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Chapter 11: Problem 37

Generally aldehydes are more susceptible to oxidation in air than are ketones. Use acetaldehyde and acetone as examples and show why ketones such as acetone are more stable than aldehydes in this respect.

Short Answer

Expert verified
Acetaldehyde, an aldehyde, is more susceptible to oxidation because it has a hydrogen atom on its carbonyl group that leaves the carbon electron-deficient and more liable to oxidation. In contrast, the carbonyl carbon in acetone, a ketone, is attached to two alkyl groups, which donate electron density, making it more stable and resistant to oxidation.

Step by step solution

01

Introduction to Aldehydes and Ketones

Both aldehydes and ketones are carbonyl compounds, meaning they contain a carbon-oxygen double bond (C=O). The key difference lies in what additional groups are attached to the carbonyl carbon. Aldehydes, such as acetaldehyde, have a hydrogen atom and an alkyl (R) group attached, while ketones, such as acetone, have two alkyl groups attached instead.
02

Understanding Oxidation

Oxidation is a reaction that involves the loss of electrons. In organic chemistry, oxidation usually involves either the addition of oxygen atoms (or removal of hydrogen atoms) to a molecule. Oxidation reactions are commonly observed with aldehydes.
03

Oxidation of Acetaldehyde

When an aldehyde like acetaldehyde is exposed to an oxidizing agent, it is readily oxidized to form a carboxylic acid. The hydrogen atom on the carbonyl group makes the carbon atom more electron-poor, rendering it more susceptible to oxidation.
04

Resilience of Acetone to Oxidation

Ketones such as acetone are more resistant to oxidation due to the absence of the hydrogen atom on the carbonyl carbon. Instead, two alkyl groups are present, which are able to donate electron density to the carbonyl carbon, thereby stabilizing it and making it less prone to oxidation.

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Most popular questions from this chapter

How many carbon-carbon sigma bonds are present in each of these molecules? (a) benzene, (b) cyclobutane, (c) 3-ethyl-2-methylpentane.

A compound has the empirical formula \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\). Upon controlled oxidation, it is converted into a compound of empirical formula \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O},\) which behaves as a ketone. Draw possible structures for the original compound and the final compound.

What are structural isomers?

2-Butanone can be reduced to 2 -butanol by reagents such as lithium aluminum hydride \(\left(\mathrm{LiAlH}_{4}\right)\). (a) Write the formula of the product. Is it chiral? (b) In reality, the product does not exhibit optical activity. Explain.

State which member of each of these pairs of compounds is the more reactive and explain why: (a) propane and cyclopropane, (b) ethylene and methane, (c) acetaldehyde and acetone.
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