Q 1 :

Given below are two statements:

Statement (I): Aminobenzene and aniline are same organic compounds.

Statement (II): Aminobenzene and aniline are different organic compounds.

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

     

  • Statement I is correct but Statement II is incorrect

     

  • Statement I is incorrect but Statement II is correct

     

  • Both Statement I and Statement II are incorrect

     

(2)

Aniline is IUPAC accepted name of aminobenzene.

 



Q 2 :

X g of ethylamine is subjected to reaction with NaNO2/HCl followed by water; evolved dinitrogen gas which occupied 2.24 L volume at STP. X is ______ ×10-1 g.                 [2024]



(45)

C2H5NH2→NaNO2+HClC2H5N2+

                    →H2OC2H5OH+N2

Moles of nitrogen

(nN2)=Volume at STP in L22.4=2.2422.4 mol=0.1 mol

From reaction it is clear that moles of ethylamine (nC2H5NH2)=nN2=0.1mol.

Mass of ethylamine = nC2H5NH2× molar mass of ethylamine

=0.1 mol×45 g mol-1=45×10-1 g



Q 3 :

X g of ethanamine was subjected to reaction with NaNO2/HCl followed by hydrolysis to liberate N2 and HCl. The HCl generated was completely neutralised by 0.2 moles of NaOH. X is ______ g.               [2024]



(9)

   HCl+NaOH→NaCl+H2O

   By stoichiometry

   nHCl=nNaOH=0.2 mol

   C2H5NH2→NaNO2+HCl

   C2H5N2+Cl-→H2OC2H5OH+N2+HCl

   nC2H5NH2=nHCl=0.2 mol

   WC2H5NH2 (X)=MC2H5NH2×nC2H5NH2=45×0.2 g=9 g

 



Q 4 :

Identify B formed in the reaction.                                           [2024]

Cl-(CH2)4-Cl→excess NH3A→NaOHB+H2O+NaCl

  •  

  • H2N-(CH2)4-NH2

     

  • ClN+H3-(CH2)4-N+H3Cl-

     

  •  

(2)

In Hoffmann ammonolysis reaction excess of NH3 forms primary amine. Since NH3 is in excess, it replaced both Cl of substrate.

 



Q 5 :

Find out the major product formed from the following reaction. [Me:-CH3]                               [2024]

  •  

  •  

  •  

  •  

(3)



Q 6 :

Phthalimide is made to undergo the following sequence of reactions.                                                     [2024]

Phthalimide→(i) KOH(ii) BenzylchlorideP

Total number of π bonds present in product 'P' is/are ______



(8)

 



Q 7 :

An amine (X) is prepared by ammonolysis of benzyl chloride. On adding p-toluenesulphonyl chloride to it, the solution remains clear. The molar mass of the amine (X) formed is ______ g mol-1.

(Given molar mass in g mol-1: C : 12, H : 1, O : 16, N : 14)                                [2024]



(287)

Ammonolysis of alkyl halides produces a mixture of primary, secondary and tertiary amines and quaternary ammonium salt. Major product formed depends upon relative amount of alkyl halide and ammonia.

As on adding p-toluenesulphonyl chloride (Hinsberg reagent) solution remains clear, this means the amine X is tertiary amine. Formula of X is (C6H5CH2)3N.
Molar mass of

X=[3×(7×12+7×1)+14] g mol-1=287 g mol-1

 



Q 8 :

Match the compounds (List–I) with the appropriate Catalyst/Reagents (List–II) for their reduction into corresponding amines.       

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

  • A–III, B–II, C–IV, D–I

     

  • A–III, B–IV, C–II, D–I

     

  • A–II, B–I, C–III, D–IV

     

  • A–II, B–IV, C–III, D–I

     

(2)

 



Q 9 :

The correct order in aqueous medium of basic strength in case of methyl substituted amines is:             [2023]
 

  • NH3>Me3N>MeNH2>Me2NH

     

  • Me2NH>Me3N>MeNH2>NH3

     

  • Me2NH>MeNH2>Me3N>NH3

     

  • Me3N>Me2NH>MeNH2>NH3

     

(3)

The observed basic strength order in aqueous medium is

Me2NH>MeNH2>Me3N>NH3



Q 10 :

Match List I with List II                   [2023]

  List I (Amines)   List II (pKb)
A. Aniline I. 3.25
B. Ethanamine II. 3.00
C. N-Ethylethanamine III. 9.38
D. N, N-Diethylethanamine IV. 3.29

Choose the correct answer from the options given below:

  • A-III, B-IV, C-II, D-I

     

  • A-III, B-II, C-IV, D-I

     

  • A-I, B-IV, C-II, D-III

     

  • A-III, B-II, C-I, D-IV

     

(1)

Greater the basic strength, smaller the pKb value.



Q 11 :

The major product ‘P’ for the following sequence of reactions is:  [2023]

  •  

  •  

  •  

  •  

(4)



Q 12 :

Reaction of propanamide with Br2/KOH (aq) produces:  [2023]

  • Propanenitrile

     

  • Ethylamine

     

  • Ethylnitrile

     

  • Propylamine

     

(2)



Q 13 :

The major product formed in the following reaction is               [2023]

  •  

  •  

  •  

  •  

(4)



Q 14 :

The major products A and B from the following reactions are:          [2023]

  •  

  •  

  •  

  •  

(4)



Q 15 :

In the reaction given below:

The product 'X' is:                       [2023]

  •  

  •  

  •  

  •  

(1)



Q 16 :

In the reaction given below:

'A' is                                                           [2023]

  •  

  •  

  •  

  •  

(4)



Q 17 :

How many of the transformations given below would result in aromatic amines?      [2023]



(3)



Q 18 :

Number of isomeric aromatic amines with molecular formula C8H11N, which can be synthesized by Gabriel Phthalimide synthesis is _________.           [2023]



(6)

DU=2C+2+N-X-H2

for C8H11N

DU=16+2+1-112=4



Q 19 :

When a concentrated solution of sulphanilic acid and 1-naphthylamine is treated with nitrous acid (273 K) and acidified with acetic acid, the mass (g) of 0.1 mole of product formed is:

(Given molar mass in g mol-1: H : 1, C : 12, N : 14, O : 16, S : 32)                  [2025]

  • 343

     

  • 330

     

  • 33

     

  • 66

     

(3)

 



Q 20 :

Compound ‘P’ undergoes the following sequence of reactions:                 [2026]

  •  

  •  

  •  

  •  

(4)



Q 21 :

The major product of the following reaction is:

  •  

  •  

  •  

  •  

(3)



Q 22 :

The number of compounds from the following which can undergo reaction with Br2/KOH (alcoholic) to give respective products and these respective products can also be obtained separately by Gabriel phthalimide reaction is:              [2026]

[IMAGE 260]

  • 5

     

  • 4

     

  • 3

     

  • 6

     

(3)

Secondary amides do not give Hoffmann bromamide degradation reaction.

[IMAGE 261]

 



Q 23 :

Which statements are True?

A. In Hoffmann bromamide degradation, 4 moles of NaOH and 2 moles of Br2 are consumed per mole of an amide.

B. Hoffmann bromamide reaction is not given by alkyl amides.

C. Primary amines can be synthesized by Hoffmann bromamide degradation.

D. Secondary amide on reaction with Br2 and NaOH will give secondary amine.

E. The by-products of Hoffmann degradation are Na2CO3, NaBr and H2O.

Choose the correct answer from the options given below:                     [2026]

  • A, C and E only

     

  • B, C and D only

     

  • C and E only

     

  • C, D and E only

     

(3)

Ar/RCONH21° amide+Br2+4NaOH→Ar/RNH21° amine+Na2CO3+2NaBr+2H2O

Secondary amides do not give this reaction



Q 24 :

Arrange the following compounds according to increasing order of boiling points.

n-C4H9OH (A), n-C4H9NH2 (B), n-C4H10 (C) and C2H5NHC2H5 (D).           [2026]

  • C<B<A<D

     

  • D<C<B<A

     

  • C<D<B<A

     

  • D<B<A<C

     

(3)

[IMAGE 264]