Problem 41
Question
\(m\)-Chlorobenzaldehyde on reaction with conc. \(\mathrm{KOH}\) 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.
Step-by-Step Solution
Verified Answer
The reaction yields (d) potassium m-chlorobenzoate and m-chlorobenzyl alcohol.
1Step 1: Identify the Reaction
The reaction of chlorobenzaldehyde with KOH is a Cannizzaro reaction, which occurs with aldehydes lacking an alpha-hydrogen. In this case, m-chlorobenzaldehyde will undergo disproportionation.
2Step 2: Understand the Cannizzaro Reaction
In the Cannizzaro reaction, one aldehyde molecule is oxidized to form a carboxylate salt, while another is reduced to form an alcohol. This is specific to aldehydes with no hydrogen on the carbon adjacent to the carbonyl group.
3Step 3: Analyze the Products
When m-chlorobenzaldehyde reacts with concentrated KOH, it forms potassium m-chlorobenzoate (by oxidation) and m-chlorobenzyl alcohol (by reduction).
4Step 4: Select the Product Pair
Based on Step 3, the reaction yields (d) potassium m-chlorobenzoate and m-chlorobenzyl alcohol. This matches the option that represents the products formed in the Cannizzaro reaction involving m-chlorobenzaldehyde.
Key Concepts
AldehydesDisproportionationChemical Reactions with KOHOrganic Chemistry
Aldehydes
Aldehydes are a class of organic compounds that contain a carbonyl group (\[ C=O \] ) with at least one hydrogen atom attached to the carbon. This makes aldehydes reactive and crucial in various chemical reactions.
They are typically represented as \[ RCHO \] , where \( R \) is a hydrocarbon group, and \( CHO \) is the aldehyde group.
Aldehydes vary in reactivity based largely on the presence of an alpha hydrogen, which can influence reaction pathways like the Cannizzaro reaction. In the Cannizzaro reaction, one molecule is oxidized while the other is reduced, showing the unique duality of aldehydes.
This reaction pathway demonstrates the diversity of chemical transformations aldehydes can undergo, making them an important topic in organic chemistry.
They are typically represented as \[ RCHO \] , where \( R \) is a hydrocarbon group, and \( CHO \) is the aldehyde group.
Aldehydes vary in reactivity based largely on the presence of an alpha hydrogen, which can influence reaction pathways like the Cannizzaro reaction. In the Cannizzaro reaction, one molecule is oxidized while the other is reduced, showing the unique duality of aldehydes.
This reaction pathway demonstrates the diversity of chemical transformations aldehydes can undergo, making them an important topic in organic chemistry.
Disproportionation
Disproportionation is a special type of redox reaction where a single species is simultaneously oxidized and reduced, resulting in two different products.
In the Cannizzaro reaction, disproportionation is observed with aldehydes that lack an alpha hydrogen.
This leads to one aldehyde molecule converting into an alcohol while another transforms into a carboxylate salt.
This reaction is particularly useful in organic synthesis for modifying simple aldehydes into valuable derivatives.
Understanding disproportionation helps chemists design reactions that selectively yield specific products from simple starting materials.
In the Cannizzaro reaction, disproportionation is observed with aldehydes that lack an alpha hydrogen.
This leads to one aldehyde molecule converting into an alcohol while another transforms into a carboxylate salt.
This reaction is particularly useful in organic synthesis for modifying simple aldehydes into valuable derivatives.
Understanding disproportionation helps chemists design reactions that selectively yield specific products from simple starting materials.
Chemical Reactions with KOH
Potassium hydroxide (KOH) is a strong base used frequently in chemical reactions. In organic chemistry, KOH can trigger various transformations due to its reactive nature.
With aldehydes, KOH can facilitate the Cannizzaro reaction by providing the basic environment needed for disproportionation to occur.
In this context, KOH acts as a catalyst, where it doesn't get consumed but enables the reaction to progress.
The reaction with KOH highlights the importance of bases in facilitating and controlling reaction pathways in organic chemistry.
With aldehydes, KOH can facilitate the Cannizzaro reaction by providing the basic environment needed for disproportionation to occur.
In this context, KOH acts as a catalyst, where it doesn't get consumed but enables the reaction to progress.
The reaction with KOH highlights the importance of bases in facilitating and controlling reaction pathways in organic chemistry.
Organic Chemistry
Organic chemistry is the study of carbon-containing compounds and their properties, structures, and reactions. It includes a vast array of reactions that transform simple molecules into complex structures.
The Cannizzaro reaction is a prime example of the nuanced transformations within organic chemistry, demonstrating concepts like oxidation, reduction, and disproportionation.
Organic chemistry provides insight into the myriad ways carbon compounds interact, forming the basis for everything from pharmaceuticals to plastics.
Mastery of concepts like the Cannizzaro reaction allows chemists to manipulate molecules intentionally and with precision, driving innovation in materials science, biochemistry, and beyond.
The Cannizzaro reaction is a prime example of the nuanced transformations within organic chemistry, demonstrating concepts like oxidation, reduction, and disproportionation.
Organic chemistry provides insight into the myriad ways carbon compounds interact, forming the basis for everything from pharmaceuticals to plastics.
Mastery of concepts like the Cannizzaro reaction allows chemists to manipulate molecules intentionally and with precision, driving innovation in materials science, biochemistry, and beyond.
Other exercises in this chapter
Problem 40
In the Cannizzaro reaction given below, $$ 2 \mathrm{PhCHO} \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{PhCH}_{2} \mathrm{OH}+\mathrm{PhCO}_{2}^{-}, $$
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