Give the IUPAC name of the primary organic product for oxidation of the following compound. Select one: a. 2-methylbutanone b. 2-methyl-3-butanone c. 3-methylbutanoic acid d. 2-methylbutanoic acid Clear my choice
The Correct Answer and Explanation is:
Correct Answer: d. 2-methylbutanoic acid
Explanation:
The oxidation of an organic compound depends on the type of functional group present. When a primary alcohol is oxidized, it forms an aldehyde, which can further oxidize to form a carboxylic acid. When a secondary alcohol is oxidized, it typically forms a ketone. Tertiary alcohols do not oxidize under mild conditions.
Consider the structure of the starting compound: assume it is 2-methylbutanol, a primary alcohol. Its structure includes a four-carbon chain (butane) with a methyl group on carbon 2 and a hydroxyl group (-OH) on the terminal carbon. This makes it a primary alcohol, and upon oxidation, the hydroxyl group is converted into a carboxylic acid (-COOH) functional group.
The result is 2-methylbutanoic acid. This molecule has:
- A straight-chain carboxylic acid backbone with four carbon atoms (butanoic acid),
- A methyl group substituted on the second carbon.
Now, analyze the other options:
- a. 2-methylbutanone: This is a ketone, likely formed from a secondary alcohol, not a primary alcohol.
- b. 2-methyl-3-butanone: This name is not correct by IUPAC standards; it seems misnamed.
- c. 3-methylbutanoic acid: This structure differs from the correct product because the methyl group is on the third carbon, not the second.
- d. 2-methylbutanoic acid: This matches the expected oxidation product of a primary alcohol with the correct position of the methyl group.
Therefore, the correct IUPAC name of the primary organic oxidation product is 2-methylbutanoic acid.
