Q 1 :

Given below are two statements:

Statement (I): The NH2 group in Aniline is ortho and para directing and a powerful activating group.

Statement (II): Aniline does not undergo Friedel-Craft's reaction (alkylation and acylation).

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

  • Both Statement I and Statement II are correct

     

  • Both Statement I and Statement II are incorrect

     

  • Statement I is incorrect but Statement II is correct

     

  • Statement I is correct but Statement II is incorrect

     

(1)

Statement (I): Due to its +M effect, -NH2 group increases electron density at ortho and para positions of the benzene ring and hence it is ortho and para directing for electrophilic aromatic substitution reactions.

Statement (II): Aniline forms salt with aluminium chloride, a Lewis acid, used as a catalyst in Friedel-Crafts reaction. Due to this, nitrogen of aniline acquires positive charge and hence acts as a strong deactivating group for further reaction. Hence, Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation).

 



Q 2 :

From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be ______ ×10-1 g.                     [2024]



(95)

   Number of moles of aniline (nC6H5NH2)=Given massMolar mass=6.5593mol

   Maximum 0.07 mol of acetanilide can be obtained from 0.07 mol of aniline.

   Maximum amount of acetanilide = number of moles × molar

   mass of acetanilide = 6.5593×135g=9.5g=95×10-1g



Q 3 :

The product A formed in the following reaction is             [2024]

  •  

  •  

  •  

  •  

(4)

 



Q 4 :

Identify the product (A) in the following reaction.               [2024]

 

  •  

  •  

  •  

  •  

(4)



Q 5 :

9.3 g of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product ‘P’. The mass of product ‘P’ obtained is 26.4 g. The percentage yield is ______ %.                    [2024]



(80)

nC6H5NH2=WC6H5NH2MC6H5NH2=9.3 g93 gmol-1=0.1 mol

nC6H3Br3NH2=WC6H3Br3NH2MC6H3Br3NH2=26.4 g330 gmol-1=0.08 mol

If the reaction was 100% complete, the moles of C6H3Br3NH2 would be 0.1 mol.

Percentage yield of reaction=Amount of product formedAmount of product expected for 100% yield×100

=0.080.1×100=80%



Q 6 :

9.3 g of pure aniline upon diazotisation followed by coupling with phenol gives an orange dye. The mass of orange dye produced (assume 100% yield/conversion) is _______ g. (nearest integer)                           [2024]



(20)

naniline=Given massMolar mass=9.3g93gmol-1=0.1mol

ndye=naniline=0.1mol

Wdye=ndye×Mdye=0.1mol×198gmol-1=19.8g20g



Q 7 :

If 279 g of aniline is reacted with one equivalent of benzenediazonium chloride, the maximum amount of aniline yellow formed will be ______ g. (nearest integer)
(consider complete conversion)                         [2024]



(591)

Number of moles of aniline (nC6H5NH2)

=Given massMolar mass=279g93g mol-1=3mol

C6H5N2C6H4NH2

M=(12×12)+(3×14)+(14×1)+(11×1)=197

Moles of yellow dye = moles of aniline = 3 mol

Mass of yellow dye = number of moles × molar mass

                              =3 mol×197 g mol-1=591 g           



Q 8 :

For reaction                   

The correct order of set of reagents for the above conversion is:             [2025]

  • Br2|FeBr3,H2O(Δ),NaOH

     

  • H2SO4,Ac2O,Br2,H2O(Δ),NaOH

     

  • Ac2O,Br2,H2O(Δ),NaOH

     

  • Ac2O,H2SO4,Br2,NaOH

     

(2)

 



Q 9 :

The sequence from the following that would result in giving predominantly 3, 4, 5-tribromoaniline is:        [2025]

  •  

  •  

  •  

  •  

(3)

 



Q 10 :

Consider the following sequence of reactions:

11.25 mg of chlorobenzene will produce ---- ×10-1 mg of product B.

(Consider the reactions result in complete conversion.)

[Given molar mass of C, H, O, N and Cl as 12, 1, 16, 14 and 35.5 g mol-1 respectively]                 [2025]



(93)

Moles of chlorobenzene (nC6H5Cl) = WC6H5ClMC6H5Cl

=11.25×10-3112.5 mol=10-4 mol

Moles of B = nC6H5Cl=10-4 mol

Mass of B=moles of B×molar mass of B=10-4×93 g

=93×10-1 mg



Q 11 :

Given below are two statements:

Statement I: Pure aniline and other arylamines are usually colourless.

Statement II: Arylamines get coloured on storage due to atmospheric reduction.

In the light of the above statements, choose the most appropriate answer from the options given below:          [2023]

  • Both Statements I and Statements II are correct

     

  • Statements I is incorrect but Statements II is correct

     

  • Both Statements I and Statements II are incorrect

     

  • Statements I is correct but Statements II is incorrect

     

(4)

Arylamines get coloured due to atmospheric oxidation.

 



Q 12 :

Compound A from the following reaction sequence is:             [2023]

  • Salicylic acid

     

  • Phenol

     

  • Benzoic acid

     

  • Aniline

     

(4)



Q 13 :

Identify X in the sequence given:

  •  

  •  

  •  

  •  

(1)



Q 14 :

Which of the following process will produce 2° amine?

  • Gabriel synthesis

     

  • Hoffmann bromamide reaction

     

  • Reduction of carbylamines

     

  • Reduction of nitro compounds

     

(3)

Reduction of RNC will produce RNHCH3.