In the given reaction `CH_(3)-cH_(2)COOHoverset((i)CH_(3)Li" (excess)"(ii)H_(3)O^(o+))rarr [X]`
[X] will be

Options

A. `CH_(3)-cH_(2)COOLi`
B. `CH_(3)-CH_(2)-overset(O)overset(||)C-CH_(3)`
C. `CH_(3)-CH_(2)-CH_(2)OH`
D. `CH_(3)-CH_(2)-CH_(2)-CH_(3)`

Answer

To solve the given reaction step by step, we will analyze the reaction of propanoic acid (CH₃CH₂COOH) with excess methyl lithium (CH₃Li) followed by hydrolysis.

### Step 1: Nucleophilic Attack
1. **Identify the Reactants**: We have propanoic acid (CH₃CH₂COOH) and methyl lithium (CH₃Li).
2. **Nucleophilic Attack**: Methyl lithium is a strong nucleophile. The lithium (Li) in CH₃Li has a partial positive charge, while the carbon in the methyl group (CH₃) has a partial negative charge. The nucleophile attacks the carbonyl carbon of the carboxylic acid (C=O) in propanoic acid.

Reaction:
\[ CH₃CH₂COOH + CH₃Li \rightarrow CH₃CH₂C(O^-)CH₃ + Li^+ \]

### Step 2: Formation of Alkoxide
3. **Formation of Alkoxide**: After the nucleophilic attack, we form an alkoxide intermediate (CH₃CH₂C(O^-)CH₃) where the carbonyl oxygen now carries a negative charge (O⁻).

### Step 3: Hydrolysis
4. **Hydrolysis**: When we add water (H₃O⁺) to the reaction, the alkoxide will be protonated to form an alcohol. The hydrolysis will convert the alkoxide into a stable alcohol.

Reaction:
\[ CH₃CH₂C(O^-)CH₃ + H₂O \rightarrow CH₃CH₂C(OH)CH₃ \]

### Step 4: Rearrangement
5. **Rearrangement**: The product formed (CH₃CH₂C(OH)CH₃) has two hydroxyl (OH) groups on the same carbon, which is not stable. Therefore, it will undergo dehydration to form a more stable compound. This leads to the formation of a ketone.

Final Product:
\[ CH₃CH₂C(=O)CH₃ \]

### Conclusion
The final product [X] after the reaction is **butan-2-one** (or methyl ethyl ketone).

### Final Answer
[X] = CH₃CH₂C(=O)CH₃ (butan-2-one)

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