The Reaction of Chloroform with Alcoholic KOH and P-Toluidine

The reaction of chloroform (CHCl3) with alcoholic potassium hydroxide (KOH) and p-toluidine (C7H9N) forms an organic compound known as p-chloroaniline (C6H4ClNH2). This reaction is a nucleophilic substitution reaction and involves the replacement of the chlorine atom in chloroform with the p-toluidine group.

Nucleophilic Substitution Reaction
A nucleophilic substitution reaction occurs when a nucleophile attacks an electron-deficient carbon atom, resulting in the displacement of a leaving group. In this case, the p-toluidine acts as the nucleophile and the chlorine atom in chloroform acts as the leaving group.

Reaction Mechanism
The reaction proceeds through the following steps:

1. Formation of the alkoxide ion: Alcoholic KOH (potassium hydroxide dissolved in alcohol) reacts with chloroform to form the alkoxide ion (CHCl2O-). This is achieved by the deprotonation of the hydrogen atom in chloroform by the hydroxide ion (OH-).

2. Nucleophilic attack: The alkoxide ion acts as a nucleophile and attacks the carbon atom in p-toluidine. The lone pair of electrons on the nitrogen atom in p-toluidine is attracted to the electron-deficient carbon atom in the alkoxide ion.

3. Formation of the intermediate: The nucleophilic attack leads to the formation of an intermediate compound in which the p-toluidine is attached to the alkoxide group. This intermediate compound is highly unstable.

4. Rearrangement: The rearrangement occurs due to the instability of the intermediate compound. The chlorine atom in chloroform is displaced from the carbon atom and forms a new bond with the nitrogen atom in p-toluidine.

5. Formation of p-chloroaniline: The final product of the reaction is p-chloroaniline, which is formed by the substitution of the chlorine atom in chloroform with the p-toluidine group.

Conclusion
In summary, the reaction of chloroform with alcoholic KOH and p-toluidine involves a nucleophilic substitution reaction. The alkoxide ion formed from alcoholic KOH attacks the carbon atom in p-toluidine, resulting in the displacement of the chlorine atom in chloroform. This reaction leads to the formation of p-chloroaniline as the final product.

P-toluidine + alc.KOH +CHCl3 ———>
. p-isocyanomethyl benzene