Question

on treating phenol with chloroform in presence of aqueous sodium hydroxide at 340k followed by acid hydrolysis gives the product.
i.write the name of the product.
ii. give the name of the reaction.​

Answer 1

Answer:

When phenol treated with chloroform in presence of aqueous sodium hydroxide at 340 K to form 2− hydroxybenzaldehyde or salicylaldehyde. 

This reaction is known as Reimer-Tiemann reaction.

Explanation:

hope it helps

Answer 2

Answer:

(i)The product formed is Salicylaldehyde.

(ii) The name of the reaction is the Reimer Tiemann reaction.

Explanation:

The reaction proceeds as follows,

$\mathrm{C_6H_5(OH)+CHCl_3+NaOH\xrightarrow[H^{+}]{340K}C_6H_4CHO-2-OH}$         (1)

Where,

C₆H₅(OH)=Phenol

CHCl₃=chloroform

NaOH=sodium hydroxide

C₆H₄CHO-2-OH=Salicylaldehyde

• The product formed is Salicylaldehyde.
• A type of substitution reaction called the Reimer-Tiemann reaction is named for the chemists Karl Reimer and Ferdinand Tiemann. The process is used to ortho-formylate C₆H₅(OH) (phenols). O-hydroxybenzaldehyde is created when phenols, or C₆H₅(OH), are treated with CHCl₃ (chloroform) in the presence of NaOH (sodium hydroxide). This is done by adding an aldehyde group (-CHO) to the ortho position of the benzene ring. The Reimer Tiemann reaction is the common name for the reaction.

#SPJ2