JEE Main 2021 Answer Key Chemistry - Morning Shift (26-02-2021) Notes | EduRev

Mock Test Series for JEE Main & Advanced 2021

JEE : JEE Main 2021 Answer Key Chemistry - Morning Shift (26-02-2021) Notes | EduRev

 Page 1


JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
PART–B : CHEMISTRY
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. Match List -I with List - II.
List - I List - II
(Ore) (Element Present)
(a) Kernite (i) Tin
(b) Cassiterite (ii) Boron
(c) Calamine (iii) Fluorine
(d) Cryolite (iv) Zinc
Choose the most appropriate answer from the
options given below:
(1) (a) ? (ii), (b) ? (i), (c) ? (iv), d ? (iii)
(2) (a) ? (iii), (b) ? (i), (c) ? (ii), d ? (iv)
(3) (a) ? (ii), (b) ? (iv), (c) ? (i), d ? (iii)
(4) (a) ? (i), (b) ? (iii), (c) ? (iv), d ? (ii)
Answer (1)
Sol. Kernite : Na
2
B
4
O
7
?4H
2
O (Boron)
Cassiterite : SnO
2
 (Tin)
Calamine : ZnCO
3
 (Zinc)
Cryolite : Na
3
AlF
6
 (Fluorine)
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
2. Identify the major products A and B
respectively in the following reactions of
phenol:
A B
(i) CHCl , NaOH
3
(ii) H O
3
+
Br in CS
22
273K
OH
(1)
OH OH
Br CHO
and
(2)
OH OH
Br
CHO
and
(3)
OH OH
Br
CHO
and
(4)
OH OH
Br
CHO
and
Answer (2)
Sol. OH
Br /CS 2 2
OH
Br
273 K
Major
OH
(i) CHCl ,NaOH 3
OH
(ii) H O
3
+
Major
B
CHO
A
So option (2) is the correct answer
3.
Hydrolysis
373K
A
(CHCl )
48 2
B
(CHO)
48
B reacts with Hydroxyl amine but does not
give Tollen’s test. Identify A and B.
(1) 2,2-Dichlorobutane and Butanal
(2) 1,1-Dichlorobutane and Butanal
(3) 1,1-Dichlorobutane and 2-Butanone
(4) 2,2-Dichlorobutane and Butan-2-one
Answer (4)
Sol. Cl Cl
Hydrolysis
373 K
O
(C H Cl )
48 2
A
(C H O)
48
B
B is a ketone, cannot give Tollen’s test.
A ? 2, 2-Dichlorobutane
B ? Butan-2-one
So correct option should be (4)
Page 2


JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
PART–B : CHEMISTRY
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. Match List -I with List - II.
List - I List - II
(Ore) (Element Present)
(a) Kernite (i) Tin
(b) Cassiterite (ii) Boron
(c) Calamine (iii) Fluorine
(d) Cryolite (iv) Zinc
Choose the most appropriate answer from the
options given below:
(1) (a) ? (ii), (b) ? (i), (c) ? (iv), d ? (iii)
(2) (a) ? (iii), (b) ? (i), (c) ? (ii), d ? (iv)
(3) (a) ? (ii), (b) ? (iv), (c) ? (i), d ? (iii)
(4) (a) ? (i), (b) ? (iii), (c) ? (iv), d ? (ii)
Answer (1)
Sol. Kernite : Na
2
B
4
O
7
?4H
2
O (Boron)
Cassiterite : SnO
2
 (Tin)
Calamine : ZnCO
3
 (Zinc)
Cryolite : Na
3
AlF
6
 (Fluorine)
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
2. Identify the major products A and B
respectively in the following reactions of
phenol:
A B
(i) CHCl , NaOH
3
(ii) H O
3
+
Br in CS
22
273K
OH
(1)
OH OH
Br CHO
and
(2)
OH OH
Br
CHO
and
(3)
OH OH
Br
CHO
and
(4)
OH OH
Br
CHO
and
Answer (2)
Sol. OH
Br /CS 2 2
OH
Br
273 K
Major
OH
(i) CHCl ,NaOH 3
OH
(ii) H O
3
+
Major
B
CHO
A
So option (2) is the correct answer
3.
Hydrolysis
373K
A
(CHCl )
48 2
B
(CHO)
48
B reacts with Hydroxyl amine but does not
give Tollen’s test. Identify A and B.
(1) 2,2-Dichlorobutane and Butanal
(2) 1,1-Dichlorobutane and Butanal
(3) 1,1-Dichlorobutane and 2-Butanone
(4) 2,2-Dichlorobutane and Butan-2-one
Answer (4)
Sol. Cl Cl
Hydrolysis
373 K
O
(C H Cl )
48 2
A
(C H O)
48
B
B is a ketone, cannot give Tollen’s test.
A ? 2, 2-Dichlorobutane
B ? Butan-2-one
So correct option should be (4)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
4. Given below are two statements : One is
labelled as Assertion A and the other is
labelled as Reason R
Assertion A : Dipole-dipole interactions are
only non-covalent interactions, resulting in
hydrogen bond formation.
Reason R : Fluorine is the most
electronegative element and hydrogen bonds
in HF are symmetrical.
In the light of the above statements, choose
the most appropriate answer from the options
given below:
(1) A is false but R is true
(2) Both A and R are true and R is the correct
explanation of A
(3) A is true but R is false
(4) Both A and R are true but R is NOT the
correct explanation of A
Answer (1)
Sol. ? Dipole - Dipole are not only the interaction
responsible for hydrogen bond formation.
Ion-dipole can also be responsible for
hydrogen bond formation.
? F is most electronegative element and
anhydrous HF in solid phase has
symmetrical hydrogen bonding
So the correct option is (1).
5. For the given reaction :
CH ? CHBr
1. NaNH 
2
CH 
3
2. Red hot iron tube, 873K 
(A)
(major product)
What is ‘A’?
(1) (2)
CH
3
HC
3
CH
3
(3) CH
3
CH
2
CH
2
NH
2
(4)
CH 
CH 
3
CH NH
2
Answer (2)
Sol. CH – CH = CH – Br
3
(i) NaNH
2
(ii) Red hot iron
      tube, 873 K
CH
3
CH
3
HC
3
Mesitylene
So the correct option should be (2).
6. The structure of Neoprene is :
(1)
HN N
NN
NHCH
2
n
NH
(2)
CH
2
CH
n
CN
(3)
CH CH      
2
CH
CH
2
CN
n
CH
2
CH
(4)            C      CH      CH
2
CH
2
Cl
n
Answer (4)
Sol. Neoprene is a polymer of monomer
chloroprene
nCH
2
C CH CH
2
Polymerisation
CH
2
C CH CH
2
n
Cl Cl
Cl Chloroprene
Neoprene
So the correct option should be (4)
7. For the given reaction :
CN
CH CH
23
Br
2
UV light
'A'
(Major product)
monobrominated
What is ‘A’?
(1)
CN
CHCH
23
Br
(2)
CN
CHCH
23
Br
(3)
CN
CHCH
23
Br
(4)
CN
CH
Br
CH
3
Answer (4)
Page 3


JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
PART–B : CHEMISTRY
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. Match List -I with List - II.
List - I List - II
(Ore) (Element Present)
(a) Kernite (i) Tin
(b) Cassiterite (ii) Boron
(c) Calamine (iii) Fluorine
(d) Cryolite (iv) Zinc
Choose the most appropriate answer from the
options given below:
(1) (a) ? (ii), (b) ? (i), (c) ? (iv), d ? (iii)
(2) (a) ? (iii), (b) ? (i), (c) ? (ii), d ? (iv)
(3) (a) ? (ii), (b) ? (iv), (c) ? (i), d ? (iii)
(4) (a) ? (i), (b) ? (iii), (c) ? (iv), d ? (ii)
Answer (1)
Sol. Kernite : Na
2
B
4
O
7
?4H
2
O (Boron)
Cassiterite : SnO
2
 (Tin)
Calamine : ZnCO
3
 (Zinc)
Cryolite : Na
3
AlF
6
 (Fluorine)
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
2. Identify the major products A and B
respectively in the following reactions of
phenol:
A B
(i) CHCl , NaOH
3
(ii) H O
3
+
Br in CS
22
273K
OH
(1)
OH OH
Br CHO
and
(2)
OH OH
Br
CHO
and
(3)
OH OH
Br
CHO
and
(4)
OH OH
Br
CHO
and
Answer (2)
Sol. OH
Br /CS 2 2
OH
Br
273 K
Major
OH
(i) CHCl ,NaOH 3
OH
(ii) H O
3
+
Major
B
CHO
A
So option (2) is the correct answer
3.
Hydrolysis
373K
A
(CHCl )
48 2
B
(CHO)
48
B reacts with Hydroxyl amine but does not
give Tollen’s test. Identify A and B.
(1) 2,2-Dichlorobutane and Butanal
(2) 1,1-Dichlorobutane and Butanal
(3) 1,1-Dichlorobutane and 2-Butanone
(4) 2,2-Dichlorobutane and Butan-2-one
Answer (4)
Sol. Cl Cl
Hydrolysis
373 K
O
(C H Cl )
48 2
A
(C H O)
48
B
B is a ketone, cannot give Tollen’s test.
A ? 2, 2-Dichlorobutane
B ? Butan-2-one
So correct option should be (4)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
4. Given below are two statements : One is
labelled as Assertion A and the other is
labelled as Reason R
Assertion A : Dipole-dipole interactions are
only non-covalent interactions, resulting in
hydrogen bond formation.
Reason R : Fluorine is the most
electronegative element and hydrogen bonds
in HF are symmetrical.
In the light of the above statements, choose
the most appropriate answer from the options
given below:
(1) A is false but R is true
(2) Both A and R are true and R is the correct
explanation of A
(3) A is true but R is false
(4) Both A and R are true but R is NOT the
correct explanation of A
Answer (1)
Sol. ? Dipole - Dipole are not only the interaction
responsible for hydrogen bond formation.
Ion-dipole can also be responsible for
hydrogen bond formation.
? F is most electronegative element and
anhydrous HF in solid phase has
symmetrical hydrogen bonding
So the correct option is (1).
5. For the given reaction :
CH ? CHBr
1. NaNH 
2
CH 
3
2. Red hot iron tube, 873K 
(A)
(major product)
What is ‘A’?
(1) (2)
CH
3
HC
3
CH
3
(3) CH
3
CH
2
CH
2
NH
2
(4)
CH 
CH 
3
CH NH
2
Answer (2)
Sol. CH – CH = CH – Br
3
(i) NaNH
2
(ii) Red hot iron
      tube, 873 K
CH
3
CH
3
HC
3
Mesitylene
So the correct option should be (2).
6. The structure of Neoprene is :
(1)
HN N
NN
NHCH
2
n
NH
(2)
CH
2
CH
n
CN
(3)
CH CH      
2
CH
CH
2
CN
n
CH
2
CH
(4)            C      CH      CH
2
CH
2
Cl
n
Answer (4)
Sol. Neoprene is a polymer of monomer
chloroprene
nCH
2
C CH CH
2
Polymerisation
CH
2
C CH CH
2
n
Cl Cl
Cl Chloroprene
Neoprene
So the correct option should be (4)
7. For the given reaction :
CN
CH CH
23
Br
2
UV light
'A'
(Major product)
monobrominated
What is ‘A’?
(1)
CN
CHCH
23
Br
(2)
CN
CHCH
23
Br
(3)
CN
CHCH
23
Br
(4)
CN
CH
Br
CH
3
Answer (4)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
Sol.
CH CH
23
Br
2
UV light
CN
CH
Br
(Major Product)
Monobrominated
CN
So the correct option should be (4)
8. The orbital having two radial as well as two
angular nodes is :
(1) 3p (2) 4d
(3) 5d (4) 4f
Answer (3)
Sol. Number of radial nodes = (n – l – 1)
Number of angular nodes = l
for 5d; n = 5, l = 2
5d orbital has two radial nodes and two
angular nodes
So, the correct option should be (3)
9. Given below are two statements:
Statement I : o-Nitrophenol is steam volatile
due to intramolecular hydrogen bonding.
Statement II : o-Nitrophenol has high melting
due to hydrogen bonding.
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 false but statement II is true
(3) Statement I is true but statement II is false
(4) Both statement I and statement II are false
Answer (3)
Sol.
O
N
O
O
H
O- Nitrophenol
?
?
?
Has intramolecular hydrogen
bonding and steam volatile
Melting point is not affected  
by intramolecular hydrogen 
bonding
–
So the correct option should be (3)
10. Compound A is used as a strong oxidizing
agent is amphoteric in nature. It is the part of
lead storage batteries. Compound A is
(1) Pb
3
O
4
(2) PbSO
4
(3) PbO (4) PbO
2
Answer (4)
Sol. PbO
2
 is strong oxidizing agent because Pb
+4
 is
not stable and can be easily reduced to Pb
+2
.
PbO
2
 is used in lead storage batteries. It is
also amphoteric in nature
So, the answer should be (4)
11. On treating a compound with warm dil. H
2
SO
4
,
gas X is evolved which turns K
2
Cr
2
O
7
 paper
acidified with dil. H
2
SO
4
 to a green compound
Y. X and Y respectively are :
(1) X = SO
3
, Y = Cr
2
O
3
(2) X = SO
3
, Y = Cr
2
(SO
4
)
3
(3) X = SO
2
, Y = Cr
2
(SO
4
)
3
(4) X = SO
2
, Y = Cr
2
O
3
Answer (3)
Sol. SO
2
 + K
2
Cr
2
O
7
 + dil. H
2
SO
4
 ? SO
3
 + Cr
2
(SO
4
)
3
(green)
12. An amine on reaction with benzenesulphonyl
chloride produces a compound insoluble in
alkaline solution. This amine can be prepared
by ammonolysis of ethyl chloride. The correct
structure of amine is :
(1)
32 2 2 3
H
CH CH CH N—CH CH
(2) CH
3
CH
2
NH
2
(3) CH
3
CH
2
CH
2
NHCH
3
(4)
NH—CHCHCH
  
22 3
Answer (1)
Sol. Given amine on reaction with PhSO
2
Cl
produces a compound insoluble in alkaline
solution it means it is a 2º amine.
? It is prepared by ammonolysis of C
2
H
5
Cl, it
must contain an ethyl group
CHCH CHN—CHCH
32 2 2 3
  
H
|
 is a 2° amine as well
as it contain ethyl group.
Page 4


JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
PART–B : CHEMISTRY
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. Match List -I with List - II.
List - I List - II
(Ore) (Element Present)
(a) Kernite (i) Tin
(b) Cassiterite (ii) Boron
(c) Calamine (iii) Fluorine
(d) Cryolite (iv) Zinc
Choose the most appropriate answer from the
options given below:
(1) (a) ? (ii), (b) ? (i), (c) ? (iv), d ? (iii)
(2) (a) ? (iii), (b) ? (i), (c) ? (ii), d ? (iv)
(3) (a) ? (ii), (b) ? (iv), (c) ? (i), d ? (iii)
(4) (a) ? (i), (b) ? (iii), (c) ? (iv), d ? (ii)
Answer (1)
Sol. Kernite : Na
2
B
4
O
7
?4H
2
O (Boron)
Cassiterite : SnO
2
 (Tin)
Calamine : ZnCO
3
 (Zinc)
Cryolite : Na
3
AlF
6
 (Fluorine)
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
2. Identify the major products A and B
respectively in the following reactions of
phenol:
A B
(i) CHCl , NaOH
3
(ii) H O
3
+
Br in CS
22
273K
OH
(1)
OH OH
Br CHO
and
(2)
OH OH
Br
CHO
and
(3)
OH OH
Br
CHO
and
(4)
OH OH
Br
CHO
and
Answer (2)
Sol. OH
Br /CS 2 2
OH
Br
273 K
Major
OH
(i) CHCl ,NaOH 3
OH
(ii) H O
3
+
Major
B
CHO
A
So option (2) is the correct answer
3.
Hydrolysis
373K
A
(CHCl )
48 2
B
(CHO)
48
B reacts with Hydroxyl amine but does not
give Tollen’s test. Identify A and B.
(1) 2,2-Dichlorobutane and Butanal
(2) 1,1-Dichlorobutane and Butanal
(3) 1,1-Dichlorobutane and 2-Butanone
(4) 2,2-Dichlorobutane and Butan-2-one
Answer (4)
Sol. Cl Cl
Hydrolysis
373 K
O
(C H Cl )
48 2
A
(C H O)
48
B
B is a ketone, cannot give Tollen’s test.
A ? 2, 2-Dichlorobutane
B ? Butan-2-one
So correct option should be (4)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
4. Given below are two statements : One is
labelled as Assertion A and the other is
labelled as Reason R
Assertion A : Dipole-dipole interactions are
only non-covalent interactions, resulting in
hydrogen bond formation.
Reason R : Fluorine is the most
electronegative element and hydrogen bonds
in HF are symmetrical.
In the light of the above statements, choose
the most appropriate answer from the options
given below:
(1) A is false but R is true
(2) Both A and R are true and R is the correct
explanation of A
(3) A is true but R is false
(4) Both A and R are true but R is NOT the
correct explanation of A
Answer (1)
Sol. ? Dipole - Dipole are not only the interaction
responsible for hydrogen bond formation.
Ion-dipole can also be responsible for
hydrogen bond formation.
? F is most electronegative element and
anhydrous HF in solid phase has
symmetrical hydrogen bonding
So the correct option is (1).
5. For the given reaction :
CH ? CHBr
1. NaNH 
2
CH 
3
2. Red hot iron tube, 873K 
(A)
(major product)
What is ‘A’?
(1) (2)
CH
3
HC
3
CH
3
(3) CH
3
CH
2
CH
2
NH
2
(4)
CH 
CH 
3
CH NH
2
Answer (2)
Sol. CH – CH = CH – Br
3
(i) NaNH
2
(ii) Red hot iron
      tube, 873 K
CH
3
CH
3
HC
3
Mesitylene
So the correct option should be (2).
6. The structure of Neoprene is :
(1)
HN N
NN
NHCH
2
n
NH
(2)
CH
2
CH
n
CN
(3)
CH CH      
2
CH
CH
2
CN
n
CH
2
CH
(4)            C      CH      CH
2
CH
2
Cl
n
Answer (4)
Sol. Neoprene is a polymer of monomer
chloroprene
nCH
2
C CH CH
2
Polymerisation
CH
2
C CH CH
2
n
Cl Cl
Cl Chloroprene
Neoprene
So the correct option should be (4)
7. For the given reaction :
CN
CH CH
23
Br
2
UV light
'A'
(Major product)
monobrominated
What is ‘A’?
(1)
CN
CHCH
23
Br
(2)
CN
CHCH
23
Br
(3)
CN
CHCH
23
Br
(4)
CN
CH
Br
CH
3
Answer (4)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
Sol.
CH CH
23
Br
2
UV light
CN
CH
Br
(Major Product)
Monobrominated
CN
So the correct option should be (4)
8. The orbital having two radial as well as two
angular nodes is :
(1) 3p (2) 4d
(3) 5d (4) 4f
Answer (3)
Sol. Number of radial nodes = (n – l – 1)
Number of angular nodes = l
for 5d; n = 5, l = 2
5d orbital has two radial nodes and two
angular nodes
So, the correct option should be (3)
9. Given below are two statements:
Statement I : o-Nitrophenol is steam volatile
due to intramolecular hydrogen bonding.
Statement II : o-Nitrophenol has high melting
due to hydrogen bonding.
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 false but statement II is true
(3) Statement I is true but statement II is false
(4) Both statement I and statement II are false
Answer (3)
Sol.
O
N
O
O
H
O- Nitrophenol
?
?
?
Has intramolecular hydrogen
bonding and steam volatile
Melting point is not affected  
by intramolecular hydrogen 
bonding
–
So the correct option should be (3)
10. Compound A is used as a strong oxidizing
agent is amphoteric in nature. It is the part of
lead storage batteries. Compound A is
(1) Pb
3
O
4
(2) PbSO
4
(3) PbO (4) PbO
2
Answer (4)
Sol. PbO
2
 is strong oxidizing agent because Pb
+4
 is
not stable and can be easily reduced to Pb
+2
.
PbO
2
 is used in lead storage batteries. It is
also amphoteric in nature
So, the answer should be (4)
11. On treating a compound with warm dil. H
2
SO
4
,
gas X is evolved which turns K
2
Cr
2
O
7
 paper
acidified with dil. H
2
SO
4
 to a green compound
Y. X and Y respectively are :
(1) X = SO
3
, Y = Cr
2
O
3
(2) X = SO
3
, Y = Cr
2
(SO
4
)
3
(3) X = SO
2
, Y = Cr
2
(SO
4
)
3
(4) X = SO
2
, Y = Cr
2
O
3
Answer (3)
Sol. SO
2
 + K
2
Cr
2
O
7
 + dil. H
2
SO
4
 ? SO
3
 + Cr
2
(SO
4
)
3
(green)
12. An amine on reaction with benzenesulphonyl
chloride produces a compound insoluble in
alkaline solution. This amine can be prepared
by ammonolysis of ethyl chloride. The correct
structure of amine is :
(1)
32 2 2 3
H
CH CH CH N—CH CH
(2) CH
3
CH
2
NH
2
(3) CH
3
CH
2
CH
2
NHCH
3
(4)
NH—CHCHCH
  
22 3
Answer (1)
Sol. Given amine on reaction with PhSO
2
Cl
produces a compound insoluble in alkaline
solution it means it is a 2º amine.
? It is prepared by ammonolysis of C
2
H
5
Cl, it
must contain an ethyl group
CHCH CHN—CHCH
32 2 2 3
  
H
|
 is a 2° amine as well
as it contain ethyl group.
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
13. Which of the following is ‘a’ FALSE statement?
(1) Carius method is used for the estimation
of nitrogen in an organic compound
(2) Phosphoric acid produced on oxidation of
phosphorus present in an organic
compound is precipitated as Mg
2
P
2
O
7
 by
adding magnesia mixture
(3) Kjeldahl’s method is used for the
estimation of nitrogen in an organic
compound
(4) Carius tube is used in the estimation of
sulphur in an organic compound
Answer (1)
Sol. Carius method is used for the estimation for
halogens and sulphur in organic compound.
14. Find A, B and C in the following reactions :
NH
3
 + A + CO
2
 ? (NH
4
)
2
CO
3
(NH
4
)
2
CO
3
 + H
2
O + B ? NH
4
HCO
3
NH
4
HCO
3
 + NaCl ? NH
4
Cl + C
(1) A – H
2
O ; B – CO
2
 ; C – NaHCO
3
(2) A – H
2
O ; B – O
2
 ; C – Na
2
CO
3
(3) A – H
2
O ; B – O
2
 ; C – NaHCO
3
(4) A – O
2
 ; B – CO
2
 ; C – Na
2
CO
3
Answer (1)
Sol. These reactions are from Solvay process
(formation of washing soda)
A = H
2
O
B = CO
2
C = NaHCO
3
15. The presence of ozone in troposphere :
(1) Protects us from greenhouse effect
(2) Protects us from the X-ray radiation
(3) Generates photochemical smog
(4) Protects us from the UV radiation
Answer (3)
Sol. Ozone in stratosphere (not in troposphere)
prevent us from U.V. radiation.
Ozone in troposphere generates
photochemical smog.
16. Given below are two statements:
Statement I : A mixture of chloroform and
aniline can be separated by simple distillation.
Statement II : When separating aniline from a
mixture of aniline and water by steam
distillation aniline boils below its boiling point.
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 false
(2) Both statement I and statement II are true
(3) Statement I is true but statement II is false
(4) Statement I is false but statement II is true
Answer (2)
Sol. ? Mixture of chloroform and aniline can be
separated by simple distillation as these
two liquids have sufficient difference in
boiling point.
Chloroform (b.p. 334 K), aniline (b.p. 457 K)
? In steam distillation, if one of the
substances is water and the other, a water
insoluble substance (like aniline) then the
mixture will boil close to but below 373 K.
17. Statements about heavy water are given below.
A. Heavy water is used in exchange reactions
for the study of reaction mechanisms.
B. Heavy water is prepared by exhaustive
electrolysis of water.
C. Heavy water has higher boiling point than
ordinary water.
D. Viscosity of H
2
O is greater than D
2
O.
Choose the most appropriate answer from the
options given below:
(1) A and B only (2) A and C only
(3) A and D only (4) A, B and C only
Answer (4)
Sol. ? Viscosity of D
2
O is greater than H
2
O.
? B.P. of D
2
O is greater than H
2
O.
18. Which of the following vitamin is helpful in
delaying the blood clotting?
(1) Vitamin B (2) Vitamin E
(3) Vitamin K (4) Vitamin C
Answer (3)
Sol. Vitamin K is helpful in delaying the blood
clotting.
Page 5


JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
PART–B : CHEMISTRY
SECTION - I
Multiple Choice Questions: This section contains 20
multiple choice questions. Each question has 4
choices (1), (2), (3) and (4), out of which ONLY ONE
is correct.
Choose the correct answer :
1. Match List -I with List - II.
List - I List - II
(Ore) (Element Present)
(a) Kernite (i) Tin
(b) Cassiterite (ii) Boron
(c) Calamine (iii) Fluorine
(d) Cryolite (iv) Zinc
Choose the most appropriate answer from the
options given below:
(1) (a) ? (ii), (b) ? (i), (c) ? (iv), d ? (iii)
(2) (a) ? (iii), (b) ? (i), (c) ? (ii), d ? (iv)
(3) (a) ? (ii), (b) ? (iv), (c) ? (i), d ? (iii)
(4) (a) ? (i), (b) ? (iii), (c) ? (iv), d ? (ii)
Answer (1)
Sol. Kernite : Na
2
B
4
O
7
?4H
2
O (Boron)
Cassiterite : SnO
2
 (Tin)
Calamine : ZnCO
3
 (Zinc)
Cryolite : Na
3
AlF
6
 (Fluorine)
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
2. Identify the major products A and B
respectively in the following reactions of
phenol:
A B
(i) CHCl , NaOH
3
(ii) H O
3
+
Br in CS
22
273K
OH
(1)
OH OH
Br CHO
and
(2)
OH OH
Br
CHO
and
(3)
OH OH
Br
CHO
and
(4)
OH OH
Br
CHO
and
Answer (2)
Sol. OH
Br /CS 2 2
OH
Br
273 K
Major
OH
(i) CHCl ,NaOH 3
OH
(ii) H O
3
+
Major
B
CHO
A
So option (2) is the correct answer
3.
Hydrolysis
373K
A
(CHCl )
48 2
B
(CHO)
48
B reacts with Hydroxyl amine but does not
give Tollen’s test. Identify A and B.
(1) 2,2-Dichlorobutane and Butanal
(2) 1,1-Dichlorobutane and Butanal
(3) 1,1-Dichlorobutane and 2-Butanone
(4) 2,2-Dichlorobutane and Butan-2-one
Answer (4)
Sol. Cl Cl
Hydrolysis
373 K
O
(C H Cl )
48 2
A
(C H O)
48
B
B is a ketone, cannot give Tollen’s test.
A ? 2, 2-Dichlorobutane
B ? Butan-2-one
So correct option should be (4)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
4. Given below are two statements : One is
labelled as Assertion A and the other is
labelled as Reason R
Assertion A : Dipole-dipole interactions are
only non-covalent interactions, resulting in
hydrogen bond formation.
Reason R : Fluorine is the most
electronegative element and hydrogen bonds
in HF are symmetrical.
In the light of the above statements, choose
the most appropriate answer from the options
given below:
(1) A is false but R is true
(2) Both A and R are true and R is the correct
explanation of A
(3) A is true but R is false
(4) Both A and R are true but R is NOT the
correct explanation of A
Answer (1)
Sol. ? Dipole - Dipole are not only the interaction
responsible for hydrogen bond formation.
Ion-dipole can also be responsible for
hydrogen bond formation.
? F is most electronegative element and
anhydrous HF in solid phase has
symmetrical hydrogen bonding
So the correct option is (1).
5. For the given reaction :
CH ? CHBr
1. NaNH 
2
CH 
3
2. Red hot iron tube, 873K 
(A)
(major product)
What is ‘A’?
(1) (2)
CH
3
HC
3
CH
3
(3) CH
3
CH
2
CH
2
NH
2
(4)
CH 
CH 
3
CH NH
2
Answer (2)
Sol. CH – CH = CH – Br
3
(i) NaNH
2
(ii) Red hot iron
      tube, 873 K
CH
3
CH
3
HC
3
Mesitylene
So the correct option should be (2).
6. The structure of Neoprene is :
(1)
HN N
NN
NHCH
2
n
NH
(2)
CH
2
CH
n
CN
(3)
CH CH      
2
CH
CH
2
CN
n
CH
2
CH
(4)            C      CH      CH
2
CH
2
Cl
n
Answer (4)
Sol. Neoprene is a polymer of monomer
chloroprene
nCH
2
C CH CH
2
Polymerisation
CH
2
C CH CH
2
n
Cl Cl
Cl Chloroprene
Neoprene
So the correct option should be (4)
7. For the given reaction :
CN
CH CH
23
Br
2
UV light
'A'
(Major product)
monobrominated
What is ‘A’?
(1)
CN
CHCH
23
Br
(2)
CN
CHCH
23
Br
(3)
CN
CHCH
23
Br
(4)
CN
CH
Br
CH
3
Answer (4)
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
Sol.
CH CH
23
Br
2
UV light
CN
CH
Br
(Major Product)
Monobrominated
CN
So the correct option should be (4)
8. The orbital having two radial as well as two
angular nodes is :
(1) 3p (2) 4d
(3) 5d (4) 4f
Answer (3)
Sol. Number of radial nodes = (n – l – 1)
Number of angular nodes = l
for 5d; n = 5, l = 2
5d orbital has two radial nodes and two
angular nodes
So, the correct option should be (3)
9. Given below are two statements:
Statement I : o-Nitrophenol is steam volatile
due to intramolecular hydrogen bonding.
Statement II : o-Nitrophenol has high melting
due to hydrogen bonding.
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 false but statement II is true
(3) Statement I is true but statement II is false
(4) Both statement I and statement II are false
Answer (3)
Sol.
O
N
O
O
H
O- Nitrophenol
?
?
?
Has intramolecular hydrogen
bonding and steam volatile
Melting point is not affected  
by intramolecular hydrogen 
bonding
–
So the correct option should be (3)
10. Compound A is used as a strong oxidizing
agent is amphoteric in nature. It is the part of
lead storage batteries. Compound A is
(1) Pb
3
O
4
(2) PbSO
4
(3) PbO (4) PbO
2
Answer (4)
Sol. PbO
2
 is strong oxidizing agent because Pb
+4
 is
not stable and can be easily reduced to Pb
+2
.
PbO
2
 is used in lead storage batteries. It is
also amphoteric in nature
So, the answer should be (4)
11. On treating a compound with warm dil. H
2
SO
4
,
gas X is evolved which turns K
2
Cr
2
O
7
 paper
acidified with dil. H
2
SO
4
 to a green compound
Y. X and Y respectively are :
(1) X = SO
3
, Y = Cr
2
O
3
(2) X = SO
3
, Y = Cr
2
(SO
4
)
3
(3) X = SO
2
, Y = Cr
2
(SO
4
)
3
(4) X = SO
2
, Y = Cr
2
O
3
Answer (3)
Sol. SO
2
 + K
2
Cr
2
O
7
 + dil. H
2
SO
4
 ? SO
3
 + Cr
2
(SO
4
)
3
(green)
12. An amine on reaction with benzenesulphonyl
chloride produces a compound insoluble in
alkaline solution. This amine can be prepared
by ammonolysis of ethyl chloride. The correct
structure of amine is :
(1)
32 2 2 3
H
CH CH CH N—CH CH
(2) CH
3
CH
2
NH
2
(3) CH
3
CH
2
CH
2
NHCH
3
(4)
NH—CHCHCH
  
22 3
Answer (1)
Sol. Given amine on reaction with PhSO
2
Cl
produces a compound insoluble in alkaline
solution it means it is a 2º amine.
? It is prepared by ammonolysis of C
2
H
5
Cl, it
must contain an ethyl group
CHCH CHN—CHCH
32 2 2 3
  
H
|
 is a 2° amine as well
as it contain ethyl group.
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
13. Which of the following is ‘a’ FALSE statement?
(1) Carius method is used for the estimation
of nitrogen in an organic compound
(2) Phosphoric acid produced on oxidation of
phosphorus present in an organic
compound is precipitated as Mg
2
P
2
O
7
 by
adding magnesia mixture
(3) Kjeldahl’s method is used for the
estimation of nitrogen in an organic
compound
(4) Carius tube is used in the estimation of
sulphur in an organic compound
Answer (1)
Sol. Carius method is used for the estimation for
halogens and sulphur in organic compound.
14. Find A, B and C in the following reactions :
NH
3
 + A + CO
2
 ? (NH
4
)
2
CO
3
(NH
4
)
2
CO
3
 + H
2
O + B ? NH
4
HCO
3
NH
4
HCO
3
 + NaCl ? NH
4
Cl + C
(1) A – H
2
O ; B – CO
2
 ; C – NaHCO
3
(2) A – H
2
O ; B – O
2
 ; C – Na
2
CO
3
(3) A – H
2
O ; B – O
2
 ; C – NaHCO
3
(4) A – O
2
 ; B – CO
2
 ; C – Na
2
CO
3
Answer (1)
Sol. These reactions are from Solvay process
(formation of washing soda)
A = H
2
O
B = CO
2
C = NaHCO
3
15. The presence of ozone in troposphere :
(1) Protects us from greenhouse effect
(2) Protects us from the X-ray radiation
(3) Generates photochemical smog
(4) Protects us from the UV radiation
Answer (3)
Sol. Ozone in stratosphere (not in troposphere)
prevent us from U.V. radiation.
Ozone in troposphere generates
photochemical smog.
16. Given below are two statements:
Statement I : A mixture of chloroform and
aniline can be separated by simple distillation.
Statement II : When separating aniline from a
mixture of aniline and water by steam
distillation aniline boils below its boiling point.
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 false
(2) Both statement I and statement II are true
(3) Statement I is true but statement II is false
(4) Statement I is false but statement II is true
Answer (2)
Sol. ? Mixture of chloroform and aniline can be
separated by simple distillation as these
two liquids have sufficient difference in
boiling point.
Chloroform (b.p. 334 K), aniline (b.p. 457 K)
? In steam distillation, if one of the
substances is water and the other, a water
insoluble substance (like aniline) then the
mixture will boil close to but below 373 K.
17. Statements about heavy water are given below.
A. Heavy water is used in exchange reactions
for the study of reaction mechanisms.
B. Heavy water is prepared by exhaustive
electrolysis of water.
C. Heavy water has higher boiling point than
ordinary water.
D. Viscosity of H
2
O is greater than D
2
O.
Choose the most appropriate answer from the
options given below:
(1) A and B only (2) A and C only
(3) A and D only (4) A, B and C only
Answer (4)
Sol. ? Viscosity of D
2
O is greater than H
2
O.
? B.P. of D
2
O is greater than H
2
O.
18. Which of the following vitamin is helpful in
delaying the blood clotting?
(1) Vitamin B (2) Vitamin E
(3) Vitamin K (4) Vitamin C
Answer (3)
Sol. Vitamin K is helpful in delaying the blood
clotting.
JEE (MAIN)-2021 : Phase-1(26-02-2021)-M
19. Which one of the following lanthanoids does not
form MO
2
?
[M is lanthanoid metal]
(1) Nd
(2) Pr
(3) Dy
(4) Yb
Answer (4)
Sol. Nd (60) = 4f
4
 6s
2
Pr (59) = 4f
3
 6s
2
Dy (66) = 4f
10
 6s
2
Yb (70) = 4f
14
 6s
2
Yb
+2
 has fully-filled 4f orbital, it will require very
large amount of energy to reach +4 oxidation
state.
20. Match List-I with List-II.
List-I List-II
Electronic ?
i
H in kJ mol
–1
configuration
of elements
(a) 1s
2
 2s
2
(i) 801
(b) 1s
2
 2s
2
 2p
4
(ii) 899
(c) 1s
2
 2s
2
 2p
3
(iii) 1314
(d) 1s
2
 2s
2
 2p
1
(iv) 1402
Choose the most appropriate answer from the
options given below :
(1) (a) ? (i), (b) ? (iii), (c) ? (iv), (d) ? (ii)
(2) (a) ? (i), (b) ? (iv), (c) ? (iii), (d) ? (ii)
(3) (a) ? (iv), (b) ? (i), (c) ? (ii), (d) ? (iii)
(4) (a) ? (ii), (b) ? (iii), (c) ? (iv), (d) ? (i)
Answer (4)
Sol. On moving left to right in periodic table,
ionisation energy increases (generally) but
group-13 elements have lesser I.E than group-
2 due to stable ns
2
 electronic configuration of
group-2 elements and group-15 elements have
greater I.E than group-16 elements due to half-
filled stable np
3
 configuration of group-15
elements.
? Overall order of I.E should be
c > b > a > d
SECTION - II
Numerical Value Type Questions: This section
contains 10 questions. In Section II, attempt any five
questions out of 10. The answer to each question is a
NUMERICAL VALUE. For each question, enter the
correct numerical value (in decimal notation,
truncated/rounded-off to the second decimal place;
e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using
the mouse and the on-screen virtual numeric keypad
in the place designated to enter the answer.
1. For a chemical reaction A + B  C + D
(
T
?
r
H = 80 kJ mol
–1
) the entropy change
T
?
r
S depends on the temperature T (in K) as
T
?
r
S = 2T (J K
–1
 mol
–1
).
Minimum temperature at which it will become
spontaneous is ____________ K. (Integer)
Answer (200)
Sol. A + B  C + D
For a reaction to be spontaneous
?
r
G < 0
or ?
r
G° < 0 (For the given case)
??
r
H° – T ?
r
S° < 0
? 80 × 1000 – T × 2T < 0
? T
2
 > 40000
T > 200
?
min
T200K ?
2. 224 mL of SO
2(g)
 at 298 K and 1 atm is passed
through 100 mL of 0.1 M NaOH solution. The
non-volatile solute produced is dissolved in 36
g of water. The lowering of vapour pressure of
solution (assuming the solution is dilute)
()
( )
°
=
2
HO
P24mmofHg is x × 10
–2
 mm of Hg, the
value of x is _________. (Integer answer)
Answer (24)
Sol.
2
SO
10.224
n 0.0092 0.01moles
0.082 298
?
?? ?
?
NaOH + SO
2
 ? NaHSO
3
3
NaHSO
n0.01 ?
33
NaHSO Na HSO
??
??
Ignoring the dissociation of 
3
HSO
?
 into H
+
 and
2
3
SO
?
van’t Hoff factor (i) = 2
3 2
2 2
3
o
NaHSO HO S
o
HO HO
NaHSO
in
PP
nin P
?
?
?
Read More
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