The product of the following reaction : `CH_(2) = CH - CCCl_(3)` + HBr to Product

Options

A. `CH_(3)- CH(Br) - CCl_(3)`
B. `CH_(2)(Br) -CH_(2) - "CCl"_(3)`
C. `CH_(2)(Br) - CH(Cl) - CHCl_(2)`
D. `CH_(3) - CH_(2) - "CCl"_(3)`

Answer

To determine the product of the reaction between `CH2=CH-CCCl3` and `HBr`, we can follow these steps:

### Step 1: Identify the Reactants
The reactants are:
- An alkene: `CH2=CH-CCCl3`
- A hydrogen halide: `HBr`

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

### Step 3: Protonation of the Alkene
The double bond in the alkene will react with HBr, leading to the formation of a carbocation. The H+ from HBr will add to one of the carbon atoms of the double bond.

### Step 4: Determine the Possible Carbocations
There are two possible sites for protonation:
1. Protonation at the terminal carbon (C1):
- This leads to the formation of a primary carbocation: `CH3-CH+-CCCl3`
2. Protonation at the middle carbon (C2):
- This leads to the formation of a secondary carbocation: `CH2+-CH-CCCl3`

### Step 5: Stability of Carbocations
The secondary carbocation is more stable than the primary carbocation due to hyperconjugation and inductive effects. Therefore, the reaction will favor the formation of the secondary carbocation.

### Step 6: Nucleophilic Attack
Once the more stable carbocation is formed, the bromide ion (Br-) from HBr will attack this carbocation:
- The nucleophile (Br-) will attack the secondary carbocation, leading to the final product.

### Step 7: Write the Product
The final product after the nucleophilic attack will be:
- `CH2Br-CH-CCCl3`

### Final Product
Thus, the product of the reaction `CH2=CH-CCCl3 + HBr` is:
- `CH2Br-CH-CCCl3`

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