Question

Why do aqueous solutions of phenol give acidic solutions?

Answer 1

Hint: The conjugate base of the phenol is Phenoxide particle. This means that it is formed from an acid that has given up its hydrogen. Therefore this acid is the phenol molecule. This implies the hydrogen of the hydroxyl leaves and solely $ {O^ - } $ remains, forming the ‘oxide ion’ a part of the Phenoxide ion.

Complete answer:
The compounds in which hydroxyl groups directly connected to an aromatic carbocyclic nucleus are referred to as phenols. On reacting with active metals like $ Na $ , $ K $ phenol form Phenoxide ion which means that phenols are acidic in nature. Moreover, Phenoxide ions are more stable as a result of resonance and also the delocalization of charge density. Therefore this shows those phenols are acidic in nature.
 $ {C_6}{H_5}OH\left( {aq} \right) + {H_2}O\left( l \right) \to {C_6}{H_5}{O^ - } + {H_3}{O^ + } $
 $ p{K_a} = 8.00 $
Phenols lose their hydrogen ions from its $ OH $ bond, as by losing the hydrogen ion; the Phenoxide ion becomes stable. Though it is a weak acid, it is in equilibrium with the Phenoxide anion $ {C_6}{H_5}{O^ - } $ which is alternatively known as Phenoxide. As compared to aliphatic compounds containing $ OH $ group, Phenoxide is more stable due to resonance stabilization of Phenoxide ion by the aromatic ring. Negative charge of oxygen atom delocalized on the ortho or para carbon atoms due to which the stability increases.
Due to the delocalization of the charge the Phenoxide ion is more stable and it is also an ambident nucleophile. Ambident nucleophiles are those nucleophiles in which there are two nucleophilic sites. In other words, the nucleophiles which have two sites by which they can attack are termed as ambident nucleophiles.

Note:
Let us substitute the ring by nitro groups which will cause the charge density of the anion to decrease thus increasing the acidic nature. $ 1,3,5 - trinitrophenol $ ; $ 1,3,5 - {\left( {{O_2}N} \right)_3}{C_6}{H_2}OH $ Which is also known as picric acid whose, $ p{K_a} = 0.38 $ which is very strong for a carbon acid.