In the given reaction
`CH_(3)-underset(CH_(3))underset("| ")"CH"-CH=CH_(2)+HCl rarr [X]`
Major product [X] will be

Options

A. 2 - chloro -3- methylbutane
B. 1 - chloro -3- methylbutane
C. 2 - chloro -2- methylbutane
D. 2 - chloropentane

Answer

To determine the major product [X] from the reaction of the given alkene with HCl, we can follow these steps:

### Step 1: Identify the Alkene
The given alkene is 2-methylpropene (CH₃-CH(CH₃)-CH=CH₂).

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

### Step 3: Determine Possible Carbocation Formation
In this case, the double bond can react in two ways:
1. The H⁺ can add to the terminal carbon (C1), leading to a primary carbocation.
2. The H⁺ can add to the second carbon (C2), leading to a secondary carbocation.

### Step 4: Stability of Carbocations
Secondary carbocations are more stable than primary carbocations. Therefore, we will favor the formation of the secondary carbocation.

### Step 5: Form the Carbocation
If H⁺ adds to C2, we form a secondary carbocation at C1. The structure will look like this:
- CH₃-CH⁺-CH₂-CH₃

### Step 6: Nucleophilic Attack by Cl⁻
The Cl⁻ ion from HCl will then attack the positively charged carbon (C1) of the carbocation.

### Step 7: Final Product Formation
The final product after the nucleophilic attack will be:
- CH₃-CH(Cl)-CH₂-CH₃

This compound can be named as 2-chloro-2-methylpropane.

### Conclusion
The major product [X] formed from the reaction of 2-methylpropene with HCl is **2-chloro-2-methylpropane**.

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