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Question: What does phenol form by reaction with \[CHC{l_3}\] in the presence of \(NaOH\) or \(KOH\)?...

What does phenol form by reaction with CHCl3CHC{l_3} in the presence of NaOHNaOH or KOHKOH?

Explanation

Solution

We have to remember that the phenol is an aromatic compound having the chemical formula C6H5OH{C_6}{H_5}OH. And this compound contains one phenyl ring which makes a bond with one hydroxyl group. And the chloroform is a chemical compound with molecular formula, CHCl3CHC{l_3}. This is a colorless liquid with a strong odor or a pungent smell. The sodium hydroxide and potassium hydroxide act as a base.

Complete answer:
We need to know that when the phenol is reacted with chloroform and potassium hydroxide which act as a base (or sodium hydroxide) then there is a formation of aldehyde which is known as salicylaldehyde with potassium chloride and water. This reaction is known as the Reimer – Tiemann reaction. Here, the aldehydic group (−CHO)\left( { - CHO} \right)attacks the ortho position of the phenol. The reaction can be written as,

Here the reaction starts with the deprotonation of CHCl3CHC{l_3} in the presence of a strong base and forms a chloroform carbanion. This will readily undergo the alpha elimination reaction which gives dichlorocarbene.
The phenol also undergoes the deprotonation by aqueous hydroxide, forming phenoxide which is negatively charged. And this negative charge will delocalize the benzene ring which becomes highly nucleophilic. Hence, there undergoes a nucleophilic attack on the dichlorocarbene and forms an intermediate which is known as dichloromethyl substituted phenol. At last, this intermediate undergoes basic hydrolysis and there is a formation of ortho – hydroxy benzaldehyde which is known as salicylaldehyde.

Note:
We have to know that when the phenol is reacted with chloroform in the presence of bases, there is a formation of salicylaldehyde by attacking the aldehydic group on the ortho position of the phenol. And this reaction is known as the Reimer – Tiemann reaction. Here, the first step is the deprotonation of chloroform. The carbene is highly electron deficient. Because, the chlorine atom has an electron – withdrawing nature.