Problem 101
Question
The reaction of benzaldehdye with alkali gives (a) benzene + benzyl alcohol (b) phenol + benzene (c) benzene + benzyl alcohol (d) benzyl alcohol + sodium benzoate
Step-by-Step Solution
Verified Answer
The reaction forms (d) benzyl alcohol + sodium benzoate.
1Step 1: Understand Cannizzaro Reaction
The reaction of benzaldehyde with an alkali is a classic example of the Cannizzaro reaction. In this reaction, a molecule of an aldehyde without an alpha hydrogen undergoes disproportionation in the presence of a strong base.
2Step 2: Identify Products of Cannizzaro Reaction
In the Cannizzaro reaction, two molecules of benzaldehyde react in the presence of a strong base such as NaOH. One molecule is reduced to benzyl alcohol, while the other is oxidized to benzoate ion.
3Step 3: Neutralize Benzoate Ion
The benzoate ion formed during the reaction is combined with the sodium ion present in NaOH, resulting in sodium benzoate.
4Step 4: Summarize Overall Reaction
The overall equation for the Cannizzaro reaction of benzaldehyde with NaOH is as follows: 2 C₆H₅CHO + NaOH → C₆H₅CH₂OH + C₆H₅COONa. This means benzyl alcohol and sodium benzoate are the final products.
Key Concepts
Benzaldehyde ReactionAldehyde DisproportionationSodium Benzoate Formation
Benzaldehyde Reaction
The reaction of benzaldehyde with a strong base is a prime example of the Cannizzaro reaction. Benzaldehyde, a member of the aldehyde family, is unique because it lacks an alpha hydrogen. This feature makes it a perfect candidate for the Cannizzaro reaction instead of undergoing typical aldehyde reactions. In this context, when benzaldehyde is treated with an alkali, such as sodium hydroxide (NaOH), it participates in a process called disproportionation.
This reaction involves two molecules of benzaldehyde interacting in order to produce two different products. Specifically:
This reaction involves two molecules of benzaldehyde interacting in order to produce two different products. Specifically:
- One molecule of benzaldehyde is reduced to form benzyl alcohol.
- Simultaneously, another molecule of benzaldehyde is oxidized to form a benzoate ion.
Aldehyde Disproportionation
Disproportionation refers to a chemical reaction where a single reactant undergoes both oxidation and reduction to form two different products. In the context of the Cannizzaro reaction with benzaldehyde, this process is pivotal. When placed in a strong alkaline solution, like NaOH, benzaldehyde molecules react with each other without the need for any additional reactants other than the base.
This disproportionation is characterized by:
This disproportionation is characterized by:
- The transfer of an electron from one benzaldehyde molecule, reducing it to benzyl alcohol.
- The concurrent oxidation of another benzaldehyde molecule to create a benzoate ion.
Sodium Benzoate Formation
Once disproportionation occurs, the oxidative half of the reaction produces a benzoate ion as an intermediate. This ion doesn’t remain alone for long. It quickly pairs up with the sodium ions present in the sodium hydroxide solution to create sodium benzoate.
Sodium benzoate is a familiar compound used widely as a preservative in the food industry due to its antimicrobial properties. Here in the Cannizzaro reaction, the formation of sodium benzoate is integral. It confirms that the oxidation part of the reaction has indeed taken place. Therefore, observing sodium benzoate among the products tells us:
Sodium benzoate is a familiar compound used widely as a preservative in the food industry due to its antimicrobial properties. Here in the Cannizzaro reaction, the formation of sodium benzoate is integral. It confirms that the oxidation part of the reaction has indeed taken place. Therefore, observing sodium benzoate among the products tells us:
- Disproportionation has been successful.
- The necessary chemical transformations have occurred.
- Both benzyl alcohol and sodium benzoate are the final products.
Other exercises in this chapter
Problem 98
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Which of the following products is formed when benzaldehyde is treated with \(\mathrm{CH}_{3} \mathrm{MgBr}\) and the addition product so obtained is subjected
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