JEE Main 2020 Chemistry January 7 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 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. Which of the following reactions are possible ? 
 fuEu eas ls dkSulh vfHkfØ;k lEHko gS \ 
 (A) 
 Cl 
+ 
Anhy. AlCl3 
? ?
  
 (B) 
 
AlCl 3 
+ Cl2(excess) 
Cl 
Cl 
Cl 
Cl 
Cl 
Cl 
Dark 
 
 (C) 
 
Anhy. AlCl 3 
? ?
+ CH2=CH –Cl 
CH=CH2 
  
 (D) 
 
Anhy. AlCl3 
? ?
+ CH2=CH –CH2 –Cl 
CH2 –CH=CH2 
  
 (1) A, B, C  (2) B, D   (3) A, C, D  (4) A, C 
Ans. (2) 
Sol. Vinyl halides and aryl halides do not give Friedel craft's reaction. 
 foukby gSykbM rFkk ,sfjy gSykbM fÝMy Øk¶V vfHkfØ;k ugha nsrs gSA 
  
2. A and B are in the given reaction ? 
 mijksDr vfHkfØ;k esa  A rFkk B gS \ 
 NH2 
Ac 2O 
CH3 
A 
Br 2 
AcOH 
B 
 
 (1) 
 NHCOCH3 
CH3 
Br 
 (2) 
 NHCOCH3 
CH3 
Br 
 (3) 
 NHCOCH3 
CH2Br 
Br 
  (4) 
 NH –COCH3 
CH2 –Br 
COCH3 
 
Ans. (1) 
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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. Which of the following reactions are possible ? 
 fuEu eas ls dkSulh vfHkfØ;k lEHko gS \ 
 (A) 
 Cl 
+ 
Anhy. AlCl3 
? ?
  
 (B) 
 
AlCl 3 
+ Cl2(excess) 
Cl 
Cl 
Cl 
Cl 
Cl 
Cl 
Dark 
 
 (C) 
 
Anhy. AlCl 3 
? ?
+ CH2=CH –Cl 
CH=CH2 
  
 (D) 
 
Anhy. AlCl3 
? ?
+ CH2=CH –CH2 –Cl 
CH2 –CH=CH2 
  
 (1) A, B, C  (2) B, D   (3) A, C, D  (4) A, C 
Ans. (2) 
Sol. Vinyl halides and aryl halides do not give Friedel craft's reaction. 
 foukby gSykbM rFkk ,sfjy gSykbM fÝMy Øk¶V vfHkfØ;k ugha nsrs gSA 
  
2. A and B are in the given reaction ? 
 mijksDr vfHkfØ;k esa  A rFkk B gS \ 
 NH2 
Ac 2O 
CH3 
A 
Br 2 
AcOH 
B 
 
 (1) 
 NHCOCH3 
CH3 
Br 
 (2) 
 NHCOCH3 
CH3 
Br 
 (3) 
 NHCOCH3 
CH2Br 
Br 
  (4) 
 NH –COCH3 
CH2 –Br 
COCH3 
 
Ans. (1) 
     
    
    
    
     
 
                       
  
 
PAGE # 2 
 
Sol. 
 
NH2 
Ac2O 
CH3 
Br2 
AcOH 
NH –C –CH3 
CH3 
NH –C –CH3 
CH3 
O O 
Br 
 
 
3.  The correct statement about gluconic acid is  
 (1) It is prepared by oxidation of glucose with HNO3 
 (2) It is obtained by partial oxidation of glucose    
(3) It is dicarboxylic acid   
(4) It forms hemiactal or acetal  
Xywdksfud vEy ds lUnHkZ esa dkSulk dFku lR; gS \  
 (1) bls HNO3 ds lkFk Xywdksl ds vkWDlhdj.k }kjk cuk;k tkrk gSA 
(2) ;g Xywdksl ds vkaf'kd vkWDlhdj.k }kjk izkIr gksrk gSA 
(3) ;g f}dkckZsfDlfyd vEy gSA 
(4) ;g gSeh,slhVsy ;k ,slhVSy cukrk gSA 
Ans. (2) 
Sol. Gluconic acid 
 
CH2 –CH –CH –CH –CH –COOH 
OH OH OH OH OH 
is obtained by partial oxidation of glucose by Tollen's 
reagent or Fehling solution or Br2,H2O.    
 Gluconic acid can not form hemiacetal or acetal   
 Xywdksfud vEy 
 
CH2 –CH –CH –CH –CH –COOH 
OH OH OH OH OH 
 VkWysu vfHkdeZd ;k Qsgfyax foy;u ;k Br2,H2O }kjk 
Xywdksl ds vkaf'kd vkWDlhdj.k }kjk izkIr gksrk gSA    
 Xywdksfud vEy gSeh,slhVsy ;k ,slhVSy ugha cukrk gSA 
  
4. Stability order of following alkoxide ions is 
 fuEufyf[kr ,YdksDlkbM vk;uksa ds LFkkf;Ro dk Øe gS % 
 
 
NO2 
O
–
 
(A) 
 
NO2 
O
–
 
(B) 
 
O2N 
O
–
 
(C) 
 
 (1) C > B > A   (2) A > C > B  (3) B > A > C  (4) C > A > B 
Ans. (1) 
Sol. When negative charge is delocalised with electron withdrawing group like (NO2) then stability increases.  
(A) Negative charge is delocalised with NO2 group  
(B) Negative charge is delocalised with carbon of alkene  
(C) Negative charge is localised  
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. Which of the following reactions are possible ? 
 fuEu eas ls dkSulh vfHkfØ;k lEHko gS \ 
 (A) 
 Cl 
+ 
Anhy. AlCl3 
? ?
  
 (B) 
 
AlCl 3 
+ Cl2(excess) 
Cl 
Cl 
Cl 
Cl 
Cl 
Cl 
Dark 
 
 (C) 
 
Anhy. AlCl 3 
? ?
+ CH2=CH –Cl 
CH=CH2 
  
 (D) 
 
Anhy. AlCl3 
? ?
+ CH2=CH –CH2 –Cl 
CH2 –CH=CH2 
  
 (1) A, B, C  (2) B, D   (3) A, C, D  (4) A, C 
Ans. (2) 
Sol. Vinyl halides and aryl halides do not give Friedel craft's reaction. 
 foukby gSykbM rFkk ,sfjy gSykbM fÝMy Øk¶V vfHkfØ;k ugha nsrs gSA 
  
2. A and B are in the given reaction ? 
 mijksDr vfHkfØ;k esa  A rFkk B gS \ 
 NH2 
Ac 2O 
CH3 
A 
Br 2 
AcOH 
B 
 
 (1) 
 NHCOCH3 
CH3 
Br 
 (2) 
 NHCOCH3 
CH3 
Br 
 (3) 
 NHCOCH3 
CH2Br 
Br 
  (4) 
 NH –COCH3 
CH2 –Br 
COCH3 
 
Ans. (1) 
     
    
    
    
     
 
                       
  
 
PAGE # 2 
 
Sol. 
 
NH2 
Ac2O 
CH3 
Br2 
AcOH 
NH –C –CH3 
CH3 
NH –C –CH3 
CH3 
O O 
Br 
 
 
3.  The correct statement about gluconic acid is  
 (1) It is prepared by oxidation of glucose with HNO3 
 (2) It is obtained by partial oxidation of glucose    
(3) It is dicarboxylic acid   
(4) It forms hemiactal or acetal  
Xywdksfud vEy ds lUnHkZ esa dkSulk dFku lR; gS \  
 (1) bls HNO3 ds lkFk Xywdksl ds vkWDlhdj.k }kjk cuk;k tkrk gSA 
(2) ;g Xywdksl ds vkaf'kd vkWDlhdj.k }kjk izkIr gksrk gSA 
(3) ;g f}dkckZsfDlfyd vEy gSA 
(4) ;g gSeh,slhVsy ;k ,slhVSy cukrk gSA 
Ans. (2) 
Sol. Gluconic acid 
 
CH2 –CH –CH –CH –CH –COOH 
OH OH OH OH OH 
is obtained by partial oxidation of glucose by Tollen's 
reagent or Fehling solution or Br2,H2O.    
 Gluconic acid can not form hemiacetal or acetal   
 Xywdksfud vEy 
 
CH2 –CH –CH –CH –CH –COOH 
OH OH OH OH OH 
 VkWysu vfHkdeZd ;k Qsgfyax foy;u ;k Br2,H2O }kjk 
Xywdksl ds vkaf'kd vkWDlhdj.k }kjk izkIr gksrk gSA    
 Xywdksfud vEy gSeh,slhVsy ;k ,slhVSy ugha cukrk gSA 
  
4. Stability order of following alkoxide ions is 
 fuEufyf[kr ,YdksDlkbM vk;uksa ds LFkkf;Ro dk Øe gS % 
 
 
NO2 
O
–
 
(A) 
 
NO2 
O
–
 
(B) 
 
O2N 
O
–
 
(C) 
 
 (1) C > B > A   (2) A > C > B  (3) B > A > C  (4) C > A > B 
Ans. (1) 
Sol. When negative charge is delocalised with electron withdrawing group like (NO2) then stability increases.  
(A) Negative charge is delocalised with NO2 group  
(B) Negative charge is delocalised with carbon of alkene  
(C) Negative charge is localised  
     
    
    
    
     
 
                       
  
 
PAGE # 3 
 
 
Sol. tc _.kkos'k bySDVªkWu vkd"khZ lewg tSls (NO2) ds lkFk foLFkkuhÑr gksrk gS] rks LFkkf;Ro c<+ tkrk gSA 
(A) _.kkos'k NO2 lewg ds lkFk foLFkkuhÑr gSA  
(B) _.kkos'k ,Ydhu ds dkcZu ds lkFk foLFkkuhÑr gSA 
(C) _.kkos'k LFkkuhÑr gSA 
 
5. 
 
HBr 
? ?
A 
Na 
Ether  
?
B 
O 
CH2 –Br 
 
 A and B are – 
 A rFkk B gS & 
 (1) 
 
& 
Br 
O 
Br 
CH2Br 
OH 
 (2) 
 
Br 
OH 
& 
Br 
CH2Br 
 
(3*) 
 
OH 
Br 
& 
OH 
CH2Br 
 (4) 
 OH 
Br 
& 
OH 
CH2Br 
 
Ans. (3) 
Sol.  
 
HBr 
? ?(SN2) ?
Na, ether 
O 
CH2 –Br 
O 
CH2 –Br 
H + 
Br 
CH2 –Br 
OH 
Br 
Wurtz reaction  
OH 
 
 
6. For the complex [Ma2b2] if M is sp
3
 or dsp
2
 hybridised respectively then total number of optical isomers 
are respectively : 
 ladqy [Ma2b2] ds fy;s ;’fn M Øe’’'k% sp
3
 ;k dsp
2 
ladfjr gS rks izdkf'kd leko;oh;ksa dh dqy la[;k Øe'k% gS &  
 (1) 1, 1    (2)  2, 1   (3) 0, 0   (4) 1, 2 
Ans. (3) 
Sol. Both will not show optical isomerism. 
 nksuksa izdkf'kd leko;ork ugha n'kkZrsA  
 
7. Bond order and magnetic nature of CN
– 
 are respectively 
 (1) 3, diamagnetic     (2) 3, paramagnetic 
  (3) 2.5, paramagnetic     (4) 2.5, diamagnetic 
 CN
– 
dk ca/k Øe rFkk pqEcdh; izÑfr Øe'k% gS& 
 (1) 3, izfrpqEcdh;  (2) 3, vuqpqEcdh;  (3) 2.5, vuqpqEcdh; (4) 2.5, izfrpqEcdh; 
Ans. (1) 
Sol. CN
–
 is a 14 electron system. 
 CN
– 
14 bysDVªkWu ra=k gSA  
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. Which of the following reactions are possible ? 
 fuEu eas ls dkSulh vfHkfØ;k lEHko gS \ 
 (A) 
 Cl 
+ 
Anhy. AlCl3 
? ?
  
 (B) 
 
AlCl 3 
+ Cl2(excess) 
Cl 
Cl 
Cl 
Cl 
Cl 
Cl 
Dark 
 
 (C) 
 
Anhy. AlCl 3 
? ?
+ CH2=CH –Cl 
CH=CH2 
  
 (D) 
 
Anhy. AlCl3 
? ?
+ CH2=CH –CH2 –Cl 
CH2 –CH=CH2 
  
 (1) A, B, C  (2) B, D   (3) A, C, D  (4) A, C 
Ans. (2) 
Sol. Vinyl halides and aryl halides do not give Friedel craft's reaction. 
 foukby gSykbM rFkk ,sfjy gSykbM fÝMy Øk¶V vfHkfØ;k ugha nsrs gSA 
  
2. A and B are in the given reaction ? 
 mijksDr vfHkfØ;k esa  A rFkk B gS \ 
 NH2 
Ac 2O 
CH3 
A 
Br 2 
AcOH 
B 
 
 (1) 
 NHCOCH3 
CH3 
Br 
 (2) 
 NHCOCH3 
CH3 
Br 
 (3) 
 NHCOCH3 
CH2Br 
Br 
  (4) 
 NH –COCH3 
CH2 –Br 
COCH3 
 
Ans. (1) 
     
    
    
    
     
 
                       
  
 
PAGE # 2 
 
Sol. 
 
NH2 
Ac2O 
CH3 
Br2 
AcOH 
NH –C –CH3 
CH3 
NH –C –CH3 
CH3 
O O 
Br 
 
 
3.  The correct statement about gluconic acid is  
 (1) It is prepared by oxidation of glucose with HNO3 
 (2) It is obtained by partial oxidation of glucose    
(3) It is dicarboxylic acid   
(4) It forms hemiactal or acetal  
Xywdksfud vEy ds lUnHkZ esa dkSulk dFku lR; gS \  
 (1) bls HNO3 ds lkFk Xywdksl ds vkWDlhdj.k }kjk cuk;k tkrk gSA 
(2) ;g Xywdksl ds vkaf'kd vkWDlhdj.k }kjk izkIr gksrk gSA 
(3) ;g f}dkckZsfDlfyd vEy gSA 
(4) ;g gSeh,slhVsy ;k ,slhVSy cukrk gSA 
Ans. (2) 
Sol. Gluconic acid 
 
CH2 –CH –CH –CH –CH –COOH 
OH OH OH OH OH 
is obtained by partial oxidation of glucose by Tollen's 
reagent or Fehling solution or Br2,H2O.    
 Gluconic acid can not form hemiacetal or acetal   
 Xywdksfud vEy 
 
CH2 –CH –CH –CH –CH –COOH 
OH OH OH OH OH 
 VkWysu vfHkdeZd ;k Qsgfyax foy;u ;k Br2,H2O }kjk 
Xywdksl ds vkaf'kd vkWDlhdj.k }kjk izkIr gksrk gSA    
 Xywdksfud vEy gSeh,slhVsy ;k ,slhVSy ugha cukrk gSA 
  
4. Stability order of following alkoxide ions is 
 fuEufyf[kr ,YdksDlkbM vk;uksa ds LFkkf;Ro dk Øe gS % 
 
 
NO2 
O
–
 
(A) 
 
NO2 
O
–
 
(B) 
 
O2N 
O
–
 
(C) 
 
 (1) C > B > A   (2) A > C > B  (3) B > A > C  (4) C > A > B 
Ans. (1) 
Sol. When negative charge is delocalised with electron withdrawing group like (NO2) then stability increases.  
(A) Negative charge is delocalised with NO2 group  
(B) Negative charge is delocalised with carbon of alkene  
(C) Negative charge is localised  
     
    
    
    
     
 
                       
  
 
PAGE # 3 
 
 
Sol. tc _.kkos'k bySDVªkWu vkd"khZ lewg tSls (NO2) ds lkFk foLFkkuhÑr gksrk gS] rks LFkkf;Ro c<+ tkrk gSA 
(A) _.kkos'k NO2 lewg ds lkFk foLFkkuhÑr gSA  
(B) _.kkos'k ,Ydhu ds dkcZu ds lkFk foLFkkuhÑr gSA 
(C) _.kkos'k LFkkuhÑr gSA 
 
5. 
 
HBr 
? ?
A 
Na 
Ether  
?
B 
O 
CH2 –Br 
 
 A and B are – 
 A rFkk B gS & 
 (1) 
 
& 
Br 
O 
Br 
CH2Br 
OH 
 (2) 
 
Br 
OH 
& 
Br 
CH2Br 
 
(3*) 
 
OH 
Br 
& 
OH 
CH2Br 
 (4) 
 OH 
Br 
& 
OH 
CH2Br 
 
Ans. (3) 
Sol.  
 
HBr 
? ?(SN2) ?
Na, ether 
O 
CH2 –Br 
O 
CH2 –Br 
H + 
Br 
CH2 –Br 
OH 
Br 
Wurtz reaction  
OH 
 
 
6. For the complex [Ma2b2] if M is sp
3
 or dsp
2
 hybridised respectively then total number of optical isomers 
are respectively : 
 ladqy [Ma2b2] ds fy;s ;’fn M Øe’’'k% sp
3
 ;k dsp
2 
ladfjr gS rks izdkf'kd leko;oh;ksa dh dqy la[;k Øe'k% gS &  
 (1) 1, 1    (2)  2, 1   (3) 0, 0   (4) 1, 2 
Ans. (3) 
Sol. Both will not show optical isomerism. 
 nksuksa izdkf'kd leko;ork ugha n'kkZrsA  
 
7. Bond order and magnetic nature of CN
– 
 are respectively 
 (1) 3, diamagnetic     (2) 3, paramagnetic 
  (3) 2.5, paramagnetic     (4) 2.5, diamagnetic 
 CN
– 
dk ca/k Øe rFkk pqEcdh; izÑfr Øe'k% gS& 
 (1) 3, izfrpqEcdh;  (2) 3, vuqpqEcdh;  (3) 2.5, vuqpqEcdh; (4) 2.5, izfrpqEcdh; 
Ans. (1) 
Sol. CN
–
 is a 14 electron system. 
 CN
– 
14 bysDVªkWu ra=k gSA  
     
    
    
    
     
 
                       
  
 
PAGE # 4 
 
 
8. Which of the following is incorrect? 
 fuEu esa ls dkSulk fodYi xyr gS \ 
 (1) 
m m m m
NaCl NaBr KCl KBr
? ? ? ?
? ? ? ? ? ? ?  (2) 
m 2 m m m
H O HCl NaOH NaCl
? ? ? ?
? ? ? ? ? ? ? 
(3) 
m m m m
Nal NaBr NaBr KBr
? ? ? ?
? ? ? ? ? ? ?  (4) 
m m m m
NaCl KCl NaBr KBr
? ? ? ?
? ? ? ? ? ? ? 
Ans. (3) 
Sol. Theory based. 
 lS)kfUrd  
 
9. 
2
Hot & conc.
NaOH Cl ? ? ? ? A + other products 
 
22
Cold& dil.
Ca(OH) Cl ? ? ? ?B + other products  
 A & B are respectively 
 (1) NaClO3, Ca(OCl)2     (2) NaClO3, Ca(ClO3)2 
  (3) NaCl, Ca(ClO3)2    (4) NaClO, Ca(ClO3)2 
 ? ? ? ?
2
Cl NaOH
z rFkk lkUn xeZ
 A + vU; mRikn 
 ? ? ? ? ? ?
2 2
Cl OH Ca
ruq B.Mk rFkk 
B + vU; mRikn 
 A rFkk B Øe'k% gS & 
 (1) NaClO3, Ca(OCl)2     (2) NaClO3, Ca(ClO3)2 
  (3) NaCl, Ca(ClO3)2    (4) NaClO, Ca(ClO3)2 
Ans. (1) 
Sol. 6NaOH + 3Cl2 ??5NaCl + NaClO3 + 3H2O 
 2Ca(OH)2 + Cl2 ??Ca(OCl)2 + CaCl2 + H2O 
 
10. There are two beakers (I) having pure volatile solvent and (II) having volatile solvent and non-volatile 
solute. If both beakers are placed together in a closed container then:  
 (1) Volume of solvent beaker will decrease and solution beaker will increase     
(2) Volume of solvent beaker will increase and solution beaker will also increase     
(3) Volume of solvent beaker will decrease and solution beaker will also decrease    
(4) Volume of solvent beaker will increase and solution beaker will decrease  
;gk¡ nks chdj gS] (I) 'kq) ok"i'khy foyk;d j[krk gS rFkk (II)  ok"i'khy foyk;d rFkk vok"i'khy foys; j[krk gSA 
;fn nksuksa chdj ,d can ik=k esa ,d lkFk mifLFkr gS] rc &  
 (1) foyk;d chdj dk vk;ru ?kVsxk rFkk foy;u chdj dk vk;ru c<sxkA    
(2) foyk;d chdj dk vk;ru c<sxk rFkk foy;u chdj dk vk;ru Hkh c<sxkA 
(3) foyk;d chdj dk vk;ru ?kVsxk rFkk foy;u chdj dk vk;ru Hkh ?kVsxkA  
(4) foyk;d chdj dk vk;ru c<sxk rFkk foy;u chdj dk vk;ru ?kVsxkA  
Ans. (1) 
Sol. There will be lowering in vapour pressure in second beaker. 
 ;gk¡ f}rh; chdj esa ok"i nkc esa voueu gksxkA  
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. Which of the following reactions are possible ? 
 fuEu eas ls dkSulh vfHkfØ;k lEHko gS \ 
 (A) 
 Cl 
+ 
Anhy. AlCl3 
? ?
  
 (B) 
 
AlCl 3 
+ Cl2(excess) 
Cl 
Cl 
Cl 
Cl 
Cl 
Cl 
Dark 
 
 (C) 
 
Anhy. AlCl 3 
? ?
+ CH2=CH –Cl 
CH=CH2 
  
 (D) 
 
Anhy. AlCl3 
? ?
+ CH2=CH –CH2 –Cl 
CH2 –CH=CH2 
  
 (1) A, B, C  (2) B, D   (3) A, C, D  (4) A, C 
Ans. (2) 
Sol. Vinyl halides and aryl halides do not give Friedel craft's reaction. 
 foukby gSykbM rFkk ,sfjy gSykbM fÝMy Øk¶V vfHkfØ;k ugha nsrs gSA 
  
2. A and B are in the given reaction ? 
 mijksDr vfHkfØ;k esa  A rFkk B gS \ 
 NH2 
Ac 2O 
CH3 
A 
Br 2 
AcOH 
B 
 
 (1) 
 NHCOCH3 
CH3 
Br 
 (2) 
 NHCOCH3 
CH3 
Br 
 (3) 
 NHCOCH3 
CH2Br 
Br 
  (4) 
 NH –COCH3 
CH2 –Br 
COCH3 
 
Ans. (1) 
     
    
    
    
     
 
                       
  
 
PAGE # 2 
 
Sol. 
 
NH2 
Ac2O 
CH3 
Br2 
AcOH 
NH –C –CH3 
CH3 
NH –C –CH3 
CH3 
O O 
Br 
 
 
3.  The correct statement about gluconic acid is  
 (1) It is prepared by oxidation of glucose with HNO3 
 (2) It is obtained by partial oxidation of glucose    
(3) It is dicarboxylic acid   
(4) It forms hemiactal or acetal  
Xywdksfud vEy ds lUnHkZ esa dkSulk dFku lR; gS \  
 (1) bls HNO3 ds lkFk Xywdksl ds vkWDlhdj.k }kjk cuk;k tkrk gSA 
(2) ;g Xywdksl ds vkaf'kd vkWDlhdj.k }kjk izkIr gksrk gSA 
(3) ;g f}dkckZsfDlfyd vEy gSA 
(4) ;g gSeh,slhVsy ;k ,slhVSy cukrk gSA 
Ans. (2) 
Sol. Gluconic acid 
 
CH2 –CH –CH –CH –CH –COOH 
OH OH OH OH OH 
is obtained by partial oxidation of glucose by Tollen's 
reagent or Fehling solution or Br2,H2O.    
 Gluconic acid can not form hemiacetal or acetal   
 Xywdksfud vEy 
 
CH2 –CH –CH –CH –CH –COOH 
OH OH OH OH OH 
 VkWysu vfHkdeZd ;k Qsgfyax foy;u ;k Br2,H2O }kjk 
Xywdksl ds vkaf'kd vkWDlhdj.k }kjk izkIr gksrk gSA    
 Xywdksfud vEy gSeh,slhVsy ;k ,slhVSy ugha cukrk gSA 
  
4. Stability order of following alkoxide ions is 
 fuEufyf[kr ,YdksDlkbM vk;uksa ds LFkkf;Ro dk Øe gS % 
 
 
NO2 
O
–
 
(A) 
 
NO2 
O
–
 
(B) 
 
O2N 
O
–
 
(C) 
 
 (1) C > B > A   (2) A > C > B  (3) B > A > C  (4) C > A > B 
Ans. (1) 
Sol. When negative charge is delocalised with electron withdrawing group like (NO2) then stability increases.  
(A) Negative charge is delocalised with NO2 group  
(B) Negative charge is delocalised with carbon of alkene  
(C) Negative charge is localised  
     
    
    
    
     
 
                       
  
 
PAGE # 3 
 
 
Sol. tc _.kkos'k bySDVªkWu vkd"khZ lewg tSls (NO2) ds lkFk foLFkkuhÑr gksrk gS] rks LFkkf;Ro c<+ tkrk gSA 
(A) _.kkos'k NO2 lewg ds lkFk foLFkkuhÑr gSA  
(B) _.kkos'k ,Ydhu ds dkcZu ds lkFk foLFkkuhÑr gSA 
(C) _.kkos'k LFkkuhÑr gSA 
 
5. 
 
HBr 
? ?
A 
Na 
Ether  
?
B 
O 
CH2 –Br 
 
 A and B are – 
 A rFkk B gS & 
 (1) 
 
& 
Br 
O 
Br 
CH2Br 
OH 
 (2) 
 
Br 
OH 
& 
Br 
CH2Br 
 
(3*) 
 
OH 
Br 
& 
OH 
CH2Br 
 (4) 
 OH 
Br 
& 
OH 
CH2Br 
 
Ans. (3) 
Sol.  
 
HBr 
? ?(SN2) ?
Na, ether 
O 
CH2 –Br 
O 
CH2 –Br 
H + 
Br 
CH2 –Br 
OH 
Br 
Wurtz reaction  
OH 
 
 
6. For the complex [Ma2b2] if M is sp
3
 or dsp
2
 hybridised respectively then total number of optical isomers 
are respectively : 
 ladqy [Ma2b2] ds fy;s ;’fn M Øe’’'k% sp
3
 ;k dsp
2 
ladfjr gS rks izdkf'kd leko;oh;ksa dh dqy la[;k Øe'k% gS &  
 (1) 1, 1    (2)  2, 1   (3) 0, 0   (4) 1, 2 
Ans. (3) 
Sol. Both will not show optical isomerism. 
 nksuksa izdkf'kd leko;ork ugha n'kkZrsA  
 
7. Bond order and magnetic nature of CN
– 
 are respectively 
 (1) 3, diamagnetic     (2) 3, paramagnetic 
  (3) 2.5, paramagnetic     (4) 2.5, diamagnetic 
 CN
– 
dk ca/k Øe rFkk pqEcdh; izÑfr Øe'k% gS& 
 (1) 3, izfrpqEcdh;  (2) 3, vuqpqEcdh;  (3) 2.5, vuqpqEcdh; (4) 2.5, izfrpqEcdh; 
Ans. (1) 
Sol. CN
–
 is a 14 electron system. 
 CN
– 
14 bysDVªkWu ra=k gSA  
     
    
    
    
     
 
                       
  
 
PAGE # 4 
 
 
8. Which of the following is incorrect? 
 fuEu esa ls dkSulk fodYi xyr gS \ 
 (1) 
m m m m
NaCl NaBr KCl KBr
? ? ? ?
? ? ? ? ? ? ?  (2) 
m 2 m m m
H O HCl NaOH NaCl
? ? ? ?
? ? ? ? ? ? ? 
(3) 
m m m m
Nal NaBr NaBr KBr
? ? ? ?
? ? ? ? ? ? ?  (4) 
m m m m
NaCl KCl NaBr KBr
? ? ? ?
? ? ? ? ? ? ? 
Ans. (3) 
Sol. Theory based. 
 lS)kfUrd  
 
9. 
2
Hot & conc.
NaOH Cl ? ? ? ? A + other products 
 
22
Cold& dil.
Ca(OH) Cl ? ? ? ?B + other products  
 A & B are respectively 
 (1) NaClO3, Ca(OCl)2     (2) NaClO3, Ca(ClO3)2 
  (3) NaCl, Ca(ClO3)2    (4) NaClO, Ca(ClO3)2 
 ? ? ? ?
2
Cl NaOH
z rFkk lkUn xeZ
 A + vU; mRikn 
 ? ? ? ? ? ?
2 2
Cl OH Ca
ruq B.Mk rFkk 
B + vU; mRikn 
 A rFkk B Øe'k% gS & 
 (1) NaClO3, Ca(OCl)2     (2) NaClO3, Ca(ClO3)2 
  (3) NaCl, Ca(ClO3)2    (4) NaClO, Ca(ClO3)2 
Ans. (1) 
Sol. 6NaOH + 3Cl2 ??5NaCl + NaClO3 + 3H2O 
 2Ca(OH)2 + Cl2 ??Ca(OCl)2 + CaCl2 + H2O 
 
10. There are two beakers (I) having pure volatile solvent and (II) having volatile solvent and non-volatile 
solute. If both beakers are placed together in a closed container then:  
 (1) Volume of solvent beaker will decrease and solution beaker will increase     
(2) Volume of solvent beaker will increase and solution beaker will also increase     
(3) Volume of solvent beaker will decrease and solution beaker will also decrease    
(4) Volume of solvent beaker will increase and solution beaker will decrease  
;gk¡ nks chdj gS] (I) 'kq) ok"i'khy foyk;d j[krk gS rFkk (II)  ok"i'khy foyk;d rFkk vok"i'khy foys; j[krk gSA 
;fn nksuksa chdj ,d can ik=k esa ,d lkFk mifLFkr gS] rc &  
 (1) foyk;d chdj dk vk;ru ?kVsxk rFkk foy;u chdj dk vk;ru c<sxkA    
(2) foyk;d chdj dk vk;ru c<sxk rFkk foy;u chdj dk vk;ru Hkh c<sxkA 
(3) foyk;d chdj dk vk;ru ?kVsxk rFkk foy;u chdj dk vk;ru Hkh ?kVsxkA  
(4) foyk;d chdj dk vk;ru c<sxk rFkk foy;u chdj dk vk;ru ?kVsxkA  
Ans. (1) 
Sol. There will be lowering in vapour pressure in second beaker. 
 ;gk¡ f}rh; chdj esa ok"i nkc esa voueu gksxkA  
     
    
    
    
     
 
                       
  
 
PAGE # 5 
 
 
11. Metal with low melting point containing impurities of high melting point can be purified by 
 (1) Zone refining   (2) Vapor phase refining  
 (3) Distillation    (4) Liquation  
 mPp xyukad dh v'kqf);ksa ;qDr U;wu xyukad okyh /kkrq fuEu }kjk 'kq) gks ldrh gS&  
 (1) {ks=k ifj'kks/ku    (2) ok"i izkoLFkk ifj'kks/ku 
 (3) vklou     (4) nzohdj.k 
Ans. (4) 
Sol. Theory based 
 lS)kfUrd 
 
12. Which of the following statements are correct ? 
(I) On decomposition of H2O2, O2 gas is released . 
(II) 2-ethylanthraquinol is used in preparation of H2O2 
(III) On heating KClO3, Pb(NO3)2, NaNO3, O2 gas is released.  
(IV) In the preparation of sodium peroxoborate, H2O2 is treated with sodium metaborate. 
 (1) I, II, IV   (2) II, III, IV  (3) I, II, III, IV  (4) I, II, III 
 fuEu esa ls dkSuls dFku lgh gS \ 
(I) H2O2 ds fo?kVu ij O2 xSl fu"dkflr gksrh gSA 
(II) 2-,fFky,UFkszD;wuksy dk mi;ksx H2O2 ds fojpu esa gksrk gSA  
(III) KClO3, Pb(NO3)2, NaNO3 dks xeZ djus ij O2 xSl fu"dkflr gksrh gSA  
(IV) lksfM;e ijvkWDlkscksjsV ds fojpu esa H2O2 dks lksfM;e esVkcksjsV ds lkFk mipkfjr fd;k tkrk gSA  
 (1) I, II, IV   (2) II, III, IV  (3) I, II, III, IV  (4) I, II, III 
Ans. (3) 
Sol. Theory based 
 lS)kfUrd 
 
13. Amongs the following which is redox reaction ? 
 (1) N2 + O2 
2000K
? ?? ? ?     (2) Formation of O3 from O2 
 (3) Reaction between NaOH and H2SO4   (4) Reaction between AgNO3 and NaCl 
 fuEu esa ls dkSulh jsMkWDl vfHkfØ;k gS \ 
 (1) N2 + O2 
2000K
? ?? ? ?     (2) O2 ls O3 dk fuekZ.k 
 (3) NaOH rFkk H2SO4 ds e/; vfHkfØ;k   (4) AgNO3 rFkk NaCl ds e/; vfHkfØ;k 
Ans. (1) 
Sol. N2  + O2 ??2NO  
 3O2 ??2O3  
2NaOH + H2SO4 ??Na2SO4 + 2H2O 
AgNO3 + NaCl ??NaNO3 + AgCl