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Chapter 28: Problem 24

Which compound does not give cannizzaro's reaction? (a) trichloro acetaldehyde (b) benzaldehyde (c) formaldehyde (d) acetaldehyde

Short Answer

Expert verified
Acetaldehyde does not give a Cannizzaro reaction.

Step by step solution

01

Identify Cannizzaro Reaction

The Cannizzaro reaction is a redox reaction wherein two molecules of an aldehyde that lacks an alpha-hydrogen undergo a disproportionation reaction in the presence of a strong base, forming an alcohol and a carboxylic acid salt. This reaction is indicative for aldehydes without alpha-hydrogens.
02

Analyze Each Compound for Alpha-Hydrogen

Assess the structure of each compound: (a) Trichloro acetaldehyde (CCl鈧僀HO) doesn't have an alpha-hydrogen due to the presence of chlorine atoms. (b) Benzaldehyde (C鈧咹鈧匔HO) does not have an alpha-hydrogen since it is attached directly to the aromatic ring. (c) Formaldehyde (HCHO) lacks any alpha-hydrogen as it only has hydrogen atoms. (d) Acetaldehyde (CH鈧僀HO) has three hydrogens attached to the alpha carbon, thus has an alpha-hydrogen.
03

Determine Cannizzaro Suitability

All compounds except for acetaldehyde lack alpha-hydrogens, making them candidates for the Cannizzaro reaction under alkaline conditions. Acetaldehyde, having an alpha-hydrogen, does not undergo this reaction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Aldehydes
Aldehydes are a class of organic compounds characterized by the presence of a carbonyl group (C=O) with the carbon atom bonded to at least one hydrogen atom. The general formula for an aldehyde is RCHO, where R is a hydrocarbon group or hydrogen. Aldehydes play a crucial role in a variety of chemical reactions, including the Cannizzaro reaction.
  • They are typically more reactive than ketones due to the presence of the hydrogen atom bonded to the carbonyl carbon.
  • The carbonyl group in aldehydes makes these compounds polar, thus more reactive with nucleophiles.
Understanding the properties and reactivity of aldehydes is important, especially when considering specific reactions like the Cannizzaro reaction. Aldehydes without alpha-hydrogens are particularly suitable for this redox reaction.
Alpha-Hydrogen
The term 'alpha-hydrogen' refers to a hydrogen atom attached to the carbon adjacent to the carbonyl group in an organic compound. The presence or absence of alpha-hydrogens plays a significant role in the reactivity of aldehydes.
  • Aldehydes with no alpha-hydrogens, like formaldehyde and benzaldehyde, are prime candidates for the Cannizzaro reaction.
  • However, if an aldehyde contains alpha-hydrogens, such as acetaldehyde, it may undergo different reactions, such as aldol condensation.
The Cannizzaro reaction specifically requires aldehydes without alpha-hydrogens because this structure prevents the compound from participating in acid-base reactions that involve the deprotonation of an alpha-hydrogen.
Redox Reaction
In a redox reaction, there is a simultaneous reduction and oxidation process occurring. These reactions are fundamental to many biochemical and industrial processes. During a redox reaction, one substance is oxidized (loses electrons), while another is reduced (gains electrons).
  • The Cannizzaro reaction is a classic example of a redox reaction; one aldehyde molecule is reduced to an alcohol, while another is oxidized to a carboxylic acid salt.
  • This type of reaction is important in organic chemistry because it allows for the transformation of simple aldehyde molecules into more complex products.
Recognizing when a reaction is a redox process helps understand the overall changes in oxidation states and the movement of electrons.
Disproportionation Reaction
A disproportionation reaction involves a single species undergoing both oxidation and reduction simultaneously, resulting in two different products. In the context of the Cannizzaro reaction, this process is evident.
  • Two molecules of an aldehyde without alpha-hydrogens react in the presence of a strong base.
  • During this reaction, one molecule is reduced to form an alcohol, and the other is oxidized to form a carboxylic acid salt.
This type of reaction is unique because the same reactant is both the oxidizing and reducing agent. Understanding disproportionation reactions is crucial for predicting product formation in complex chemical transformations, especially those involving reactive intermediates.

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Most popular questions from this chapter

Which of the following undergoes aldol condensation? 1\. Acetaldehyde 2\. Propionaldehyde 3\. Benzaldehyde 4\. Trideuteroacetaldehyde. (a) 1,2 (b) \(1,2,3\) (c) \(1,2,4\) (d) 2,4

During reduction of carbonyl compounds by hydrazine and KOH, the first intermediate formed is (a) \(\mathrm{RCH}=\mathrm{NH}\) (b) \(\mathrm{RCONH}_{2}\) (c) \(\mathrm{RCH}=\mathrm{NNH}_{2}\) (d) \(\mathrm{RC} \equiv \mathrm{N}\)

Match the following: List I (Reactants) 1\. \(\mathrm{HC} \equiv \mathrm{CH}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{Hg}^{2 *}}{\longrightarrow}\) 2\. \(\mathrm{PhCH}_{3} \frac{1 . \mathrm{Cl}_{2}, \Delta}{2 . \mathrm{H}_{2} \mathrm{O} / \mathrm{OH}^{-}}\) 3\. \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{CO}+\mathrm{H}_{2} \stackrel{\left[\mathrm{Co}(\mathrm{CO})_{4}\right]_{2}}{\longrightarrow}\) 4\. \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{O}_{2} \frac{\mathrm{PdCl}_{2}, \mathrm{CuCl}_{2}}{\mathrm{H}_{2} \mathrm{O}}\) List II (Products) (i) \(\mathrm{CH}_{3} \mathrm{CHO}\) (ii) \(\mathrm{PhCHO}\) (iii) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) The correct matching is: \(\begin{array}{llll}1 & 2 & 3 & 4\end{array}\) (a) (iii) (ii) (i) (i) (b) (i) (i) (ii) (iii) (c) (ii) (iii) (i) (i) (d) (i) (ii) (iii) (i)

On strong heating, ammonium acetate gives (a) urea (b) formamide (c) acetamide (d) methyl cyanide

Which of the following will give yellow precipitate with \(\mathrm{I}_{2} / \mathrm{NaOH} ?\) 1\. \(\mathrm{ICH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\) 2. \(\mathrm{CH}_{3} \mathrm{COOCOCH}_{3}\) 3\. \(\mathrm{CH}_{3} \mathrm{CONH}_{2}\) 4\. \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CH}_{3}\) (a) 1,3 (b) 3,4 (c) 1,2 (d) 1,4
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