Consider the following reaction sequence ,
`CH_(3)CH(OH)CH_(3) overset("Conc" H_(2)SO_(4))underset("Heat")(to) X underset(H_(2)O) overset("dil." H_(2)SO_(4))(to) Y`
X and Y in the reaction respectively are

Options

A. `CH_3 CH = CH_(2) , CH_(3)CH(OH)CH_(3)`
B. `CH_(3)CH = CHCH_(3) , CH_(3)CH_(2)CH_(2)OH`
C. `CH_(3)CH_(2)CH = CH_(2) , CH_(3) , CH_(2)CH_(2) OH`
D. `CH_(3) CH_(2) CH_(2) CH_(3) , CH_(3)CH(OH)CH_(2)CH_(3)`

Explanation

To solve the reaction sequence given in the question, we will break it down step by step.

### Step 1: Identify the starting compound
The starting compound is 2-propanol, which can be represented as:
\[ \text{CH}_3\text{CH(OH)CH}_3 \]

### Step 2: React with concentrated sulfuric acid (H₂SO₄)
When 2-propanol reacts with concentrated sulfuric acid and heat, it undergoes dehydration. The mechanism involves the following:
- The hydroxyl group (–OH) is protonated by H⁺ from H₂SO₄, making it a better leaving group.
- Water (H₂O) is eliminated, forming a carbocation.

The reaction can be represented as:
\[ \text{CH}_3\text{CH(OH)CH}_3 \xrightarrow{\text{Conc. H}_2\text{SO}_4, \text{Heat}} \text{CH}_3\text{C}^+\text{HCH}_3 \]

### Step 3: Formation of the alkene
The carbocation formed can lose a hydrogen atom to form an alkene. The possible product is propene:
\[ \text{CH}_3\text{C}^+\text{HCH}_3 \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 \]

Thus, the product X is:
\[ \text{X} = \text{CH}_3\text{CH}=\text{CH}_2 \quad (\text{Propene}) \]

### Step 4: React with dilute sulfuric acid (H₂SO₄)
Now, we take the alkene (propene) and react it with dilute sulfuric acid. In this case, the reaction involves hydration:
- The H⁺ from H₂SO₄ adds to one of the carbons of the double bond, following Markovnikov's rule, which states that the hydrogen will add to the carbon with more hydrogens already attached.
- Water (H₂O) acts as a nucleophile and attacks the carbocation formed.

The reaction can be represented as:
\[ \text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow{\text{dil. H}_2\text{SO}_4} \text{CH}_3\text{CH(OH)CH}_3 \]

### Step 5: Identify the final product Y
The final product Y is 2-propanol:
\[ \text{Y} = \text{CH}_3\text{CH(OH)CH}_3 \]

### Conclusion
Thus, the products X and Y in the reaction sequence are:
- **X**: Propene (\( \text{CH}_3\text{CH}=\text{CH}_2 \))
- **Y**: 2-Propanol (\( \text{CH}_3\text{CH(OH)CH}_3 \))

### Summary of the Answer
- **X**: Propene (\( \text{CH}_3\text{CH}=\text{CH}_2 \))
- **Y**: 2-Propanol (\( \text{CH}_3\text{CH(OH)CH}_3 \))