Question

Identify the major product formed when propene is treated with hydrogen chloride followed by Swarts reaction.

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

HCl adds to alkenes according to Markovnikov's rule. Swarts reaction then replaces halogen atoms such as Cl or Br with fluorine to form alkyl fluorides.

Answer 1

The Correct Option is C

Step 1: Addition of HCl to propene.
Propene is \[ CH_3-CH=CH_2 \] When HCl is added, the reaction follows Markovnikov's rule.
Hydrogen attaches to the carbon atom having more hydrogen atoms, while chlorine attaches to the more substituted carbon atom.
Therefore, \[ CH_3-CH=CH_2 + HCl \rightarrow CH_3-CH(Cl)-CH_3 \] The product formed is \[ 2\text{-chloropropane} \]

Step 2: Apply Swarts reaction.
Swarts reaction converts alkyl chlorides or alkyl bromides into alkyl fluorides using metallic fluorides such as: \[ AgF,\ Hg_2F_2,\ SbF_3,\ CoF_2 \] Thus, \[ CH_3-CH(Cl)-CH_3 \xrightarrow{\text{Swarts reaction}} CH_3-CH(F)-CH_3 \]

Step 3: Identify the final product.
The product obtained is \[ 2\text{-fluoropropane} \]

Step 4: Final conclusion.
Hence, the major product formed is \[ \boxed{\text{2-Fluoropropane}} \] Therefore, the correct option is \[ \boxed{(3)} \]