`{:(" "CH_(3)),(" |"),(CH_(3)CH_(2)-C-CH-CH_(3)underset("heat")overset(H_(2)SO_(4))(rarr)underset("Major")(rarr)),(" | |"),(" "CH_(3)""OH):}`
What is the major product P in the above reaction?

Options

A. `{:(""CH_(3)" "CH_(3)),( "| |"),(CH_(2)-CH_(2)-CH_(3)-CH=CH_(2)):}`
B. `CH_(3)-overset(CH_(3))overset("| ")(CH)-overset(CH_(3))overset("| ")(CH)-CH=CH_(2)`
C. `{:(" "CH_(3)),(" |"),(CH_(3)CH_(2)-C-CH=CH_(2)),(" |"),(" "CH_(3)):}`
D. `{:(" "CH_(3)),(" |"),(CH_(3)CH_(2)-C=C-CH_(3)),(" |"),(" "CH_(3)):}`

Answer

To solve the given problem, we need to analyze the reaction step by step. The reaction involves the dehydration of an alcohol in the presence of sulfuric acid (H₂SO₄) and heat, leading to the formation of an alkene. Let's break it down:

### Step 1: Identify the Reactant
The reactant is a compound with the following structure:
- CH₃-CH₂-C(OH)-CH-CH₃
This can be rewritten as:
- CH₃-CH₂-C(OH)-CH(CH₃)-CH₃

### Step 2: Protonation of the Alcohol
In the presence of sulfuric acid (H₂SO₄), the hydroxyl group (OH) of the alcohol gets protonated to form a better leaving group:
- CH₃-CH₂-C(OH₂⁺)-CH(CH₃)-CH₃

### Step 3: Elimination of Water
The protonated alcohol (OH₂⁺) can lose a water molecule (H₂O), leading to the formation of a carbocation. The structure now looks like this:
- CH₃-CH₂-C⁺-CH(CH₃)-CH₃
This is a secondary carbocation.

### Step 4: Carbocation Rearrangement
To stabilize the carbocation, a 1,2-methyl shift occurs. The methyl group from the adjacent carbon migrates to the carbocation center:
- CH₃-CH⁺-C(CH₃)-CH₂-CH₃
This results in a more stable tertiary carbocation.

### Step 5: Deprotonation to Form Alkene
Finally, a proton (H⁺) is removed from the adjacent carbon to form a double bond, resulting in the major product:
- CH₃-CH=C(CH₃)-CH₂-CH₃

### Step 6: Identify the Major Product
The major product P is:
- CH₃-CH=C(CH₃)-CH₂-CH₃
This is an internal alkene, which is typically more stable than terminal alkenes.

### Conclusion
Thus, the major product P in the reaction is:
- **CH₃-CH=C(CH₃)-CH₂-CH₃**