Step 1: Understanding the Question:
We are given the product of a reaction and the reagents used. We need to deduce the structure of the starting material P.
Step 2: Key Formula or Approach:
The reagents are 1. NaOCl (sodium hypochlorite) and 2. H\(_3\)O\(^+\) (acid workup). This set of reagents is used for the haloform reaction.
The haloform reaction is a characteristic reaction of methyl ketones (compounds containing the R-C(=O)-CH\(_3\) group) or alcohols that can be oxidized to methyl ketones.
The reaction converts a methyl ketone into a carboxylate salt and a haloform (in this case, chloroform, CHCl\(_3\)). The subsequent acid workup protonates the carboxylate to give a carboxylic acid.
\[ \text{R-C(=O)CH}_3 \xrightarrow{\text{1. NaOCl}} \text{R-COO}^- + \text{CHCl}_3 \xrightarrow{\text{2. H}_3\text{O}^+} \text{R-COOH} \]
Step 3: Detailed Explanation:
We can identify the structure of the starting material P by working backward from the product.
The given product is an \(\alpha,\beta\)-unsaturated carboxylic acid.
\[ \text{Product} = \text{CH}_3\text{-CH=C(CH}_3)\text{-COOH} \]
This product has the structure R-COOH, where the 'R' group is:
\[ \text{R} = \text{CH}_3\text{-CH=C(CH}_3)\text{-} \]
Since the haloform reaction converts a starting material of the form R-C(=O)CH\(_3\) into R-COOH, the starting material P must have had this 'R' group attached to a methyl ketone group (-C(=O)CH\(_3\)).
Let's construct the structure of P:
\[ \text{P} = \underbrace{\text{CH}_3\text{-CH=C(CH}_3)\text{-}}_{\text{R group}} \underbrace{\text{C(=O)CH}_3}_{\text{Methyl ketone group}} \]
The starting material P is 3,4-dimethylpent-3-en-2-one.
Step 4: Final Answer:
Comparing this deduced structure with the options, it matches the structure in option (A).