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When phenol is treated with CHCl3 and NaOH, the product formed is [2002]
- a)Benzaldehyde
- b)Salicylaldehyde
- c)Salicylic acid
- d)Benzoic acid
Correct answer is option 'B'. Can you explain this answer?
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When phenol is treated with CHCl3 and NaOH, the product formed is [200...
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When phenol is treated with CHCl3 and NaOH, the product formed is [200...
Explanation:
When phenol is treated with CHCl3 (chloroform) and NaOH (sodium hydroxide), the product formed is salicylaldehyde. Let's understand the reaction step by step:
Step 1: Formation of chloroform:
When chloroform (CHCl3) reacts with sodium hydroxide (NaOH), it undergoes a series of reactions to form sodium formate (HCOONa) and dichlorocarbene (CCl2), which is a reactive intermediate.
CHCl3 + NaOH → HCOONa + CCl2
Step 2: Reaction with phenol:
The formed dichlorocarbene (CCl2) reacts with phenol (C6H5OH) to form the desired product, salicylaldehyde.
CCl2 + C6H5OH → C6H4(OH)CHO
Here, the -CHO group indicates the presence of an aldehyde functional group, and the -OH group indicates the presence of a hydroxyl group. Therefore, the product formed is salicylaldehyde.
Summary:
- When phenol is treated with chloroform and sodium hydroxide, chloroform reacts with sodium hydroxide to form dichlorocarbene as a reactive intermediate.
- The formed dichlorocarbene then reacts with phenol to form salicylaldehyde.
- This reaction is a useful method for the synthesis of salicylaldehyde from phenol.
When phenol is treated with CHCl3 (chloroform) and NaOH (sodium hydroxide), the product formed is salicylaldehyde. Let's understand the reaction step by step:
Step 1: Formation of chloroform:
When chloroform (CHCl3) reacts with sodium hydroxide (NaOH), it undergoes a series of reactions to form sodium formate (HCOONa) and dichlorocarbene (CCl2), which is a reactive intermediate.
CHCl3 + NaOH → HCOONa + CCl2
Step 2: Reaction with phenol:
The formed dichlorocarbene (CCl2) reacts with phenol (C6H5OH) to form the desired product, salicylaldehyde.
CCl2 + C6H5OH → C6H4(OH)CHO
Here, the -CHO group indicates the presence of an aldehyde functional group, and the -OH group indicates the presence of a hydroxyl group. Therefore, the product formed is salicylaldehyde.
Summary:
- When phenol is treated with chloroform and sodium hydroxide, chloroform reacts with sodium hydroxide to form dichlorocarbene as a reactive intermediate.
- The formed dichlorocarbene then reacts with phenol to form salicylaldehyde.
- This reaction is a useful method for the synthesis of salicylaldehyde from phenol.
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When phenol is treated with CHCl3 and NaOH, the product formed is [2002]a)Benzaldehydeb)Salicylaldehydec)Salicylic acidd)Benzoic acidCorrect answer is option 'B'. Can you explain this answer?
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