Question

1-phenyl-2-chloropropane on treating with alc. KOH gives mainly.
A. 1-phenyl propene
B. 2-phenyl propene
C. 1-phenylpropan-2-ol
D. 1-phenyl-propan-1-ol

Answer 1

Hint: In the given reaction, dehydrohalogenation is taking place which means the chloride ion is removed which further acts as a nucleophile and attacks the metal cation to form salt. In this reaction the alcoholic potassium hydroxide acts as a base. The main product formed will be an unsaturated hydrocarbon compound.

Complete step by step answer:
In the given reaction, a dehydrohalogenation reaction is taking place. In this reaction, the alcoholic KOH is acting as a base where potassium carries a positive charge and the hydroxyl group carries the negative charge. The hydroxyl anion will abstract the hydrogen present at the alpha position of 1-phenyl-2-chloropropane to eliminate water. After the removal of the hydrogen the charge will shift to form a double bond and chlorine will be removed as chloride anion. The chloride anion will further act as a nucleophile and will attack the positive charged potassium to form potassium chloride. The resulting main product which will be formed is 1-phenyl propene
The reaction between 1-phenyl-2-chloropropane with alc. KOH is shown below.

In this reaction, 1-phenyl-2-chloropropane reacts with alcoholic potassium hydroxide to form 1-phenyl propene.
Therefore, the correct option is A.

Note: The reaction is a type of elimination reaction, as a saturated compound is converted to an unsaturated compound. If in place of alcoholic potassium hydroxide, aqueous potassium hydroxide solution was used, the resulting product which will be formed is alcohol.