JEE Main 2021 Answer Key Chemistry - Evening Shift (18-03-2021) JEE Notes | EduRev

Mock Test Series for JEE Main & Advanced 2021

JEE : JEE Main 2021 Answer Key Chemistry - Evening Shift (18-03-2021) JEE Notes | EduRev

 Page 1


 
 
18
th
 March. 2021 | Shift 2 
SECTION - A 
1. 
 
dilNaOH
?? ? ? ? ? “X” 
H ,Heat
?
? ? ? ? ? ? ”Y” 
Consider the above reaction, the product ‘X’ and ‘Y’ respectively are : 
(1)   ,  
(2)  ,   
(3)   ,  
(4)  ,   
Ans. (2) 
Sol. 
 
 
O 
dil NaOH 
(Aldol condensation) 
O 
OH 
Heat 
O 
 
 
2. The charges on the colloidal CdS sol and TiO
2
 sol are, respectively : 
 (1) positive and negative  (2) negative and negative 
 (3) negative and positive  (4) positive and positive 
Ans. (3) 
Sol. CdS ? Sulphide sol. ? Negative sol. 
 TiO
2 
? Oxide sol. ? Positive sol. 
 
O O
OH
O
OH
O
O O
OH
O O
OH
O
Page 2


 
 
18
th
 March. 2021 | Shift 2 
SECTION - A 
1. 
 
dilNaOH
?? ? ? ? ? “X” 
H ,Heat
?
? ? ? ? ? ? ”Y” 
Consider the above reaction, the product ‘X’ and ‘Y’ respectively are : 
(1)   ,  
(2)  ,   
(3)   ,  
(4)  ,   
Ans. (2) 
Sol. 
 
 
O 
dil NaOH 
(Aldol condensation) 
O 
OH 
Heat 
O 
 
 
2. The charges on the colloidal CdS sol and TiO
2
 sol are, respectively : 
 (1) positive and negative  (2) negative and negative 
 (3) negative and positive  (4) positive and positive 
Ans. (3) 
Sol. CdS ? Sulphide sol. ? Negative sol. 
 TiO
2 
? Oxide sol. ? Positive sol. 
 
O O
OH
O
OH
O
O O
OH
O O
OH
O
 
 
3. The oxide that shows magnetic property is : 
 (1) SiO
2   
(2) Na
2
O  (3) Mn
3
O
4
  (4) MgO 
Ans. (3) 
Sol. Mn
3
O
4
 is paramagnetic due to presence of unpaired electrons.  
 
4. Given below are two statements : 
 Statement I : Bohr’s theory accounts for the stability and line spectrum of Li
+
 ion. 
 Statement II : Bohr’s theory was unable to explain the splitting of spectral lines in the presence of a magnetic 
field. 
 In the light of the above statements, choose the most appropriate answer from the options given below : 
 (1) Both statement I and statement II are true. 
 (2) Statement I is true but statement II is false. 
 (3) Statement I is false but statement II is true. 
 (4) Both statement I and statement II are false. 
Ans. (3) 
Sol. S-1 ? false 
 S-2 ? True 
 Hence option 3 
 
5. Match List-I with List-II : 
  List-I   List-II 
 (1) Mercury  (i) Vapour phase refining 
 (2) Copper  (ii) Distillation Refining 
 (3) Silicon  (iii) Electrolytic Refining 
 (4) Nickel  (iv) Zone Refining 
 Choose the most appropriate answer from the option given below : 
 (1) (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)   (2) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii) 
 (3) (a)-(ii), (b)-(iv), (c)-(iii),(d)-(ii)   (4) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) 
Ans. (4) 
Sol. Theory based 
 
6. Match List-I with List-II : 
  List-I   List-II 
 (Class of Chemicals)  (Example) 
 (a) Antifertility drug  (i) Meprobamate 
 (b) Antibiotic   (ii) Alitame 
 (c) Tranquilizer   (iii) Norethindrone 
 (d) Artificial Sweetener  (iv) Salvarsan 
 Options : 
 (1) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)   (2) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) 
 (3) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)   (4) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) 
Ans. (4) 
Sol. (a) Antifertility drug   ??  Norethindrone 
 (b) Antibiotic    ??  Salvarsan 
 (c) Tranquilizer   ??  Meprobamate 
 (d) Artificial sweetener  ??  Alitame 
Page 3


 
 
18
th
 March. 2021 | Shift 2 
SECTION - A 
1. 
 
dilNaOH
?? ? ? ? ? “X” 
H ,Heat
?
? ? ? ? ? ? ”Y” 
Consider the above reaction, the product ‘X’ and ‘Y’ respectively are : 
(1)   ,  
(2)  ,   
(3)   ,  
(4)  ,   
Ans. (2) 
Sol. 
 
 
O 
dil NaOH 
(Aldol condensation) 
O 
OH 
Heat 
O 
 
 
2. The charges on the colloidal CdS sol and TiO
2
 sol are, respectively : 
 (1) positive and negative  (2) negative and negative 
 (3) negative and positive  (4) positive and positive 
Ans. (3) 
Sol. CdS ? Sulphide sol. ? Negative sol. 
 TiO
2 
? Oxide sol. ? Positive sol. 
 
O O
OH
O
OH
O
O O
OH
O O
OH
O
 
 
3. The oxide that shows magnetic property is : 
 (1) SiO
2   
(2) Na
2
O  (3) Mn
3
O
4
  (4) MgO 
Ans. (3) 
Sol. Mn
3
O
4
 is paramagnetic due to presence of unpaired electrons.  
 
4. Given below are two statements : 
 Statement I : Bohr’s theory accounts for the stability and line spectrum of Li
+
 ion. 
 Statement II : Bohr’s theory was unable to explain the splitting of spectral lines in the presence of a magnetic 
field. 
 In the light of the above statements, choose the most appropriate answer from the options given below : 
 (1) Both statement I and statement II are true. 
 (2) Statement I is true but statement II is false. 
 (3) Statement I is false but statement II is true. 
 (4) Both statement I and statement II are false. 
Ans. (3) 
Sol. S-1 ? false 
 S-2 ? True 
 Hence option 3 
 
5. Match List-I with List-II : 
  List-I   List-II 
 (1) Mercury  (i) Vapour phase refining 
 (2) Copper  (ii) Distillation Refining 
 (3) Silicon  (iii) Electrolytic Refining 
 (4) Nickel  (iv) Zone Refining 
 Choose the most appropriate answer from the option given below : 
 (1) (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)   (2) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii) 
 (3) (a)-(ii), (b)-(iv), (c)-(iii),(d)-(ii)   (4) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) 
Ans. (4) 
Sol. Theory based 
 
6. Match List-I with List-II : 
  List-I   List-II 
 (Class of Chemicals)  (Example) 
 (a) Antifertility drug  (i) Meprobamate 
 (b) Antibiotic   (ii) Alitame 
 (c) Tranquilizer   (iii) Norethindrone 
 (d) Artificial Sweetener  (iv) Salvarsan 
 Options : 
 (1) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)   (2) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) 
 (3) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)   (4) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) 
Ans. (4) 
Sol. (a) Antifertility drug   ??  Norethindrone 
 (b) Antibiotic    ??  Salvarsan 
 (c) Tranquilizer   ??  Meprobamate 
 (d) Artificial sweetener  ??  Alitame 
 
 
18
th
 March. 2021 | Shift 2 
7. Main Products formed during a reaction of 1-methoxy naphthalene with hydroiodic acid are : 
 (1) 
 
  (2) 
 
 (3)    (4)  
Ans. (4) 
Sol. 
 
 
OCH
3
 
H 
O 
H 
CH
3
 
I
–
 
OH 
+ CH
3
–I 
 
 
 
8.  
 
Consider the given reaction, percentage yield of : 
 (1) A > C > B  (2) B > C > A  (3) C > B > A  (4) C > A > B 
Ans. (3) 
Sol. 
 Order of % yield ?  
 
NH
2
 
NO
2
 
NH
2
 NH
2
 
> > 
NO
2
 
NO
2
 
 
NH
2 
HNO 3,H 2SO 4 
288 K
 
NH
2 
NO
2 
A
 
NH
2 
NO
2 
B
 
NH
2 
NO
2 
C
 
+
 
+
 
OH 
and CH
3
I 
I 
and CH
3
OH 
OH 
and CH
3
OH 
I
and CH
3
I
Page 4


 
 
18
th
 March. 2021 | Shift 2 
SECTION - A 
1. 
 
dilNaOH
?? ? ? ? ? “X” 
H ,Heat
?
? ? ? ? ? ? ”Y” 
Consider the above reaction, the product ‘X’ and ‘Y’ respectively are : 
(1)   ,  
(2)  ,   
(3)   ,  
(4)  ,   
Ans. (2) 
Sol. 
 
 
O 
dil NaOH 
(Aldol condensation) 
O 
OH 
Heat 
O 
 
 
2. The charges on the colloidal CdS sol and TiO
2
 sol are, respectively : 
 (1) positive and negative  (2) negative and negative 
 (3) negative and positive  (4) positive and positive 
Ans. (3) 
Sol. CdS ? Sulphide sol. ? Negative sol. 
 TiO
2 
? Oxide sol. ? Positive sol. 
 
O O
OH
O
OH
O
O O
OH
O O
OH
O
 
 
3. The oxide that shows magnetic property is : 
 (1) SiO
2   
(2) Na
2
O  (3) Mn
3
O
4
  (4) MgO 
Ans. (3) 
Sol. Mn
3
O
4
 is paramagnetic due to presence of unpaired electrons.  
 
4. Given below are two statements : 
 Statement I : Bohr’s theory accounts for the stability and line spectrum of Li
+
 ion. 
 Statement II : Bohr’s theory was unable to explain the splitting of spectral lines in the presence of a magnetic 
field. 
 In the light of the above statements, choose the most appropriate answer from the options given below : 
 (1) Both statement I and statement II are true. 
 (2) Statement I is true but statement II is false. 
 (3) Statement I is false but statement II is true. 
 (4) Both statement I and statement II are false. 
Ans. (3) 
Sol. S-1 ? false 
 S-2 ? True 
 Hence option 3 
 
5. Match List-I with List-II : 
  List-I   List-II 
 (1) Mercury  (i) Vapour phase refining 
 (2) Copper  (ii) Distillation Refining 
 (3) Silicon  (iii) Electrolytic Refining 
 (4) Nickel  (iv) Zone Refining 
 Choose the most appropriate answer from the option given below : 
 (1) (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)   (2) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii) 
 (3) (a)-(ii), (b)-(iv), (c)-(iii),(d)-(ii)   (4) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) 
Ans. (4) 
Sol. Theory based 
 
6. Match List-I with List-II : 
  List-I   List-II 
 (Class of Chemicals)  (Example) 
 (a) Antifertility drug  (i) Meprobamate 
 (b) Antibiotic   (ii) Alitame 
 (c) Tranquilizer   (iii) Norethindrone 
 (d) Artificial Sweetener  (iv) Salvarsan 
 Options : 
 (1) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)   (2) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) 
 (3) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)   (4) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) 
Ans. (4) 
Sol. (a) Antifertility drug   ??  Norethindrone 
 (b) Antibiotic    ??  Salvarsan 
 (c) Tranquilizer   ??  Meprobamate 
 (d) Artificial sweetener  ??  Alitame 
 
 
18
th
 March. 2021 | Shift 2 
7. Main Products formed during a reaction of 1-methoxy naphthalene with hydroiodic acid are : 
 (1) 
 
  (2) 
 
 (3)    (4)  
Ans. (4) 
Sol. 
 
 
OCH
3
 
H 
O 
H 
CH
3
 
I
–
 
OH 
+ CH
3
–I 
 
 
 
8.  
 
Consider the given reaction, percentage yield of : 
 (1) A > C > B  (2) B > C > A  (3) C > B > A  (4) C > A > B 
Ans. (3) 
Sol. 
 Order of % yield ?  
 
NH
2
 
NO
2
 
NH
2
 NH
2
 
> > 
NO
2
 
NO
2
 
 
NH
2 
HNO 3,H 2SO 4 
288 K
 
NH
2 
NO
2 
A
 
NH
2 
NO
2 
B
 
NH
2 
NO
2 
C
 
+
 
+
 
OH 
and CH
3
I 
I 
and CH
3
OH 
OH 
and CH
3
OH 
I
and CH
3
I
 
 
 
9. An organic compound “A” on treatment with benzene sulphonyl chloride gives compound B. B is soluble in 
dil. NaOH solution. Compound A is : 
 (1) C
6
H
5
–N–(CH
3
)
2    
(2) C
6
H
5
–NHCH
2
CH
3
 
 (3) 
   
 (4) C
6
H
5
–CH
2 
NH CH
3 
Ans. (3) 
Sol. 
  
 
 
C
6
H
5
–CH–NH
2
 
CH
3
 
C
6
H
5
SO
2
Cl 
–HCl 
C
6
H
5
SO
2
NHCH–C
6
H
5
 
CH
3
 
(ppt) (Acidic in 
nature) 
NaOH 
–H
2
O 
C
6
H
5
–CH–NSO
2
C
6
H
5
 
CH
3
 
Na
+
 
Soluble 
 
 
10. The first ionization energy of magnesium is smaller as compound to that of elements X and Y, but higher 
than that of Z. The elements X, Y and Z, respectively are : 
 (1) argon, lithium and sodium 
 (2) chlorine, lithium and sodium 
 (3) neon, sodium and chlorine 
 (4) argon, chlorine and sodium 
Ans. (4) 
Sol. Order of I.E. 
 3rd period ? Na < Al < Mg < Si < S < P < Cl < Ar 
 
11. In the following molecule, 
 
Hybridisation of Carbon a, b and c respectively are : 
(1) sp
3
, sp
2
, sp
2  
(2) sp
3
, sp
2
, sp
  
(3) sp
3
, sp, sp   (4) sp
3
, sp, sp
2 
Ans. (1) 
Sol. a ?? sp
3
  
 b ?? sp
2
  
c ?? sp
2
  
 
 
C=C–O
H
3
C
a
H
b
c 
H 
C
6
H
5
–CH–NH
2 
CH
3
Page 5


 
 
18
th
 March. 2021 | Shift 2 
SECTION - A 
1. 
 
dilNaOH
?? ? ? ? ? “X” 
H ,Heat
?
? ? ? ? ? ? ”Y” 
Consider the above reaction, the product ‘X’ and ‘Y’ respectively are : 
(1)   ,  
(2)  ,   
(3)   ,  
(4)  ,   
Ans. (2) 
Sol. 
 
 
O 
dil NaOH 
(Aldol condensation) 
O 
OH 
Heat 
O 
 
 
2. The charges on the colloidal CdS sol and TiO
2
 sol are, respectively : 
 (1) positive and negative  (2) negative and negative 
 (3) negative and positive  (4) positive and positive 
Ans. (3) 
Sol. CdS ? Sulphide sol. ? Negative sol. 
 TiO
2 
? Oxide sol. ? Positive sol. 
 
O O
OH
O
OH
O
O O
OH
O O
OH
O
 
 
3. The oxide that shows magnetic property is : 
 (1) SiO
2   
(2) Na
2
O  (3) Mn
3
O
4
  (4) MgO 
Ans. (3) 
Sol. Mn
3
O
4
 is paramagnetic due to presence of unpaired electrons.  
 
4. Given below are two statements : 
 Statement I : Bohr’s theory accounts for the stability and line spectrum of Li
+
 ion. 
 Statement II : Bohr’s theory was unable to explain the splitting of spectral lines in the presence of a magnetic 
field. 
 In the light of the above statements, choose the most appropriate answer from the options given below : 
 (1) Both statement I and statement II are true. 
 (2) Statement I is true but statement II is false. 
 (3) Statement I is false but statement II is true. 
 (4) Both statement I and statement II are false. 
Ans. (3) 
Sol. S-1 ? false 
 S-2 ? True 
 Hence option 3 
 
5. Match List-I with List-II : 
  List-I   List-II 
 (1) Mercury  (i) Vapour phase refining 
 (2) Copper  (ii) Distillation Refining 
 (3) Silicon  (iii) Electrolytic Refining 
 (4) Nickel  (iv) Zone Refining 
 Choose the most appropriate answer from the option given below : 
 (1) (a)-(ii), (b)-(iii), (c)-(i), (d)-(iv)   (2) (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii) 
 (3) (a)-(ii), (b)-(iv), (c)-(iii),(d)-(ii)   (4) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) 
Ans. (4) 
Sol. Theory based 
 
6. Match List-I with List-II : 
  List-I   List-II 
 (Class of Chemicals)  (Example) 
 (a) Antifertility drug  (i) Meprobamate 
 (b) Antibiotic   (ii) Alitame 
 (c) Tranquilizer   (iii) Norethindrone 
 (d) Artificial Sweetener  (iv) Salvarsan 
 Options : 
 (1) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)   (2) (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i) 
 (3) (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)   (4) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii) 
Ans. (4) 
Sol. (a) Antifertility drug   ??  Norethindrone 
 (b) Antibiotic    ??  Salvarsan 
 (c) Tranquilizer   ??  Meprobamate 
 (d) Artificial sweetener  ??  Alitame 
 
 
18
th
 March. 2021 | Shift 2 
7. Main Products formed during a reaction of 1-methoxy naphthalene with hydroiodic acid are : 
 (1) 
 
  (2) 
 
 (3)    (4)  
Ans. (4) 
Sol. 
 
 
OCH
3
 
H 
O 
H 
CH
3
 
I
–
 
OH 
+ CH
3
–I 
 
 
 
8.  
 
Consider the given reaction, percentage yield of : 
 (1) A > C > B  (2) B > C > A  (3) C > B > A  (4) C > A > B 
Ans. (3) 
Sol. 
 Order of % yield ?  
 
NH
2
 
NO
2
 
NH
2
 NH
2
 
> > 
NO
2
 
NO
2
 
 
NH
2 
HNO 3,H 2SO 4 
288 K
 
NH
2 
NO
2 
A
 
NH
2 
NO
2 
B
 
NH
2 
NO
2 
C
 
+
 
+
 
OH 
and CH
3
I 
I 
and CH
3
OH 
OH 
and CH
3
OH 
I
and CH
3
I
 
 
 
9. An organic compound “A” on treatment with benzene sulphonyl chloride gives compound B. B is soluble in 
dil. NaOH solution. Compound A is : 
 (1) C
6
H
5
–N–(CH
3
)
2    
(2) C
6
H
5
–NHCH
2
CH
3
 
 (3) 
   
 (4) C
6
H
5
–CH
2 
NH CH
3 
Ans. (3) 
Sol. 
  
 
 
C
6
H
5
–CH–NH
2
 
CH
3
 
C
6
H
5
SO
2
Cl 
–HCl 
C
6
H
5
SO
2
NHCH–C
6
H
5
 
CH
3
 
(ppt) (Acidic in 
nature) 
NaOH 
–H
2
O 
C
6
H
5
–CH–NSO
2
C
6
H
5
 
CH
3
 
Na
+
 
Soluble 
 
 
10. The first ionization energy of magnesium is smaller as compound to that of elements X and Y, but higher 
than that of Z. The elements X, Y and Z, respectively are : 
 (1) argon, lithium and sodium 
 (2) chlorine, lithium and sodium 
 (3) neon, sodium and chlorine 
 (4) argon, chlorine and sodium 
Ans. (4) 
Sol. Order of I.E. 
 3rd period ? Na < Al < Mg < Si < S < P < Cl < Ar 
 
11. In the following molecule, 
 
Hybridisation of Carbon a, b and c respectively are : 
(1) sp
3
, sp
2
, sp
2  
(2) sp
3
, sp
2
, sp
  
(3) sp
3
, sp, sp   (4) sp
3
, sp, sp
2 
Ans. (1) 
Sol. a ?? sp
3
  
 b ?? sp
2
  
c ?? sp
2
  
 
 
C=C–O
H
3
C
a
H
b
c 
H 
C
6
H
5
–CH–NH
2 
CH
3
 
 
18
th
 March. 2021 | Shift 2 
12. In the reaction of hypobromite with amide, the carbonyl carbon is lost as : 
 (1) HCO
3
–  
(2) CO
3
2–
  (3) CO
2
   (4) CO 
Ans. (2) 
Sol. CO
3
2–
 
 
13. The oxidation states of nitrogen in NO, NO
2
, N
2
O and NO
3
–
 are in the order of : 
 (1) NO
2
 > NO
3
–
 > NO > N
2
O   (2) N
2
O > NO
2
 > NO > NO
3
– 
 (3) NO
3
–
 > NO
2
 > NO > N
2
O   (4) NO > NO
2
 > N
2
O > NO
3
– 
Ans. (3) 
Sol. O.S. of 'N' 
 NO ? +2 
 NO
2
 ? ?+4 
 N
2
O ? +1 
 NO
3
–
 ? ?+5 
 Decreasing order of ox. state of 'N' is as follows 
 NO
3
–
 > NO
2
 > NO > N
2
O 
 
14. Match List-I and List-II : 
  List-I   List-II 
 (a) Be    (i) treatment of cancer 
 (b) Mg    (ii) extraction of metals 
 (c) Ca    (iii) incendiary bombs and signals 
 (d) Ra    (iv) windows of X-ray tubes 
     (v) bearings for motor engines 
 Choose the most appropriate answer from the option given below : 
 Options : 
 (1) (a)-(iii), (b)-(iv), (c)-(ii), (d)-(v) 
 (2) (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii) 
 (3) (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i) 
 (4) (a)-(iii), (b)-(iv), (c)-(v), (d)-(ii) 
Ans. (3) 
Sol. Fact (NCERT) 
 Due to radioactive nature Ra - is used in treatment of cancer. 
 
15. Deficiency of vitamin K causes: 
 (1)  Cheilosis 
 (2) Increase in blood clotting time 
 (3) Increase in fragility of RBC’s 
 (4) Decrease in blood clotting time 
Ans. (2) 
Sol. Deficiency of vitamin "K" causes ? ?in blood clotting time. 
 
 
 
 
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