1-phenyl-2-chloropropane on treating with alc. KOH gives mainly :

Question

Accepted Answer

1-phenyl-2-chloropropane on treating with alcoholic KOH undergoes a dehydrohalogenation reaction. This is a type of elimination reaction where the halogen (chlorine in this case) and a hydrogen atom are removed from adjacent carbon atoms, resulting in the formation of a double bond.

1. **Explanation for Option 1 (1-phenylpropene):**
   - In the presence of alcoholic KOH, 1-phenyl-2-chloropropane undergoes an E2 elimination reaction. The chlorine atom is removed from the second carbon, and a hydrogen atom is removed from the adjacent carbon (the first carbon in this case), leading to the formation of a double bond between the first and second carbon atoms. This results in the formation of 1-phenylpropene. The reaction favors the formation of the more stable alkene, which in this case is 1-phenylpropene due to the presence of the phenyl group that stabilizes the double bond through resonance.

2. **Explanation for why other options are incorrect:**
   - **Option 2 (2-phenylpropene):** This compound would require the double bond to form between the second and third carbon atoms. However, the elimination reaction in this scenario favors the formation of the double bond between the first and second carbon atoms due to the stability provided by the phenyl group. Therefore, 2-phenylpropene is not the major product.
   - **Option 3 (1-phenylpropane-2-ol):** This compound would result from a substitution reaction where the chlorine atom is replaced by a hydroxyl group. However, alcoholic KOH promotes elimination reactions rather than substitution, especially in the presence of a secondary alkyl halide like 1-phenyl-2-chloropropane.
   - **Option 4 (1-phenylpropane-1-ol):** Similar to Option 3, this would require a substitution reaction where the chlorine atom is replaced by a hydroxyl group at the first carbon. Again, alcoholic KOH favors elimination over substitution in this context, making this option incorrect.

In conclusion, the reaction of 1-phenyl-2-chloropropane with alcoholic KOH predominantly leads to the formation of 1-phenylpropene due to the mechanism of E2 elimination and the stabilization provided by the phenyl group.