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Chapter 11: Problem 39

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.

Short Answer

Expert verified
The original compound might be a pentanol and the final compound might be a pentanone.

Step by step solution

01

Determine the Molecular Structures

Start with the empirical formula of the original compound, \(\mathrm{C}_{5} \mathrm{H}_{12}\mathrm{O}\). This suggests a molecule with a chain of five carbon atoms, twelve hydrogen atoms, and one oxygen atom. A possible molecular structure would be a pentanol, with the hydroxyl group attached to one of the end carbon atoms.
02

Oxidation

Oxidation in organic compounds generally involves the loss of hydrogen. In this case, during the reaction, the empirical formula changes to \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}\), which suggests that two hydrogen atoms have been removed from the original compound.
03

Final Compound Structure

The final compound behaves as a ketone, so a carbonyl group is present. Ketones are characterized by the presence of a carbon-oxygen double bond (\(\mathrm{C}=\mathrm{O}\)) within the carbon chain. A possible molecular structure for the final compound would be a pentanone, where the \(\mathrm{C}=\mathrm{O}\) group is in the middle of the carbon chain.

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Most popular questions from this chapter

Sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) adds to the double bond of alkenes as \(\mathrm{H}^{+}\) and \(-\mathrm{OSO}_{3} \mathrm{H}\). Predict the products when sulfuric acid reacts with (a) ethylene and (b) propene.

When a mixture of methane and bromine vapor is exposed to light, this reaction occurs slowly: $$\mathrm{CH}_{4}(g)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Br}(g)+\mathrm{HBr}(g)$$Suggest a mechanism for this reaction.

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