Question

When phenol reacts with chloroform and an alkali like$NaOH$, the product formed is salicylaldehyde. If carbon tetrachloride is used in place of chloroform, the product obtained is:

Answer 1

This reaction is called Riemer Tiemann reaction. In this, there is ortho-formylation occurring which takes place after the formation of phenoxide ion. In this type of reaction, there are two products that are formed one is major and the other is minor. The major one will be para and the product which is minor will be para.

Complete step-by-step answer:
In this reaction two products are formed as when the o-position is occupied, then the process of formylation occurs at para position. This gives out salicylaldehyde along with sodium chloride and water as the byproducts. This reaction is used in the conversion of phenol to an o-hydroxy benzaldehyde with the help of chloroform, a base, and acid.

The process begins with the abstraction of the ${H^ + }$ ion form chloroform with the base to form a trichlorocarban ion. Then phenol gets converted to the ion phenoxide which is caused by the acidic proton being removed by$- OH$due to the presence of $NaOH$which is a strong alkali. It loses a chloride ion and form dichlorocarbene which is neutral in nature. The phenol reagent is made to get deprotonated with the help of base and then carbene is attacked by it.

The reaction for the same is as follows:
$phenol + CHC{l_3} + 3NaOH \to salicylaldehyde + 3NaCl + 2{H_2}O$

Now if there was the presence of carbon tetrachloride instead of chloroform then the product now formed would be salicylic acid. The reaction for the same is as follows:

$phenol + CC{l_4} + 4NaOH \to salicylic acid + 4NaCl + 2{H_2}O$

Note: Carbene is a very good electrophile thus it is responsible for the activation of the aromatic ring. This causes the resonance effect to be happy which is an electron donating process and this makes phenoxide ion’s ortho and para positions electron rich. The salicylic acid is also called $2 -$hydroxybenzoic acid.