JEE Main 2021 Chemistry March 16 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


 
 
16
th
 MARCH 2021 | Shift 2
1. 
 
Cl 
A 
Cl 
 
 Identify the reagent(s) 'A' and condition(s) for the reaction 
 (1) A = HCl; Anhydrous AlCl
3
 
 (2) A = HCl, ZnCl
2 
 (3) A = Cl
2
, dark, Anhydrous AlCl
3 
 (4) A = Cl
2
; UV light 
Ans. (4) 
Sol. 
 
 
Cl
2
 
h ? 
Cl 
Cl
2
 
h ? 
Cl 
Cl 
 
 
2. The INCORRECT statement regarding the structure of C
60
 is: 
 (1) It contains 12 six-membered rings and 24 five-membered rings. 
 (2) Each carbon atom forms three sigma bonds. 
 (3) The five-membered rings are fused only to six-membered rings.  
 (4) The six-membered rings are fused to both six and five-membered rings. 
Ans. (1) 
Sol. it contain 12 five membered ring & 20 six membered ring 
 
3. Match List-I with List-II: 
  List-I       List-II 
  Test/Reagents/Observation(s)   Species detected 
 (a) Lassaigne's Test    (i) Carbon 
 (b) Cu(II) oxide     (ii) Sulphur 
 (c) Silver nitrate     (iii) N, S, P and halogen 
 (d) The sodium fusion extract gives black (iv) Halogen Specifically 
  precipitate with acetic acid & lead acetate 
The correct match is: 
 (1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 (2) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) 
 (3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 (4) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
Ans. (1) 
Sol. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
Page 2


 
 
16
th
 MARCH 2021 | Shift 2
1. 
 
Cl 
A 
Cl 
 
 Identify the reagent(s) 'A' and condition(s) for the reaction 
 (1) A = HCl; Anhydrous AlCl
3
 
 (2) A = HCl, ZnCl
2 
 (3) A = Cl
2
, dark, Anhydrous AlCl
3 
 (4) A = Cl
2
; UV light 
Ans. (4) 
Sol. 
 
 
Cl
2
 
h ? 
Cl 
Cl
2
 
h ? 
Cl 
Cl 
 
 
2. The INCORRECT statement regarding the structure of C
60
 is: 
 (1) It contains 12 six-membered rings and 24 five-membered rings. 
 (2) Each carbon atom forms three sigma bonds. 
 (3) The five-membered rings are fused only to six-membered rings.  
 (4) The six-membered rings are fused to both six and five-membered rings. 
Ans. (1) 
Sol. it contain 12 five membered ring & 20 six membered ring 
 
3. Match List-I with List-II: 
  List-I       List-II 
  Test/Reagents/Observation(s)   Species detected 
 (a) Lassaigne's Test    (i) Carbon 
 (b) Cu(II) oxide     (ii) Sulphur 
 (c) Silver nitrate     (iii) N, S, P and halogen 
 (d) The sodium fusion extract gives black (iv) Halogen Specifically 
  precipitate with acetic acid & lead acetate 
The correct match is: 
 (1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 (2) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) 
 (3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 (4) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
Ans. (1) 
Sol. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 
 
  
4. 
 
CN 
OCH 3 
6 5
3
(i)C H MgBr Ether
(1.0equivalent),dry
Major Pr oduct (ii) H O
X
?
? ? ? ? ? ? ? ? ? ? ? 
 The structure of X is: 
 (1)
 
O 
OCH 3 
C 6H 5 
   (2)
 
OCH 3 
NH 2 
 
 (3)
 
O 
C 6H 5 
C 6H 5 
   (4)
 
C 6H 5 
NH 2 
 
Ans. (1) 
Sol. 
 
 
C ? ?N 
C
6
H
5
–MgBr 
CH–CH
3
 
OCH
3
 
C=NMgBr 
CH–CH
3
 
OCH
3
 
C
6
H
5
 
CH–CH
3
 
OCH
3
 
C–C
6
H
5
 
O 
H
3
O
+
 
 
 
 
Page 3


 
 
16
th
 MARCH 2021 | Shift 2
1. 
 
Cl 
A 
Cl 
 
 Identify the reagent(s) 'A' and condition(s) for the reaction 
 (1) A = HCl; Anhydrous AlCl
3
 
 (2) A = HCl, ZnCl
2 
 (3) A = Cl
2
, dark, Anhydrous AlCl
3 
 (4) A = Cl
2
; UV light 
Ans. (4) 
Sol. 
 
 
Cl
2
 
h ? 
Cl 
Cl
2
 
h ? 
Cl 
Cl 
 
 
2. The INCORRECT statement regarding the structure of C
60
 is: 
 (1) It contains 12 six-membered rings and 24 five-membered rings. 
 (2) Each carbon atom forms three sigma bonds. 
 (3) The five-membered rings are fused only to six-membered rings.  
 (4) The six-membered rings are fused to both six and five-membered rings. 
Ans. (1) 
Sol. it contain 12 five membered ring & 20 six membered ring 
 
3. Match List-I with List-II: 
  List-I       List-II 
  Test/Reagents/Observation(s)   Species detected 
 (a) Lassaigne's Test    (i) Carbon 
 (b) Cu(II) oxide     (ii) Sulphur 
 (c) Silver nitrate     (iii) N, S, P and halogen 
 (d) The sodium fusion extract gives black (iv) Halogen Specifically 
  precipitate with acetic acid & lead acetate 
The correct match is: 
 (1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 (2) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) 
 (3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 (4) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
Ans. (1) 
Sol. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 
 
  
4. 
 
CN 
OCH 3 
6 5
3
(i)C H MgBr Ether
(1.0equivalent),dry
Major Pr oduct (ii) H O
X
?
? ? ? ? ? ? ? ? ? ? ? 
 The structure of X is: 
 (1)
 
O 
OCH 3 
C 6H 5 
   (2)
 
OCH 3 
NH 2 
 
 (3)
 
O 
C 6H 5 
C 6H 5 
   (4)
 
C 6H 5 
NH 2 
 
Ans. (1) 
Sol. 
 
 
C ? ?N 
C
6
H
5
–MgBr 
CH–CH
3
 
OCH
3
 
C=NMgBr 
CH–CH
3
 
OCH
3
 
C
6
H
5
 
CH–CH
3
 
OCH
3
 
C–C
6
H
5
 
O 
H
3
O
+
 
 
 
 
 
 
16
th
 MARCH 2021 | Shift 2
5. Ammonolysis of Alkylhalides followed by the treatment with NaOH solution can be used to 
prepare primary, secondary and tertiary amines. The purpose of NaOH in the reaction is: 
 (1) to remove basic impurities 
 (2) to activate NH
3
used in the reaction 
 (3) to increase the reactivity of alkyl halide 
 (4) to remove acidic impurities 
Ans. (4) 
Sol.  
 
 
 
 
R–X 
NH
3
 
–HX 
R-NH
2
 
R– X 
–HX 
R
2
NH 
–HX R–X 
R–X 
R
3
N 
R
4
NX 
 
 
During the reaction HX (acid) is form 
 Hence, we use NaOH to remove this acidic impurities 
 
6. Arrange the following metal complex/compounds in the increasing order of spin only magnetic 
moment. Presume all the three, high spin system. 
 (Atomic numbers Ce = 58, Gd = 64 and Eu = 63) 
 (a) (NH
4
)
2
[Ce(NO
3
)
6
]  (b)Gd(NO
3
)
3
 and (c)Eu(NO
3
)
3 
 Answer is: 
 (1)(a)<(c)<(b)    (2)(a)<(b)<(c) 
 (3)(c)<(a)<(b)    (4)(b)<(a)<(c) 
Ans. (1) 
Sol. (NH
4
)
2
 [Ce(NO
3
)
6
]  (n = 0)   ? ? ? ?= 0 B.M 
 Eu (NO
3
)
3
   (n = 6)   ?  ? = 6.93 B.M 
 Gd(NO
3
)
3
   (n = 7)   ?  ? = 7.94 B.M 
 
Page 4


 
 
16
th
 MARCH 2021 | Shift 2
1. 
 
Cl 
A 
Cl 
 
 Identify the reagent(s) 'A' and condition(s) for the reaction 
 (1) A = HCl; Anhydrous AlCl
3
 
 (2) A = HCl, ZnCl
2 
 (3) A = Cl
2
, dark, Anhydrous AlCl
3 
 (4) A = Cl
2
; UV light 
Ans. (4) 
Sol. 
 
 
Cl
2
 
h ? 
Cl 
Cl
2
 
h ? 
Cl 
Cl 
 
 
2. The INCORRECT statement regarding the structure of C
60
 is: 
 (1) It contains 12 six-membered rings and 24 five-membered rings. 
 (2) Each carbon atom forms three sigma bonds. 
 (3) The five-membered rings are fused only to six-membered rings.  
 (4) The six-membered rings are fused to both six and five-membered rings. 
Ans. (1) 
Sol. it contain 12 five membered ring & 20 six membered ring 
 
3. Match List-I with List-II: 
  List-I       List-II 
  Test/Reagents/Observation(s)   Species detected 
 (a) Lassaigne's Test    (i) Carbon 
 (b) Cu(II) oxide     (ii) Sulphur 
 (c) Silver nitrate     (iii) N, S, P and halogen 
 (d) The sodium fusion extract gives black (iv) Halogen Specifically 
  precipitate with acetic acid & lead acetate 
The correct match is: 
 (1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 (2) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) 
 (3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 (4) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
Ans. (1) 
Sol. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 
 
  
4. 
 
CN 
OCH 3 
6 5
3
(i)C H MgBr Ether
(1.0equivalent),dry
Major Pr oduct (ii) H O
X
?
? ? ? ? ? ? ? ? ? ? ? 
 The structure of X is: 
 (1)
 
O 
OCH 3 
C 6H 5 
   (2)
 
OCH 3 
NH 2 
 
 (3)
 
O 
C 6H 5 
C 6H 5 
   (4)
 
C 6H 5 
NH 2 
 
Ans. (1) 
Sol. 
 
 
C ? ?N 
C
6
H
5
–MgBr 
CH–CH
3
 
OCH
3
 
C=NMgBr 
CH–CH
3
 
OCH
3
 
C
6
H
5
 
CH–CH
3
 
OCH
3
 
C–C
6
H
5
 
O 
H
3
O
+
 
 
 
 
 
 
16
th
 MARCH 2021 | Shift 2
5. Ammonolysis of Alkylhalides followed by the treatment with NaOH solution can be used to 
prepare primary, secondary and tertiary amines. The purpose of NaOH in the reaction is: 
 (1) to remove basic impurities 
 (2) to activate NH
3
used in the reaction 
 (3) to increase the reactivity of alkyl halide 
 (4) to remove acidic impurities 
Ans. (4) 
Sol.  
 
 
 
 
R–X 
NH
3
 
–HX 
R-NH
2
 
R– X 
–HX 
R
2
NH 
–HX R–X 
R–X 
R
3
N 
R
4
NX 
 
 
During the reaction HX (acid) is form 
 Hence, we use NaOH to remove this acidic impurities 
 
6. Arrange the following metal complex/compounds in the increasing order of spin only magnetic 
moment. Presume all the three, high spin system. 
 (Atomic numbers Ce = 58, Gd = 64 and Eu = 63) 
 (a) (NH
4
)
2
[Ce(NO
3
)
6
]  (b)Gd(NO
3
)
3
 and (c)Eu(NO
3
)
3 
 Answer is: 
 (1)(a)<(c)<(b)    (2)(a)<(b)<(c) 
 (3)(c)<(a)<(b)    (4)(b)<(a)<(c) 
Ans. (1) 
Sol. (NH
4
)
2
 [Ce(NO
3
)
6
]  (n = 0)   ? ? ? ?= 0 B.M 
 Eu (NO
3
)
3
   (n = 6)   ?  ? = 6.93 B.M 
 Gd(NO
3
)
3
   (n = 7)   ?  ? = 7.94 B.M 
 
 
 
7. Identify the elements X and Y using the ionisation energy values given below: 
 Ionization energy (kJ/mol) 
  1
st
 2
nd
 
 X 495 4563 
 Y 731 1450   
 (1) X = F;  Y = Mg   (2) X = Mg; Y = F 
 (3) X = Na;  Y = Mg   (4) X = Mg;  Y = Na 
Ans. (3) 
Sol. 2
nd
 I. E of Alkali metals is higher than their respective period. 
 
 
8. The INCORRECT statements below regarding colloidal solutions is: 
 (1) A colloidal solution shows colligative properties. 
 (2) An ordinary filter paper can stop the flow of colloidal particles. 
 (3) A colloidal solution shows Brownian motion of colloidal particles. 
 (4) The flocculating power of Al
3+
 is more than that of Na
+
. 
Ans. (2) 
Sol. Colloidal solutions can pass through ordinary filter paper but cannot pass through special filter 
 collodial solution coated paper. 
 
9. The characteristics of elements X, Y and Z with atomic numbers, respectively, 33, 53 and 83 
 are: 
 (1) X and Z are non-metals and Y is a metalloid. 
 (2) X and Y are metalloids and Z is a metal 
 (3) X, Y and Z are metals. 
 (4) X is a metalloid, Y is a non-metal and Z is a metal. 
Ans. (4) 
Sol. Atomic No.  Element 
 (1) 33     ? ? ? ? As (Metalloid) 
 (2) 53    ? ? ? ? I (Non metal) 
 (3) 83    ? ? ? ? Bi (Metal) 
 
10. The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H
3
PO
3
 solution and 
100 mL of 2 M H
3
PO
2
 solution, respectively, are: 
 (1) 100 mL and 50 mL   (2) 50 mL and 50 mL 
 (3) 100 mL and 100 mL   (4) 100 mL and 200 mL 
Ans. (4) 
Sol. (1) 2NaOH  +  H
3
PO
3
  ??  Na
2
HPO
3  
+  2H
2
O
 
  
100m mole  50m mole 
  100m mole = M × V
ml
 
  100m mole = 1 × V
ml 
  
V
ml
 = 100 ml 
 (2) NaOH  + H
3
PO
2
  ??  NaH
2
PO
2
 + H
2
O 
  200m mole  200m mole 
  200m mole = M × V
ml
 
  V
ml
 = 200 ml 
 
 
Page 5


 
 
16
th
 MARCH 2021 | Shift 2
1. 
 
Cl 
A 
Cl 
 
 Identify the reagent(s) 'A' and condition(s) for the reaction 
 (1) A = HCl; Anhydrous AlCl
3
 
 (2) A = HCl, ZnCl
2 
 (3) A = Cl
2
, dark, Anhydrous AlCl
3 
 (4) A = Cl
2
; UV light 
Ans. (4) 
Sol. 
 
 
Cl
2
 
h ? 
Cl 
Cl
2
 
h ? 
Cl 
Cl 
 
 
2. The INCORRECT statement regarding the structure of C
60
 is: 
 (1) It contains 12 six-membered rings and 24 five-membered rings. 
 (2) Each carbon atom forms three sigma bonds. 
 (3) The five-membered rings are fused only to six-membered rings.  
 (4) The six-membered rings are fused to both six and five-membered rings. 
Ans. (1) 
Sol. it contain 12 five membered ring & 20 six membered ring 
 
3. Match List-I with List-II: 
  List-I       List-II 
  Test/Reagents/Observation(s)   Species detected 
 (a) Lassaigne's Test    (i) Carbon 
 (b) Cu(II) oxide     (ii) Sulphur 
 (c) Silver nitrate     (iii) N, S, P and halogen 
 (d) The sodium fusion extract gives black (iv) Halogen Specifically 
  precipitate with acetic acid & lead acetate 
The correct match is: 
 (1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 (2) (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii) 
 (3) (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv) 
 (4) (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii) 
Ans. (1) 
Sol. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)  
 
 
  
4. 
 
CN 
OCH 3 
6 5
3
(i)C H MgBr Ether
(1.0equivalent),dry
Major Pr oduct (ii) H O
X
?
? ? ? ? ? ? ? ? ? ? ? 
 The structure of X is: 
 (1)
 
O 
OCH 3 
C 6H 5 
   (2)
 
OCH 3 
NH 2 
 
 (3)
 
O 
C 6H 5 
C 6H 5 
   (4)
 
C 6H 5 
NH 2 
 
Ans. (1) 
Sol. 
 
 
C ? ?N 
C
6
H
5
–MgBr 
CH–CH
3
 
OCH
3
 
C=NMgBr 
CH–CH
3
 
OCH
3
 
C
6
H
5
 
CH–CH
3
 
OCH
3
 
C–C
6
H
5
 
O 
H
3
O
+
 
 
 
 
 
 
16
th
 MARCH 2021 | Shift 2
5. Ammonolysis of Alkylhalides followed by the treatment with NaOH solution can be used to 
prepare primary, secondary and tertiary amines. The purpose of NaOH in the reaction is: 
 (1) to remove basic impurities 
 (2) to activate NH
3
used in the reaction 
 (3) to increase the reactivity of alkyl halide 
 (4) to remove acidic impurities 
Ans. (4) 
Sol.  
 
 
 
 
R–X 
NH
3
 
–HX 
R-NH
2
 
R– X 
–HX 
R
2
NH 
–HX R–X 
R–X 
R
3
N 
R
4
NX 
 
 
During the reaction HX (acid) is form 
 Hence, we use NaOH to remove this acidic impurities 
 
6. Arrange the following metal complex/compounds in the increasing order of spin only magnetic 
moment. Presume all the three, high spin system. 
 (Atomic numbers Ce = 58, Gd = 64 and Eu = 63) 
 (a) (NH
4
)
2
[Ce(NO
3
)
6
]  (b)Gd(NO
3
)
3
 and (c)Eu(NO
3
)
3 
 Answer is: 
 (1)(a)<(c)<(b)    (2)(a)<(b)<(c) 
 (3)(c)<(a)<(b)    (4)(b)<(a)<(c) 
Ans. (1) 
Sol. (NH
4
)
2
 [Ce(NO
3
)
6
]  (n = 0)   ? ? ? ?= 0 B.M 
 Eu (NO
3
)
3
   (n = 6)   ?  ? = 6.93 B.M 
 Gd(NO
3
)
3
   (n = 7)   ?  ? = 7.94 B.M 
 
 
 
7. Identify the elements X and Y using the ionisation energy values given below: 
 Ionization energy (kJ/mol) 
  1
st
 2
nd
 
 X 495 4563 
 Y 731 1450   
 (1) X = F;  Y = Mg   (2) X = Mg; Y = F 
 (3) X = Na;  Y = Mg   (4) X = Mg;  Y = Na 
Ans. (3) 
Sol. 2
nd
 I. E of Alkali metals is higher than their respective period. 
 
 
8. The INCORRECT statements below regarding colloidal solutions is: 
 (1) A colloidal solution shows colligative properties. 
 (2) An ordinary filter paper can stop the flow of colloidal particles. 
 (3) A colloidal solution shows Brownian motion of colloidal particles. 
 (4) The flocculating power of Al
3+
 is more than that of Na
+
. 
Ans. (2) 
Sol. Colloidal solutions can pass through ordinary filter paper but cannot pass through special filter 
 collodial solution coated paper. 
 
9. The characteristics of elements X, Y and Z with atomic numbers, respectively, 33, 53 and 83 
 are: 
 (1) X and Z are non-metals and Y is a metalloid. 
 (2) X and Y are metalloids and Z is a metal 
 (3) X, Y and Z are metals. 
 (4) X is a metalloid, Y is a non-metal and Z is a metal. 
Ans. (4) 
Sol. Atomic No.  Element 
 (1) 33     ? ? ? ? As (Metalloid) 
 (2) 53    ? ? ? ? I (Non metal) 
 (3) 83    ? ? ? ? Bi (Metal) 
 
10. The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M H
3
PO
3
 solution and 
100 mL of 2 M H
3
PO
2
 solution, respectively, are: 
 (1) 100 mL and 50 mL   (2) 50 mL and 50 mL 
 (3) 100 mL and 100 mL   (4) 100 mL and 200 mL 
Ans. (4) 
Sol. (1) 2NaOH  +  H
3
PO
3
  ??  Na
2
HPO
3  
+  2H
2
O
 
  
100m mole  50m mole 
  100m mole = M × V
ml
 
  100m mole = 1 × V
ml 
  
V
ml
 = 100 ml 
 (2) NaOH  + H
3
PO
2
  ??  NaH
2
PO
2
 + H
2
O 
  200m mole  200m mole 
  200m mole = M × V
ml
 
  V
ml
 = 200 ml 
 
 
 
 
16
th
 MARCH 2021 | Shift 2
11. Which of the following reduction reaction CANNOT be carried out with coke?  
 (1) Fe
2
O
3
? Fe     (2)ZnO ? Zn 
 (3) Al
2
O
3
? Al     (4) Cu
2
O ? Cu 
Ans. (3) 
Sol. Al is extracted by electrolytic reduction of Al
2
O
3
 
 
12. An unsaturated hydrocarbon X on ozonolysis gives A. Compound A when warmed with 
ammonical silver nitrate forms a bright silver mirror along the sides of the test tube. The 
unsaturated  hydrocarbon X is: 
 (1) CH
3
–C ?C–CH
3
    (2)
3 3
3 3
CH C C CH
| |
CH CH
? ? ? 
 (3) HC ?C–CH
2
–CH
3
    (4) 
 
CH 3—C= 
CH 3 
 
Ans. (3) 
Sol.  
 
 
 
 
CH
3
CH
2
?CH 
(i) O
3
 
(ii) H
2
O 
CH
3
CH
2
COOH + H–C–OH 
O 
[Ag(NH
3
)
2
]
+
 
Tollen's 
reagent 
 
HCOOH CO
2
 + H
2
O+2Ag 
 
 
13. Statement-I: Sodium hydride can be used as an oxidising agent. 
 Statement-II: The lone pair of electrons on nitrogen in pyridine makes it basic: 
 Choose the CORRECT answer from the options given below: 
 (1) Statement I is true but statement II is false 
 (2) Both statement I and statement II are false 
 (3) Both statement I and statement II are true 
 (4) Statement I is false but statement II is true 
Ans. (4) 
Sol. ? ?NaH is used as reducing agent. 
? ? ? ? ?The ?p on nitrogen in pyridine makes it basic 
 
 
N 
 
 
14. Which of the following polymer is used in the manufacture of wood laminates? 
 (1) Melamine formaldehyde resin  (2)cis-poly isoprene 
 (3) Phenol and formaldehyde resin  (4) Urea formaldehyde resin 
Ans. (1) 
Sol. Melamine formaldehyde resin is used in the manufacture of wood laminates.