What is the major product that will be formed in the following reaction ?
`CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-CH=CH_2+HCl to`

Options

A. `CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-undersetunderset(CH_3)(|)CH-CH_3`
B. `CH_3-undersetunderset(Cl)(|)oversetoverset(CH_3)(|)C-undersetunderset(CH_3)(|)CH-CH_3`
C. `CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-CH_2-CH_2-Cl`
D. `CH_3-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-undersetunderset(Cl)(|)CH-CH_2-CH_2-undersetunderset(Cl)(|)CH-undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-CH_3`

Answer

To determine the major product formed in the reaction of the given alkene with HCl, we can follow these steps:

### Step 1: Identify the Reactants
The reactant is an alkene:
\[ \text{CH}_3-\text{C}(\text{CH}_3)(\text{CH}_3)-\text{C}=\text{CH}_2 \]
This is a substituted alkene with a double bond between the last two carbon atoms.

### Step 2: Understand the Reaction Mechanism
When alkenes react with hydrogen halides (like HCl), they undergo an electrophilic addition reaction. The double bond acts as a nucleophile and attacks the electrophile (H⁺ from HCl), leading to the formation of a carbocation.

### Step 3: Protonation of the Alkene
The double bond between the carbon atoms will react with HCl. The H⁺ ion will add to one of the carbon atoms involved in the double bond. The more substituted carbon (the one with more alkyl groups) will preferentially form the carbocation due to stability reasons (Markovnikov's rule).

### Step 4: Formation of the Carbocation
In this case, the double bond is between the terminal carbon (C1) and the adjacent carbon (C2). When H⁺ adds to C1, it creates a carbocation at C2, which is more stable because it is a tertiary carbocation (due to the three alkyl groups attached).

### Step 5: Nucleophilic Attack by Chloride Ion
The Cl⁻ ion from HCl will then attack the positively charged carbocation. This will result in the formation of the major product.

### Step 6: Write the Major Product
The major product formed will be:
\[ \text{CH}_3-\text{C}(\text{CH}_3)(\text{CH}_3)-\text{C}(\text{Cl})-\text{CH}_2 \]
This can be simplified to:
\[ \text{CH}_3-\text{C}(\text{Cl})(\text{CH}_3)-\text{C}(\text{CH}_3)-\text{CH}_2 \]

### Final Product
Thus, the major product of the reaction is:
\[ \text{CH}_3-\text{C}(\text{Cl})(\text{CH}_3)-\text{C}(\text{CH}_3)-\text{CH}_2 \]