehyde having no \(\alpha\) - hydrogen (b) Step-1 is the rate determining step (c) Step-2 involves hydride shift (d) In crossed Cannizzaro reaction involving formaldehyde, it always undergoes oxidation.

Short answer

(Select all that apply) a) The first step involves the attack of a hydroxide ion on the carbonyl carbon of an aldehyde having no α-hydrogen. b) The rate-determining step is the nucleophilic attack of the hydroxide ion on the carbonyl carbon. c) A hydride shift occurs between two aldehyde molecules during the reaction. d) In a crossed Cannizzaro reaction involving formaldehyde, formaldehyde always undergoes oxidation. Answer: a, b, c, d

Step-by-step solution

Attack of hydroxide ion on the carbonyl carbon

In the Cannizzaro reaction, the first step involves the attack of a hydroxide ion (OH-) on the carbonyl carbon (C=O) of an aldehyde having no α-hydrogen. The hydroxide ion acts as a nucleophile and forms a tetrahedral intermediate. Statement (a) is correct.

The rate-determining step

Step 1, where the nucleophilic attack of the hydroxide ion on the carbonyl carbon takes place, is the slowest step of the reaction and thus is the rate-determining step. Statement (b) is correct.

Hydride shift

In the Cannizzaro reaction, there is a hydride shift (H-) from one molecule of the aldehyde to another molecule during the reaction. This step involves the transfer of a hydride ion from one molecule to another, leading to the oxidation of one aldehyde and the reduction of another. Statement (c) is correct.

Formaldehyde in crossed Cannizzaro reaction

In a crossed Cannizzaro reaction, two different aldehydes are involved, one with no α-hydrogen (like formaldehyde). When formaldehyde is involved in the crossed Cannizzaro reaction, it always undergoes oxidation due to its high electrophilic nature, being the most reactive aldehyde in this reaction. Statement (d) is correct. Based on the analysis of each statement, all the given options (a), (b), (c), and (d) are correct about the Cannizzaro reaction.

Key concepts

Nucleophilic Attack

In the chemistry of organic reactions, a nucleophilic attack is a fundamental step that plays a crucial role. It involves a nucleophile, which is a chemical species rich in electrons, targeting a positively charged or partially positive atom to form a chemical bond. During the Cannizzaro reaction, this nucleophile is the hydroxide ion (OH-). It approaches the carbonyl carbon of an aldehyde that lacks alpha-hydrogen atoms, due to the partial positive charge of this carbon atom in the carbonyl group. The hydroxide ion effectively donates a pair of electrons to form a new bond with the carbon, generating a tetrahedral intermediate. This step is not only crucial for the Cannizzaro reaction but is a cornerstone of many organic reactions involving carbonyl compounds.

Understanding nucleophilic attack is essential as it often governs the reactivity and outcome of a synthesis reaction. In the context of the Cannizzaro reaction, recognizing the electrophilic nature of the carbonyl carbon provides insight into the mechanism's initiation.

Rate-Determining Step

The rate-determining step (RDS) is the slowest step in a reaction mechanism and thus determines the overall rate at which a reaction progresses. In the Cannizzaro reaction, the rate-determining step is the initial nucleophilic attack where hydroxide ions react with the carbonyl carbon of the aldehyde. Due to the formation of a high-energy intermediate and the need to overcome energy barriers associated with this process, this step requires more time than subsequent steps in the reaction.

The significance of identifying the RDS lies in its utility for optimizing reaction conditions, understanding kinetics, and designing chemical synthesis pathways. For students and chemists, knowing the RDS is critical for controlling the rate and efficiency of a chemical reaction.

Hydride Shift

A hydride shift is another fundamental concept in organic chemistry, and it involves the movement of a hydride ion, which is a hydrogen atom bearing an extra electron, resulting in a negative charge. During the Cannizzaro reaction, after the initial nucleophilic attack, a crucial step is the transfer of the hydride ion from the alkoxide ion of one aldehyde molecule to the carbonyl carbon of another. This hydride transfer is what leads to one molecule being reduced to an alcohol while the other is oxidized to a carboxylate ion.

The hydride shift is integral to understanding how electron density moves within a molecule, and it explains how new bonds are created while others are broken. Additionally, it can be a stereospecific process, influencing the stereochemistry of the products formed. This understanding is pivotal for predicting reaction outcomes and designing molecules with specific orientations.

Crossed Cannizzaro Reaction

In a crossed Cannizzaro reaction, two different aldehydes are reacted together, specifically one with and one without an alpha-hydrogen. This type of reaction can lead to complex product mixtures unless one of the reactants is significantly more reactive. Formaldehyde is an ideal candidate for oxidation in such reactions due to its high reactivity. When mixed with another less reactive aldehyde, formaldehyde will preferentially be oxidized due to its lack of electron-donating alpha-hydrogens, which makes it more susceptible to the nucleophilic attack by the hydroxide ion.

Understanding the dynamics of crossed Cannizzaro reactions allows chemists to apply this method selectively, avoiding undesired side reactions and improving yields of the desired products. Knowing how to manage the reactivity of different aldehydes is essential for successful synthetic applications of the crossed Cannizzaro reaction.

Formaldehyde Oxidation

The formaldehyde oxidation in the context of the Cannizzaro reaction refers to the transformation of formaldehyde, which undergoes oxidation to form formate ion. Since formaldehyde does not contain any alpha-hydrogen atoms, it cannot undergo enolization, making it highly susceptible to nucleophilic attack. In the presence of a base such as hydroxide, formaldehyde quickly reacts to form the formate ion, completing the oxidation half of the Cannizzaro reaction.

Grasping the concept of formaldehyde oxidation is essential due to its implications in various chemical syntheses and industrial applications. Formaldehyde is a fundamental building block in organic chemistry, and its behavior during oxidation reactions is critical for students to understand the transformation of simple carbonyl compounds into more complex molecules.