Q 1 :

Phenolic group can be identified by a positive:                   [2024]

  • Phthalein dye test

     

  • Lucas test

     

  • Tollen's test

     

  • Carbylamine test

     

(1)

Phthalein dye test is a test for phenolic group. For example phenol condense with phthalic anhydride in the presence of concentrated H2SO4 to give phenolphthalein which gives a dark pink color with NaOH solution.

 



Q 2 :

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R:

Assertion A: Aryl halides cannot be prepared by replacement of hydroxyl group of phenol by halogen atom.

Reason R: Phenols react with halogen acids violently.

In the light of the above statements, choose the most appropriate from the options given below:                [2024]

  • A is false but R is true

     

  • Both A and R are true and R is the correct explanation of A

     

  • A is true but R is false

     

  • Both A and R are true but R is NOT the correct explanation of A

     

(3)

Due to resonance, there is partial double bond character in O-H bond of phenol. This makes the O-H bond difficult to break. Hence it does not react with halogen acids to give aryl halides.

 



Q 3 :

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R:

Assertion A: pKa value of phenol is 10.0 while that of ethanol is 15.9.

Reason R: Ethanol is stronger acid than phenol. 

In the light of the above statements, choose the correct answer from the options given below:             [2024]

  • A is true but R is false.

     

  • Both A and R are true but R is NOT the correct explanation of A.

     

  • Both A and R are true and R is the correct explanation of A.

     

  • A is false but R is true.

     

(1)

Phenols are stronger acids than alcohols because phenoxide ion is resonance stabilised whereas there is no resonance in alkoxide ion.

 



Q 4 :

Given below are two statements:

Statement I: Bromination of phenol in solvent with low polarity such as CHCl3 or CS2 requires Lewis acid catalyst.

Statement II: The Lewis acid catalyst polarises the bromine to generate Br+.

In the light of the above statements, choose the correct answer from the options given below:                    [2024]

  • Statement I is true but Statement II is false.

     

  • Both Statement I and Statement II are true.

     

  • Both Statement I and Statement II are false.

     

  • Statement I is false but Statement II is true.

     

(4)

The usual halogenation of benzene takes place in the presence of a Lewis acid, such as FeBr3, which polarises the halogen molecule. In case of phenol, the polarisation of bromine molecule takes place even in the absence of Lewis acid. It is due to the highly activating effect of –OH group attached to the benzene ring, due to which charge density on ring increases and polarization of Br2 occurs even in absence of Lewis acid.

 



Q 5 :

Match List - I with List - II                               [2024]

Choose the correct answer from the options given below:

  • (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

     

  • (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

     

  • (A)-(II), (B)-(I), (C)-(III), (D)-(IV)

     

  • (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

     

(4)

(A) Kolbe reaction in presence of NaOH;CO2 and HCl.

(B) Reimer-Tiemann reaction

(C) Oxidation of phenol

(D) Williamson ether synthesis

 



Q 6 :

Phenol treated with chloroform in the presence of sodium hydroxide, which is further hydrolyzed in the presence of an acid, results in                 [2024]

  • Benzene-1,2-diol

     

  • Benzene-1,3-diol

     

  • 2-Hydroxybenzaldehyde

     

  • Salicylic acid

     

(3)

This is Reimer-Tiemann formylation reaction. The product is 2-hydroxybenzaldehyde (salicylaldehyde).

 



Q 7 :

Salicylaldehyde is synthesized from phenol, when reacted with.                        [2024]

  •  

  • CO2,NaOH

     

  • CCl4,NaOH

     

  • HCCl3,NaOH

     

(4)

1.  Formyl chloride (HCOCl) is unstable and does not exist, hence it cannot be used as a reagent.

2.  This is Kolbe's reaction.

3.  This is Reimer-Tiemann carboxylation reaction.

4.  This is Reimer-Tiemann formylation reaction.

 



Q 8 :

Match List - I with List - II.                                                                        [2024]

  List - I (Reactants)   List - II (Products)
(A) Phenol, Zn/Δ (I) Salicylaldehyde
(B) Phenol, CHCl3, NaOH, HCl (II) Salicylic acid
(C) Phenol, CO2, NaOH, HCl (III) Benzene
(D) Phenol, Conc. HNO3 (IV) Picric acid

 

Choose the correct answer from the options given below:

  • (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

     

  • (A)-(III), (B)-(I), (C)-(II), (D)-(IV)

     

  • (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

     

  • (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

     

(2)

(B) This is Reimer-Tiemann formylation reaction

(C) This is Kolbe’s reaction

(D) Nitration of phenol



Q 9 :

Which one of the following reactions is NOT possible?                   [2024]

  •  

  •  

  •  

  •  

(3)

 



Q 10 :

Given below are two statements:                                                                                        [2024]

Statement I: Picric acid is 2, 4, 6 - trinitrotoluene.
Statement II: Phenol - 2, 4 - disulphonic acid is treated with Conc. HNO3 to get picric acid.

In the light of the above statements, choose the most appropriate answer from the options given below:

  • Both Statement I and Statement II are incorrect

     

  • Statement I is correct but Statement II is incorrect

     

  • Both Statement I and Statement II are correct

     

  • Statement I is incorrect but Statement II is correct

     

(4)

Picric acid is prepared by treating phenol first with concentrated sulphuric acid which converts it to phenol-2,4-disulphonic acid. Because of the –M effect of the SO3H group, electron density on the ring decreases and hence ring is less prone to getting oxidized. Phenol-2,4-disulphonic acid is then treated with concentrated nitric acid to get 2, 4, 6-trinitrophenol.

 



Q 11 :

In Reimer-Tiemann reaction, phenol is converted into salicylaldehyde through an intermediate. The structure of the intermediate is ______.            [2024]

  •  

  •  

  •  

  •  

(1)

 



Q 12 :

Match List-I with List-II.                                                  [2024]

Choose the correct answer from the options given below:

  • (A) - (III), (B) - (II), (C) - (I), (D) - (IV)

     

  • (A) - (IV), (B) - (II), (C) - (III), (D) - (I)

     

  • (A) - (I), (B) - (IV), (C) - (II), (D) - (III)

     

  • (A) - (II), (B) - (IV), (C) - (I), (D) - (III)

     

(4)


(A) This reaction is diazotization.
(B) This is oxidation of phenol.
(C) This is Reimer Tiemann formylation reaction.
(D) This is Kolbe’s reaction.

 



Q 13 :

Consider the given chemical reaction sequence:                                                  [2024]

Total sum of oxygen atoms in Product A and Product B are ______. 



(14)

Sum of oxygen atoms in A and B = 7 + 7 = 14.



Q 14 :

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

(Given molar mass in g mol-1 of C, H, O, Br are 12, 1, 16, 80 respectively)                    [2025]

  • 6.0 g

     

  • 20.44 g

     

  • 10.22 g

     

  • 4.0 g

     

(3)

Moles of phenol=Given massMolar mass=294mol

By stoichiometry of the reaction,

Moles of Br2=3×Moles of phenol=3×294mol

Mass of bromine=Moles of Br2×Molar mass of Br2

Mass of bromine=3×294 mol×160 g mol-1=10.21 g



Q 15 :

Identify the products [A] and [B], respectively in the following reaction:               [2025]

  •  

  •  

  •  

  •  

(1)

 



Q 16 :

The steam volatile compounds among the following are:

Choose the correct answer from the options given below:             [2025]

  • (A) and (B) only

     

  • (A), (B) and (C) only

     

  • (B) and (D) only

     

  • (A) and (C) only

     

(1)

(A) and (B) are steam volatile because of intramolecular hydrogen bonding.

 



Q 17 :

Benzene is treated with oleum to produce compound (X) which when further heated with molten sodium hydroxide followed by acidification produces compound (Y). The compound Y is treated with zinc metal to produce compound (Z).

Identify the structure of compound (Z) from the following option.             [2025]

  •  

  •  

  •  

  •  

(2)

 



Q 18 :

In the cumene to phenol preparation in presence of air, the intermediate is                      [2023]

  •  

  •  

  •  

  •  

(4)



Q 19 :

The increasing order of pK for the following phenols is

(A) 2,4-Dinitrophenol
(B) 4-Nitrophenol
(C) 2,4,5-Trimethylphenol
(D) Phenol
(E) 3-Chlorophenol  

Choose the correct answer from the option given below:          [2023]

  • (A), (E), (B), (D), (C)

     

  • (C), (E), (D), (B), (A)

     

  • (C), (D), (E), (B), (A)

     

  • (A), (B), (E), (D), (C)

     

(4)

Increasing order of pKa means most acidic will be placed first.

–M groups at ortho and para position increases acidic strength in phenol hence (A) is the most acidic.

+M, hyperconjugation, +I decreases Ka value of the compound hence (C) is least acidic.



Q 20 :

The correct order of pK values for the following compounds is:  [2023]

  • a > b > c > d

     

  • b > a > d > c

     

  • b > d > a > c

     

  • c > a > d > b

     

(3)

pKa1Ka ; Ka increases by electron withdrawing group.

Ka=c>a>d>b,  pKa=b>d>a>c



Q 21 :

An organic compound 'A' with empirical formula C6H6O gives sooty flame on burning. Its reaction with bromine solution in low polarity solvent results in high yield of B. B is      [2023]

  •  

  •  

  •  

  •  

(3)



Q 22 :

A compound ‘X’ when treated with phthalic anhydride in presence of concentrated H2SO4 yields ‘Y’. ‘Y’ is used as an acid/base indicator. ‘X’ and ‘Y’ are respectively  [2023]

  • Toluidine, Phenolphthalein

     

  • Salicylaldehyde, Phenolphthalein

     

  • Carbolic acid, Phenolphthalein

     

  • Anisole, Methyl orange

     

(3)