`F_3C-CH=CH_2 overset(HBr)(rarr) (A) overset(Nal) underset(dry . "acetone")(rarr) (B) `

Options

A. `F_3C-CH(Br)-CH_3 and F_3C-CH(I)-CH_3`
B. `F_3C-CH_2-CH_2Br and F_3C-CH_2-CH_2 I`
C. `BrF_2C-CH=CH_2 and IF_2C-CH=CH_2`
D. `BrF_2C-CH_2-CH_2Br and IF_2C-CH_2-CH_2I`

Answer

To solve the given question step by step, we will analyze the reactions of the compound \( F_3C-CH=CH_2 \) with \( HBr \) and then with \( NaI \) in dry acetone.

### Step 1: Reaction of \( F_3C-CH=CH_2 \) with \( HBr \)

1. **Identify the Reactants**: We have an alkene \( F_3C-CH=CH_2 \) and \( HBr \).
2. **Markovnikov's Rule**: According to Markovnikov's rule, when \( HBr \) adds to the alkene, the hydrogen atom (H) will attach to the carbon with more hydrogen atoms (the less substituted carbon), and the bromine (Br) will attach to the carbon with fewer hydrogen atoms (the more substituted carbon).
3. **Addition of HBr**:
- The double bond between \( CH \) and \( CH_2 \) will break.
- The hydrogen from \( HBr \) will add to the \( CH_2 \) carbon, and the bromine will add to the \( CH \) carbon.
4. **Product A**: The resulting product after this addition is:
\[ A = F_3C-CH(Br)-CH_3 \]

### Step 2: Reaction of Product A with \( NaI \) in Dry Acetone

1. **Identify the Reactants**: The product \( A \) is \( F_3C-CH(Br)-CH_3 \) and it reacts with \( NaI \) in dry acetone.
2. **Finkelstein Reaction**: This is a nucleophilic substitution reaction where iodide ion (\( I^- \)) from \( NaI \) will replace the bromine atom.
3. **Better Leaving Group**: Bromine is a better leaving group than fluorine, so the iodide ion will replace the bromine atom.
4. **Product B**: The resulting product after this substitution is:
\[ B = F_3C-CH(I)-CH_3 \]

### Final Products

- **Product A**: \( F_3C-CH(Br)-CH_3 \)
- **Product B**: \( F_3C-CH(I)-CH_3 \)

### Conclusion

After analyzing the reactions, we can conclude that:
- The correct answer for product A is \( F_3C-CH(Br)-CH_3 \).
- The correct answer for product B is \( F_3C-CH(I)-CH_3 \).