Consider the following reaction sequence
`H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH=CH_(2) underset(2." "KOH-EtOH, Heat)overset(1." "HBr)to`
the major product formed is

Options

A. `H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-underset(OH)underset(|)(C)H-CH_(3)`
B. `H_(3)C-underset(OE t)underset(|)overset(CH_(3))overset(|)(C)-CH(CH_(3))_(2)`
C. `H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH=CH_(2)`
D. `H_(3)C-underset(CH_(3))underset(|)(C)=underset(CH_(3))underset(|)(C)-CH_(3)`

Answer

To solve the given reaction sequence, we will follow these steps:

### Step 1: Identify the Starting Material
The starting material is 2-methyl-2-pentene, which can be represented as:
\[ \text{H}_3C-\text{C}(\text{CH}_3)-\text{C}(\text{CH}_3)=\text{CH}_2 \]

### Step 2: Reaction with HBr
When 2-methyl-2-pentene reacts with HBr, the double bond will react with HBr to form a carbocation. The HBr can add to the double bond in two ways:
1. H adds to the less substituted carbon (1° carbocation).
2. H adds to the more substituted carbon (2° carbocation).

However, the more stable carbocation will be favored. In this case, the more stable 2° carbocation will form:
\[ \text{CH}_3-\text{C}^+(\text{CH}_3)-\text{CH}_2-\text{Br} \]

### Step 3: Carbocation Rearrangement
The formed 2° carbocation can undergo a rearrangement (methyl shift) to form a more stable 3° carbocation:
\[ \text{CH}_3-\text{C}^+(\text{CH}_2-\text{CH}_3)-\text{CH}_3 \]

### Step 4: Nucleophilic Attack
The bromide ion (Br⁻) will then attack the carbocation to form the product:
\[ \text{CH}_3-\text{C}(\text{Br})(\text{CH}_2-\text{CH}_3)-\text{CH}_3 \]

### Step 5: Reaction with KOH (Ethanolic)
The next step involves treating the product with KOH in ethanol, which acts as a strong base. The base will abstract a proton from the β-carbon (the carbon adjacent to the carbon bearing the Br). According to Zaitsev's rule, the base will preferentially remove the hydrogen from the carbon that has the most substituents (the carbon with fewer hydrogens).

In this case, the elimination will occur as follows:
1. The base abstracts a hydrogen from the β-carbon that has fewer hydrogens.
2. The Br leaves, resulting in the formation of a double bond.

### Step 6: Final Product
The final product after elimination will be:
\[ \text{CH}_3-\text{C}(\text{CH}_3)=\text{C}(\text{CH}_2-\text{CH}_3) \]
This is a more stable alkene.

### Summary of Major Product
The major product formed from the reaction sequence is:
\[ \text{CH}_3-\text{C}(\text{CH}_3)=\text{C}(\text{CH}_2-\text{CH}_3) \]

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