Give reasons for the following:
(a) Protanation of Phenols is difficult whereas ethanol easily undergoes protonation.
(b) Boiling point of ethanol is higher than that of dimethyl ether.
(c) Anisole on reaction with HI gives phenol and $C H_{3} - I$ as main products and not iodobenzene and $C H_{3} O H$ .

Open in App

Solution

(a) In phenols, the lone pair of electrons on the oxygen atom are delocalised over the benzene ring due to resonance and hence are not easily available for protonation. In contrast, in ethanol, the lone pairs on oxygen atom are localised due to the absence of resonance and hence are easily available for protonation.

(b) Ethanol has intermolecular H-bonding due to the presence of a hydrogen attached to the electron negative oxygen atom.
Consequently, a large amount of energy is required to break these hydrogen bonds. Therefore, the boiling point of ethanol is higher than that of dimethyl ether.

(c) Protonation of anisole gives methyl phenyloxonium ion,

$C_{6} H_{5} - + O | H - C H_{3}$

In this ion, the bond between O and $C H_{3}$ is weaker than the bond between $O - C_{6} H_{5}$ which has partial double bond character.

Therefore, attack by $I^{-}$ ion exclusively breaks the weaker the $O - C H_{3}$ bond forming methyl ionide and phenol.

Suggest Corrections

Similar questions

Q. Give reasons for the following:
(a) Protanation of Phenols is difficult whereas ethanol easily undergoes protonation.
(b) Boiling point of ethanol is higher than that of dimethyl ether.
(c) Anisole on reaction with HI gives phenol and

C H_{3} - I

as main products and not iodobenzene and

C H_{3} O H

Give reasons for the following:

(i) o-nitrophenol is more acidic than o-methoxyphenol.

(ii) Butan-1-ol has a higher boiling point than diethyl ether.

(iii) $(C H_{3})_{3} C - O - C H_{3}$ on reaction with HI gives $(C H_{3})_{3} C I -$ and $C H_{3} - O H$ as the main products and not $(C H_{3})_{3}$ C–OH and $C H_{3} - I$ .

Q. Phenyl methyl ether (anisole) reacts with HI to give phenol and methyl iodide and not iodobenzene and methyl alcohol because___________________.

Q. (a) How do you convert the following :
(i) Phenol to Anisole
(ii) Ethanol to Propan-2-ol
(b) Write mechanism of the following reaction :

C_{2} H_{5} O H H_{2} S O_{4} / 443 K - ------- \to C H_{2} = C H_{2} + H_{2} O

Q. Give reasons of the following.
(i) Phenol is more acidic than methanol.
(ii) The C-O-H bond angle in alcohols is slightly less than the tetrahedral angle

(109^{\circ} 28^{^{'}})

.
(iii)

(C H_{3})_{3} C - O - C H_{3}

on reaction with HI gives

(C H_{3})_{3} C - I a n d C H_{3} - O H

as the main products and not

(C H_{3})_{3} C - O H a n d C H_{3} - I

Join BYJU'S Learning Program

Give reasons for the following: (a) Protanation of Phenols is difficult whereas ethanol easily undergoes protonation. (b) Boiling point of ethanol is higher than that of dimethyl ether. (c) Anisole on reaction with HI gives phenol and CH3−I as main products and not iodobenzene and CH3OH.