`H_(3)C underset(CH_(3))underset(|) overset(CH_(3)) overset(|) (--) overset(C_(2)H_(5))overset(|)CH-underset(CH_(3))underset(|)(CH)-OH underset(/_\) overset(H^(o+))toP` Product mixture, the product's mixture contains

Options

A. `H_(3)C-underset(CH_(3))underset(|)(CH)-underset(OH)underset(|)(CH)-overset(CH_(3)) overset(|)(CH)-C_(2)H_(50`
B. `H_(3)C-underset(CH_(3))underset(|)overset(CH_(3)) overset(|)(C)-underset(C_(2)H_(5))underset(|)(C)=underset(CH_(3))underset(|)(C)-H`
C. `H_(3)C-underset(CH_(3))underset(|)overset(CH_(3)) overset(|)(C)-underset(C_(2)H_(5))underset(|)(C)=underset(CH_(3))underset(|)(C)-CH_(3)`
D. `H_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-underset(C_(2)H_(5))underset(|)(CH)-overset(H) overset(|)(C)=CH_(2)`

Explanation

To solve the problem, we need to analyze the given structure and the reaction conditions. The structure represents a long-chain alcohol that will undergo protonation and dehydration in the presence of concentrated sulfuric acid and heat.

### Step-by-Step Solution:

1. **Identify the Structure**:
The given structure is a long-chain alcohol with a hydroxyl group (-OH) attached to the third carbon of a carbon chain. The carbon chain has three methyl groups (CH₃) attached to the first carbon, an ethyl group (C₂H₅) attached to the second carbon, and the third carbon has a hydroxyl group.

2. **Protonation of the Hydroxyl Group**:
When the alcohol is treated with concentrated sulfuric acid, the hydroxyl group (-OH) gets protonated to form -OH₂⁺. This increases the leaving ability of the water molecule.

3. **Formation of Carbocation**:
Upon heating, water (H₂O) is eliminated, leading to the formation of a carbocation. The carbocation formed here is a secondary carbocation because it is attached to two other carbon atoms.

4. **Carbocation Rearrangement**:
To stabilize the carbocation, a hydride shift can occur. A hydrogen atom from an adjacent carbon can shift to the carbocation carbon, converting the secondary carbocation into a tertiary carbocation, which is more stable.

5. **Elimination Reaction**:
The tertiary carbocation can now undergo elimination to form an alkene. In this step, a hydrogen atom is removed from one of the adjacent carbons, resulting in the formation of a double bond.

6. **Identify the Major Product**:
The major product will be the most stable alkene formed from this elimination reaction. In this case, the product will have a double bond between the carbons adjacent to the original carbocation.

7. **Conclusion**:
The major product formed from the dehydration of the alcohol is an alkene with the structure that reflects the rearrangement and elimination steps described.

### Final Product:
The major product P in the mixture is an alkene formed after dehydration of the alcohol.