Identify the major product formed when propene is treated with hydrogen chloride followed by Swarts reaction.
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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.
Step 1: Write the structure of propene and plan the two steps. Propene: $CH_3{-}CH{=}CH_2$. Step 1 is HCl addition (Markovnikov's rule). Step 2 is Swarts reaction to convert $-Cl$ to $-F$. Step 2: Apply Markovnikov's rule for HCl addition to propene. Markovnikov's rule: $H^+$ adds to the carbon with more hydrogens (less substituted), and $Cl^-$ adds to the more substituted carbon. $C_3$ ($=CH_2$) has 2 H atoms (less substituted), and $C_2$ ($=CH-$) has 1 H atom (more substituted). So $H^+$ goes to $C_3$ and $Cl^-$ goes to $C_2$. Step 3: Draw the product of HCl addition. \[ CH_3{-}CH{=}CH_2 + HCl \rightarrow CH_3{-}CHCl{-}CH_3 \] Product: 2-chloropropane (secondary alkyl chloride). Step 4: Explain Swarts reaction. Swarts reaction converts alkyl chlorides or bromides into alkyl fluorides using metallic fluorides ($AgF$, $Hg_2F_2$, $SbF_3$, or $CoF_2$). The fluoride ion displaces the chloride via nucleophilic substitution. This is the only practical method for making secondary alkyl fluorides since direct fluorination is too violent. Step 5: Apply Swarts reaction to 2-chloropropane. \[ CH_3{-}CHCl{-}CH_3 \xrightarrow{AgF\ \text{or}\ SbF_3} CH_3{-}CHF{-}CH_3 \] The chlorine at $C_2$ is replaced by fluorine to give 2-fluoropropane. Step 6: Identify the major product. The sequence (Markovnikov HCl addition then Swarts fluorination) gives 2-fluoropropane as the major product. Step 7: State the final answer. \[ \boxed{\text{2-Fluoropropane}\ (CH_3{-}CHF{-}CH_3)} \]
