The electrophile involved in the reaction is.
${C_6}{H_5}OH + CHC{l_3} + NaOH \to {C_6}{H_4}(CHO)OH$
A) Dichloromethyl cation
B) Dichlorocarbene
C) Trichloromethyl anion
D) Formyl cation
Answer
585.9k+ views
Hint: In order to solve this problem, first we need to know what an electrophile is. Electrophile is a molecule that forms a bond to its reaction partner which is the nucleophile by accepting both the bonding electrons from that reaction partner. Electrophilic reagents are said to be Lewis acids. Carbon dioxide is said to be an example of an electrophile.
Complete answer:
In this reaction, phenol reacts with chloroform and sodium hydroxide to give benzoic acid.
${C_6}{H_5}OH + CHC{l_3} + NaOH \to {C_6}{H_4}(CHO)OH$
The reaction for the electrophile can be given by.
$NaOH + CC{l_3}^ - \to :CC{l_2} + NaC{l^ - } + {H_2}O$
Additional information:
The mechanism of the Reimer – Tiemann reaction is as follows. Chloroform which is on the reaction side is deprotonated by a strong base which can be sodium hydroxide or potassium hydroxide to form the chloroform carbanion which will quickly alpha-eliminate to give dichlorocarbene.
The hydroxide can also deprotonate the phenol to give a negatively charged phenoxide. The negative charge is delocalized into the aromatic ring, making it more nucleophilic. Nucleophilic attack on the dichlorocarbene gives an intermediate dichloromethyl substituted phenol. After hydrolysis, the desired product is formed.
Hence the correct option is B which is dichlorocarbene.
Note: Dichlorocarbene is the reactive intermediate with the chemical formula $CC{l_2}$ . It is a common intermediate which is being produced from chloroform. Salicylaldehyde can be prepared from phenol and chloroform by heating with either potassium hydroxide or sodium hydroxide this reaction is called the Reimer - Tiemann reaction. Reimer–Tiemann reaction is a chemical reaction which is used for the formation of phenols at the ortho position.
Complete answer:
In this reaction, phenol reacts with chloroform and sodium hydroxide to give benzoic acid.
${C_6}{H_5}OH + CHC{l_3} + NaOH \to {C_6}{H_4}(CHO)OH$
The reaction for the electrophile can be given by.
$NaOH + CC{l_3}^ - \to :CC{l_2} + NaC{l^ - } + {H_2}O$
Additional information:
The mechanism of the Reimer – Tiemann reaction is as follows. Chloroform which is on the reaction side is deprotonated by a strong base which can be sodium hydroxide or potassium hydroxide to form the chloroform carbanion which will quickly alpha-eliminate to give dichlorocarbene.
The hydroxide can also deprotonate the phenol to give a negatively charged phenoxide. The negative charge is delocalized into the aromatic ring, making it more nucleophilic. Nucleophilic attack on the dichlorocarbene gives an intermediate dichloromethyl substituted phenol. After hydrolysis, the desired product is formed.
Hence the correct option is B which is dichlorocarbene.
Note: Dichlorocarbene is the reactive intermediate with the chemical formula $CC{l_2}$ . It is a common intermediate which is being produced from chloroform. Salicylaldehyde can be prepared from phenol and chloroform by heating with either potassium hydroxide or sodium hydroxide this reaction is called the Reimer - Tiemann reaction. Reimer–Tiemann reaction is a chemical reaction which is used for the formation of phenols at the ortho position.
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