Page 1
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 1
Date : 2
nd
September 2020
Time : 2 : 00 pm - 5 : 00 pm
Subject : Chemistry
1. Cast iron is used for the manufacture of :
(1) Wrought iron and steel (2) Wrought iron and pig iron
(3) Wrougth iron, pig iron and steel (4) Pig iron, scrap iron and steel
Sol. 1
Refer topic metallurgy
2. The shape/structure of [XeF
5
]
—
and XeO
3
F
2
, respectively, are :
(1) Pentagonal planar and trigonal bipyramidal
(2) Trigonal bipyramidal and trigonal bipyramidal
(3) Octahedral and square pyramidal
(4) Trigonal bipyramidal and pentagonal planar
Sol. 1
[XeF
5
]
–
5BP + 2LP = 7VSEP ? sp
3
d
3
hybridisation
XeO
3
F
2
5BP + 0LP = 5VSEP ? sp
3
d hybridisation
3. Simplified absorption spectra of three complexes ((i), (ii) and (iii)) of M
n+
ion are pro-
vided below; their ?
max
values are marked as A, B and C respectively. The correct match
between the complexes and their ?
max
values is :
A
B
C
Absorption
Wave length (nm)
(i) [M(NCS)
6
]
(–6+n)
(ii) [MF
6
]
(–6+n)
(iii) [M(NH
3
)
6
]
n+
(1) A-(i), B-(ii), C-(iii) (2) A-(iii), B-(i), C-(ii)
(3) A-(ii), B-(iii), C-(i) (4) A-(ii), B-(i), C-(iii)
Sol. 2
? =
? ? absorbedf max
hc
?
A ? NH
3
comp (iii)
B ? NCS comp (i)
C ? F
–
comp (ii)
using spectrochemical series of ligand
F
–
< NCS
–
< NH
3
order of ?+e
crystal field spliting energy
So. NH
3
complex ? A
F
–
complex – C
NCS
–
complex ? B
Page 2
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 1
Date : 2
nd
September 2020
Time : 2 : 00 pm - 5 : 00 pm
Subject : Chemistry
1. Cast iron is used for the manufacture of :
(1) Wrought iron and steel (2) Wrought iron and pig iron
(3) Wrougth iron, pig iron and steel (4) Pig iron, scrap iron and steel
Sol. 1
Refer topic metallurgy
2. The shape/structure of [XeF
5
]
—
and XeO
3
F
2
, respectively, are :
(1) Pentagonal planar and trigonal bipyramidal
(2) Trigonal bipyramidal and trigonal bipyramidal
(3) Octahedral and square pyramidal
(4) Trigonal bipyramidal and pentagonal planar
Sol. 1
[XeF
5
]
–
5BP + 2LP = 7VSEP ? sp
3
d
3
hybridisation
XeO
3
F
2
5BP + 0LP = 5VSEP ? sp
3
d hybridisation
3. Simplified absorption spectra of three complexes ((i), (ii) and (iii)) of M
n+
ion are pro-
vided below; their ?
max
values are marked as A, B and C respectively. The correct match
between the complexes and their ?
max
values is :
A
B
C
Absorption
Wave length (nm)
(i) [M(NCS)
6
]
(–6+n)
(ii) [MF
6
]
(–6+n)
(iii) [M(NH
3
)
6
]
n+
(1) A-(i), B-(ii), C-(iii) (2) A-(iii), B-(i), C-(ii)
(3) A-(ii), B-(iii), C-(i) (4) A-(ii), B-(i), C-(iii)
Sol. 2
? =
? ? absorbedf max
hc
?
A ? NH
3
comp (iii)
B ? NCS comp (i)
C ? F
–
comp (ii)
using spectrochemical series of ligand
F
–
< NCS
–
< NH
3
order of ?+e
crystal field spliting energy
So. NH
3
complex ? A
F
–
complex – C
NCS
–
complex ? B
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 2
4. The correct observation in the following reactions is :
Gly cosidic bond Seliwanoff 's
Cleavage reagent
(Hydrolysis)
Sucrose A B ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
(1) Formation of red colour (2) Formation of blue colour
(3) Formation of violet colour (4) Gives no colour
Sol. 1
Glycosidic bond Seliwanoff 's
Cleavage reagent
(Hydrolysis)
Sucrose Glucose Fructose RedColour ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
5. The results given in the below table were obtained during kinetic studies of the follow-
ing reaction :
2A + B ? C + D
Experiment
[A]/
molL
–1
[B]/
molL
–1
Initial rate/
molL
–1
min
–1
I 0.1 0.1 6.00 × 10
–3
II 0.1 0.2 2.40 × 10
–2
III 0.2 0.1 1.20 × 10
–2
IV X 0.2 7.20 × 10
–2
V 0.3 Y 2.88 × 10
–1
X and Y in the given table are respectively :
(1) 0.4, 0.4 (2) 0.3, 0.4 (3) 0.4, 0.3 (4) 0.3, 0.3
Sol. 2
2A + B
? ? ? ?
C + D
Exp. (I) 6 × 10
–3
= K (0.1)
p
(0.1)
q
(II) 2.4 × 10
–2
= K (0.1)
p
(0.2)
q
(III) 1.2 × 10
–2
= K (0.2)
p
(0.1)
q
exp(I)
exp(II)
1
4
=
q
1
2
? ?
? ?
? ?
? q = 2
Exp.(I)
Exp.(III)
1
2
=
p
1
2
? ?
? ?
? ?
? p = 1
exp. (I) ? exp (IV)
–2
–2
0.6 10
7.2 10
?
?
=
1 2
0.1 0.1
.
x 0.2
? ? ? ?
? ? ? ?
? ? ? ?
1 0.1 1
–
12 x 4
?
[x] = 0.3
exp (I) ? exp(V)
–2
–1
0.6 10
2.88 10
?
?
=
2 1
0.1 0.1
0.3 y
? ? ? ?
?
? ? ? ?
? ? ? ?
–2
2
2
1 1 10
y 0.16
48 3 y
? ? ? ?
y = 0.4
Ans(2)
Page 3
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 1
Date : 2
nd
September 2020
Time : 2 : 00 pm - 5 : 00 pm
Subject : Chemistry
1. Cast iron is used for the manufacture of :
(1) Wrought iron and steel (2) Wrought iron and pig iron
(3) Wrougth iron, pig iron and steel (4) Pig iron, scrap iron and steel
Sol. 1
Refer topic metallurgy
2. The shape/structure of [XeF
5
]
—
and XeO
3
F
2
, respectively, are :
(1) Pentagonal planar and trigonal bipyramidal
(2) Trigonal bipyramidal and trigonal bipyramidal
(3) Octahedral and square pyramidal
(4) Trigonal bipyramidal and pentagonal planar
Sol. 1
[XeF
5
]
–
5BP + 2LP = 7VSEP ? sp
3
d
3
hybridisation
XeO
3
F
2
5BP + 0LP = 5VSEP ? sp
3
d hybridisation
3. Simplified absorption spectra of three complexes ((i), (ii) and (iii)) of M
n+
ion are pro-
vided below; their ?
max
values are marked as A, B and C respectively. The correct match
between the complexes and their ?
max
values is :
A
B
C
Absorption
Wave length (nm)
(i) [M(NCS)
6
]
(–6+n)
(ii) [MF
6
]
(–6+n)
(iii) [M(NH
3
)
6
]
n+
(1) A-(i), B-(ii), C-(iii) (2) A-(iii), B-(i), C-(ii)
(3) A-(ii), B-(iii), C-(i) (4) A-(ii), B-(i), C-(iii)
Sol. 2
? =
? ? absorbedf max
hc
?
A ? NH
3
comp (iii)
B ? NCS comp (i)
C ? F
–
comp (ii)
using spectrochemical series of ligand
F
–
< NCS
–
< NH
3
order of ?+e
crystal field spliting energy
So. NH
3
complex ? A
F
–
complex – C
NCS
–
complex ? B
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 2
4. The correct observation in the following reactions is :
Gly cosidic bond Seliwanoff 's
Cleavage reagent
(Hydrolysis)
Sucrose A B ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
(1) Formation of red colour (2) Formation of blue colour
(3) Formation of violet colour (4) Gives no colour
Sol. 1
Glycosidic bond Seliwanoff 's
Cleavage reagent
(Hydrolysis)
Sucrose Glucose Fructose RedColour ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
5. The results given in the below table were obtained during kinetic studies of the follow-
ing reaction :
2A + B ? C + D
Experiment
[A]/
molL
–1
[B]/
molL
–1
Initial rate/
molL
–1
min
–1
I 0.1 0.1 6.00 × 10
–3
II 0.1 0.2 2.40 × 10
–2
III 0.2 0.1 1.20 × 10
–2
IV X 0.2 7.20 × 10
–2
V 0.3 Y 2.88 × 10
–1
X and Y in the given table are respectively :
(1) 0.4, 0.4 (2) 0.3, 0.4 (3) 0.4, 0.3 (4) 0.3, 0.3
Sol. 2
2A + B
? ? ? ?
C + D
Exp. (I) 6 × 10
–3
= K (0.1)
p
(0.1)
q
(II) 2.4 × 10
–2
= K (0.1)
p
(0.2)
q
(III) 1.2 × 10
–2
= K (0.2)
p
(0.1)
q
exp(I)
exp(II)
1
4
=
q
1
2
? ?
? ?
? ?
? q = 2
Exp.(I)
Exp.(III)
1
2
=
p
1
2
? ?
? ?
? ?
? p = 1
exp. (I) ? exp (IV)
–2
–2
0.6 10
7.2 10
?
?
=
1 2
0.1 0.1
.
x 0.2
? ? ? ?
? ? ? ?
? ? ? ?
1 0.1 1
–
12 x 4
?
[x] = 0.3
exp (I) ? exp(V)
–2
–1
0.6 10
2.88 10
?
?
=
2 1
0.1 0.1
0.3 y
? ? ? ?
?
? ? ? ?
? ? ? ?
–2
2
2
1 1 10
y 0.16
48 3 y
? ? ? ?
y = 0.4
Ans(2)
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 3
6. Match the type of interaction in column A with the distance dependence of their inter-
action energy in column B :
A B
(I) ion-ion (a)
1
r
(II) dipole-dipole (b)
2
1
r
(III) London dispersion (c)
3
1
r
(d)
6
1
r
(1) (I)-(a), (II)-(b), (III)-(d) (2) (I)-(a), (II)-(b), (III)-(c)
(3) (I)-(b), (II)-(d), (III)-(c) (4) (I)-(a), (II)-(c), (III)-(d)
Sol. 4
ion - ion ?
1
r
dipole – dipole ?
3
1
r
Londong dispersion ?
6
1
r
7. The major product obtained from E
2
– elimination of 3-bromo-2-fluoropentane is :
(1)
CH CH CH=C–F
3 2
CH
3
(2)
CH CH –CH–CH=CH
3 2 2
Br
(3)
CH –CH –C=C–CH
3 2 3
Br
(4)
CH –CH=CH–CH–CH
3 3
F
Sol. 1
8. Consider the reaction sequence given below :
Br
OH
H O
2
OH + Br ....(1)
rate = k[t-BuBr]
OH
C H OH
2 5 H C
2
CH
3
CH
3
+ HOH + Br ....(2)
rate = k[t-BuBr][OH ]
Which of the following statements is true :
(1) Changing the concentration of base will have no effect on reaction (1).
(2) Doubling the concentration of base will double the rate of both the reactions.
(3) Changing the base from
OH
?
to
OR
?
will have no effect on reaction (2).
(4) Changing the concentration of base will have no effect on reaction (2).
Page 4
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 1
Date : 2
nd
September 2020
Time : 2 : 00 pm - 5 : 00 pm
Subject : Chemistry
1. Cast iron is used for the manufacture of :
(1) Wrought iron and steel (2) Wrought iron and pig iron
(3) Wrougth iron, pig iron and steel (4) Pig iron, scrap iron and steel
Sol. 1
Refer topic metallurgy
2. The shape/structure of [XeF
5
]
—
and XeO
3
F
2
, respectively, are :
(1) Pentagonal planar and trigonal bipyramidal
(2) Trigonal bipyramidal and trigonal bipyramidal
(3) Octahedral and square pyramidal
(4) Trigonal bipyramidal and pentagonal planar
Sol. 1
[XeF
5
]
–
5BP + 2LP = 7VSEP ? sp
3
d
3
hybridisation
XeO
3
F
2
5BP + 0LP = 5VSEP ? sp
3
d hybridisation
3. Simplified absorption spectra of three complexes ((i), (ii) and (iii)) of M
n+
ion are pro-
vided below; their ?
max
values are marked as A, B and C respectively. The correct match
between the complexes and their ?
max
values is :
A
B
C
Absorption
Wave length (nm)
(i) [M(NCS)
6
]
(–6+n)
(ii) [MF
6
]
(–6+n)
(iii) [M(NH
3
)
6
]
n+
(1) A-(i), B-(ii), C-(iii) (2) A-(iii), B-(i), C-(ii)
(3) A-(ii), B-(iii), C-(i) (4) A-(ii), B-(i), C-(iii)
Sol. 2
? =
? ? absorbedf max
hc
?
A ? NH
3
comp (iii)
B ? NCS comp (i)
C ? F
–
comp (ii)
using spectrochemical series of ligand
F
–
< NCS
–
< NH
3
order of ?+e
crystal field spliting energy
So. NH
3
complex ? A
F
–
complex – C
NCS
–
complex ? B
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 2
4. The correct observation in the following reactions is :
Gly cosidic bond Seliwanoff 's
Cleavage reagent
(Hydrolysis)
Sucrose A B ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
(1) Formation of red colour (2) Formation of blue colour
(3) Formation of violet colour (4) Gives no colour
Sol. 1
Glycosidic bond Seliwanoff 's
Cleavage reagent
(Hydrolysis)
Sucrose Glucose Fructose RedColour ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
5. The results given in the below table were obtained during kinetic studies of the follow-
ing reaction :
2A + B ? C + D
Experiment
[A]/
molL
–1
[B]/
molL
–1
Initial rate/
molL
–1
min
–1
I 0.1 0.1 6.00 × 10
–3
II 0.1 0.2 2.40 × 10
–2
III 0.2 0.1 1.20 × 10
–2
IV X 0.2 7.20 × 10
–2
V 0.3 Y 2.88 × 10
–1
X and Y in the given table are respectively :
(1) 0.4, 0.4 (2) 0.3, 0.4 (3) 0.4, 0.3 (4) 0.3, 0.3
Sol. 2
2A + B
? ? ? ?
C + D
Exp. (I) 6 × 10
–3
= K (0.1)
p
(0.1)
q
(II) 2.4 × 10
–2
= K (0.1)
p
(0.2)
q
(III) 1.2 × 10
–2
= K (0.2)
p
(0.1)
q
exp(I)
exp(II)
1
4
=
q
1
2
? ?
? ?
? ?
? q = 2
Exp.(I)
Exp.(III)
1
2
=
p
1
2
? ?
? ?
? ?
? p = 1
exp. (I) ? exp (IV)
–2
–2
0.6 10
7.2 10
?
?
=
1 2
0.1 0.1
.
x 0.2
? ? ? ?
? ? ? ?
? ? ? ?
1 0.1 1
–
12 x 4
?
[x] = 0.3
exp (I) ? exp(V)
–2
–1
0.6 10
2.88 10
?
?
=
2 1
0.1 0.1
0.3 y
? ? ? ?
?
? ? ? ?
? ? ? ?
–2
2
2
1 1 10
y 0.16
48 3 y
? ? ? ?
y = 0.4
Ans(2)
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 3
6. Match the type of interaction in column A with the distance dependence of their inter-
action energy in column B :
A B
(I) ion-ion (a)
1
r
(II) dipole-dipole (b)
2
1
r
(III) London dispersion (c)
3
1
r
(d)
6
1
r
(1) (I)-(a), (II)-(b), (III)-(d) (2) (I)-(a), (II)-(b), (III)-(c)
(3) (I)-(b), (II)-(d), (III)-(c) (4) (I)-(a), (II)-(c), (III)-(d)
Sol. 4
ion - ion ?
1
r
dipole – dipole ?
3
1
r
Londong dispersion ?
6
1
r
7. The major product obtained from E
2
– elimination of 3-bromo-2-fluoropentane is :
(1)
CH CH CH=C–F
3 2
CH
3
(2)
CH CH –CH–CH=CH
3 2 2
Br
(3)
CH –CH –C=C–CH
3 2 3
Br
(4)
CH –CH=CH–CH–CH
3 3
F
Sol. 1
8. Consider the reaction sequence given below :
Br
OH
H O
2
OH + Br ....(1)
rate = k[t-BuBr]
OH
C H OH
2 5 H C
2
CH
3
CH
3
+ HOH + Br ....(2)
rate = k[t-BuBr][OH ]
Which of the following statements is true :
(1) Changing the concentration of base will have no effect on reaction (1).
(2) Doubling the concentration of base will double the rate of both the reactions.
(3) Changing the base from
OH
?
to
OR
?
will have no effect on reaction (2).
(4) Changing the concentration of base will have no effect on reaction (2).
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 4
Sol. 1
9. The size of a raw mango shrinks to a much smaller size when kept in a concentrated
salt solution. Which one of the following process can explain this ?
(1) Diffusion (2) Osmosis
(3) Reverse osmosis (4) Dialysis
Sol. 2
Theoritical
Ans. Osmosis
Option (2)
10. If you spill a chemical toiled cleaning liquid on your hand, your first aid would be :
(1) Aqueous NH
3
(2) Aqueous NaHCO
3
(3) Aqueous NaOH (4) Vinegar
Sol. 2
Fact
11. Arrange the followig labelled hydrogens in decreasing order of acidity :
NO
2
H –O
O– H
COO H
a
b
c
d
(1) b > a > c > d (2) b > c > d > a
(3) c > b > d > a (4) c > b > a > d
Sol. 2
Order of acidic strength
12. An organic compound ‘A’ (C
9
H
10
O) when treated with conc. HI undergoes cleavage to
yield compounds ‘B’ and ‘C’. ‘B’ gives yellow precipitate with AgNO
3
where as ‘C’
tautomerizes to ‘D’. ‘D’ gives positive iodoform test. ‘A’ could be :
(1)
CH –O–CH=CH
2 2 (2)
O–CH=CH
2 H C
3
(3)
O–CH –CH=CH
2 2 (4)
O–CH=CH=CH
3
Page 5
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 1
Date : 2
nd
September 2020
Time : 2 : 00 pm - 5 : 00 pm
Subject : Chemistry
1. Cast iron is used for the manufacture of :
(1) Wrought iron and steel (2) Wrought iron and pig iron
(3) Wrougth iron, pig iron and steel (4) Pig iron, scrap iron and steel
Sol. 1
Refer topic metallurgy
2. The shape/structure of [XeF
5
]
—
and XeO
3
F
2
, respectively, are :
(1) Pentagonal planar and trigonal bipyramidal
(2) Trigonal bipyramidal and trigonal bipyramidal
(3) Octahedral and square pyramidal
(4) Trigonal bipyramidal and pentagonal planar
Sol. 1
[XeF
5
]
–
5BP + 2LP = 7VSEP ? sp
3
d
3
hybridisation
XeO
3
F
2
5BP + 0LP = 5VSEP ? sp
3
d hybridisation
3. Simplified absorption spectra of three complexes ((i), (ii) and (iii)) of M
n+
ion are pro-
vided below; their ?
max
values are marked as A, B and C respectively. The correct match
between the complexes and their ?
max
values is :
A
B
C
Absorption
Wave length (nm)
(i) [M(NCS)
6
]
(–6+n)
(ii) [MF
6
]
(–6+n)
(iii) [M(NH
3
)
6
]
n+
(1) A-(i), B-(ii), C-(iii) (2) A-(iii), B-(i), C-(ii)
(3) A-(ii), B-(iii), C-(i) (4) A-(ii), B-(i), C-(iii)
Sol. 2
? =
? ? absorbedf max
hc
?
A ? NH
3
comp (iii)
B ? NCS comp (i)
C ? F
–
comp (ii)
using spectrochemical series of ligand
F
–
< NCS
–
< NH
3
order of ?+e
crystal field spliting energy
So. NH
3
complex ? A
F
–
complex – C
NCS
–
complex ? B
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 2
4. The correct observation in the following reactions is :
Gly cosidic bond Seliwanoff 's
Cleavage reagent
(Hydrolysis)
Sucrose A B ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
(1) Formation of red colour (2) Formation of blue colour
(3) Formation of violet colour (4) Gives no colour
Sol. 1
Glycosidic bond Seliwanoff 's
Cleavage reagent
(Hydrolysis)
Sucrose Glucose Fructose RedColour ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
5. The results given in the below table were obtained during kinetic studies of the follow-
ing reaction :
2A + B ? C + D
Experiment
[A]/
molL
–1
[B]/
molL
–1
Initial rate/
molL
–1
min
–1
I 0.1 0.1 6.00 × 10
–3
II 0.1 0.2 2.40 × 10
–2
III 0.2 0.1 1.20 × 10
–2
IV X 0.2 7.20 × 10
–2
V 0.3 Y 2.88 × 10
–1
X and Y in the given table are respectively :
(1) 0.4, 0.4 (2) 0.3, 0.4 (3) 0.4, 0.3 (4) 0.3, 0.3
Sol. 2
2A + B
? ? ? ?
C + D
Exp. (I) 6 × 10
–3
= K (0.1)
p
(0.1)
q
(II) 2.4 × 10
–2
= K (0.1)
p
(0.2)
q
(III) 1.2 × 10
–2
= K (0.2)
p
(0.1)
q
exp(I)
exp(II)
1
4
=
q
1
2
? ?
? ?
? ?
? q = 2
Exp.(I)
Exp.(III)
1
2
=
p
1
2
? ?
? ?
? ?
? p = 1
exp. (I) ? exp (IV)
–2
–2
0.6 10
7.2 10
?
?
=
1 2
0.1 0.1
.
x 0.2
? ? ? ?
? ? ? ?
? ? ? ?
1 0.1 1
–
12 x 4
?
[x] = 0.3
exp (I) ? exp(V)
–2
–1
0.6 10
2.88 10
?
?
=
2 1
0.1 0.1
0.3 y
? ? ? ?
?
? ? ? ?
? ? ? ?
–2
2
2
1 1 10
y 0.16
48 3 y
? ? ? ?
y = 0.4
Ans(2)
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 3
6. Match the type of interaction in column A with the distance dependence of their inter-
action energy in column B :
A B
(I) ion-ion (a)
1
r
(II) dipole-dipole (b)
2
1
r
(III) London dispersion (c)
3
1
r
(d)
6
1
r
(1) (I)-(a), (II)-(b), (III)-(d) (2) (I)-(a), (II)-(b), (III)-(c)
(3) (I)-(b), (II)-(d), (III)-(c) (4) (I)-(a), (II)-(c), (III)-(d)
Sol. 4
ion - ion ?
1
r
dipole – dipole ?
3
1
r
Londong dispersion ?
6
1
r
7. The major product obtained from E
2
– elimination of 3-bromo-2-fluoropentane is :
(1)
CH CH CH=C–F
3 2
CH
3
(2)
CH CH –CH–CH=CH
3 2 2
Br
(3)
CH –CH –C=C–CH
3 2 3
Br
(4)
CH –CH=CH–CH–CH
3 3
F
Sol. 1
8. Consider the reaction sequence given below :
Br
OH
H O
2
OH + Br ....(1)
rate = k[t-BuBr]
OH
C H OH
2 5 H C
2
CH
3
CH
3
+ HOH + Br ....(2)
rate = k[t-BuBr][OH ]
Which of the following statements is true :
(1) Changing the concentration of base will have no effect on reaction (1).
(2) Doubling the concentration of base will double the rate of both the reactions.
(3) Changing the base from
OH
?
to
OR
?
will have no effect on reaction (2).
(4) Changing the concentration of base will have no effect on reaction (2).
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 4
Sol. 1
9. The size of a raw mango shrinks to a much smaller size when kept in a concentrated
salt solution. Which one of the following process can explain this ?
(1) Diffusion (2) Osmosis
(3) Reverse osmosis (4) Dialysis
Sol. 2
Theoritical
Ans. Osmosis
Option (2)
10. If you spill a chemical toiled cleaning liquid on your hand, your first aid would be :
(1) Aqueous NH
3
(2) Aqueous NaHCO
3
(3) Aqueous NaOH (4) Vinegar
Sol. 2
Fact
11. Arrange the followig labelled hydrogens in decreasing order of acidity :
NO
2
H –O
O– H
COO H
a
b
c
d
(1) b > a > c > d (2) b > c > d > a
(3) c > b > d > a (4) c > b > a > d
Sol. 2
Order of acidic strength
12. An organic compound ‘A’ (C
9
H
10
O) when treated with conc. HI undergoes cleavage to
yield compounds ‘B’ and ‘C’. ‘B’ gives yellow precipitate with AgNO
3
where as ‘C’
tautomerizes to ‘D’. ‘D’ gives positive iodoform test. ‘A’ could be :
(1)
CH –O–CH=CH
2 2 (2)
O–CH=CH
2 H C
3
(3)
O–CH –CH=CH
2 2 (4)
O–CH=CH=CH
3
JEE Main 2020 Paper
2
nd
September 2020 | (Shift-2), Chemistry Page | 5
Sol. 1
CH – O – CH = CH
2 2
Conc. HI
CH – I + HO – CH = CH
2 2
AgNO
3
yellow ppt
tautomerize
CH – CH = O
3
(+ve iodoform)
B
C
D
13. Two elements A and B have similar chemical properties. They don’t form solid
hydrogencarbonates, but react with nitrogen to form nitrides. A and B, respectively,
are :
(1) Na and Ca (2) Cs and Ba
(3) Na and Rb (4) Li and Mg
Sol. 4
LiHCO
3
& Mg(HCO
3
)
2
does not exist in solid form but both forms nitrides with nitrogen
gas
14. The number of subshells associated with n = 4 and m = –2 quantum numbers is :
(1) 4 (2) 8 (3) 2 (4) 16
Sol. 3
n = 4
?
= 0 m = 0
?
= 1 m = – 1, 0, +1
?
= 2 m = – 2, + 2, –1, +1, 0
?
= 3 m 3, 2, 1,0 ? ? ? ?
Ans. ‘2’ Subshells
Option (3)
15. The major product of the following reaction is :
OH
NO
2
CH
3
conc. HNO + conc. H SO
3 2 4
(1)
OH
NO
2
H C
3
O N
2
(2)
OH
NO
2
H C
3
NO
2
NO
2
(3)
OH
NO
2
H C
3
NO
2
(4)
OH
NO
2
H C
3
NO
2
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