Question

The product can also be obtained by the reaction of which of the following?
 \[Sodium{\text{ }}phenoxide\xrightarrow[{\Pr essure}]{{C{O_2}}}\,A\,\xrightarrow{{{H_3}{O^ + }}}\,o - hydroxybenzoic\,acid\]
A.$Phenol\, + \,chloroform\, + alkali$
B.$Phenol\, + \,pyrene\, + alkali$
C.$Phenol\, + \,acetyl\,chloride\, + AlC{l_3}$
D.$Phenol\, + \,methyl\,chloride\, + AlC{l_3}$

Answer 1

Hint:The conversion of sodium phenoxide to o-hydroxybenzoic acid is known as Kolbe’s reaction. The o-hydroxybenzoic acid is also called salicylic acid. From the above option try to solve each with a mechanism. The starting reactant can be phenol, benzene ring having OH group by the reaction of phenol with a variety of reagent as given above we can make products as aldehyde also.

Complete step-by-step answer:The reaction of phenol with chloroform and alkali is a very famous named reaction known as Reimer-Tiemann reaction. In this reaction we get salicylaldehyde which is also called as $2 - hydroxybenzaldehyde$ . Let’s start the reaction by phenol and other reagents.
A.

Next we have the reaction between phenol with pyrene in alkali, use of alkali is done to make the medium basic hence, in this reaction the products we get are salicylic acid that is \[o - hydroxybenzoic\,acid\] .
B.

When the reaction takes place between phenol and acetyl chloride in the presence of $AlC{l_3}$ then we get an acetyl group at ortho and para positions. The activity is due to the high electron density of phenol at ortho position and para position hence in the reaction we get these two substitutions. This reaction is called acetylation reaction.
C.

Last reaction is the reaction of methylation. It means there is an introduction of methyl groups at ortho and para position because of high electron density at these two positions. The para product is found to be major. It means when reactions get completed the amount of para product is high.
D.

Among all the reactions we get the same product as \[o - hydroxybenzoic\,acid\] or we can say it as salicylic acid only in option B.

Option B is correct.

Note:While solving these types of questions try to write the reaction taking place in all four options. By this it's easier to choose the correct option. There are two reactions, acetylation and alkylation in case of acetylation an acetyl group gets attached at ortho and para position while in case of alkylation an alkyl group gets attached to these positions.