`CH_(3)-CH=CH_(2)+HIrarr'X'`. Here product 'X' is

Options

A. `CH_(3)CH_(2)CH_(2)I`
B. `CH_(3)-CHI-CH_(3)`
C. `CH_(3)CH_(2)CH_(3)`
D. `CI_(3)-CH_(2)-CI_(3)`

Answer

To determine the product 'X' formed when propene (CH₃-CH=CH₂) reacts with hydroiodic acid (HI), we can follow these steps:

### Step 1: Identify the reactants
The reactants are propene (CH₃-CH=CH₂) and hydroiodic acid (HI).

### Step 2: Understand the reaction mechanism
When an alkene reacts with HI, the reaction proceeds via electrophilic addition. The double bond in the alkene acts as a nucleophile and attacks the electrophilic hydrogen (H⁺) from HI.

### Step 3: Apply Markovnikov's Rule
Markovnikov's Rule states that when HX (where X is a halogen) adds to an unsymmetrical alkene, the hydrogen atom will attach to the carbon with the greater number of hydrogen atoms (the more substituted carbon), while the halide (I⁻) will attach to the carbon with fewer hydrogen atoms.

In propene:
- The double bond is between the first and second carbon atoms.
- The first carbon (C1) has 3 hydrogen atoms (CH₃).
- The second carbon (C2) has 1 hydrogen atom (CH).

According to Markovnikov's Rule:
- H⁺ will attach to C1 (the more substituted carbon).
- I⁻ will attach to C2 (the less substituted carbon).

### Step 4: Write the product
After applying Markovnikov's Rule:
- The product formed will be CH₃-CH(I)-CH₃, which is 2-iodopropane.

### Step 5: Conclusion
The product 'X' formed from the reaction of propene with HI is 2-iodopropane (CH₃-CH(I)-CH₃).

### Final Answer
The product 'X' is 2-iodopropane (CH₃-CH(I)-CH₃).

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