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 problem step by step, we will analyze the reaction of propanoic acid (CH₃CH₂COOH) with excess methyl lithium (CH₃Li) and then hydrolyze the product.

### Step 1: Identify the starting material
The starting material is propanoic acid, which has the structure:
\[ \text{CH}_3\text{CH}_2\text{COOH} \]

### Step 2: Reaction with methyl lithium (CH₃Li)
Methyl lithium is a strong nucleophile. When propanoic acid reacts with CH₃Li, the nucleophile (CH₃⁻) will attack the carbonyl carbon (C=O) of the carboxylic acid. This results in the formation of a lithium carboxylate intermediate.

The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{Li} \rightarrow \text{CH}_3\text{CH}_2\text{C(OH)(CH}_3)\text{O}^- \text{Li}^+ \]

### Step 3: Formation of the lithium carboxylate
After the nucleophilic attack, we have a lithium carboxylate:
\[ \text{CH}_3\text{CH}_2\text{C(OH)(CH}_3)\text{O}^- \text{Li}^+ \]

### Step 4: Second addition of methyl lithium
Since the problem states that there is an excess of CH₃Li, we can add another equivalent of methyl lithium. The lithium carboxylate will undergo another nucleophilic attack by CH₃Li at the carbonyl carbon, leading to the formation of a tertiary alcohol.

The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{C(OH)(CH}_3)\text{O}^- \text{Li}^+ + \text{CH}_3\text{Li} \rightarrow \text{CH}_3\text{CH}_2\text{C(CH}_3)(OH)\text{O}^- \text{Li}^+ \]

### Step 5: Hydrolysis
The final step involves hydrolyzing the lithium salt with an acid (H₃O⁺). This will convert the lithium carboxylate into the corresponding alcohol.

The hydrolysis can be represented as:
\[ \text{CH}_3\text{CH}_2\text{C(CH}_3)(OH)\text{O}^- \text{Li}^+ + \text{H}_3\text{O}^+ \rightarrow \text{CH}_3\text{CH}_2\text{C(CH}_3)(OH) + \text{Li}^+ + \text{OH}^- \]

### Step 6: Final product
After hydrolysis, the final product (X) will be:
\[ \text{CH}_3\text{CH}_2\text{C(OH)(CH}_3) \]
This is 2-methyl-1-propanol.

### Summary
The final product [X] is 2-methyl-1-propanol (or tert-butyl alcohol).

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