`(CH_(3))_(2)CHOH overset("Mild")underset("oxidation[O]")rarr (X) overset((i)CH_(3)Mgl)underset((ii)"Hydrolysis")rarr (Y)`
In the above sequence of reaction, (Y) is:

Options

A. Isobutyl alcohol
B. n-Butyl alcohol
C. Tertiary butyl alcohol
D. Isobutylene

Explanation

To solve the given question, we will analyze the reactions step by step.

### Step 1: Identify the starting compound
The starting compound is `(CH₃)₂CHOH`, which is a secondary alcohol known as isopropyl alcohol or 2-propanol.

### Step 2: Mild oxidation of the secondary alcohol
When a secondary alcohol undergoes mild oxidation, it typically forms a ketone. In this case, the oxidation of isopropyl alcohol will yield acetone (propan-2-one). Therefore, we can denote this product as (X).

**Reaction:**
\[ (CH₃)₂CHOH \xrightarrow{\text{Mild Oxidation}} \text{Acetone (X)} \]

### Step 3: Reaction of acetone with methyl magnesium iodide
Next, we react acetone (X) with methyl magnesium iodide (CH₃MgI), which is a Grignard reagent. In this reaction, the nucleophilic methyl group from the Grignard reagent attacks the carbonyl carbon of acetone.

**Reaction:**
\[ \text{Acetone (X)} + \text{CH₃MgI} \rightarrow \text{(CH₃)₂C(OH)CH₃} \text{ (intermediate)} \]

### Step 4: Hydrolysis of the intermediate
After the Grignard reaction, the intermediate formed is a tertiary alcohol. Upon hydrolysis (adding water), this intermediate will yield a tertiary alcohol known as tert-butyl alcohol (or 2-methylpropan-2-ol).

**Reaction:**
\[ \text{(CH₃)₂C(OH)CH₃} \xrightarrow{\text{Hydrolysis}} \text{tert-Butyl Alcohol (Y)} \]

### Conclusion
Thus, the final product (Y) after the hydrolysis step is tert-butyl alcohol.

### Final Answer
(Y) is tert-butyl alcohol.

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