`CH_3CH_2CH_2OHoverset(Conc.H_2SO_4)rarrXoverset(Cl_2,hv)rarrY,X and Y` are

Options

A. `X=CH_3CH_2CH_3,Y=CH_3CH_2CH_2Cl`
B. `X=CH_3CH=CH_2,Y=CH_2ClCH=CH_2`
C. `X=CH_2=CH_2,Y=CH_3CH_2Cl`
D. `X=CH_3CH_2CH_3,Y=CH_3CH=CH_2`

Answer

To solve the problem, we need to identify the compounds X and Y formed from the reaction of 1-butanol (CH3CH2CH2OH) with concentrated sulfuric acid and chlorine in the presence of light.

### Step-by-Step Solution:

1. **Identify the Starting Material:**
The starting material is 1-butanol (CH3CH2CH2OH).

2. **Reaction with Concentrated Sulfuric Acid:**
When 1-butanol reacts with concentrated sulfuric acid (H2SO4), it undergoes dehydration. The acid donates a proton (H+) to the hydroxyl group (-OH), converting it into a better leaving group (water, H2O).

The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{H}_2\text{SO}_4 \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2^+ + \text{H}_2\text{O} \]

3. **Formation of Carbocation:**
The dehydration leads to the formation of a carbocation. In this case, the primary carbocation (CH3CH2CH2+) is formed, which is not very stable. However, it can rearrange to a more stable secondary carbocation by hydride shift.

The rearrangement can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2^+ \rightarrow \text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_3 \]

4. **Formation of Alkene (X):**
The more stable carbocation can lose a proton (H+) to form an alkene. The resulting alkene from 1-butanol after dehydration is butene (specifically, 2-butene).

The reaction can be represented as:
\[ \text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_3 \rightarrow \text{CH}_3\text{CH}=\text{CH}\text{CH}_3 + \text{H}^+ \]
Thus, X = 2-butene (CH3CH=CHCH3).

5. **Reaction with Chlorine (Cl2) in the Presence of Light:**
The next step involves the reaction of 2-butene with chlorine (Cl2) in the presence of light (hv). This reaction leads to the formation of free radicals.

The chlorine molecule (Cl2) dissociates into two chlorine radicals (Cl•) in the presence of light:
\[ \text{Cl}_2 \xrightarrow{\text{hv}} 2 \text{Cl}• \]

6. **Formation of Product Y:**
The chlorine radical can add to the double bond of 2-butene, resulting in the formation of a vicinal dichloride. The addition of Cl• to the double bond leads to the formation of 1,2-dichlorobutane.

The reaction can be represented as:
\[ \text{CH}_3\text{CH}=\text{CH}\text{CH}_3 + \text{Cl}• \rightarrow \text{CH}_3\text{CH}(\text{Cl})\text{CH}(\text{Cl})\text{CH}_3 \]
Thus, Y = 1,2-dichlorobutane (CH3CHClCHClCH3).

### Final Answer:
- X = 2-butene (CH3CH=CHCH3)
- Y = 1,2-dichlorobutane (CH3CHClCHClCH3)