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m-Chlorobenzaldehyde on reaction with conc. KOH at room temperature gives (a) potassium m-chlorobenzoate and m-hydroxybenzaldehyde (b) m-hydroxybenzaldehyde and m-chlorobenzyl alcohol (c) m-chlorobenzyl alcohol and m-hydroxybenzyl alcohol (d) potassium m-chlorobenzoate and m-chlorobenzyl alcohol

Short Answer

Expert verified
The correct option is (d) potassium m-chlorobenzoate and m-chlorobenzyl alcohol.

Step by step solution

01

Understand the Reaction

The reaction mentioned is an example of the Cannizzaro reaction, which involves base-induced disproportionation of an aromatic aldehyde to form a carboxylic acid salt and an alcohol.
02

Identify the Reactants

The reactant in this scenario is m-chlorobenzaldehyde, which is an aromatic aldehyde substituted with a chlorine atom at the meta position.
03

Determine the Reaction Products

In Cannizzaro reaction, one molecule of the aldehyde is oxidized to a carboxylic acid (which in the presence of KOH forms a carboxylate salt), and another molecule is reduced to an alcohol.
04

Apply to m-Chlorobenzaldehyde

When m-chlorobenzaldehyde undergoes the Cannizzaro reaction: - One molecule of m-chlorobenzaldehyde is reduced to m-chlorobenzyl alcohol. - Another molecule is oxidized to m-chlorobenzoic acid, which forms potassium m-chlorobenzoate with KOH.

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Key Concepts

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

m-Chlorobenzaldehyde
m-Chlorobenzaldehyde is a specialized type of aldehyde that belongs to the category of aromatic aldehydes. It features a benzene ring with an aldehyde group \((\text{CHO})\). The key distinguishing factor of m-chlorobenzaldehyde is the chlorine atom attached to the third carbon position of the benzene ring. This positioning is referred to as the "meta" position. This chlorine substitution significantly affects the chemical behavior and reactivity of the aldehyde, especially in the context of reactions like the Cannizzaro reaction.

In organic chemistry, knowing the structure of your reactants is essential for predicting reaction outcomes. With m-chlorobenzaldehyde, the meta-positioned chlorine impacts both steric and electronic properties. This not only influences the reaction rate but also plays a role in stabilizing the reaction intermediates.
Disproportionation Reaction
The Cannizzaro reaction is a classic example of a disproportionation reaction in organic chemistry. In a disproportionation reaction, a single reactant undergoes simultaneous oxidation and reduction to form two different products. For the Cannizzaro reaction, it's notable because it doesn't require any external oxidizing or reducing agents besides the presence of a strong base like \((\text{KOH})\).

During the Cannizzaro reaction of m-chlorobenzaldehyde, one molecule is reduced to form m-chlorobenzyl alcohol. At the same time, another molecule is oxidized to become m-chlorobenzoic acid, which further reacts with \((\text{KOH})\) to form its salt.
  • This process vividly demonstrates how complex transformations can happen within a single reaction.
  • The absence of external agents simplifies the reaction mechanics considerably.
Understanding disproportionation is key to mastering reactions involving aldehydes.
Aromatic Aldehyde
Aromatic aldehydes, like m-chlorobenzaldehyde, are characterized by the combination of an aromatic ring and an aldehyde functional group. This fusion of structural elements renders aromatic aldehydes significantly stable and reactive compared to their non-aromatic counterparts. The aromatic ring provides a stable backbone due to resonance stabilization, while the aldehyde group offers a reactive site for chemical reactions. This duality is what allows aromatic aldehydes to participate readily in reactions like the Cannizzaro reaction.

Understanding the unique properties of aromatic aldehydes, such as their electronic makeup and stability, is imperative for grasping how and why they react as they do in various conditions. It is this remarkable nature that allows them to undergo specific transformations efficiently in the presence of a strong base.
Potassium m-Chlorobenzoate
Potassium m-chlorobenzoate is the salt form of m-chlorobenzoic acid. This compound is one of the primary products of the Cannizzaro reaction involving m-chlorobenzaldehyde in the presence of concentrated \(\text{KOH}\). When m-chlorobenzaldehyde molecules undergo oxidation, the resulting carboxylic acid (\

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Most popular questions from this chapter

Trihloroacetaldehyde was subjected to cannizzaro's reaction by using \(\mathrm{NaOH}\). The mixture of the products contains sodium trichloroacetate and another co mpound. The other compound is: (a) Trichloromethanol (b) \(2,2,2\)-Trichloropropanol (c) Chloroform (d) \(2,2,2\)-Trichloroethanol

When \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{COOH}\) is reduced with \(\mathrm{LiAlH}_{4}\) the compound obtained will be (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{COOH}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH}\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}^{3}-\mathrm{CHO}\)

From which of the following tertiary butyl alcohol is obtained by the action of methyl magnesium iodide? (a) \(\mathrm{CO}_{2}\) (b) \(\mathrm{HCHO}\) (c) \(\mathrm{CH}_{3} \mathrm{CHO}\) (d) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\)

Which of the following reactions, yield a product with a three membered ring? (a) \(\mathrm{CH}_{3}-\mathrm{C}(\mathrm{O})-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl} \stackrel{\mathrm{KOH}, \mathrm{H}_{2} \mathrm{O}}{\longrightarrow}\) (b) \(\mathrm{PhCHO}+\mathrm{Br}-\mathrm{CH}_{2}-\mathrm{C}(\mathrm{O})-\mathrm{OEt}\) \(\mathrm{t}-\mathrm{BuO}^{-/} \mathrm{t}-\mathrm{BuOH}\) (c) \(\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3} \stackrel{\mathrm{mCPBA}}{\longrightarrow}\) (d) 3- bromobutan \(-2\) - ol \({ }^{-\mathrm{OH} / \mathrm{H}_{2} \mathrm{O}}\)

Here the strongest acid is (a) \(\mathrm{CH}_{3}-\mathrm{COOH}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{COOH}\) (c) \(\mathrm{Cl}-\mathrm{CH}_{2}-\mathrm{COOH}\) (d) \(\mathrm{Br}-\mathrm{CH}_{2}-\mathrm{COOH}\)

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