`CH_(3)-CH=CH-CH_(2)-CH_(3) overset(HI)to underset("major")A`
Compound A is

Options

A. `CH_(3)-CH_(2)-underset(|)underset(|)CH-CH_(2)-CH_(3)`
B. `CH_(3)-underset(|)underset(|)CH-CH_(2)-CH_(2)-CH_(3)`
C. `CH_(3)-underset(|)underset(|)overset(CH_(2))overset(|)C-CH_(2)-CH_(3)`
D. `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH_(2)I`

Answer

To solve the problem, we need to determine the major product (Compound A) formed when 1-pentene (CH₃-CH=CH-CH₂-CH₃) reacts with hydrogen iodide (HI).

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is 1-pentene (CH₃-CH=CH-CH₂-CH₃) and it reacts with hydrogen iodide (HI).

2. **Protonation of the Alkene**:
The double bond in 1-pentene is attacked by the proton (H⁺) from HI. This can occur at either of the carbon atoms involved in the double bond:
- If the first carbon (C1) of the double bond is protonated, we get a carbocation at C2.
- If the second carbon (C2) of the double bond is protonated, we get a carbocation at C1.

3. **Formation of Carbocations**:
- **Carbocation from C1 protonation**:
CH₃-CH(+)-CH₂-CH₂-CH₃ (2° carbocation at C2)
- **Carbocation from C2 protonation**:
CH₃-CH₂-CH(+)-CH₂-CH₃ (2° carbocation at C1)

4. **Stability of Carbocations**:
- The stability of carbocations can be evaluated based on hyperconjugation and inductive effects.
- The carbocation formed by protonating C1 has 3 alpha hydrogens available for hyperconjugation.
- The carbocation formed by protonating C2 has 2 alpha hydrogens available for hyperconjugation.
- Therefore, the carbocation at C2 (from protonation at C1) is more stable due to having more alpha hydrogens (3 vs. 2).

5. **Nucleophilic Attack by Iodide Ion**:
The more stable carbocation (at C2) will be attacked by the iodide ion (I⁻) to form the final product.

6. **Final Product (Compound A)**:
The major product (Compound A) is formed as follows:
- The structure of Compound A will be CH₃-CH₂-CH(I)-CH₂-CH₃.

### Conclusion:
The major product (Compound A) formed from the reaction of 1-pentene with HI is **CH₃-CH₂-CH(I)-CH₂-CH₃**.

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