Consider the given reaction Compound 'X' is
`CH_3-CH=CH_2+HBr to X`

Options

A. 1-bromopropane
B. 2-bromopropane
C. 1, 2-dibromopropane
D. 1,3-dibromopropane

Answer

To solve the reaction of the compound \( CH_3-CH=CH_2 \) with \( HBr \) to find compound \( X \), we can follow these steps:

### Step 1: Identify the Reactants
The reactant is an alkene, specifically propene, which has the structure:
\[ CH_3-CH=CH_2 \]

### Step 2: Understand the Reaction with HBr
When propene reacts with hydrogen bromide (\( HBr \)), the double bond between the carbon atoms will break. This reaction is an example of an electrophilic addition reaction.

### Step 3: Break Down the Reaction
The \( HBr \) molecule can dissociate into \( H^+ \) and \( Br^- \). The double bond in propene acts as a nucleophile and attacks the \( H^+ \) ion.

### Step 4: Formation of Carbocations
When the double bond breaks, two possible carbocations can form:
1. A primary carbocation at the end carbon (less stable).
2. A secondary carbocation at the middle carbon (more stable).

Given that the secondary carbocation is more stable, it will be the favored pathway for the reaction.

### Step 5: Attack of Bromide Ion
Once the secondary carbocation is formed, the bromide ion (\( Br^- \)) will attack the positively charged carbon atom. This leads to the formation of the final product.

### Step 6: Write the Product
The product formed from this reaction is:
\[ CH_3-CHBr-CH_3 \]
This compound is known as 2-bromopropane.

### Conclusion
Thus, the compound \( X \) formed from the reaction of propene with \( HBr \) is:
\[ \text{Compound } X = 2-\text{bromopropane} \]

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