Chapter 19: Problem 440

In alkaline solution, \(4-\) methy1-4-hydroxy-2-pentanone is partly converted into acetone. Show all steps of a likely mechanism. What does this reaction amount to?

Short Answer

Expert verified

In an alkaline solution, 4-methyl-4-hydroxy-2-pentanone undergoes a series of reactions involving deprotonation, intramolecular rearrangement, and protonation to form an unstable intermediate, 5-hydroxy-3-methyl-2-pentanone. This intermediate then decomposes to produce acetone and other products. The reaction amounts to a decomposition of 4-methyl-4-hydroxy-2-pentonanone, with acetone being a major product.

Step by step solution

Identify the starting compound

First, let's identify the starting compound, 4-methyl-4-hydroxy-2-pentanone. This is a ketone with a hydroxy group and a methyl group on the same carbon.

Deprotonation of the hydroxy group

In an alkaline solution, the hydroxyl group (OH) in 4-methyl-4-hydroxy-2-pentonanone can be deprotonated due to the presence of OH- ions. This will result in the formation of a negatively charged oxygen atom (O-): \( \text{4-methyl-4-hydroxy-2-pentonanone} + OH^{-} \rightarrow \text{4-methyl-4-oxo-2-pentonanoate} + H_2O \) The resulting species is called 4-methyl-4-oxo-2-pentonanoate, which has a negative charge on the oxygen atom.

Intramolecular rearrangement

Now that we have formed a negatively charged oxygen atom in 4-methyl-4-oxo-2-pentonanoate, the oxygen will act as a nucleophile and attack the carbonyl carbon. This will form an oxyanion, which will then push the double bond electrons on the carbonyl carbon to the nearby methyl group, causing a fundamental shift in the structure: \( \text{4-methyl-4-oxo-2-pentonanoate} \rightarrow \text{Enolate intermediate} \)

Protonation of the enolate intermediate

In the next step, the enolate intermediate formed in step 3 can capture an available proton (H+) from the solvent to form a new hydroxy group adjacent to the newly formed double bond: \( \text{Enolate intermediate} + H^+ \rightarrow \text{5-hydroxy-3-methyl-2-pentanone} \)

Formation of acetone

The formed 5-hydroxy-3-methyl-2-pentanone is unstable and undergoes a decomposition reaction. This produces acetone, which is one of the final products of this reaction: \( \text{5-hydroxy-3-methyl-2-pentanone} \rightarrow \text{acetone} + \text{other products} \)

Conclusion

Through the steps described above, the mechanism depicts how 4-methyl-4-hydroxy-2-pentonanone is partly converted into acetone in an alkaline solution. This reaction amounts to a decomposition of 4-methyl-4-hydroxy-2-pentonanone, as it is converted into other products, with acetone being one of the major products.

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In alkaline solution, \(4-\) methy1-4-hydroxy-2-pentanone is partly converted into acetone. Show all steps of a likely mechanism. What does this reaction amount to?

Short Answer

Step by step solution

Identify the starting compound

Deprotonation of the hydroxy group

Intramolecular rearrangement

Protonation of the enolate intermediate

Formation of acetone

Conclusion

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