Q: Phenol is heated with a solution of mixture of KBr and KBrO3. The major product obtained in the above reaction is

Br2 is formed by a redox reaction. 5Br−+BrO3−+6H+⟶3Br2+3H2O −OH group is the activating group and thus activates the benzene towards electrophilic substitution reaction at o-and p-positions giving yellowish white precipitate of 2,4,6-tribromophenol.

Q: The structure of the compound that gives a tribromo derivative on treatment with bromine water is

m-cresol due to phenoxide ion in H2O solvent, gives tribromoderivative at all ortho and para positions.

Q: Replacement of Cl of chlorobenzene to give phenol requires drastic conditions but chlorine of 2,4-dinitrochlorobenzene is readily replaced. This is because

(d) NO2 withdraws electrons from ortho/para positions.

Q: What amount of bromine will be required to convert 2 g of phenol into 2,4,6-tribromophenol?

94gC6H5OH+3×160g3Br2⟶2, 4, 6,−tribromophenol94 g of phenol require Br2=3×160 g2 g of phenol will require Br2=943×160×2=10.22 g

Question 1

Explain why is ortho nitrophenol more acidic than ortho methoxyphenol?

Accepted Answer

In ortho nitro phenol, electron withdrawing nitro group is present in ortho position, the electron density of O−H bond is reduced.This makes removal of proton easy. After deprotonation, o-nitrophenoxide ion is resonance stabilized. Hence, ortho nitrophenol is a stronger acid.In ortho methoxyphenol, electron releasing methoxy group is present in ortho position, the electron density of O−H bond is increased.This makes removal of proton difficult. Hence, ortho methoxyphenol is less acidic.

Question 2

Phenol on treatment with CO2​ in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with (CH3​CO)2​O in the presence of catalytic amount of H2​SO4​ produces:

Accepted Answer

The very first reaction in the above road map looks like Kolbe's reaction which results to salicylic acid as

The salicylic acid with acetic anhydride [(CH3​CO)2​O] in the presence of catalytic amount of conc. H2​SO4​ undergoes acylation to produce aspirin as

Aspirin is a non-narcotic analgesic (Pain killer).

Question 3

Phenol is heated with a solution of mixture of KBr and KBrO3​. The major product obtained in the above reaction is

Accepted Answer

Br2​ is formed by a redox reaction.
5Br−+BrO3−​+6H+⟶3Br2​+3H2​O
−OH group is the activating group and thus activates the benzene towards electrophilic substitution reaction at o-and p-positions giving yellowish white precipitate of 2,4,6-tribromophenol.

Question 4

The structure of the compound that gives a tribromo derivative on treatment with bromine water is

Accepted Answer

m-cresol due to phenoxide ion in H2​O solvent, gives tribromoderivative at all ortho and para positions.

Question 5

O2​ gas is passed into hot cumene. The product is then treated with aq.H2​SO4​. The final product(s) formed is/are

Accepted Answer

(c) phenol and acetone

Question 6

Replacement of Cl of chlorobenzene to give phenol requires drastic conditions but chlorine of 2,4-dinitrochlorobenzene is readily replaced. This is because

Accepted Answer

(d) NO2​ withdraws electrons from ortho/para positions.

Question 7

What amount of bromine will be required to convert 2 g of phenol into 2,4,6-tribromophenol?

Accepted Answer

94gC6​H5​OH​+3×160g3Br2​​⟶2, 4, 6,−tribromophenol94 g of phenol require Br2​=3×160 g2 g of phenol will require Br2​=943×160​×2=10.22 g

Question 8

Benzoquinone is prepared by reaction of phenol with:

Accepted Answer

Oxidation of phenol with chromic acid produces a conjugated diketone known as benzoquinone.
Chromic acid is also called tetrachromic acid or chromic (IV) acid.
It is usually a mixture made by adding concentrated H2​SO4​ to a dichromate.

Question 9

The major product(s) of the following reaction is/are

Accepted Answer

-OH group is activating group and is 0 - and ρ-directingAlso, −SO3​H is a better leaving group and is knocked out by Br−.

Question 10

In the following reaction, the product (s) formed is/are

Accepted Answer

Phenolic compounds in alkaline solution react with chloroform (CHCl3​) at a temperature lower than that of CHCl3​ to form ortho-isomer as the major product (due to greater stability resulting from intramolecular hydrogen bonding).Maior as stable due to intramolecular H-bondingThus, (b) and (d) are correct.

Question 11

Complete the following with appropriate structures :

Accepted Answer

Reimer-Tiemann reaction

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