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Reaction of phenol with chloroform and sodium hydroxide to give o-hydroxybenzaldehyde involves the formation of
- a)Dichlorocarbene
- b)Trichlorocarbene
- c)Chlorine atoms
- d)Chlorone molecules
Correct answer is option 'A'. Can you explain this answer?
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Reaction of phenol with chloroform and sodium hydroxide to give o-hydr...
Formation of o-hydroxybenzaldehyde from phenol, chloroform and sodium hydroxide involves the formation of dichlorocarbene.
Explanation:
The reaction of phenol with chloroform and sodium hydroxide is known as the Reimer-Tiemann reaction. In this reaction, phenol reacts with chloroform and sodium hydroxide to give ortho-hydroxybenzaldehyde as the main product.
The mechanism of this reaction involves the formation of a reactive intermediate called dichlorocarbene. This intermediate is formed by the reaction of chloroform with sodium hydroxide. The reaction can be summarized as follows:
Step 1: Formation of dichlorocarbene
In the first step, chloroform reacts with sodium hydroxide to form dichlorocarbene. This is an electrophilic and highly reactive intermediate.
CHCl3 + NaOH → Cl2C: + NaCl + H2O
Step 2: Attack of phenoxide ion on dichlorocarbene
In the second step, the phenol is deprotonated by sodium hydroxide to form the phenoxide ion. The phenoxide ion then attacks the dichlorocarbene intermediate to form the ortho-hydroxybenzaldehyde.
C6H5O- + Cl2C: → C6H4(OH)CHO + Cl-
In this reaction, the phenoxide ion acts as a nucleophile while the dichlorocarbene acts as an electrophile. The nucleophilic attack of the phenoxide ion on the electrophilic dichlorocarbene leads to the formation of the desired product.
Conclusion:
Thus, the formation of o-hydroxybenzaldehyde from phenol, chloroform, and sodium hydroxide involves the formation of dichlorocarbene as an intermediate. The reaction mechanism is important for understanding the formation of other compounds using the Reimer-Tiemann reaction.
Explanation:
The reaction of phenol with chloroform and sodium hydroxide is known as the Reimer-Tiemann reaction. In this reaction, phenol reacts with chloroform and sodium hydroxide to give ortho-hydroxybenzaldehyde as the main product.
The mechanism of this reaction involves the formation of a reactive intermediate called dichlorocarbene. This intermediate is formed by the reaction of chloroform with sodium hydroxide. The reaction can be summarized as follows:
Step 1: Formation of dichlorocarbene
In the first step, chloroform reacts with sodium hydroxide to form dichlorocarbene. This is an electrophilic and highly reactive intermediate.
CHCl3 + NaOH → Cl2C: + NaCl + H2O
Step 2: Attack of phenoxide ion on dichlorocarbene
In the second step, the phenol is deprotonated by sodium hydroxide to form the phenoxide ion. The phenoxide ion then attacks the dichlorocarbene intermediate to form the ortho-hydroxybenzaldehyde.
C6H5O- + Cl2C: → C6H4(OH)CHO + Cl-
In this reaction, the phenoxide ion acts as a nucleophile while the dichlorocarbene acts as an electrophile. The nucleophilic attack of the phenoxide ion on the electrophilic dichlorocarbene leads to the formation of the desired product.
Conclusion:
Thus, the formation of o-hydroxybenzaldehyde from phenol, chloroform, and sodium hydroxide involves the formation of dichlorocarbene as an intermediate. The reaction mechanism is important for understanding the formation of other compounds using the Reimer-Tiemann reaction.
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Reaction of phenol with chloroform and sodium hydroxide to give o-hydroxybenzaldehyde involves the formation ofa)Dichlorocarbeneb)Trichlorocarbenec)Chlorine atomsd)Chlorone moleculesCorrect answer is option 'A'. Can you explain this answer?
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