JEE Exam > JEE Notes > Mock Tests for JEE Main and Advanced 2025 > JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions
JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
PAGE # 1
PART : CHEMISTRY
SECTION – 1 : (Maximum Marks : 80)
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. V 34 . 0 º E
Cu / Cu
2 ? ?
V 522 . 0 º E
Cu / Cu
? ?
? ? ?
Cu / Cu
2 º E ?
(1) 0.158V (2) –0.158 V (3) 0.182 V (4) –0.182 V
Ans. (1)
Sol.
Cu
2+
º
1
E
Cu
+
0.522
Cu
0.34
2 × 0.34 = 522 . 0 1 E
º
1
? ?
522 . 0 – 68 . 0 E
º
1
?
158 . 0 E
º
1
?
2. Correct order of electron gain enthalpy (kJ/mole) of F, Cl, Br, I
F, Cl, Br, I dh bySDVªkWu xzg.k ,UFkSYih (kJ/eksy) dk lgh Øe gS \
(1) –296, – 325, – 333, – 349
(2) –333, – 349, – 325, –296
(3) –349, – 333, – 325, – 296
(4) –325, – 333, – 349, – 296
Ans. (2)
3. Arrange the following in order of their pKb value
fuEu dks buds pKb eku ds Øe O;ofLFkr dhft, &
(A)
N NH
N
(B) CH3 –NH –CH3 (C) CH3 –CH=NH
(1) A > B > C (2) B > A > C (3) C > B > A (4) B > C > A
Ans. (3)
Sol. Option "A" represent Guanadine, the conjugate acid of which is resonance stabilised. The option 'B' is
aliphatic amine, here the 'N' is sp
3
whereas in option 'C' the 'N' is sp
2
, hence B is more basic than C.
fodYi "A" xqokusfMu izdkj dk ;kSfxd iznf'kZr djrk gS ftldk la;qXeh vEy vuqukn }kjk LFkk;hd`r gSA fodYi (B)
,fyQsfVd ,ehu iznf'kZr djrk gSA ftlesa 'N' sp
3
ladfjr gS tcfd fodYi 'C' esa 'N' sp
2
ladfjr gSA vr% B, C dh
rqyuk esa vfèkd {kkjh; gksxkA
Page 2
PAGE # 1
PART : CHEMISTRY
SECTION – 1 : (Maximum Marks : 80)
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. V 34 . 0 º E
Cu / Cu
2 ? ?
V 522 . 0 º E
Cu / Cu
? ?
? ? ?
Cu / Cu
2 º E ?
(1) 0.158V (2) –0.158 V (3) 0.182 V (4) –0.182 V
Ans. (1)
Sol.
Cu
2+
º
1
E
Cu
+
0.522
Cu
0.34
2 × 0.34 = 522 . 0 1 E
º
1
? ?
522 . 0 – 68 . 0 E
º
1
?
158 . 0 E
º
1
?
2. Correct order of electron gain enthalpy (kJ/mole) of F, Cl, Br, I
F, Cl, Br, I dh bySDVªkWu xzg.k ,UFkSYih (kJ/eksy) dk lgh Øe gS \
(1) –296, – 325, – 333, – 349
(2) –333, – 349, – 325, –296
(3) –349, – 333, – 325, – 296
(4) –325, – 333, – 349, – 296
Ans. (2)
3. Arrange the following in order of their pKb value
fuEu dks buds pKb eku ds Øe O;ofLFkr dhft, &
(A)
N NH
N
(B) CH3 –NH –CH3 (C) CH3 –CH=NH
(1) A > B > C (2) B > A > C (3) C > B > A (4) B > C > A
Ans. (3)
Sol. Option "A" represent Guanadine, the conjugate acid of which is resonance stabilised. The option 'B' is
aliphatic amine, here the 'N' is sp
3
whereas in option 'C' the 'N' is sp
2
, hence B is more basic than C.
fodYi "A" xqokusfMu izdkj dk ;kSfxd iznf'kZr djrk gS ftldk la;qXeh vEy vuqukn }kjk LFkk;hd`r gSA fodYi (B)
,fyQsfVd ,ehu iznf'kZr djrk gSA ftlesa 'N' sp
3
ladfjr gS tcfd fodYi 'C' esa 'N' sp
2
ladfjr gSA vr% B, C dh
rqyuk esa vfèkd {kkjh; gksxkA
PAGE # 2
4. 1-Methylethylene oxide
HBr
excess
? ? ? ? ? X, Product 'X' will be –
1-esfFky,sfFkyhu vkWDlkbM vkf/kD; ? ? ? ? ? ?
vkf/kD; HBr
X, mRikn 'X' gksxk &
(1)
Br
Br
(2)
Br
Br
(3)
Br Br (4)
Br
Ans. (2)
Sol.
Br
Br
O
HBr
O
H
? ?
Br
–
Br
OH
HBr
5. Correct order of Intermolecular forces
(1) Dipole – Dipole > Ion – Ion > Dipole – Ion
(2) Ion – Ion > Dipole – Dipole > Dipole – Ion
(3) Dipole – Ion > Dipole – Dipole > Ion – Ion
(4) Ion – Ion > Dipole – Ion > Dipole – Dipole
vUrjvkf.od cyksa dk lgh Øe gS &
(1) f}/kzqo – f}/kzqo > vk;u – vk;u > f}/kzqo – vk;u
(2) vk;u – vk;u > f}/kzqo – f}/kzqo > f}/kzqo – vk;u
(3) f}/kzqo – vk;u > f}/kzqo – f}/kzqo > vk;u – vk;u
(4) vk;u – vk;u > f}/kzqo – vk;u > f}/kzqo – f}/kzqo
Ans. (4)
6. Hex-3-ynal
4
(1)NaBH
? ??? ? ?
3
(2)PBr
? ?? ? ?
(3)Mg/Ether
? ?? ?? ? ?
2
(4)CO
? ?? ? ?
(5)H
?
? ? ? ? ? (X), formed product X will be:
gsDl-3-vkbuSy
4
(1)NaBH
? ??? ? ?
3
(2)PBr
? ?? ? ?
(3)Mg/Ether
? ?? ?? ? ?
2
(4)CO
? ?? ? ?
(5)H
?
? ? ? ? ? (X), izkIr mRikn X gksxk %
(1)
O
C –OH
(2)
C –OH
O
H
H
H
(3)
O
OH
(4)
O
OH
Ans. (1)
Sol.
CHO
4
NaBH
? ?? ? ?
OH
3
PBr
? ? ? ?
Br
Mg/Ether
COOH
H
?
? ? ? ?
O
C –O
2
CO
? ? ? ? MgBr
Page 3
PAGE # 1
PART : CHEMISTRY
SECTION – 1 : (Maximum Marks : 80)
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. V 34 . 0 º E
Cu / Cu
2 ? ?
V 522 . 0 º E
Cu / Cu
? ?
? ? ?
Cu / Cu
2 º E ?
(1) 0.158V (2) –0.158 V (3) 0.182 V (4) –0.182 V
Ans. (1)
Sol.
Cu
2+
º
1
E
Cu
+
0.522
Cu
0.34
2 × 0.34 = 522 . 0 1 E
º
1
? ?
522 . 0 – 68 . 0 E
º
1
?
158 . 0 E
º
1
?
2. Correct order of electron gain enthalpy (kJ/mole) of F, Cl, Br, I
F, Cl, Br, I dh bySDVªkWu xzg.k ,UFkSYih (kJ/eksy) dk lgh Øe gS \
(1) –296, – 325, – 333, – 349
(2) –333, – 349, – 325, –296
(3) –349, – 333, – 325, – 296
(4) –325, – 333, – 349, – 296
Ans. (2)
3. Arrange the following in order of their pKb value
fuEu dks buds pKb eku ds Øe O;ofLFkr dhft, &
(A)
N NH
N
(B) CH3 –NH –CH3 (C) CH3 –CH=NH
(1) A > B > C (2) B > A > C (3) C > B > A (4) B > C > A
Ans. (3)
Sol. Option "A" represent Guanadine, the conjugate acid of which is resonance stabilised. The option 'B' is
aliphatic amine, here the 'N' is sp
3
whereas in option 'C' the 'N' is sp
2
, hence B is more basic than C.
fodYi "A" xqokusfMu izdkj dk ;kSfxd iznf'kZr djrk gS ftldk la;qXeh vEy vuqukn }kjk LFkk;hd`r gSA fodYi (B)
,fyQsfVd ,ehu iznf'kZr djrk gSA ftlesa 'N' sp
3
ladfjr gS tcfd fodYi 'C' esa 'N' sp
2
ladfjr gSA vr% B, C dh
rqyuk esa vfèkd {kkjh; gksxkA
PAGE # 2
4. 1-Methylethylene oxide
HBr
excess
? ? ? ? ? X, Product 'X' will be –
1-esfFky,sfFkyhu vkWDlkbM vkf/kD; ? ? ? ? ? ?
vkf/kD; HBr
X, mRikn 'X' gksxk &
(1)
Br
Br
(2)
Br
Br
(3)
Br Br (4)
Br
Ans. (2)
Sol.
Br
Br
O
HBr
O
H
? ?
Br
–
Br
OH
HBr
5. Correct order of Intermolecular forces
(1) Dipole – Dipole > Ion – Ion > Dipole – Ion
(2) Ion – Ion > Dipole – Dipole > Dipole – Ion
(3) Dipole – Ion > Dipole – Dipole > Ion – Ion
(4) Ion – Ion > Dipole – Ion > Dipole – Dipole
vUrjvkf.od cyksa dk lgh Øe gS &
(1) f}/kzqo – f}/kzqo > vk;u – vk;u > f}/kzqo – vk;u
(2) vk;u – vk;u > f}/kzqo – f}/kzqo > f}/kzqo – vk;u
(3) f}/kzqo – vk;u > f}/kzqo – f}/kzqo > vk;u – vk;u
(4) vk;u – vk;u > f}/kzqo – vk;u > f}/kzqo – f}/kzqo
Ans. (4)
6. Hex-3-ynal
4
(1)NaBH
? ??? ? ?
3
(2)PBr
? ?? ? ?
(3)Mg/Ether
? ?? ?? ? ?
2
(4)CO
? ?? ? ?
(5)H
?
? ? ? ? ? (X), formed product X will be:
gsDl-3-vkbuSy
4
(1)NaBH
? ??? ? ?
3
(2)PBr
? ?? ? ?
(3)Mg/Ether
? ?? ?? ? ?
2
(4)CO
? ?? ? ?
(5)H
?
? ? ? ? ? (X), izkIr mRikn X gksxk %
(1)
O
C –OH
(2)
C –OH
O
H
H
H
(3)
O
OH
(4)
O
OH
Ans. (1)
Sol.
CHO
4
NaBH
? ?? ? ?
OH
3
PBr
? ? ? ?
Br
Mg/Ether
COOH
H
?
? ? ? ?
O
C –O
2
CO
? ? ? ? MgBr
PAGE # 3
7.
Cl N2 N
CH3
CH3
+ SO3
? ?
Na
? ?
'X'
, formed product 'X' is used as:
(1) Lab test of Phenol
(2) Acid base titration indicator
(3) Food colouring
(4) It can be used as an alternative to detect amino acid in place of ninhydrin
Cl N2 N
CH3
CH3
+ SO3
? ?
Na
? ?
'X'
, izkIr mRikn 'X' dk mi;ksx fd;k tkrk gS \
(1) fQukWy ds iz;ksx'kkyk ijh{k.k esa
(2) vEy {kkj vuqekiu eas lwpd ds :i esa
(3) [kk| inkFkksaZ dksa jaxhu djus esa
(4) uhugkbMªhu ds LFkku ij ,ehuksa vEy ds fuèkkZj.k gsrq bldk mi;ksx fd;k tkrk gS
Ans. (2)
Sol.
+ N2
H3C
SO3
? ?
Methyl orange
N
H3C
N=N
H3C
SO3 N
H3C
Na
?
Methyl orange is used as an indicator in acid base titration.
esfFky vkWjsat dk mi;ksx vEy {kkj vuqekiu eas lwpd ds :i esa fd;k tkrk gSA
8. In which of the following Saytzeff product will not be formed as major product ?
fuEu eas ls dkSulh vfHkfØ;k esa eq[; mRikn lSRtSQ mRikn ugha gksxk \
(A)
OH
H
+
? ?
(B)
Br
alc. KOH
? ?
(C)
Br
t BuO
–
K
+
? ?
(D)
H ? ?
O
OH
(1) Only dsoy C (2) Only dsoy B (3) A, C & D (4) B & D
Ans. (1)
Sol. (1)
OH
H
+
? ?
(2)
Br
alc. KOH
? ?
(3)
Br
t BuO
–
K
+
? ?
(4)
H ? ?
O
OH
H
O
Page 4
PAGE # 1
PART : CHEMISTRY
SECTION – 1 : (Maximum Marks : 80)
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. V 34 . 0 º E
Cu / Cu
2 ? ?
V 522 . 0 º E
Cu / Cu
? ?
? ? ?
Cu / Cu
2 º E ?
(1) 0.158V (2) –0.158 V (3) 0.182 V (4) –0.182 V
Ans. (1)
Sol.
Cu
2+
º
1
E
Cu
+
0.522
Cu
0.34
2 × 0.34 = 522 . 0 1 E
º
1
? ?
522 . 0 – 68 . 0 E
º
1
?
158 . 0 E
º
1
?
2. Correct order of electron gain enthalpy (kJ/mole) of F, Cl, Br, I
F, Cl, Br, I dh bySDVªkWu xzg.k ,UFkSYih (kJ/eksy) dk lgh Øe gS \
(1) –296, – 325, – 333, – 349
(2) –333, – 349, – 325, –296
(3) –349, – 333, – 325, – 296
(4) –325, – 333, – 349, – 296
Ans. (2)
3. Arrange the following in order of their pKb value
fuEu dks buds pKb eku ds Øe O;ofLFkr dhft, &
(A)
N NH
N
(B) CH3 –NH –CH3 (C) CH3 –CH=NH
(1) A > B > C (2) B > A > C (3) C > B > A (4) B > C > A
Ans. (3)
Sol. Option "A" represent Guanadine, the conjugate acid of which is resonance stabilised. The option 'B' is
aliphatic amine, here the 'N' is sp
3
whereas in option 'C' the 'N' is sp
2
, hence B is more basic than C.
fodYi "A" xqokusfMu izdkj dk ;kSfxd iznf'kZr djrk gS ftldk la;qXeh vEy vuqukn }kjk LFkk;hd`r gSA fodYi (B)
,fyQsfVd ,ehu iznf'kZr djrk gSA ftlesa 'N' sp
3
ladfjr gS tcfd fodYi 'C' esa 'N' sp
2
ladfjr gSA vr% B, C dh
rqyuk esa vfèkd {kkjh; gksxkA
PAGE # 2
4. 1-Methylethylene oxide
HBr
excess
? ? ? ? ? X, Product 'X' will be –
1-esfFky,sfFkyhu vkWDlkbM vkf/kD; ? ? ? ? ? ?
vkf/kD; HBr
X, mRikn 'X' gksxk &
(1)
Br
Br
(2)
Br
Br
(3)
Br Br (4)
Br
Ans. (2)
Sol.
Br
Br
O
HBr
O
H
? ?
Br
–
Br
OH
HBr
5. Correct order of Intermolecular forces
(1) Dipole – Dipole > Ion – Ion > Dipole – Ion
(2) Ion – Ion > Dipole – Dipole > Dipole – Ion
(3) Dipole – Ion > Dipole – Dipole > Ion – Ion
(4) Ion – Ion > Dipole – Ion > Dipole – Dipole
vUrjvkf.od cyksa dk lgh Øe gS &
(1) f}/kzqo – f}/kzqo > vk;u – vk;u > f}/kzqo – vk;u
(2) vk;u – vk;u > f}/kzqo – f}/kzqo > f}/kzqo – vk;u
(3) f}/kzqo – vk;u > f}/kzqo – f}/kzqo > vk;u – vk;u
(4) vk;u – vk;u > f}/kzqo – vk;u > f}/kzqo – f}/kzqo
Ans. (4)
6. Hex-3-ynal
4
(1)NaBH
? ??? ? ?
3
(2)PBr
? ?? ? ?
(3)Mg/Ether
? ?? ?? ? ?
2
(4)CO
? ?? ? ?
(5)H
?
? ? ? ? ? (X), formed product X will be:
gsDl-3-vkbuSy
4
(1)NaBH
? ??? ? ?
3
(2)PBr
? ?? ? ?
(3)Mg/Ether
? ?? ?? ? ?
2
(4)CO
? ?? ? ?
(5)H
?
? ? ? ? ? (X), izkIr mRikn X gksxk %
(1)
O
C –OH
(2)
C –OH
O
H
H
H
(3)
O
OH
(4)
O
OH
Ans. (1)
Sol.
CHO
4
NaBH
? ?? ? ?
OH
3
PBr
? ? ? ?
Br
Mg/Ether
COOH
H
?
? ? ? ?
O
C –O
2
CO
? ? ? ? MgBr
PAGE # 3
7.
Cl N2 N
CH3
CH3
+ SO3
? ?
Na
? ?
'X'
, formed product 'X' is used as:
(1) Lab test of Phenol
(2) Acid base titration indicator
(3) Food colouring
(4) It can be used as an alternative to detect amino acid in place of ninhydrin
Cl N2 N
CH3
CH3
+ SO3
? ?
Na
? ?
'X'
, izkIr mRikn 'X' dk mi;ksx fd;k tkrk gS \
(1) fQukWy ds iz;ksx'kkyk ijh{k.k esa
(2) vEy {kkj vuqekiu eas lwpd ds :i esa
(3) [kk| inkFkksaZ dksa jaxhu djus esa
(4) uhugkbMªhu ds LFkku ij ,ehuksa vEy ds fuèkkZj.k gsrq bldk mi;ksx fd;k tkrk gS
Ans. (2)
Sol.
+ N2
H3C
SO3
? ?
Methyl orange
N
H3C
N=N
H3C
SO3 N
H3C
Na
?
Methyl orange is used as an indicator in acid base titration.
esfFky vkWjsat dk mi;ksx vEy {kkj vuqekiu eas lwpd ds :i esa fd;k tkrk gSA
8. In which of the following Saytzeff product will not be formed as major product ?
fuEu eas ls dkSulh vfHkfØ;k esa eq[; mRikn lSRtSQ mRikn ugha gksxk \
(A)
OH
H
+
? ?
(B)
Br
alc. KOH
? ?
(C)
Br
t BuO
–
K
+
? ?
(D)
H ? ?
O
OH
(1) Only dsoy C (2) Only dsoy B (3) A, C & D (4) B & D
Ans. (1)
Sol. (1)
OH
H
+
? ?
(2)
Br
alc. KOH
? ?
(3)
Br
t BuO
–
K
+
? ?
(4)
H ? ?
O
OH
H
O
PAGE # 4
9. Match the column
Column-I Column-II
(A) Thiamine (P) Scurvy
(B) Riboflavin (Q) Beri Beri
(C) Pyridoxine (R) Cheilosis
(D) Ascorbic acid (S) Convulsions
fuEu dk feyku dhft, &
dkWye-I dkWye-II
(A) Fkk;kehu (P) LdohZ
(B) jkbcks¶ysfou (Q) csjh csjh
(C) ikbjhMksDlhu (R) dhyksfll (Cheilosis)
(D) ,LdksfcZd vEy (S) ejksM+ iM+uk (Convulsions)
(1) A ??Q B ??R C ??P D ??S
(2) A ??Q B ??R C ??S D ??P
(3) A ??R B ??Q C ??S D ??P
(4) A ??R B ??Q C ??P D ??S
Ans. (2)
Sol.
Vitamins Deficiency Diseases
Vitamin B1 (Thiamine) Beri Beri
Vitamin B2 (Riboflavin) Cheilosis
Vitamin B6 (Pyridoxine) Convulsions
Vitamin C (Ascorbic acid) Scurvy
foVkehu ghurk tfur jksx
foVkehu B1 (Fkk;kehu) csjh csjh
foVkehu B2 (jkbcks¶ysfou) dhyksfll
foVkehu B6 (ikbjhMksDlhu) ejksM+ iM+uk
foVkehu C (,LdksfcZd vEy) LdohZ
10. Atomic radius of Ag is similar to
Ag dh ijek.kq f=kT;k fdlds leku gksrh gS \
(1) Cu (2) Hg (3) Au (4) Ni
Ans. (3)
11. Correct IUPAC name of ? ? ? ? Cl ) NH CH ( Cl NH Pt
2 3 2 3
is:
(1) Bis(ammine)chlorido(methylamine)platinum(II) chloride
(2) Diamminechlorido(methylamine)platinum(II) chloride
(3) Diamminechlorido(aminomethyl)platinum(II) chloride
(4) Chloridobis(ammine)methylamineplatinum(II) chloride
? ? ? ? Cl ) NH CH ( Cl NH Pt
2 3 2 3
dk lgh IUPAC uke gS %
(1) fcl(,sEehu)DyksjkbMksa(esfFky,sehu)IysfVue(II) DyksjkbM
(2) MkbZ,sEehuDyksjkbMks(esfFky,sehu)IysfVue(II) DyksjkbM
(3) MkbZ,sEehuDyksjkbMks(,sehuksesfFky)IysfVue(II) DyksjkbM
(4) DyksjkbMksfcl(,sEehu)esfFky,sehuIysfVue(II) DyksjkbM
Ans. (2)
Page 5
PAGE # 1
PART : CHEMISTRY
SECTION – 1 : (Maximum Marks : 80)
Straight Objective Type (lh/ks oLrqfu"B izdkj)
This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4) for its
answer, out of which Only One is correct.
bl [k.M esa 20 cgq&fodYih iz'u gSaA izR;sd iz'u ds 4 fodYi (1), (2), (3) rFkk (4) gSa] ftuesa ls flQZ ,d lgh gSA
1. V 34 . 0 º E
Cu / Cu
2 ? ?
V 522 . 0 º E
Cu / Cu
? ?
? ? ?
Cu / Cu
2 º E ?
(1) 0.158V (2) –0.158 V (3) 0.182 V (4) –0.182 V
Ans. (1)
Sol.
Cu
2+
º
1
E
Cu
+
0.522
Cu
0.34
2 × 0.34 = 522 . 0 1 E
º
1
? ?
522 . 0 – 68 . 0 E
º
1
?
158 . 0 E
º
1
?
2. Correct order of electron gain enthalpy (kJ/mole) of F, Cl, Br, I
F, Cl, Br, I dh bySDVªkWu xzg.k ,UFkSYih (kJ/eksy) dk lgh Øe gS \
(1) –296, – 325, – 333, – 349
(2) –333, – 349, – 325, –296
(3) –349, – 333, – 325, – 296
(4) –325, – 333, – 349, – 296
Ans. (2)
3. Arrange the following in order of their pKb value
fuEu dks buds pKb eku ds Øe O;ofLFkr dhft, &
(A)
N NH
N
(B) CH3 –NH –CH3 (C) CH3 –CH=NH
(1) A > B > C (2) B > A > C (3) C > B > A (4) B > C > A
Ans. (3)
Sol. Option "A" represent Guanadine, the conjugate acid of which is resonance stabilised. The option 'B' is
aliphatic amine, here the 'N' is sp
3
whereas in option 'C' the 'N' is sp
2
, hence B is more basic than C.
fodYi "A" xqokusfMu izdkj dk ;kSfxd iznf'kZr djrk gS ftldk la;qXeh vEy vuqukn }kjk LFkk;hd`r gSA fodYi (B)
,fyQsfVd ,ehu iznf'kZr djrk gSA ftlesa 'N' sp
3
ladfjr gS tcfd fodYi 'C' esa 'N' sp
2
ladfjr gSA vr% B, C dh
rqyuk esa vfèkd {kkjh; gksxkA
PAGE # 2
4. 1-Methylethylene oxide
HBr
excess
? ? ? ? ? X, Product 'X' will be –
1-esfFky,sfFkyhu vkWDlkbM vkf/kD; ? ? ? ? ? ?
vkf/kD; HBr
X, mRikn 'X' gksxk &
(1)
Br
Br
(2)
Br
Br
(3)
Br Br (4)
Br
Ans. (2)
Sol.
Br
Br
O
HBr
O
H
? ?
Br
–
Br
OH
HBr
5. Correct order of Intermolecular forces
(1) Dipole – Dipole > Ion – Ion > Dipole – Ion
(2) Ion – Ion > Dipole – Dipole > Dipole – Ion
(3) Dipole – Ion > Dipole – Dipole > Ion – Ion
(4) Ion – Ion > Dipole – Ion > Dipole – Dipole
vUrjvkf.od cyksa dk lgh Øe gS &
(1) f}/kzqo – f}/kzqo > vk;u – vk;u > f}/kzqo – vk;u
(2) vk;u – vk;u > f}/kzqo – f}/kzqo > f}/kzqo – vk;u
(3) f}/kzqo – vk;u > f}/kzqo – f}/kzqo > vk;u – vk;u
(4) vk;u – vk;u > f}/kzqo – vk;u > f}/kzqo – f}/kzqo
Ans. (4)
6. Hex-3-ynal
4
(1)NaBH
? ??? ? ?
3
(2)PBr
? ?? ? ?
(3)Mg/Ether
? ?? ?? ? ?
2
(4)CO
? ?? ? ?
(5)H
?
? ? ? ? ? (X), formed product X will be:
gsDl-3-vkbuSy
4
(1)NaBH
? ??? ? ?
3
(2)PBr
? ?? ? ?
(3)Mg/Ether
? ?? ?? ? ?
2
(4)CO
? ?? ? ?
(5)H
?
? ? ? ? ? (X), izkIr mRikn X gksxk %
(1)
O
C –OH
(2)
C –OH
O
H
H
H
(3)
O
OH
(4)
O
OH
Ans. (1)
Sol.
CHO
4
NaBH
? ?? ? ?
OH
3
PBr
? ? ? ?
Br
Mg/Ether
COOH
H
?
? ? ? ?
O
C –O
2
CO
? ? ? ? MgBr
PAGE # 3
7.
Cl N2 N
CH3
CH3
+ SO3
? ?
Na
? ?
'X'
, formed product 'X' is used as:
(1) Lab test of Phenol
(2) Acid base titration indicator
(3) Food colouring
(4) It can be used as an alternative to detect amino acid in place of ninhydrin
Cl N2 N
CH3
CH3
+ SO3
? ?
Na
? ?
'X'
, izkIr mRikn 'X' dk mi;ksx fd;k tkrk gS \
(1) fQukWy ds iz;ksx'kkyk ijh{k.k esa
(2) vEy {kkj vuqekiu eas lwpd ds :i esa
(3) [kk| inkFkksaZ dksa jaxhu djus esa
(4) uhugkbMªhu ds LFkku ij ,ehuksa vEy ds fuèkkZj.k gsrq bldk mi;ksx fd;k tkrk gS
Ans. (2)
Sol.
+ N2
H3C
SO3
? ?
Methyl orange
N
H3C
N=N
H3C
SO3 N
H3C
Na
?
Methyl orange is used as an indicator in acid base titration.
esfFky vkWjsat dk mi;ksx vEy {kkj vuqekiu eas lwpd ds :i esa fd;k tkrk gSA
8. In which of the following Saytzeff product will not be formed as major product ?
fuEu eas ls dkSulh vfHkfØ;k esa eq[; mRikn lSRtSQ mRikn ugha gksxk \
(A)
OH
H
+
? ?
(B)
Br
alc. KOH
? ?
(C)
Br
t BuO
–
K
+
? ?
(D)
H ? ?
O
OH
(1) Only dsoy C (2) Only dsoy B (3) A, C & D (4) B & D
Ans. (1)
Sol. (1)
OH
H
+
? ?
(2)
Br
alc. KOH
? ?
(3)
Br
t BuO
–
K
+
? ?
(4)
H ? ?
O
OH
H
O
PAGE # 4
9. Match the column
Column-I Column-II
(A) Thiamine (P) Scurvy
(B) Riboflavin (Q) Beri Beri
(C) Pyridoxine (R) Cheilosis
(D) Ascorbic acid (S) Convulsions
fuEu dk feyku dhft, &
dkWye-I dkWye-II
(A) Fkk;kehu (P) LdohZ
(B) jkbcks¶ysfou (Q) csjh csjh
(C) ikbjhMksDlhu (R) dhyksfll (Cheilosis)
(D) ,LdksfcZd vEy (S) ejksM+ iM+uk (Convulsions)
(1) A ??Q B ??R C ??P D ??S
(2) A ??Q B ??R C ??S D ??P
(3) A ??R B ??Q C ??S D ??P
(4) A ??R B ??Q C ??P D ??S
Ans. (2)
Sol.
Vitamins Deficiency Diseases
Vitamin B1 (Thiamine) Beri Beri
Vitamin B2 (Riboflavin) Cheilosis
Vitamin B6 (Pyridoxine) Convulsions
Vitamin C (Ascorbic acid) Scurvy
foVkehu ghurk tfur jksx
foVkehu B1 (Fkk;kehu) csjh csjh
foVkehu B2 (jkbcks¶ysfou) dhyksfll
foVkehu B6 (ikbjhMksDlhu) ejksM+ iM+uk
foVkehu C (,LdksfcZd vEy) LdohZ
10. Atomic radius of Ag is similar to
Ag dh ijek.kq f=kT;k fdlds leku gksrh gS \
(1) Cu (2) Hg (3) Au (4) Ni
Ans. (3)
11. Correct IUPAC name of ? ? ? ? Cl ) NH CH ( Cl NH Pt
2 3 2 3
is:
(1) Bis(ammine)chlorido(methylamine)platinum(II) chloride
(2) Diamminechlorido(methylamine)platinum(II) chloride
(3) Diamminechlorido(aminomethyl)platinum(II) chloride
(4) Chloridobis(ammine)methylamineplatinum(II) chloride
? ? ? ? Cl ) NH CH ( Cl NH Pt
2 3 2 3
dk lgh IUPAC uke gS %
(1) fcl(,sEehu)DyksjkbMksa(esfFky,sehu)IysfVue(II) DyksjkbM
(2) MkbZ,sEehuDyksjkbMks(esfFky,sehu)IysfVue(II) DyksjkbM
(3) MkbZ,sEehuDyksjkbMks(,sehuksesfFky)IysfVue(II) DyksjkbM
(4) DyksjkbMksfcl(,sEehu)esfFky,sehuIysfVue(II) DyksjkbM
Ans. (2)
PAGE # 5
12. Vapour pressure of pure CS2 and CH3COCH3 are 512 mm of Hg and 312 mm of Hg respectively. Total
vapour pressure of mixture is 600 mm of Hg then find incorrect statement:
(1) B A
B B
A A
? ? ? ?
?
?
?
? ? ?
? ? ?
(2) Does not obey Raoult's law
(3) Endothermic solution
(4) after adding 100 ml of each, then net volume is less than 200 ml
'kq) CS2 rFkk CH3COCH3 dk ok"i nkc Øe'k% 512 mm Hg rFkk 312 mm Hg gSA buds feJ.k dk dqy ok"i nkc
600 mm Hg gS rc vlR; dFku dh igpku dhft, &
(1) B A
B B
A A
? ? ? ?
?
?
?
? ? ?
? ? ?
(2) jkmYV fu;e dh vuqikyuk ugha djrk gSA
(3) Å"ek'kks"kh foy;u
(4) izR;sd ds 100 ml feykus ij ifj.kkeh vk;ru 200 ml ls de ik;k tkrk gSA
Ans. (4)
Sol. Above mixture of liquids show positive deviation from Raoult's Law
mijksDr feJ.k jkmYV fu;e ls /kukRed fopyu n'kkZrk gSA
13. Purest form of commercial iron is:
(1) Wrought iron (2) Cast iron (3) Pig iron (4) Iron scrap + pig iron
vkS|ksfxd ykSgs dk 'kq)r~e :i gS &
(1) fiVok ykSgk (Wrought iron) (2) <yok ykSgk (Cast iron)
(3) dPpk ykSgk (Pig iron) (4) jn~nh ykSgk + dPpk ykSgk (Iron scrap + pig iron)
Ans. (1)
Sol. Purest form is wrought iron.
vkS|ksfxd ykSgs dk 'kq)r~e :I fiVok ykSgk (Wrought iron) gSA
14.
COOH
Cl
(x)
OH
Cl
(y)
NH2
Cl
(z)
Mixture of above three organic compound was subjected to aq NaHCO3 and followed by dil NaOH.
compounds which will be soluble in given solvent will be :
(1) x in aq NaHCO3 and y in dil. NaOH.
(2) x in aq NaHCO3 and z in dil. NaOH.
(3) y in aq NaHCO3 and x in dil. NaOH.
(4) y is aq NaHCO3 and z in dil. NaOH.
Read More
|
357 docs|148 tests
|
Related Exams
About this Document
|
4.91/5 Rating |
|
Nov 22, 2024 Last updated |
Document Description: JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions for JEE 2024 is part of Mock Tests for JEE Main and Advanced 2025 preparation.
The notes and questions for JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions have been prepared according to the JEE exam syllabus. Information about JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions covers topics
like and JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions Example, for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions.
Introduction of JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions in English is available as part of our Mock Tests for JEE Main and Advanced 2025
for JEE & JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions in Hindi for Mock Tests for JEE Main and Advanced 2025 course.
Download more important topics related with notes, lectures and mock test series for JEE
Exam by signing up for free. JEE: JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025
Description
Full syllabus notes, lecture & questions for JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 - JEE | Plus excerises question with solution to help you revise complete syllabus for Mock Tests for JEE Main and Advanced 2025 | Best notes, free PDF download
Information about JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions
In this doc you can find the meaning of JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions defined & explained in the simplest way possible. Besides explaining types of
JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions theory, EduRev gives you an ample number of questions to practice JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions tests, examples and also practice JEE
tests
|
Explore Courses for JEE exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
past year papers
,Objective type Questions
,practice quizzes
,MCQs
,Exam
,Free
,study material
,JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025
,Important questions
,Summary
,JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025
,Previous Year Questions with Solutions
,Viva Questions
,video lectures
,ppt
,Semester Notes
,JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025
,shortcuts and tricks
,Sample Paper
,mock tests for examination
,Extra Questions
;
Additional Information about JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions for JEE Preparation
JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions Free PDF Download
The JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions is an invaluable resource that delves deep into the core of the JEE exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions now and kickstart your journey towards success in the JEE exam.
Importance of JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions
The importance of JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions cannot be overstated, especially for JEE aspirants.
This document holds the key to success in the JEE exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions Notes
JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions Notes on EduRev are your ultimate resource for success.
JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions JEE Questions
The "JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions JEE Questions" guide is a valuable resource for all aspiring students preparing for the
JEE exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions on the App
Students of JEE can study JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of JEE Main 2020 Chemistry January 7 Shift 1 Paper & Solutions is prepared as per the latest JEE syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup