classnote-508568e7ede56.pdf

IIT-JEE ChEmistry by N.J. sir

ORGANIC ChemIstRy

S . J .

REACTION MECHANISM

N

r i

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1


ORGANIC chemistry

DPP NO-01

Time: 15 minutes

Q.1

(b)

CH3

_____

CH3—CH2—CH2 

(o)

 _ _ _ _ _ _ _

(c)

H

CH—OH 2

_____



CH3— C — C = O: _ _ _ _ _ _ _ _ _ :

(a)

(p)



_______ _

CH3 (d)

CH3—CH2—CH2—C—CH2

 _ _ _ _ _

CH2 (q)

 

_________

(r)



_________

(s)



CH3 (e)

(f)



CH2

________

 

_________

S . J . `

(g)

Ph—CH2—CH2 CH3

 _ _ _ _

(h)

CH3—CH—CH2—CH2



Ph Me

(i)

Ph CH3–O

C—CH2—CH2—CH2

(l)



OH

(m)



CH3 Me (n)

_ __ _ _ _ _ _

N

CH3—CH—CH2—OH

(t)

 

_________

(u)



_______ _

(v)

 

_________

___

 _ _ _

Ph

(k)

________

____

Ph—C— CH—CH2—CH2  D

(j)

r i

OH

(w)

 _ _ _ _ _ _

_________

 _ _ _ _ _ _ _ _ _

OH (x)



________

`

CH3—C— C—et _ _ _ _ _ _ _ _ _ OH


2


ORGANIC chemistry

DPP NO-02

Time: 15 minutes

Dehydration of Alcohol Conc.H2SO4 CH3—CH=CH2 Q.11 Q.1 CH3—CH2—CH2—OH  

H /   _ _ _ _ _ _ _ OH 

170C

Mechanism:– (1)

Q.8

 CH3 — CH2 — CH2 — O  — H

+ H+

CH 3 — CH2 — CH2 — rds

 CH—CH—CH 3

(2)

3, 3–Dimethyl–butan–2–ol loses a molecule of water in the presence of concentrated sulphuric acid to give tetramethylethylene as a major product. Suggest a suitable mechanism.

Electrophilic addition:–   CH3 — CH2 — CH2  + 

3

 CH3 — CH — CH3 + :O:

H

C=C

ENu  

C—C E

CH3—CH=CH2+ H3O+ Mechanism:–

H Energy profile diagram:—

C=C

r i Nu

+ E  —C — C—

E

Reaction

Q.2 OH

OH

/ _ _ _ _ _ _ _  

H / _ _ _ _ _ _ _ OH 

Q.4 Q.5

OH OH

N

H  _ _ _ _ _ _ _   OH

Q.8

OH

(b)

—C — C—

CH = CH2

CH = CH2

(iv) CH = CH2

CN

(ii) CH = CH2

O – Me

NH2 (v)

(iv)

(iii)

COOH

(c) (i) (d)

H /   _ _ _ _ _ _ _ 

(iii)

(i) CH = CH2

Conc.H2SO4  _ _ _ _ _ _ _ 

OH

(ii)

NO2

H /   _ _ _ _ _ _ _ 

Q.9

Q.10

Nu

E Compare rate of electrophilic addition on alkenes:– (i)

Conc.H2SO4  _ _ _ _ _ _ _  Conc.H2SO4  _ _ _ _ _ _ _ 

OH

Q.6

Q.1

Nu

(a)

H

Q.3

Q.7

S . J .

H /   _ _ _ _ _ _ _ OH  H /   _ _ _ _ _ _ _ 

Q.1



:

E

(ii)

Ph — CH = CH2 (i)

(iii)

(iv)

(v)

PH — CH = CH — Ph (ii)


3

HI

Q.1

CH2 = CH2  _ _ _ _ _ _ _ _ _ _

Q.2

KBr CH2 = CH2 

__________

HI __________  KBr

Q.3

Q.15

HCl(2eq.) CH3 — C  CH   __________

Q.16

HCl  _ _ _ _ _ _ _ _ _ _ 

Q.17

dil.H2SO4   __________

H /H2O   __________

Q.18 Q.4

HI __________ C = CH2  KBr

HI  _ _ _ _ _ _ _ _ _ _  KF

Q.5

Q.19

H /R–OH  _ _ _ _ _ _ _ _ _ _ 

Q.20

H /H2O   __________

Me HI  _ _ _ _ _ _ _ _ _ _ 

Q.6

Q.21

S . J . CH = CH2

HCl  

Q.7

__________

O

Q.8

HI  _ _ _ _ _ _ _ _ _ _ 

D /D2O   __________

Q.22

H+

Q.23

HI  _ _ _ _ _ _ _ _ _ (Ph)2 CH—CH—CH=CH2 

Me Q.10

Q.11

Q.12

N

HBr

C = CH—Ph 

__________

Q.13

HCl  _ _ _ _ _ _ _ _ _ _ CH2 = CH — Cl 

Q.14

HCl(2eq.) CH  CH   __________

H __________

Q.25

HCl  _ _ _ _ _ _ _ CH  C — CH2 — CH = CH2 

Q.26

CH  C — CH = CH2

Q.27

H HO—CH2—CH2—CH2—CH = CH2  ___ MeOH

HCl  _ _ _ _ _ _ _ _ _ _ 

HBr __________ 

O

HBr __________ 

Q.24

Q.9

r i

HCl  _ _ _ _ _ _ _ _ _ _ 

D

H

HCl  

__________

OH

Q.28



H /MeOH  _ _ _ _ _ _ CH2= CH—CH—CH=CH2 

H  _ _ _ _ _ _ _ _ _ _ 

Q.29 O H


4


ORGANIC chemistry

DPP NO-03

Time: 15 minutes

Q.1 O

CH3—CH—CH—CH3

H   CH3—CH2—C—CH3 

Q.3

OH OH H H

H   

OH OH

OH Mechanism:– (1) CH3 —CH—CH—CH 3

+H

+

+  H2 O + CH 3—CH—CH—CH 3

: :O–H O–H

H   

O–H

: O—H +

CHO 3

CH3 CH 3

S . J .

NaNO2 + HCl  H—O—N=O + NaCl

(2)

H — O — N = O + H H2O +

:

(1)

N=O +

H

H

:

:

:

N = O  R—N — N = O  R—N—N = O—H

N

:

H

H

: :

R—N= N—O—H

H

:

:

 N2  R  R — N  N :  R—N = N

+ O

:

H

H   

NaNO2    R — N  N :   R  N2 HCl

 R — NH2

Mechanism:–

R—N: +

H   

Diazotization of primary amine:–

H  Ph— C — CH 2  

OH OH

(3)

H   

H

Q.6

Q.7

Ph

H

H

OH

OH OH

H  Et — C — C —Et   OH OH

Q.2

C—C—

OH

O—H

r i

OH OH

+  CH3 — CH 2—C—CH 3  CH3 — CH 2—C—CH 3

Q.1

CH 3

Q.4

Q.5 s

OH

CH3

 R—N = N—O:

H

H 

H


5

Demjanov Reaction:– OH NH2

OH

O

NaNO2 CH3 — CH—CH—CH3   CH 3 — C—CH 2—CH3 HCl

SbCl5 __________   

Cl O—H

OH

NH2

NaNO2  CH3 — CH—CH—CH3 + N2 CH3 — CH—CH—CH3  HCl

 OH

O H

OH Br AlBr3   

Q.6

OH

NH2

NaNO2   __ —OCH3 HCl

CH — CH — NH2

Q.2

OH

NH 2

NaNO2   HCl

__________ Q.8

NaNO2   __________ HCl

Q.3

DPP NO-04 Non Classical Carbocation:–

Q.4

Nu

E—Nu   —C — C— E

Mechanism:–  —C — C—

:

C=C E

+ Nu

E

Nu

N Nu

—C — C—

Q.2

Q.3

aq.AgNO3   __________ 

2 CH2 = CH2   _______ Br2 _ _ _ _ _ _ _ CH2 = CH2 O H H

ORGANIC chemistry Time: 15 minutes

Br2  _ _ _ _ _ _ _ CCl4

Q.5

ICl  _ _ _ _ _ _ _ CH3 — CH = CH2 

Q.6

HOCl  _ _ _ _ _ _ _ CH3 — CH = CH2 

Q.7

Cl2  _ _ _ _ _ _ _ CCl4

Q.8

Br2  _ _ _ _ _ _ _ _ _ CCl4

Q.9

Br2 /H2O   _______

Q.10

Br2  _______ H2O

E

Br

OH

S . J .

IIT-JEE ChEmistry by N.J. sir C=C

O—H

I

OH

r i

AgI    __________

Q.7 I

Q.1

__________

+ CH3 — C—CH2—CH3  CH3 — C—CH2—CH3 

Q.1

(1)

NaNO2   _______ HCl

CH3 — CH —CH—

Q.5

Mechanism:– (1) OH

Q.4

NH2

Q.11

Br2 /CCl4  _______ (1eq.)

Br2  _ _ _ _ _ _ _ CCl4


6

H N

Q.12 Q.13

Diel’s Alder Reaction:–

Br2 (CCl4 )   _______ (1eq.)

+ ||



HCl  _ _ _ _ _ _ CH3 — CH = CH2 



COOH

Q.1

CH3Cl / AlCl3

+

 COOH

COOH

CH3Cl Q.2 H

Q.14

H–O

CH3

______

HCl  _ _ _ _ _ _ 

CH3

Q.13

The reaction of propene with HOCl proceeds via the addition of [IIT '2001] Q.5 (A) H+ in first step + (B) Cl in first step (C) OH– in first step Q.6 (D) Cl+ and OH– in single step B OH

Q.18

Ans. Q.23

x



Q.4

H 2O

NO NO

(C) CH3 – CH2 – CH Cl

+

N

Q.26

The number of stereoisomers bromination of trans-2-butene is (A) 1 (B) 2 (C) 3 A

Write down the structures of the stereoismers formed when cis-2-butene is reacted with bromine. [IIT '1995]

Ans.

H Br

Me

Me

H Br





+



+



O

Q.10

+

O

obtained by [IIT '2007] (D) 4

Q.5

Br + Br H H

+

(D) CH2 – CH2 – CH2 Cl NO

A



+

Q.9

r i 

S . J .

NO Cl

Ans.

Me



+

Br2   5 compounds of (mixture) Q.7 molecular formula C4H8Br2 [IIT '2003] Number of compounds in X will be : (A) 2 (B) 3 (C) 4 (D) 5 B Q.8 CH3 – CH = CH2 + NOCl  P [IIT '2006] Identify the adduct. (A) CH3 – CH – CH2 (B) CH3 – CH – CH2 H  

Cl

Ans.

+ NO2

Q.15

Ans.



HOOC Q.3

O—

+



O

Q.11

Q.12

O

+



+



Me


7


ORGANIC chemistry

DPP NO- 05

Time: 15 minutes

Fill in the blanks:– O

Q.1

CH3 OH OH H

H  

CH3 OH OH H

+ H+  H



Mechanism :– (i)

(ii)

O H 



Intermediatecarbocation  Rearrargemet  





Intermediatecarbocation 

Re arrangedcarbocation O

(iii)

+

 H +



Re arrangedcarbocation (iv)

[H ] = - - - - - - - - - - - - - - t

(v)

Name of Reaction is - - - - - - - - - - - - - - - -

Q.2

OH

OH

C

C

S . J . 

CH3 — O

CH3

H

Mechanism : – CH3 — O

OH

OH

C

C

H  Product 

CH3 +

r i

H  Intermediate Carbocation



H

Intermediate Carbocation  rearrangedCarbocation

Q.3

(i)



(ii)

RearrangedCarbocation  H+ Pr oduct 

PH

NH2

OH

C

C

H



NaNO2 |HCl.   Pr oduct



H

NH2 PH

N

CH3

Mechanism : – (i)



C H

OH

C

CH3

NaNO2 |HCl.   ---- ---------

H

Rearrangement

H+ + - - - - - - - - - - -  - - - - - - - - - - - None of Reaction - - - - - - - - - - - - - -


8

CH3

Q.4

CH3

CH3

C° +

CH3

CH3

CH3

C

Q.6

           Q.5 CH3 — C°H2 + CH3 — C°H2            (Major Pr oduct) (Major Pr oduct)

C°

+

C

     Q.7 (explain)



Ph

H

H

C

C

CH 3

CH3

O +

CH = CH 2  H / Ph

H

Mechanism : – (i) Ph

H

H

C

C

CH 3

CH 3

  

+H CH = CH2

(Intermediate1 )

(Rearrangement) Nucleophile  Pr oduct   Intermediate3  (Intermediate2 )





HCl | CCl 4

A

Q.8

HBr CCl4

B

Ph—CH= CH 2

D

HCl R–O–O–R/CCl 4

Ph

S . J .

C

C

HBr (R–O–O–R)/CCl 4

C

CH3

Q.9



CH3

CH3

C

CH2

Br /h

aq.KOH

2 2  A   CH3 

CH3

C

CH2

r i CH3

OH

H

Conc. H2SO4

HBr(R–O–O–R) C   B CCl4

A & C are Idenitcal / Isomers / Positon Isomers.

  Q.10

Ph

Ph

O

C

C

NaOH OH  electrolysis

Ph

Anode (I)

 

N

_ _ _ _ _ _ _ _ _ e— +

Ph

Cathode

O

O°

(i)

2H2O + 2e— _ _ _ _ _ _

Ph

Ph

O

C

C

(ii)

Ph

(iii)

Intermediate - - - - - - - - - - - 

Ph

O°

CO2 +         Intermediate


9


ORGANIC chemistry

DPP NO- 06 1.

Time: 15 minutes

Which of following carbocation will undergo rearrangement? + + CH2 CH2 (a)

(b)

(c)

(f)

Ph — C — Ph CH2 — CH3

+

(g)

H   

(A) , Major product (A) is (major)

+

+ +

CH2 — CH2 — OH

3.

(d) + +

(e)

Ph — CH — Ph

(h)

(A) O

r i

Ph — CH — Ph

Ph +

CH — CH3

(i) CH2 — O — CH2 — CH2 — CH2 (B)

+

(j)

Ph — CH — Ph + CH2

+

+

(k)

(l)

S . J . (C)

CH3

(m)

Ph

+

Ph — C — OH

Ph

+

(n) CH2 — CH2 — CH2

(o)

+ CH

(q)

(r) CH3CH2CH2+ 

(t) (CH3)3 C CHCH3 CH2—OH 2.

CH2 — CH3

+ (p) CH2 — C =O

CH3 — CH2 — O



CH — CH2

(D)

4.

OH

H   

Reaction-1

(s) (CH3)2 CH CHCH3

N

(u) (CH3CH2)3CCH2+

H  (A)  

H   

Reaction-2

OH

(A) an heating isomerizes to (B). What is the structure of (B). CH3 OH

Reaction-3

(A)

Sum of -hydrogen (A + B + C) is.

(B)

(C)

(D)

5.

OH

(A)

OH

(A) (major)

(A) (major)

H   

(A) (major)

H  (x)  


10

(B)

OH

OH

H  (y)  

H  (B)  

Reaction-2

OH H  (z)  

(C)

OH

H  (p)  

(D)

Reaction-3

H  (C)  

OH Total number of products obtained in above reactions including minor products is (include stereoisomer) x y z p

6.

In which of following reaction rearrangement take place with change is carbon skeleton. CH3 + (A) CH — C — CH (B) CH3 — CH2 CH2  3 2

H  (D)  

Reaction-4

Sum of -hydrogen is 8.

+ CH3 — CH — CH2 — CH2

CH2—OH

S . J . OH

(B)

CH3

(D) CH3 — CH  — CH3

r i A+B+C+D=

In which of following reaction resonance stabilized product will form. (A)

CH3

(C)

OH

CH3

7.

Sum of -hydrogen in major product of the reaction. OH

Reaction-1

(C)

OH

H    H   

H   

(D) All

H  (A)  

N


11


ORGANIC chemistry

DPP NO- 07 Q.1

Time: 15 minutes OH

Compare rate of dehydration (acid-catalyzed)

OH

OH



H  A+B 

(h)



(i)

H OH

H CH3

(a)

H  A+B 

CH3

OH

OH



H  A+B 

(j) OH

(b)

OH

OH

(c)

OH

(d)

OH

OH

H2 SO4  

(m)

OH

CH3

(e)

S . J . OH

(n)

Q.2

Q.3

Predict the major product dehydration of alcohols (A) 2-pentanol (B) 1-methyl cyclopentanol (C) 2-methyl cyclohexanol (D) 2, 2-dimethyl-1-propanol

of

(o)

(b)

H SO

N

— CH2 — OH

H PO



H  

CH3

(f)

(r)

CH3



H  

OH

(s)

Q.4

OH

H3PO4  

Write – Mechanism 

(a)

H   

OH

H2 SO4  

(b)

H

CH3—CH2—CH — C —CH3   OH

(g)

Cl

H3PO4  

3 4 — CH — CH2 — CH3  

OH

(e)

D D

OH CH3 D H D  

2 4  CH3—CH2—C — CH—CH3 

OH

(d)



(q)

CH3 CH3

OH



H  

H  

(p)



(c)

H2 SO4  

CD3 OH

H2 SO4 

OH

OH

acid-catalyzed

Identify-Product (a)

KHSO4   170 C

H2 SO4  

(l) OH

OH

r i

OH

(k)

OH

(c) 18 + H2 O 



H OH  



CH2 — CH — CH2 — CH2 — CH2 OH 18

OH

H   

CH 3

O


12

Q.5

HCl  

(a)

CH2

Me

Me

Predict Major Product (i)

HCl

 

(b)

(j)

HCl  

(c)

HCl  

(d)

2-methyl propene

(e)

Hl 1-methyl cyclohexene  



(k)

OH

OH

(l)

OH

OH

(n)

(o)

CD3

(b)

D (q) D

OH

OH

(p)

(c)

S . J . (r)

(d)

OH

OH OH

OH

r i (m)

Ans.1 (a)

D

(s)

Cl

Cl

(e)

Ans.5 (a)

(b)

(c)

(d)

C

Ans.2 (A)

Ans.3 (a)

(B)

(C)

(b)

(D)

C—C=C—C

(c)

(e)

(f)

(h)

+

(d)

N (g)

Cl

Cl I

(e)


13


ORGANIC chemistry

DPP NO- 08 Q.1

Time: 15 minutes

Identify major products: CH3 HgSO4 (a)   CH3 — C — C CH H2 SO4

(d)

H Ph

CH3 HgSO 2

(c)

CH3

(f)

4

Q.3

C

CH

(d) C

HgSO4   H2 SO4

Ph

D–Cl (x) D3O+ (y)

HgSO4   H2 SO4

CC



H3 O  

HBr   CCl4

(e)

4 CH3 — C  CH   H SO

(b)

H C=C

What will be major – product obtained from addition of HBr to each of the following compounds. (a) CH3 — CH2 — CH = CH2

r i

CH3

CH H3 O



(e)



(f)

HgSO4 H2 SO4

(b)

CH2

(c)

CH3 CH3 — C  CH   C

CH H3 O

(d)



(h)

H3 O l-phenyl cyclohexene  

(i)

H3 O l-methyl cyclopetene  



HBr  

(j)

(e)

H2C = C

(f)

CH3 — CH = CH — CH3

(g)

HO

(h)

CH3 — CH — CH2 — CH = CH2

4

Find total product in following reaction ? (including stereoisomer) 

(a)

OH

H   (x) products 

N Br2

HBr CCl4

  (y) (products) (Markonikoff products)

(i)



CCl4

OH (c)

(j)

O

O

1.

(a)

(b)

O

H HBr   (A)   (B)  CCl4

Br2

CH3

Ans.

OH (b)

CH = CH2

(k)

(z) (products)

CCl4

— CH2 — CH3

CH3

HBr Ph — CH2 — CH = CH2   CCl

(k) Q.2

S . J . CH3





(g)

CH3 — CH = C — CH3

(c)

Me

(C)



H   


14

O

O 3.

(d)

*

(a)

O

Br

(c)

CHD2

(g)

(e)

OH

Ph

OH

2.

M. I.

(a) x  3 ; y 2 ; z 5 (b) A 1 ; B 1 ; C 2 (c) 3 (d) 2 (e) 4 (f) x 4 ; y 4

HO

M. I.

* Br

Br (h)

(j)

*

HO Br

r i

M. I. (i)

S . J . (k)

N

Br

(g)

Br

(k)

*

(f)

Br

(i)

Br (j)

(d)

Br

(f)

(h)

Br

Br

(e)

O

M. I. (b)

Br

Me

M. I.

Br


15


ORGANIC chemistry

DPP NO- 09 Q.1

Time: 15 minutes

How will prepare following compound using alkene as the starting material ? OH

Ans. 1.

(a)

(b)

OH

Br (c)



H3 O  

(a)

(b)



H3 O  

or

(d)

or Br CH3

(e)

Br

(f) CH3 — O — C — CH3 (c)

CH3

OCH3

Q.2

or

OH (i)

(g)

Identify reactant in following reaction (a) (b)

Br2 CCl4

Br

Br

2 (B)   CCl 4

Br

(d)

HBr (C)   CCl 4

Br

HBr (D)   CCl 4

CH3

(e)

HBr (E)   CH 3 — C — CH2 — Br CCl 4

CH3

O H O

(f)

3 (F)  

(g)

(G)   racemic mixture of 2, 3 – CCl

HBr

4

Write – Mechanism

(a)


or

(e)

MeOH 

(f)


(g)

(h)

N

dibromobutane. Q.3

(d)

+ Mirror Image

Br

(c)

S . J .

(A)   meso – 2, 3 – dibromobutane

2.



H   

r i HBr   CCl4

or

or

or



H3 O  

MeOH 

(a) A  Trans but-2-ene (b) B cyclohexene (c) C  cyclohexene (d) D 2-Methylpropene (e) E can not prepared by alkene (f) F Propyne (g) G Cis but 2-ene

OH

H3 O

(b) Ph—CH2—CH2—OH  Ph — CH — CH3


16

IIT-JEE Chemistry by N.J. sir

ORGANIC chemistry

DPP NO-10

Time: 15 minutes

OZONOLYSIS:

The reaction of alkene of alkene with ozone (O3) followed by hydrolysis is known as ozonolysis. It is two types : (I) Reductive ozonolysis  In presence of reducing agent (II) Oxidative ozonolysis In presence of oxidizing agent Reducing agents: Zn, H2O or Zn, CH3COOH or (CH3)2S or (Ph)3 P etc. Oxidising agents: H2O2 or R – C – O – O – H or Ag2 O etc. || O

R

R’ C=C

R

H

O

R

O3 step I –70°C

R R’ Zn/H2O Reductive C = O + R’– C – H C H ozonolysis R O O

C R O

Oxidative H O 2 2 ozonolysis Example 1:

R – C – R + R’ – C – OH O

Mechanism:

R

O

C

+

C R’

O

S . J . R

R

R

O

+

+

R’

O +

C=O

H

H

R—C—O

O

C

R

R’ — C — O

R

R

O

C

O

O

H

R

R—C—O

O+

r i

[SCT-Cut the double bond and paste two oxygen atoms and vice versa]

O

O

R’ O=C

C

R – C – R + R’ – C – H

R’

H

H

O

O

Note : In case of oxidative ozonolysis aldehyde (not ketone) further undergoes oxidation which gives acid as product. Q.1

N

Give the product of the following reaction. [ 7 × 2 = 14] (i)

(i)O

3  H2C = CH2  (ii)Zn /H O

(vii)

H2C = CH — CH2 – CH = CH – CH3

(i)O3   (ii)Zn /H2O

2

(i)O

3  CH3 — CH = CH2  (ii)Zn /H O 2

CH3

(iii)

Q.2

Find out the structure of reactant. [11 × 2 = 22] (i)

(i)O

(i)O

3  CH – CH – X  C–H 3 2 (ii)Zn /H O 2

3  CH3 – C = CH2  (ii)Zn /H O 2

(iv)

(ii)

(i)O3   (ii)Zn /H2O

(v)

(iii)

(i)O

3   (ii)Zn /H O

(vi)

O X

H

(i)O3   (ii)Zn /H2O

H

(i)O

3  X  (ii)Zn /H O 2

O

O+

O+

O

(i)O3   (ii)Zn /H2O

H

(ii)

2

H

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17

(iv)

+

2

(v)

O

O

(i)O

3  X  (ii)Zn /H O

H

H

2

(i)O

3  X  (ii)Zn /H O

O

2

How many species will be formed. Q.4

O

(vi)

O /

3   Zn /H O

(iii)

(i)O3 X   (ii)Zn/H O

CH3

+

H 3C

2

C=O +

H C=O

How many initial ozonoids are possible in given reaction. (i)

H

(i)O / 

3 CH2 = CH2   (ii)Zn /H O

2

O

O

(vii)

(i)O

3  X  (ii)Zn /H O

(ii)

2

2

(iii)

O

(viii)

X

(i)O3 O   (ii)Zn /H2O

O

C10H12

+ HCHO

Q.9

O

O

X

(i)O

3   (ii)Zn /H O

+

2

C12H18 O

(x)

(i)O

2

(X)

O

(xi)

X

(i)O

+ HCHO

3  (ii)Zn/H O 2

C12H18

O

Q.3

Ans.

Q.21

3 C H O C6H4  3 2 3 (ii)Zn /H O

Ans.

Give the ozonolysis product of the following. H O ()

3  3 (i) X  Zn O

C

H

CH3 – CH =

(ii)

N O

3 ?  Zn /H O

2

How many species will be formed.

r i

Only mole of the compound A (molecular formula C8H12), incapable of showing stereoisomerism, reacts with only one mole of H2 on hydrogenation over Pd. A undergoes ozonolysis to give a symmetrical diketone B(C8H12O2). What are the structure of A and B? O (A)

(B)

O If after complete ozonolysis of one mole of monomer of natural polymer gives two moles of CH3 CH2O and one mole of O = C – CH = O. Identify the monomer and draw the all-cis structure of natural polymer. [IIT '2005] CH3 (a) CH2 = C – CH = CH2 H CH3 (b) C = C CH2)n (CH2 OH

Q.22

O

(i)O /  3 CH2   (ii)Zn /H O 2

S . J .

O + HCHO

O=C=O+

O

(ix)

(i)O / 

3 CH3 – CH = CH – CH3   (ii)Zn /H O

H ,

 X

(i ) O

3   Y

(ii) Zn / Cl3COOH

Identify X and Y.

Ans.

O , (Y) CH3 – C – (CH2)4 – CH = O

(X) CH3


18


ORGANIC chemistry

DPP NO-11 O

Time: 15 minutes CHO

(C)

Q.1

(D)

CHO CHO

O

is obtained by the ozonolysis of following?

Q.5

This compound on ozonolysis gives which of the following compounds O

(A)

(B)

(C)

(D)

O

O

r i O

O

(I) O

(II)

O

OHC—C—C—CH2—C—CHO O

O

O

O

Q.2

H

O3 ?   O ZnH2O H O

S . J . Q.6

(B)

(C)

(D)

CH3 | H  CH3 — C — CH3   | OH

O3 (A)   (P) + HCHO Zn,H O 2

N

Product (P) is

Q.7

O || (A) CH3 — CH2 — CHO (B) CH3 — C — CH3 (C) CH — CH — OH (D) CH3 — CHO 3 | CH3

Q.4

The reactants that lead to product (a) and (b) on ozonolysis are

Q.8

(A)

O

(B) I, III, IV (D) I, II, IV

Which of the following will give three different compounds on ozonolysis Me

(A)

(B)

(C)

(D)

Me

Which one of the following compounds gives acetone (CH3)2C=O as one of the products of its ozonolysis? (A)

(B)

(C)

(D)

ozonolysis

Sentene  O

Which is the correct structure of Sentene.

HCHO

CHO

(a)

OHC—C—C—CH2—CH = O

O

O

O

(IV)

(A) I, II, III & IV (C) I, II, III

O O

(A)

Q.3

(III)

Which starting material should be used to produce the compound shown below?

(A)

(B)

(C)

(D)

(b) (B) Q.9

In which of following reaction product formed is aromatic


19

OH Q.10 (A)

HBr  

(B)

HF    KH   

(C)

Propane reacts with chlorine in sunlight to give two products. 1-chloropropane is obtained in 44% yield and 2-chloropropane is obtained in 56% yield of the total product. 2-Methylpropane reacts with chlorine under same conditions to produce 1-chloro-2methylpropane 66% and 2-chloro-2-methylpropane 33% What will be the percent yield (X) of the major product obtained when 1,3,5-trimethylclohexane is treated with Cl2 in similar conditions. (Round answer to nearest integer)

OH H2SO4  

(D)

HO OH

IIT-JEE ChEmistry by N.J. sir DPP NO-12

Time: 15 minutes

Free Radicals:– Wurtz Reaction:– Na

S . J . Cl

Cl

Mechanism:– (Free radical Mechanism) (1) Na Na+ + e–

(3)

Na  ________ _ ether

Q.8

R — X  R— R ether

 R — X + e–  R + X   R + R R — R

:

(2)

:

Q.12

(3)

  R + R — X R — R + X

:

R — X + 2e–  R+ X

Na  _______ ether

Br

Q.11

(2)

Cl

Q.9

Q.10

(Ionic Mechanism) (1) Na Na+ + e–

Na

N

Q.1

CH3 — Cl  _______ ether

Q.2

Et — Cl  _______ ether

Q.3

CH3 — Cl + Et — Cl  _______ ether

Na

Cl

Na

Na

Na THF

Na  _ _ (Wurtz fitting) Cl + Cl — CH3  THF

Cl

I

Q.5 Q.6 Q.7

Br

Q.16

Cl Ag/Powder  _______ H—C—Cl  Cl

Q.17

_______ CH2—CH=CH—CH2  ether

Na

Br

Br

I

Na

_______ Br  ether

Cl Br Br

Na  _______ ether

Cl Zn  _______ CH2—CH2 

Br

(Ph)3 CI  _______ ether

Na  _______ ether

Na _ _____ Cl  ether

Ph — Cl   _ _ _ _ _ _ _ (fitting Reaction)

Q.13

Q.15

Na

_______ Cl  ether

Cl

Q.14

Q.4

r i

ORGANIC chemistry

Na  _______ ether

Q.18

Na  _______ ether

Br


20


ORGANIC chemistry

DPP NO-13 Q.1.

Q.2.

Q.3

Time: 15 minutes

On chlorination, an equimolar mixture of ethane and neopentane yields neopentyl chloride and ethyl chloride in o o the ratio 2.3 : 1. How does the reactivity of 1 hydrogen in neopentane compare with the of a 1 hydrogen in ethane? Give product(s) in each of the following reactions Br2 / hv CH3 CH CH2 CH2 CH3  (A) CH3 Bromination of methane is slowed down by addition of fairly large amount of HBr. (a) Suggest a possible explanation for this. (b) Account for the fact that addition of HCl does not have a similar effect upon chlorination of CH4.

r i

Q.4.

Which of the following compounds on gentle heating, will undergo homolytic bond cleavage easily: O (A) (B) (CH3)3C ─ OC(CH3)3 (CH3)2C O C CH3 (C) (CH3)3C ─ C6H5 (D) (CH3)3C ─O─O─C(CH3)3

Q.5.

Amongst the following, the most basic compound is: (A) C6H5NH2 (B) p-NO2─C6H4NH2 (C) m-NO2─C6H4NH2 Which of the following is most acidic?

Q.6

(A)



NH3



(B)

(C)

CH3 NH3

NH3

(D)

Write mechanism of following reactions? (a)

Na

Cl 

-

Et 2O

HC

(b) Q.8.

S . J . 

NH2

Q.7.

(D) C6H5CH2NH2

Na

Ph3 C─Cl 

Ph2 C

Et 2O

CH2

CH2

CPh2

Arrange the following in increasing order of stability. (a)

,

,

,

,

(regarding stability of free radical) CH3 (b)

CH2

Q.9.

N

CH2 , CH2

CH

CH

2

CH

CH2 ,

CH2

,

Supply the structure and type of intermediate species designated by   (a) CH3 CH ─CH3 + H+  ? + H2O (b) CH3CH2 ─ N ═ N ─ CH2CH3  ? + N2 |

OH

Q.10.

2



(c)

CH3CHI2 + Zn   ? + ZnI2

(e)

CH3CH2Cl + AlCl3   AlCl4– + ?

(d)

CH3C ≡ CH + NaNH2   ? N a + NH3

Name any organic compound which on electrolysis give H2 on both the electrodes.

Q.11. Given structure of major product formed by electrolysis of following salts. (a)

(b) COONa

COOK

COONa

COOK

(c) COONa

COOK


21

CH3

Q.12

CH3

CH3

C° +

C°            (Major Pr oduct) CH3

CH3

CH3

Q.13

CH3 — C°H2 + CH3 — C°H2            (Major Pr oduct)

Q.6

C

+



C

 (explain)

H

H

O +

Q.14

Ph

C

C

CH = CH 2

CH3

CH 3

 H / Ph

Mechanism : – (ii)

Ph

H

H

C

C

CH 3

CH 3

H

S . J .            (Intermediate1 )

CH = CH 2 + H

r i

(Rearrangement)

Nucleophile  Pr oduct   Intermediate3  (Intermediate2 )





HCl | CCl4

HBr CCl4

Q.15 A

N

Ph—CH2 = CH 2

D

HCl R–O–O–R/CCl4

CH3

Q.16

CH3

C

CH2

B

HBr (R–O–O–R)/CCl4

Br /h



C

CH3

aq.KOH

2 2  A   CH3 

CH3

C

CH2

CH3

OH

H

Conc. H2SO4 HBr(R–O–O–R) C   B CCl4

A & C are Idenitcal / Isomers / Positon Isomers.

 

 


22

Q.17

Ph

Ph

O

C

C

NaOH

OH  electrolysis

Ph

Anode

Cathode Ph

_ _ _ _ _ _ _ _ _ e— +

(I)

(ii)

Ph

Ph

O

C

C

Ph

O

C

C

2H2O + 2e— _ _ _ _ _ _

(i)

O°

Ph

O°

CO2 +        

Intermediate

Ph

(iii)

Intermediate - - - - - - - - - - - 

IIT-JEE ChEmistry by N.J. sir DPP NO- 14

S . J . COOH

Kolbe Electrolysis:–

O NaOH  R—R electrolysis

+

CO2 + H 2

COOH

Cathode

Anode

Mechanism:– (1)

R — C —O—H

 + OH

O

H

  R — C —O—

+ O

Q.6

Anode:–

O

O

(2)

KOH _______  electrolysis

– R — C —O—  R — C — O + e O

N

 R — C — O  R + CO2    R + R R — R 

Cathode:– (1)

2H2O + 2e— 2O H + H2

Q.1

electrolysis CH3 COOK  _______

Q.2

NaOH Et — COOH  _______

electrolysis _ COONa 

_ _

NaOOC

Q.8



electrolysis COONa  __ _ _ _

NaOOC

Q.7

H

(2)

COOH

Q.5

HOOC

O

(1)

Time: 15 minutes

NaOH _______  electrolysis

Q.4

R — C —OH

r i

ORGANIC chemistry

electrolysis _ _ _ _ __ _ SO3 K 

KO3S

COOH

Q.9

SO3H

NaOH  electrolysis

_______

COOH

Q.10

NaOH  electrolysis

_______

COOH

SO3K electrolysis  _ _ _ _ _ _ _

Q.11

electrolysis

NaOH _ _ _ _ COOH 

Q.3

HOOC

electrolysis

SO3K

Q.12

KO3S

C

electrolysis 

C

___

SO3K


23


ORGANIC chemistry

DPP NO- 15 Q.1

Time: 15 minutes

Matrix Reactions

(A)

Number of dimerization product (excluding stereoisomers)

Na CH3 — CH2 — Cl   Dry ether

(P)

1

(Q)

3

(R)

6

(S)

None

Cl Na   Dry ether

(B)

(C)

14

Na H2C = CH — CH2 — Cl   Dry ether

Q.2

Cl  * 14 Na  + CH3 — Cl (H2 C  CH2 )

*

(D)

Dry ether

Br

Identify major products. CH2Br

(1)

+ 4 Na 

AlHg   C2H5OH

(2)

Na   DE

14

(7)

I I I+

Cu  120260C

I

Cl

(4)

(6)

Br

O

(3)

S . J . Cl

Br

r i

Na   DE

(5)

Br

(8)

N

Na   Dry ether


(9)


Cl

Na   DE

Br


24


ORGANIC chemistry

DPP NO- 16 Q.1

Time: 15 minutes Cl

Identify the major product:CH2 — I

(12)

Na +   Dry ether

(1)


CH3 — CH2 — Cl + Br

Br

Br

(2)

Cl

(13) Na   Dry ether

+

2.

Br

Br

CH2

CH2

(1)

E

rxn - coordinate

Br

Se  (B) Na   (A)  Dry ether

(4)

Br

Br

Br Br

(2)

(3)

(4)

(6)

CH2 — Br

(7)

Na   (A) + (B); Dry ether

CH2 — Br

3.

N

Na H2C*=CH—CH2—Cl  A + B + C Dry ether

Br


(9)

Cl

(1) (2) (3) (4)

Cl


(10) Cl


(5) (6)

E

rxn - coordinate

E

Cl  + H — CH3  H — Cl +  CH3

Bond energy

(A) & (B) are isomer.

(11)

CH3  + CH3   CH3 — CH3

Na   Naphthacene Dry ether Br

(8)

hv  2Cl Cl — Cl 

S . J .


(5)

r i

Which of following is correct matching of energy profile diagram?   CH3  Cl  CH3 — Cl


(3)


rxn - coordinate

E rxn - coordinate

KJ    KJ   H  Cl  432   CH  H  440  mole   3  mole 

When pentane is heated to a very high temperatue, radical reactions take palce that produce (among other products) methane, ethane, propane, and butane. This type of change is called thermal cracking. Among the reactions that take place are the following:CH3CH2CH2CH2CH3  CH3  + CH3CH2CH2CH2  CH3CH2CH2CH2CH3  CH3CH2  + CH3CH2CH2  CH3  + CH3   CH3CH3 CH3  + CH3CH2CH2CH2CH3  CH4 + CH3CH2CH2CH2CH2  CH3  + CH3CH2   CH3CH2CH3 CH3CH2  + CH3CH2   CH3CH2CH2CH3 (a) For which of these reactions would you expect Eact to equal zero? (b) To be greter than zero? (c) To equal H°?


25

IIT-JEE ChEmistry by N.J. sir DPP NO-17 1.

ORGANIC chemistry Time: 15 minutes

Compound

Number of monochloroproduct

Number of monocloroproduct (excluding stereoisomer)

1. 2. 3. 4. 5. 6.

7.

8.

9. CH2

2. 1.

Compound

S . J .

Number of Dichloroproduct (including stereoisomer)

1.

1-chlorobutane

2.

R-2-chlorobutane

3.

3-chloropentane

4.

R-2-chloropentane

5.

S-2-chlorobutane

6.

R & S-2-chloropentane

7.

R & S-2-chloro butane

N

r i

Optically active product


26


ORGANIC ChemIstRy

Question bank on

r i

GRIGNARD'S REAGENT

H N, RC

O

3

N

O RCOR, H2

HCOOEt

SiCl 4

l2 PbC

Cl 2 Zn

H2 O ,

S,

2

CH

H

R–Mg–X

O +

gB r

2

O +

O , H 2O – CH 2 CH 2

H

O , H2 HO RC

O2, H2O,

O +

O +

O ,H

O

CdCl 2

S . J . –X

2 CH H= –C 2 H CIC R CICH 2O R–Mg–X PhC H2 C l CI– CN

R

Cl 2

Br2

I2

?


27

GRIGNARD’S REAGENT Q.1

The order of reactivity of alkyl halide in the reaction R – X + Mg  RMgX is (B) (A) RI > RBr > RCl (B) RCl> RBr > RI (C) RBr > RCl > RI (D) RBr > RI > RCl

Q.2 Mg Br–CH2–CC–CH2–Br  BrMg–CH2–CC–CH2 –MgBr

O || PhMgBr(excess) + CH3 – C – Cl 

 H

(C)

O O || || CH3MgBr(excess)+ CH3 – C – O  C  CH3

(D)

O || CH3MgBr (excess) + Cl  C  O  Et   H

(excess) Et 2O

  H

Mg(1 eq.) Et2O

Product

Cl

Q.9

The major product is (A) Br –Mg–CH2 –C  C –CH2–Br (B) Cyclobutyne (C) –(CH2–C  C –CH2)n – (D) CH2 = C = C =CH2 Q.3

Q.4

1 equivalent Mg

Br

 ether

D

Br

(D) None of these

D

S . J .

Which of the following reacts with Grignard reagent to give alkane? (A) nitro ethane (B) acetyl acetone (C) acetaldehyde (D) acetone

r i Y ; Y is

(B)

D D

(C)

D O

2  

Cl

(A)

On conversion into Grignard followed by treatment with ethanol, how many alkyl halides (excluding Q.10 stereoisomers) would yield 2- methyl butane (A) 2 (B) 3 (C) 4 (D) 5

X

Compounds are shown with the no. of RMgX required for complete reaction, select the incorrect option (A) CH3COOC2H5 1 (B) CH3COCl

2

(C) HOCH2COOC2H5

3

OH

Q.5

How many litres of methane would be produced when 0.595 g of CH3MgBr is treated with excess of C4H9NH2 (A) 0.8 litre (B) 0.08 litre (C) 0.112 litre (D) 1.12 litre

(D)

CHO COOC2H5

Q.11

Q.6

How many litres of ethene would be produced when 2.62 g of vinyl magnesium bromide is treated with 224 ml of ethyne at STP (A) 0.224 litre (B) 0.08 litre (C) 0.448 litre (D) 1.12 litre MgBr

Q.7

OH

N

(B)

(C)

H H

H

C = O (II)

CH3

(A) I > II > III > IV (C) II > I > IV > III

C = O (III)

CH3 CH3

C = O(IV)

Me3C

C=O

Me3C

(B) IV > III> II> I (D) III > II> I > IV

Carbonyl compound (X) + Grignard reagent (Y)  OH

OH

(A)

What will be the order of reactivity of the following carbonyl compound with Grignard’s reagent? (I)

Q.12

A

+

4

O–Ph

(D)

Ph X, Y will be-

Me Et

O O || || Et – C – Ph, MeMgBr Me – C – Ph,Et MgBr (A) (B) Q.8

(A)

In which of the following reactions 3º alcohol will be obtained as a product. O || MgBr (excess) H – C – Cl +   H

O O || || (D) Me – C – Et,Ph MgBr (D) Et – C – Ph,Et MgBr


28

Q.13

(i) Mg (R) – 2–Bromooctane   X; X is

C6H13 *

(A) CH3

COOH

(B) HOOC

(D) None of these

(A)

(B)

CH 2CH=O

(C)

CH=O

(D)

CH–CH3 | CH=O

Br

1. Mg/ether Product (s) + 2. CH3CHCH2CH 3. H3O | || OH O

(B) RMgX + C2H5OH  Alkane

Select the product from the following

(C) RMgX + CH3CH2Cl  Alkene

I.

(D) RMgX + Cl

O

Ether

The number of consumed per

moles mole

of of

COOEt

is

(A) 4

(B) 2

(C) 3

Q.20

Br

Dry Ether



O || CH3–C–CH3 NH4Cl

N

H  

X (Major)

End product of above reaction is

(A)

CH 2 || CH2= CH – CH2– C – CH 3

(B) H2C  CH – CH  C – CH3 | CH3

OH | H C  CH – CH – C – CH3 (C) 2 2 | CH3 2–OH (D) H2C = CH– CH2–CH–CH |

2CH3MgBr  A. Product A formed

(A) is ethyl acetate (B) further react with CH3MgBr/H2O+ to give acetone (C) further react with CH3Mg Br/H2O+ to give t- butyl alcohol (D) Can give pinacol when treated with Mg followed by H2O Order of rate of reaction of following compound with phenyl magnesium bromide is O || Me – C – Cl I

(A) I > II > III (C) III > I > II

Q.22



O || C2H5 O – C – OC2H5

S . J . (D) 1

Select the correct statement : (A) 1, 4- dibromobutane react with excess of magnesium in ether to generate di-Grignard reagent. (B)1, 2- dichlorocyclohexane treated with excess of Q.21 Mg in ether produces cyclohexene. (C) Vicinal dihalides undergo dehalogenation to give alkene when heated with Zn dust or Mg. (D) 1, 3- dichloropropane by treatment with Zn dust or Mg forms cyclopropane. Mg 2 CH3–CH = CH2   

r i

II. CH 3 CHCH 2CH III. CH 3 CHCH 2CH | || | | OH OH O OH (A) III (B) I, III (C) I, II (D) II, III

grignard reagent the compound

O

Q.17

H3 O

O || C–CH3

In which one of the following reaction products are Q.19 not correctly matched in (A) RMg X + CO2   Carboxylic acid ( 2) H

HO

Q.16

followed   

Product

CH3

H

(C) A and B both

Q.15

*

OEt OEt

OEt Ethyl ortho formate

C6H13

H

Q.14

MgBr + H –C

Q.18

(ii) CO2 (iii) H

O || Me – C – O – Et III

(B) II > III > I (D) II > I > III

Select the correct order of decreasing reactivity of the following compounds towards the attack of Grignard reagent (I) Methyl benzoate (II) Benzaldehyde (III) Benzoylchloride (IV) Acetophenone (A) II > III > I > IV (B) I > II > III > IV (C) III > II > IV > I (D) II > IV > I > III O

Q.23

CH3 MgX NH4 Cl

(A) Enantiomer (C) Meso Q.24

O || Me – C – H II

Product is (B) Diastereisomer (D) Achiral

Nucleophilic addition of Grignard reagent cannot occur in (A)

CH3

(C)

O O O O || || || || CH 3– C– C–CH 3 (B) CH3– C– CH2–C–CH3 O O O || || CH 3– C– CH 2–CH 2–C–CH 3 (D) O


29

Q.25

CH3 CCH2CH2 CH2 Cl CH3MgBr A, A is || O CH3 | (A) CH3 CCH2CH2CH2Cl | OH

C

O

Q.26

Q.30

(D)

  A,  (ii) H3O

O

O

HO OEt

(A) CH3CCH2CH2COCH2CH3 (B) CH3CCH2CH2CCH 3 CH3 O

OH

O

OH

Cl

NH4 Cl

Identify P1 & P2.

(D) CH3CCH2CH2CCH3

O H 3C

Q.27

CH3

(A) CH4

(B)

S . J .

CH CHO

Mg 2 3 PhCH3   (A)  (B)   (C) h

NH4 Cl

ether

OH CH2 CH

CH3

(A) CH3

CH3

(C)

OH CH

(B)

Q.33

CH CH3

r i HO Me

(C)

(D) O

14

Mg NaHCO (i) CO  (A)   (C) gas  (B)  3

2

(ii) H /H2 O

(i) CH ONH

3 2 2CH3MgBr   

(ii) H

CH3

CH3

following

Product C is (A) CO (B) 14CO2 (C) CO2 14 (D) A mixtue of CO2 and CO2

CH3

OH C

(D)

HO

Br

Q.32

CH3

the

2CH3MgX   P 1 + P2

Q.31

CH3 H 3C

RMgX (2 moles) 

Deprotonation will occur from positions: (A) 1, 2 (B) 1, 3 (C) any two positions (D) 1, 4

A formed in this reaction is

(C)

OH

C

(i) CH3MgBr(one mole)

O

CH3

CH (4)

O

OH

CH3

,

(2) OH

H (1) S

H3C

CH 3CCH 2CH2COCH 2CH 3

O

CH3

CH3

O (3) H

O

CH3

O

O

O CH3

HO

(D)

OH

(B) CH3 CCH 2CH2CH2CH3 ||

CH3

CH3

CH3 ,

(C)

H3C

(C)

HO

O

(A) CH3 – O – NH – CH3 (B) CH3 – NH – CH3 (C) CH3 – NH2 (D) CH3 – OH OH

O

Q.28

Select the correct order of reactivity towards Q.34 Grignard reagent for nucleophilic attack. O O || || (A) R  C  R  R  C  H (B) Cl – CH 2  C – H  CH 3CH 2 – C  H || || O O

N

O ||

O || NO2 < CH3 – C – O

(C) CH3–C–O

Q.35

(ii) H / H2O

CH3

(A) The product is optically active (B) The product contains plane of symmetry (C) The product shows geometrical isomerism (D) The product shows optical isomerism Which of the following is incorrect. O

O CH3MgX

  CH –C – OC H

C

(A)

O O || || (D) R  C  OR  R  C  NR 2

Cl

3

OC2 H5

2

5

(1 eq)

O

OC2H5 C2H5MgX

O

(B) CH3– C – OC2H5   CH3–C – OC2H5 OC2H5

(i) CH3MgBr / CuCl

  (X) Major + (Y) 

Q.29

(i) CH MgBr

3   (A) 



(X) and (Y) respectively are

H3 O (C) CH3MgX + C = S   CH3–C – SH

O

(A)

, CH3

CH3

S

S

(ii) H2O / H

OH

(1 eq)

OH

(B)

, CH3

CH3

OH

O

O H3 O

(D) CH3MgX + C = O   CH3–C – OH


30

Q.36

Which reaction gives 1° aromatic amine as major Q.41 product. Br

(ii)NH4Cl

NH4 Cl

(A) 1°ROH (C) 3°ROH

NH3 Mg/ ether    

(A)

(i)CH CN

RMgX 3 RMgX   (A)   (B) will be

(B) 2°ROH (D) Alkene

Br NH3 Mg/ ether    

(B)

Q.42

F Br

Hg(OAc )2  

(D) HO–CH2–CH2–CH2–CH2– OH

CH3MgBr + CH2 = CH – C – H

H O

3  

Product (1, 4

addition). It is

Q.43

OH

(A) CH2 CH

C CH3

O

(A) CH4 + IMgO

OH

C– CH3

(D) None

O

(B) CH3–O

O

C– CH3

S . J . OMgI

(i) PhMgBr   Product

Q.38

(C) CH3–C

(ii) NH4 Cl

OH

CH3

Me

MgI

Produtcs in this reaction will be (A) Stereoisomers (B) Enantiomer (C) Diastereomers (D) Geometrical isomers

(D) CH3O

Q.44

O

CH3MgBr (1 eq.)  ?

O

C–CH3 r

1 + Ph Mg Br   Ph CH2CH2OH

(i)

r

2 + Ph Mg Br   Ph CH2CH2CH2OH

(ii)

O

The product is:

(A) r2 > r1 (C) r1 = r2

OH

O

O

(A)

O

O

O

Q.39

r i

The reaction of 1 mole each of p-hydroxy acetophenone and methyl magnesium iodide will give

(B) CH2 CH = CH – CH3

H

(C) CH3CH2CH2CHO

O

N (B)

O

O

OH O

(C)

Q.40

OH

CH3

NH3 /NaBH4

O

Q.37

(B) CH3– CH–CH2–CH3

CH3 CH – CH–CH3 3 (C)

F

(D) Ph

(ii)H2O

O (A) CH3– CH–CH2OH

NH3 Mg/ ether    

(C)

(i)CH3MgCl  

CH3 – CH – CH2

Q.45

(B) r1 > r2 (D) r1 = 2r2

How many moles of Grignard reagent will be required by one mole of given compound? SH

HO

O C

OEt

C

Cl

O

(D)

(i) Br

CH2

(A) 7

(ii) CH3MgBr (2equi.)

CH3

(B)

CH3

CH3

(C)

Q.46

O

O

O

O

Cl

2 CH2= C = O   C4H8O

(A)

CH2

CH3

(B) 6

(C) 8

(D) 5

Consider the given organometallic compound. (I) (CH3)2Hg (II) (CH3)2Zn (III) (CH3)2Mg (IV) CH3Li The correct decreasing order of ionic character is (A) I > II > III > IV (B) II > I > III > IV (C) I > III > II > IV (D) IV > III > II > I

(D) All of these

CH2


31

(i) CH3MgBr

O

Q.47

CH3CH = CH –C –CH3

For Q. no. 48 to Q.no.50 Consider the given reaction and answer the following questions

P

+

(ii) H3O

(i) CuI CH3MgBr (ii) H3O+

COOCH3

Q

MeMgBr   Products

O

OH

O

(A) P is CH3CH = CH –C –Me Q is CH3CH – CH2 –C –CH3

O

OH

(B) P is CH3CH– CH2 –C –CH3 Q is CH3CH = CH –C –Me CH3

Me

Q.48

No. of RMgX consumed in the reaction is (A) 4 (B) 5 (C) 6 (D) 7

Q.49

How many product will be formed in given reaction (excluding stereo) (A) 2 (B) 3 (C) 4 (D) 5

Q.50

Which of the following reaction will give the same Hydrocarbon formed as one of the product in the above reaction. (A) EtMgBr + Me – OH  (B) PhMgBr + Me – OH  (C) MeMgBr + Ph – OH  (D) MeMgBr + CH3 – CHO 

(C) P is CH3 CH=CH–C(CH3)2 Q is (CH3)2CHCH2C –CH3 OH

O

(D) P is (CH3 )2CHCH2C –CH3 Q is CH3CH=CH–C(CH3)2 O

O

O

CH3

Me

(excess )

OCH3 SH

OH

S . J . ANSWER KEY

R – CH2 – CH – Me

R–Mg–X

HO

O +

O

R – CH2 – CH2 –OH

H N, RC

O

3

R4Si

R

SiCl4

2

l2 PbC

HC OO Et

R – CHO

Cl 2 Zn

PbR4

–X

2 CH H= –C 2

H CIC CICH2OR R–Mg–X CIC OO Et CI– CN

R

N 5 C 20 C,D 35 B 50 C

R2CHOH

O

H2 =C CH CH 2 R– OR RCH2

R–COOEt

I2

Br2

R–CN

R2

R – C– R

4 A,B 19 C 34 B,C 49 C

OH R 3C

2

Cl 2

OH

3 C 18 C 33 C,D 48 C

R, RCO

R–R

Zn

R2– C= O

2 D 17 B 32 C 47 C

O , H2 HO RC HO

HC

O +

H gB r

RSH

2

O +

R

R2Hg

OH CH 2

R–OH

HO 2 ,

O

O +

R – C– OH

Q.No. 1 Ans. A Q.No. 16 Ans. A,C,D Q.No. 31 Ans. A,B Q.No. 46 Ans. D

HO ,H

O

R 2Cd

H H 2O RC

S,

RH

l CdC 2

R – C– SH

RH

O2, H2O,

RSO2H S

r i

RCl

RBr

R–I

R–C

R

6 C 21 A 36 B

7 A 22 C 37 C

8 9 B,C,D D 23 24 A B,D 38 39 A,C,D D

10 A 25 C 40 A

11 A 26 C 41 C

12 A,B,C 27 A,B,C 42 B

13 C 28 B 43 A

14 C 29 B 44 B

15 A 30 A 45 A


32


ORGANIC chemistry

DPP NO- 01

Time: 15 minutes

Explain:

I

CH2N2 

CH2 = CH2

CH2N2 

CH  CH

2.

C

H

H

:

N

H

H

N

Characteristics :– 1. The reaction is stereospecific in nature

H

Q.1

Simon Smith Reaction:– CH2I2 CH2 = CH2   CH2 — CH 2

CH2I2   Zn

CH2

OH

Zn  Zn+2 + Ze—

1)

2e– I — CH 2 : – + I — CH2 — I 

1.

Zn+2

I

I

4.

–

OH

S . J . C—OH

IIT-JEE ChEmistry by N.J. sir H

SOCl 2

O

:

R

N

3.

N +

N

–

+ HCl

S=O

Cl

H

O—S—O + Cl

N :

Cl

Cl

R

N

R — :O

—

R—O—S—O

—

:

H

+

S=O

R

H N +

Cl :O: +

H

R

Cl + — O—S—O

H

Cl

S=O Cl

O

–

2.

H

S O +

O

—

Cl

O

O

–

O +

R — O — S — Cl

O Cl – R

+



Cl R

+ Cl

+ HCl 

Cl

+

Cl

R — O — S — Cl

 R— Cl + SO2

Machanism:– 1.

Cl

Cl

Time: 15 minutes

This reaction leads to inversion of Configuration. Darzen’s Process.

Cl

H

(Kolbe’ Schmidt Reaction)

ORGANIC chemistry

+ Cl

:O: +

OH

H

R — Cl + SO2  + HCl 

Mechanism: – 1. R

O C

O

DPP NO- 02

r i OH

CO2  NaOH high P high T

+

CH2 — Zn I (Carbenoid)

CH2I2   Zn

2.

CH2I2   Zn

3.

Zn/ Cu

Mechanism:–

(No bond rotation)

C—C

Zn–I

C

N:



:

1.

C=C

2.

– O — S — Cl

R – Cl + Cl — S

—

:

Cl

+

S O–H +

O : —H

– So2 + Cl

H— Cl + 

SO2




33

Note:

the reaction leads to selection of configuration Retention of configuration. For example:— O

Q.1

H

H :

CH3

C — CH

H

SOCl2

OH

C — CH = C = O

A

CH3

C Ph

Ph

CH3 SOCl2

Ph

:

B

N

Arndt Eistert Reaction:–

Q.2

CH2N2 R — C — CH 2 – N 2 SOCl2 R — C — Cl R — C — OH  

SOCl2

C — OH

 Ag2O/

A

CH3

CH 3

R — CH 2 – C OH

B

OH

SOCl2

O

A

R—C H

CH 3

OH

:

H

N

SOCl2

B

Nucleophilic addition elimination:— O

O

N:

1. SOCl 2 2. CH 2N 2

H

R—CH = C = O

O

3. Ag2 O/ O 4. H H

S . J .

(Overall)

R — CH 2 – C

OH

Characteristics:— 1. The reaction is known as homologation 2. The reaction occurs with relation of configuration.

–

+

CH2N2 R — C — CH 2 — N R — C — Cl 

r i

H

:

N

SOCl2

Ph

O

O Ph

Q.3

O

O

O

H

+ Cl

Ex.1

Mechanism:— + CH2 = N = N:

:

1.

N:

H

Ag2O -- - R — C — CH= N =N:  

+

–

H OH

---

–

COOH

O – + R — C — Cl + :CH 2— N

–O + R — C — CH 2 — N

N:

Cl

O + R — C — CH 2— N

Wolf Rearrangement:— O

+ R — C — CH 2 — N

N

Ag O/ 

2 N: 

H

Mechanism:— 1.

2.

N:

1. SOCl 2

2. CH 2N 2 2. Ag 2O  4. H2O

–

N:

+ Cl

Nucleophilic addition:—

1.

OH

OH

–

O

+ R — C — CH — N

+ Ag2O/  N:  R — C — CH — –N

N:

:

O

O – R — CH – C OH O

–

H

O

OH /H O

2 R — CH = C = O  R — CH 2 – C

:

1.

H—OH  R — CH – C  2



+OH

–

OH

O

:

2.

R — C — CH

+N

ROH/RO —

R — CH = C = O

– NH2 / NH 3

2

R — C — CH  O = C = CH — R.

3.

R — CH = C = O

4.

R — CH = C = O

:

:

O

2.

O

O

– + :CH 2 —N

R—NH2


34

Diazo methane Acid-Base Reaction:– 1.

O 5.

CH2N2 + HCl CH3Cl + N2

CH2N2 Et—C—H  - - - - - - - + - - - - - - - - ---

 2.

CH3 O

R — OH + CH2N2 

6.

1. SOCl 2

CH3 — CH2 — C — C — OH

2. CH 2N 2 3. Ag2 O/ O 4. H H

O

3.

+ CH2N2 

R— C

--------

O—H

7. 4.

3. Ag 2O/ COOH 4. H2 O

+ CH2N2 

CH 3

8.

O OH

Homologation Reaction:– O

S . J . H

CH3 O

O 2.

CH2N2 R—C—R 

- - - - - - - - + - - -- - - - - - - -

CH2N2 

O

O

CH2N2 

DPP NO- 03 Nitrenes :– Fill in the blanks:–

12.

CH2N2 Et—C—CH3  - - - -- - - - + - - - - - - - - ---

- - - - - - - + - - - - - - - - ---

N

+ R — N = N = N:

:

:

(a)

ORGANIC chemistry

O

h  –  

(d)

:

h  R — S – N = N = N: –  

R—O—C—N—O—S— O

(e)

H— N = N = N: – +

:

h   

h  

– h  R—C— N = N = N:  + O - - - - - - - - - - - + - - - - - - - - - - - - - - (gas)

:

:

–

B

- - - - - - - - - - - + - - - - - - - - - - - - - - (gas) (f)

R — O — C – N = N = N: + O

O

- - - - - - - - + - - - - - - - - - - (gas)

+ O - - - - - - - - - - - + - - - - - - - - - - - - - - (gas)

(c)

H

NO2

:

O

Time: 15 minutes

- - - - - - - - - - - + - - - - - - - - - - - - - - (gas)

- - - - - - - - - - - + - - - - - - - - - - - - - - (gas)

(b)

--------------

2. CH 2N 2 3. Ag2O/ 4. OH/H 2O

CH2N2 Et—C—H  - - - - - - - - + - - - - - - - - - ---

IIT-JEE ChEmistry by N.J. sir Q.1

1. SOCl 2

O

- - - - - - - + - - - -- - - - - -- -

11.

4.

C—OH

H

Cl

O

3.

10.

r i

--------------

COOH 2. CH 2N 2 3. Ag2 O/ 4. RO/H 2O

S

CH2N2 R—C—H  R—C—CH2 — H + R—C—CH2

--------------

:

1.

1. SOCl 2

9.

O

O

1. SOCl 2

O

2. CH 2N 2 CH2 —C—Cl 3. Ag2O/ 4. OH/H 2O

+ CH2N2 

CH3 — C

--------

:

5.

1. SOCl 2 2. CH 2N 2

No 2

O = C—OH


35

Fill in the blanks:–

h

H — N = C = O  

(g)

O

- - - - - - - - - - - + CO _ _ _ _

D

KOH  R — NH2  Br2

R—C—N D

Mechanism:– (h)

OH

R — N — O — S —Ph

–

1.

H O ----------- -

O

D

R— C — N

O

–

:

O

H–O–D + R — C — N: – D

+ OH D

NO2

Br—Br

OH N — Cl  - - - - - - - - - - - O R — C — N—Br :–

H–O–D +

R—C— N — Cl

(j)

OH  

----------- -

H

O

– OH

:

H

– Br + _ _ _ _ _ _ _

Hoffmann Bromanide Reaction:–

O

NaOH R – NH2 + KBr + k2CO3  R — C — NH 2 + Br2

2.

S . J . R—N=C=O

Mechanism:– 1. O

O

R—C—N—H

+

:

:

–

R — C — N: – +

OH

H

H

O – R—N—C

–

: :

OH

r i

Wolf Rearrangement

H

O

R—N=C=O

H

O

R—N—C

:

(i)

OH

O



—

CO2 + _ _ _ _ _ _ _ _ _ _ _ _

H

H2 O

Br —Br OH

O + R — C — N: –

H

O

–

:

O

:

H

R— C — N— H

Br

Br

–

r.d.s. R—N=C=O

2

OH

R— N=C=O

Q.1

N

–

15

(a)

R—N—C

H

H

R — NH 2 + OH

O = C — NH 2 Br2 /KOH   

+

- - - + - -- -

O

(b)

Br2 /KOH    ------ ---- -R — C — N — Ph H

O

—

Q.3 O

(a)

O

(iv)

R—N—C H

H

C — NH 2

D

O—H

:

: R — N + CO 2

(ii) O

Fill in the blanks:– O = C — NH 2

O

C — NH 2

CH3

(iii)

O



O

C — NH 2

O CH3

C — NH 2 Cl

Q.2

–

Compare rate of reaction:– O

(a)

: :

: + Br

–

(i) O

O R — C — N:

+ Br

R — NH 2 + OH

CH2 — CH 2 — C — NH 2

Br2 /KOH    - - -- -

–


36

(b)

O = C — NH 2 Me

Et

Br2 /KOH    - - - -- - - - - - - - -

Mechanism:– 1. R

R +

H

C=N

H l

O

(c)

R

N–H

C=N

OH

R

l

O +

Br2 /KOH    ------- ---- --

H r.d.s.

H

R l

O

Br2   ------ --KOH

(d) Ph

H O

CONH 2 H

(e)

Br2   ------ ---- -KOD/D2O

CH3

(g)

R —C—ND 2

C=N

Ph C=N

C —NH 2

CH3

(i)

-----------

CH3

C=N

OH

C

OH

N

N O

C=N

Q.5

C —NH 2

Br2  - - - - - - - - - - NaOH

Q.6

O

C=N

l

R —C — NH —R

(a)

Catalyst:– O

Et

OH

Ph C=N

H2SO4 Ph – C  N  

H

S — Cl

O

O

Migration of R is always from anti position.

H  - - - - - - - - - -  

C

Q.7 Explain:–

OH

H2SO4, SOCl2, SO3, PCl5, P2O3, PCl3, BF3,

H CH 3

Beckmann’s Rearrangement:—

H   (Catalyst)

OH

C=N

Br

R

H  - - - - - - - 

O2 N

CH3

2 ------ ---- -NH2 —C—NH2  NaOH

l

PCl5   

OH

CH3

O

R

-----------

N

KOBr  - - - - - - - - - - - 

CH3

(k)

S

O

CH3 —C — C —NH 2

(j)

PCl5  

CH3

CH3

O

-----------

C

KOBr  - - - - - - - - - 

Q.4

Br

H2SO4 

OH

CH3

Q.3

(h)

OH

S . J .

Br2   ------ ---- -KOH

O

r i

H2SO4 Ph — C —NH — CH 3 

Ph

Q.2

O

–

O

Q.1

R —C—NH 2

+ OH

H

OH

O (f)

Rl —C = N

H

NaOCl  - - - - - - - - - - - 

Et—C—NH2

O

R

Tautomer

Rl —C = N — R

O

: :

R —C = N +

OH

Ph

(b)

C=N HO— CH2

H2SO4 Ph–C  N+ H  

C


H

37


ORGANIC chemistry

DPP NO- 04

Time: 15 minutes Schmidt Reaction:–

O

H2SO4 R — C — O—H + HN3  R—NH2 + CO2 + N2.

:

:

Curtius reaction:– O O NaN3  R—C—N=N=N R — C — Cl  O NH2 –NH2  R — C — O CH 2 — CH 3  O O – HNO2 R — C — NH — NH 2  R — C — N = N = N:

+ O

Mechanism:–

1. NaN3 R—N=C=O OH/H2O   R— NH2  R — C — Cl   2. 

R — C — OH

Imp. O

:

R—N=C=O

H

+

O

N2

:



H OH / 

R —NH2 + CO2

Et O C — OH

H

H

R — C — N = N = N: H

S . J . –H+



H2O   _ _ _ _ _ _ _  _______  

Q.2

H

:

R — C — N = N = N: :

— C — Cl

O

+

O

:

Q.1

O

NaN3   _______

O

R—C

H—N = N = N:

NH–R

HOH  R — NH + CO R — N = C = O  2 2 

+H

O

R/ —OH  R—NH — C —OR 1 R — N = C = O  O R—NH2 R — N = C = O   R—N— C H

r i O

NaN3 SOCl2 _ _ _ _ _   

O

CH3

 _ _ _ _ _  _ __ __ _  _ _ _ _ _ H2O/  CH 3 Q.3

O C

_____ R

N

NaN

O

1. NH NH 2. HNO2 H/

O CH3 —O—

2

4

O

CH3—

C—OH

  N3H(H2SO4 )/ 

_ _ _ _ _ + _ _ _ _ _ _ +_ _ _ _ _ _

2 2 CH2 — C—OC2H5  

O

3

_ _ _ _ _ + _ _ _ _ _ _ +_ _ _ _ _ _

Q.2

H _____

 ____ H _ __ __ _ 

Q.5

 CH2— C—OH  N H(H SO )

 3  _ __ __ _  Cl 

O Q.4

Q.1

Schimidt miscellaneous:–

O ____

O

H R — C — R  R — C — N —R N3H/  H

1. NH2 –NH2 _ __  C—OCH 2CH 3  2. HNO2

 _ _ _ _ _ H O H/ _ _ _ _ _ 


38

Mechanism:– 1.

+

R

+H

R

C

:

O

H—N = N = N: R

O

R—C—N—O—C—R H

OH

O

H

–

O

: :

C

R

OH

:

O

:

Mechanism:– 1.

+ OH H

O + R—C—N— O—C—R –

R—C—R :

H—N = N = N: OH

OH R—C

+

N2

H—N = N = N: :

:

H +

H + R—C—N—R

O

H

Q.2

Lossen’s Rearrangement:– O

R — C — N — O — C— R H

—

+O —C—R

O O KOH  _ _ _ CH2 — C— N — O — C — Ph  

Q.1

R

O

O

O

R—C—N

R—C—R

N

H

O R—N = C = O

: :

– R — NH 2+ CO 2

OH

H

O

–

R—NH2+ R — C — O—

N

O

C— N — O — C — CH3

S . J .

r i CH3

KOH   


39


ORGANIC chemistry

r i S . J .

Reaction

Intermediates

N


40

Q.1

Q.2

EXERCISE – I 2-Chloropentane on halogenation with chlorine gives 2,3-dichloropentane. What will be the structure of free radical species formed in the reaction? (A) Planar (B) Trigonal planar (C) Square planar (D) Pyramidal Decreasing order of nucleophilicity following nucleophile is : (1) CH3O

of

(B) H

O

O

(C)

the

H

O

OH

(D)





O

O

(2) C N O S

(3)



r i S . J .

(B) 1 > 2 > 4> 3 (D) 1 > 2 > 3 > 4

(A)

The correct order of rate of Wurtz reaction. Na (I) CH2– F CH – CH 2

ether

Na

CH2– Cl

(III)

(B)

CH2 – CH2

ether

Q.8

What will be the major product, when 2-methyl butane undergoes bromination in presence of light? (A) 1-bromo-2-methyl butane (B) 2- bromo-2-methyl butane (C) 2- bromo-3-methyl butane (D) 1-bromo-3-methyl butane

Q.9

Which can not be the possible product of the given reaction

CH2 – CH2

(A) I > II > III > IV (B) II > I > III > IV (C) IV > III > II > I (D) In all rate of Wurtz reaction is same Q.4

CH3 – CH –CO2K

electrolysis

(A) (Major)

CH3 – CH –CO2K

Major product (A) of above reaction

Q.5

(A)

(B)

(C)

(D)

Mg H2O

P1



P2

(A) It is a spiro compound (B) It is a Ketone (C) It can show tautomerism (D) It is an alkene Q.6

Consider the following reactionO

OEt

Na/ether

FeSO4 + H2O2

liq. NH3

Fenton's reagent

the major product P is: O

(A)

OH

CCl4, 

Product (s)

O

N H

Br2

CH3 – CH2 – C – OAg

(A) CH3 – CH2 – Br (B) CH3 – CH2 –C– O – CH2 – CH3

Which of the following is not correct about P2. O

(C)

(D) None of these

Na

Na ether

CH2– I

2

CH2 – CH2

ether

CH2– Br

(IV)

(X)

X is :

(A) 4 > 3 > 2 > 1 (C) 2 > 1 > 3 > 4

(II)

Na/ether

CCl4. Peroxide

O

Q.3

NBS

Q.7

(4) CH3 C O 2 

O

O

(C) CH3 – CH2 – CH2 – CH3 (D) CH3 – CH2 – CH3

Q.10

Pick the correct statement for monochlorination of R-secbutyl Bromide. Me H

Br

Cl2 300°C

Et P

(A) There are five possible product; four are optically active one is optically inactive (B) There are five possible product; three are optically inactive & two are optically active (C) There are five possible product; two are optically inactive & three are optically active (D) None of these


41

Q.11

Correct order of rate of photochlorination for Q.16 following compounds is

(A) I is more soluble than bromocyclopropane (B) I gives pale yellow ppt. on addition with AgNO3 (C) I is having lower dipole moment than bromocyclopropane Br (D) I is more ionic than

CH3 C

CD3 – CD3 CH3

CH3 – CH3

CH3

CH3

(I) (A) II < I < III (C) III < I < II Q.12

(II)

(III) (B) I < II < III (D) II < III < I

Which one of the following carbocation would you expect to rearrange.

KNH2

(A)

?

(A)

(C)

(D)

r i S . J . NH2

*

*

(B)

(C)

Q.18

How many, 1,2-Shifts are involved during the course of following reaction: OH

*

(D) (B) and (C) both

conc. H2 SO4

Which of the following carbocation is most stable? CH3

(A) 1

(B) 2

(C) 3

(D) 4

CH3

+ C

CH3

H3C

+ C

(B)

CH3

CH3 + C

CH3

(D)

H

CH3

Q.19

+



(X)

OH

Product (X) is:

C

CH3



OH

CH3

HO

CH3

CH3

H3C

OH

(C)

(B)

CH3 

 



NH2

NH2

(A)

CH3

CH3

Cl NH3

Q.13

Q.17

Product can be

*

, which is not the correct statement

Br

(A)

(B)

CH3

Q.14

Which carbocation is least likely to form as an intermediate? 

(A) (C 6H5 )3 C

(B)

(C)



Q.20





(C)

Q.15

(D) CH2  CH

For the reactions (I)

Cl

(II)

Cl

(III) (IV)

(D)

N

(I)

OH

(III)

OH



+ Cl , H1°

(A) I < II < III < IV (C) I < III < IV < II

 + Cl , H2° 

Among the given dehydration order is:

compounds,

the

(II)

OH

(IV)

OH

correct

(B) II < III < IV < I (D) I < II < III = IV

OH 

CH2Cl

CH2 + Cl , H3°

Cl

+ Cl , H4°

The correct decreasing order of enthalpies of reaction for producing carbocation is: (A) H1º > H2º > H3º > H4º (B) H4º > H1º > H2º > H3º (C) H3º > H2º> H1º > H4º (D) H2º> H1º> H4º > H3º

Q.21

+

H

P. The product P is

5°C

(A)

(B)

(C)

(D)

OH


42

Q.22

Rate of dehydration when given compounds are treated with conc. H2SO4. Q.26

OH

H+ 

OH

CH2OH

(P)

CH3

(Q)

(A)

(B)

(C)

(D)

OH

OH CH3

CH3

(R)

(S)

(A) P > Q > R > S (C) R > Q > P > S

(B) Q > P > R > S (D) R > Q > S > P

Me

Q.27

Conc.H2SO4

Me

A



OH H2SO4

Q.23

OH

OH

Product 'A' is-

X



r i S . J . (A)

O

Me (B)

Me

Me

Me

O

'X' is

Me

OH

OH

(A)

(C)

(B)

C –Me

Me

(D)

Me

O

O

O

Q.28

OH

(C)

How many products are obtained in the given reaction: Et

(D)

Ph

HO

OH

OH

Et

OH

Q.24

O

CH3 H+ CH3 O

Q.29

(B) O

Q.25

CH3

N (D) OH

CH2 O CH3

CH3

Ph

(B) 2

H2SO4

OH



Ph

(C) 3

H3O

+

CH3 – CH – C – CH2 – NH2

CH3

CH3

HO

HO

(D) 4

CH3

(A)

(C)

+

Ph

CH3

CH3

CH3

(A) 1

major product is

CH3

HNO2

(X)

CH3 CH3

(major) Major product of above reaction is:

CH3

CH3

(A)

CH3

CH3

CH –C –CH2 – OH

CH3

A, A is

(B)

OH

CH3

(C)

(D) OH

OH

OH

HO

(A)

Q.30

Which will dehydrate at fastest rate by H3PO4 : (A) 2-methyl butane-2-ol (B) 3-methyl butane-2-ol (C) Butane-1-ol (D) 2-methyl butane-1-ol

Q.31

What is the order of reactivity with HBr.

(B)

CH3 OH

CH3 OH

OH

OH

(C)

(a)

(D) OH

OH

(b)

(c)

CH3O

CH3O

OH

(A) a > b > c (C) c > b > a

(B) b > a > c (D) b > c > a


43

CH2 = CH3

Q.32

HBr CCl4

Cl

H

Q.38

Stability order of following singlet halocarbene is (A) CF2 > CCl2 > CBr2 > CI2 (B) Cl2 > CBr2 > CCl2 > CF2 (C) CCl2 > CF2 > CBr2 > CI2 (D) CF2 > CI2 > CCl 2 > CBr2

Q.39

Trans-Butene-2  3  Product

CH3

What is stereochemistry of product? (A) Racemic mixture (B) Optically inactive (C) Diastereomers (D) Meso product Q.33

CHCl / KOH Solvent

In the given reaction :

Cl

H Br2

N

(A) CH3

[X]

[X] is : H N Br

(A) Br

H

(C)

Br

Br2

  Products

3

Cl3C –CH = CH2

(A)

(C)

H3C – CH = CH2

1

The order of rate of reaction of following towards carbylamine reaction: O – CH3

N

Q.37

CH2 = CH – CH3 Mg / ether

NH2

NaOH

(i) CHCl / KOH

CaO / 

(ii) H3O 

 Y.    X   3 

2

COOH

(D) None of these

2

OMe

OMe

(A)

NBS

14

NH2

OMe

O 1

NH2

(iv)

Identify X and Y :

(B) CH3 – C – CBr3

(C) CH3 – CH2 – C– CI3

(iii)

NH2

1

2

(ii)

NO2

CH3

(A) (ii) > (i) > (iii) > (iv) (B) (ii) > (iii) > (i) > (ii) (C) (iv) > (i) > (iii) > (ii) (D) None

O

(A) CH3 – C – CCl3

(D)

(i)

Which of the following compounds yield most Q.42 stable carbanion after rupture of (C1 – C2) bond: O

(B)

Cl

1

2

(I) (II) In addition of HOBr to (I) and (II) (A) Br is at C2 in both cases (B) Br is at C2 in II and at C1 in I (C) Br is at C1 in II and C2 in I (D) Br is at C1 in both cases Q.36

CHCl Br

 2  Product

Br

Q.41

Q.35

H

Br

(A) it is optically active (B) it is racemic mixture (C) it is a resolvable mixture (D) it is a mixture of erythro compounds 1

H

(D) Both (A) and (B)

Cl

CCl 4

2

CH3

Product is :

Br

Select the incorrect statement about the product mixture in the following reaction:

3

Cl

KOH

Br

Q.34

CH3

Q.40

N

(D)

N

H

H

H

r i S . J . (C)

N

(B)

Cl

(B) CH3

CH3

H

Br

Cl

Cl

CHO

CCl4 / Peroxide

CHO

CH3

OMe

D O

   2 Product(s) Product(s) is/are : 14 14 (A) CH2 = CH–CH2–D (B) CH2 = CH – CH2 –D (C) Both of these (D) None of these

OMe

(B) COONa



OMe

OMe

OMe

OMe

(C)

(D) CHO

CHO


44

OH

R

MgBr

(i) CH MgBr CHBr3 P1   P3  P2  3 

Q.43



KOH

(ii)H / 

O

HN

3   P. Identify P.

Q.47

H2SO4

OMe

R

P1, P2 and P3 are OMe

OMe

(A)

OMe

R

N–H

(A)

O

(B)

O R

CHOH

OH

(C)

Me OMe

OMe

(B)

OMe

Me2CH Me

CHO

CHOH

conc .H SO

2 4     A



r i S . J . (A) Me2CH – NH – C – Me

OMe

OMe

OH C=N

Major product of this reaction is

Me OMe

(D) Both (A) and (C)

O NH

Q.48 OH

NH

(C)

O

OMe

OMe

(B) Me – NH – C – CHMe2

CH= CH2

CHO

OH

O

OMe

(D)

Me

CHO

(C) Me – C – CH = N – Me

CH= CH2

OH

(C) CH2 = C –CH = N –Me

h / 

Q.44

   P.

Me

N3

'P' is: (A)

OH

(B)

Q.49

•• N

N=N

(C)

Chloride

H

O

(D)

N=N

P  Toluene sulphonyl

      P. 'P' is

N

O

(A)

N=N

O

O

(B)

NH

O

NH

NH2

Q.45

NH

Br

N

(C)

KOH

CH3CH2CH2NH2. Compound (X) is O

(A)

CH3

(B)

(C)

OH Cl

Q.46

(D)

O

Cl

COOD

Q.50



 product is

OD

O

(A)

(B) C2H5

C – NH2

C2H5

O

OD

(B)

CH2OD

(C)

O

O

(C) Me – C – NH – Me

HN

C2H5

CHO

O

O

NH

O

C – Cl

Which of the following can not give Hoffmann's bromamide reaction:

(A) Me – C – NH – Br

O

(D)

O

CH3

Cl

O

O

3 2 (X) C4H7OCl   C4H9ON   

(D) C2H5

(D)

C2H5

N–H O


45

Q.1

EXERCISE – II Match the column : Column-I Column-II (A) Electrophile (P) CH3 – CH = CH – CH3 (B) Nucleophile (Q) :CHCl

CH3

(C) Lewis acid (R) CH3 – CH2 (D) Lewis base (S) Guanidine

(A)

Select the correct statements : (A) Protonation increases electrophilic nature of carbonyl group – – (B) CF3SO3 is better leaving group than CH3SO3 (C) Benzyl carbocation is stabilized by resonance

(B)

CH3 H3C –CH – CH –CH = CH2 CH3

(C)

H3C –CH –CH2 –CH = CH – CH3 CH3

(D) H3C –CH –CH = CH – CH2 – CH3

is unstable

r i S . J . CH3

OH

Q.3

CH3 H3C –CH = CH – CH CH3

OH

(D) CCl3CH

Which of the following can be produced by Wurtz reaction in good yield. (A) (B)

Q.8

Select correct statement about the product (P) of the reaction : H

Me

(C)

Q.5

(D)

Select true statement(s) : (A) Instead of radical substitution, cyclopropane undergoes electrophilic addition reactions in sun light. (B) In general, bromination is more selective than chlorination Q.9 (C) The2,4,6-tri-tert, butylphenoxy radical is resistant to dimerization (D) The radical-catalysed chlorination, ArCH3  ArCH2Cl, occurs faster when Ar= phenyl than when Ar= p-nitrophenyl. Q.10 Choose all alkane that give only one monochloro derivative upon reaction with chlorine in sun light. (A)

(B)

N

(C)

(D)

NBS

Q.6

(A) P is optically inactive due to internal compensation (B) P is optically inactive due to the presence of plane of symmetry in the molecule (C) The structure of P can have three optical isomers possible (D) P can have four possible optical isomers. Products formed when HCl adds to 2,4-hexadiene is: (A) 4-chloro-2-hexene (B) 2-chloro-3-hexene (C) 2-chloro-4-hexene (D) 1-chloro-2-hexene HBr In the given reaction C7H12 (A)  

as major product

(B)

CH3

(D)

Which

of

following

reaction

CCl4

(B)

Br

Br

Br

(D) Br

Q.7

Br

H3C –CH2 –CH2 –CH2 OH

conc . H SO

4   2 X+ Y

H

D

(i) NaCN

   (ii) H

CH3

(C)

(C)

are

CHO

Br2  CH3 

(A)

(B)

Br

product

diastereomer of each other.

Br

(A)

CH2

CH =CH2 CH3

Q.11

CH3 Br

(A) can be

(C)

HBr

Br

H

(A)

  (X)+(Y) enantiomeric pair. CCl4 / h

Br / CCl

4 2  P

Me

Q.4

Product(s)

Product(s) are :



Q.2

(CH3 –CH)2CuLi

NBS

 CCl4 / peroxide

CH3

CH3 H

H

HBr



C=C

CCl4

(D) CH3 –CH – CH = CH –Ph

HCl

 peroxide

Et




46

Q.12

Which of the following can be formed during this Q.16

Which of the following will give cyclic products

reaction?

upon being heated or being treated by an acid? O

H3O 

(A)

  

OH

OH

OH OH

(B) (A)

OH

O

(B) OH

OH

OH

O

(C) (C)

(D)

r i S . J . O

O

   A

(ii) ROH



(iii) RNH2

P1, P2 and P3 are : (A) MeO NH2

(B) MeO

Q.17

(C)

O

(D)

Which of the following reaction is not representing Q.18 major product. O

OH +

H 

••

CH3

C – NH – CH3

CH3 H

(B)

14

Cl CH3

O

(C) Ph – C – NH 2 O

(D)

14

Ph–Li

C= C

3-hydroxy propanoic acid forms Lactide on heating

NH – C – NHR

C= N

succinic acid forms succinic anhydride on heating

N= C = O

MeO

(A)

Select the correct statements.

(B)

O

Q.15

P3

propanoic acid on heating

NH – C – OR

(C) MeO

Q.14

OH

P2

(A) methyl malonic acid is converted into

O

(D)

O

P1

NaN3

C – Cl

MeO

OH

(D)

(i) H2O

O

Q.13

OH

OH

CC

N Br2

CH3 – C –CH2COOH

forms acetone on heating Soda lim e



C5H8O4(A)  C4H8O2(B)    (C) 

C is a hydrocarbon occupying 0.509 litre per g at NTP approximately. Hence A and B are: (A) methyl malonic acid, propanoic acid (B) succinic acid, succinic anhydride (C) Dimethylmalonic acid, 2-Methylpropanoic acid (D) Ethyl Malonic acid, Butanoic acid

CH3

Q.19

Which of the following reaction is not incorrectly formulated.  SO Cl

2 (A) CH2=CH–CH3  2   CH2Cl–CHCl– CH3

UV light

CH

Ph –NH2

KOH

(B) HC  CH + CH2N2  HC

HN3 H2 SO4

In which of following compound chiral center is

N

HC

NH2

N H photo

(C) (CH3)3CH+Cl2    (CH3)3C–Cl ha log enation

as major product

not affected on heating. (A)

CH3 – CH – COOH OH

(B)

O HOOC CH3

HCl

(D) CH3 – C  C – H  CH3 – CH – CH2 Cl

Cl

(C) CH3 – CH – CH2 – COOH (D) CH3 – CHCOOH OH

CH2COOH


47

D

KNH

2    ?

Q.20

Cl

Q.21

NH3 ,33C

D

Cl

KNH

2    ?

liq. NH3

NH2

D

(A)

H2N

(B)

NH2

(A)

(B)

NH2 NH2

NH2

(C)

NH2

(C)

(D)

D

D

(D) D

NH2

Q.22

This question consist of two statements, printed as assertion and reason, while answering this question you are required to choose any one of the following responses. (A) If assertion is true but the reason is false. (B) If assertion is false but the reason is true. (C) If both assertion and reason are true and the reason is a correct explanation of assertion (D) If both assertion and reason are true but reason is not a correct explanation of assertion NO2

Assertion :

r i S . J .

PCl5 

C= N OH OMe

O2N

C– NH

OMe

O

OMe

Reason :

Migratory aptitude of

group is greater than migratory aptitude of

group during

NO2

cation rearrangements. Q.23

Each of the compounds in Column A is subjected to further chlorination. Match the following for them. Column-A Column-B (A) CHCl2 – CH2 – CH3 (P) Optically active original compound (B) CH2Cl – CHCl – CH3 (Q) Only one trichloro product (C) CH2Cl – CH2 – CH2–Cl (R) Three trichloro product (D) CH3 – CCl2 – CH3 (S) Four trichloro product Cl Cl

(E) CH3 – C – C – CH3

N

CH3 CH3

Q.24

(T)

Atleast one of the trichloro product is optically active

(U)

Two trichloro products

Column-I and Column-II contains four entries each. Entries of column-I are to be matched with some entries of column-II. One or more than one entries of column-I may have the matching with the same entries of column-II and one entry of column-I may have one or more than one matching with entries of column-II. Column-I Column-II (Reaction) (Type of intermediate formed) HO

(A) Ph–CHCl 2  (A)

(P)

Carbocation

(Q)

Carbanion

(R)

Free- radical

(S)

Carbene



Na

(B) R – Br   dry ether

O (i) Mg / H O

(C) CH –C –CH   2  3 3 (ii) H SO  2

CH3 OH

(D)

H

 

4,




48

Q.25

Column-I

Column-II O

(A) Caprolactum formation take place in

HN

3  

(P)

H2SO 4

N – OH

(B) Beckmann rearrangement is

H

(Q)

  

OH

(C) Schmidt reaction is

(i) CHCl / HO 

  3 

(R)

(ii) H

OH

r i S . J . Ph

(D) Reaction in which number of carbon increases

PCl

5  

C= N

(S)

CH3

EXERCISE – III Q.1

(ii) Ph –CH = CH2 + N2

Identify P1 to P8. CH3 – CH = CH – CH2

CH3

CH2 D

HBr

CH3

CH3 – CH = CH – CH2 – CH – CH2

HBr

Q.5

 

T

P1 + P2 + P3 + P4

CH3 – CH = CH – CH2 – CH – CH2 – CH –CH3 D

T

P1 + P2 + P3 + P4 + P5 + P6 + P7 + P8

HBr

 

Compare the rate of decarboxylation in sodalime Q.6 process for the following. (i) (a) HC  C – COOH (b) CH2 = CH –COOH (c) CH3 – CH2 – COOH (ii) (a) CH2 – CH2 – COOH (b) F–CH2 –CH2 – COOH

(c)

N

(a)

Formulate the reactions between but-1-ene in presence of small amount of benzoyl peroxide and (i) CCl4 (ii) CBrCl3 Give your reasons. (b) The dichlorocarbene reacts with phenol in base where as it doesn't reacts with benzene explain. Give the product of the following reaction. O

Mg

(i)

  A H2O

(ii)

CH = O

Mg

  B

CH = O

(c) Cl – CH2CH2 –COOH (iii) (a) CH3 – CH – COOH (b) CH3 – CH – COOH

H2O

O

(iii)

Cl

F

Mg

Me –C –Et

H2O

(iv)

Which compound is more stable explain. 



Q.7 

(a) CH3 – C – CH = N = N and CH3 – CH = N =

Ph –C –Me



MeOCH Cl BuLi

CHBr

3   ? t

BuOK

H SO 

Br

H SO

2 4 (B) 2    (D) = ?  (C) 

CCl 4



Give product and suggest mechanism for these reactions. (i) ?   2 

Mg

H2O

4   2   D

Identify missing products in the given reaction sequence. Br2 / h aq. KOH (a) CH3 – CH2 – CH3   (A)   

N

Q.4



O

CH3 – CH – COOH

O

H SO

4   2  C

CH3

Q.3



 B. Find

CH2 CH2

out A & B.

T

D

(iii) Cl3C – C – OH + Na  A

  P1 + P2

H

NO2

Zn

O

H

H

Q.2



CO2Et  ?

O C – Cl

(b)

NaN

Ag O

3 2     (P)    (Q)



H2O

NaNO 2

 (R)   (S) HCl


49

Q.8

Find out the total number of products (including stereo) in the given reaction:

OH CH3

CH3

NBS,CCl

4     Products.

•

• • X CH2 – CH2 XCH2 – C H2 + H –X

heat

CH3

• XCH2 CH3 + X

conc .

+ H2O

   H2SO 4

HCl HBr HI

–67 –25.1 +46

Q.10

Addition of small amount of (C2H5)4 Pb to a mixture of methane and chlorine, starts the Q.16 reaction at 140°C instead of the usual minimum 250°C. Why?

Q.12

1,2-Dimethylcyclohexene Isopropylidenecyclopentane

(a) Write a reasonable and detailed mechanism for the following transformation.



Q.11

C(CH3)2

+

 

2,2-Dimethylcyclohexanol

With the help of following data show HBr exhibits Q.15 the peroxide effect. –1 –1 H1°/ kJ mol H2°/kJ mol

H – X X + CH2 = CH2

H

CH3

Peroxide , 

Q.9

CH3

H

+ 12.6 – 50.2 –117.1

OH

(b)

HO

OH

H  / HOH

   

r i S . J .

On chlorination, an equimolar mixture of ethane and neopentane yields neopentyl chloride and ethyl chloride in the ratio 2.3 : 1. How does the reactivity of 1° hydrogen in neopentane compare with that of a 1° hydrogen in ethane?

It required 0.7 g of a hydrocarbon (A) to react completely with Br2(2.0 g) and form a non resolvable product. On treatment of (A) with HBr it Q.17 yielded monobromo alkane (B). The same compound (B) was obtained when (A) was treated with HBr in presence of peroxide. Write down the structure formula of (A) and (B) and explain the reactions involved.

- Terpeniol

Geraniol

Assuming that cation stability governs the barrier for protonation in H–X additions, predict which compound in each of the pairs in parts (a) and (b) will be more rapidly hydrochlorinated in a polar solvent. (I) (II) (a) CH2 = CH2 or (b)

or

Choose the member of the following pairs of unsaturated hydrocarbons that is more reactive towards acid-catalysed hydration and predict the regiochemistry of the alcohols formed from this compound. (a) or

Q.13

Give product(s) in each of the following reactions.

Br / h (a) CH3 –CH – CH2 – CH2 – CH3 2 (A)(major)

CH3

(C H CO) O

+ NBS 65 2 (B) S

(b)

or

(i)

CH3

(b)

N

h

CH3

(c)

or

CH3

(ii)

Give product in the following reaction. (i) NaNO 2

CH3 – C – O –Cl /

(d)C6H5–CH2–CH2–CH3

(ii)

(i)

(c) CH3–CH2–CH=CH2+Me3COCl   (C) + (D) Q.18

Q.14

(ii)

(i)

NH2

(E)(major)

We saw that acid-catalyzed dehydration of 2,2dimethyl-cyclohexanol afforded 1,2dimethylcyclohexene. To explain this product we must write a mechanism for the reaction in which a methyl shift transforms a secondary carbocation to a tertiary one. Another product of the dehydration of 2,2-dimethylcyclohexanol is isopropylidenecyclopentane. Write a mechanism to rationalize its formation.

(ii)

HCl

NH2

A

NaNO2 HCl

NH2

CH2NH2

(iii)

NaNO2 HCl

(v)

NaNO2

OH NH2

B

HCl

C

(iv)

NaNO2 HCl

D

E


50

Q.19

What are the products of the following reactions? (B) CH3CH2CHBr

(a) PhCH = CHCH3 + HBr  A H3C

(b)

CH3 C=C

H3C

+ HI  B

Peroxide

(c)

Q.5

CH3

+ HCl  D CH2CH3

Complete following reaction: HCl

H

Br2 CCl4

(b)

CH2

CH3 CS2

Reaction of R–CO – NH2 with a mixture of Br2 and KOH gives R – NH2 as the main product. The intermediates involved in this reaction are : Q.8 [JEE 1992] (A) R –CO –NHBr (B) RNHBr (C) R – N = C = O (D) R.CO.NBr2 Which one of the following has the smallest heat Q.9 of hydrogenation per mole? [JEE 1993] (A) 1-Butene (B) trans-2-Butene (C) cis-2-Butene (D) 1,3-Butadiene In the following compounds : OH

N OH

The reaction of CH3–CH=CH HBr gives : (A) CH3CHBrCH2

OD

The correct order of basicities of the following compounds is: [JEE 2001] NH

CH3CH2NH2

NH2

(1)

(2)

2

(CH3)2NH

O

CH3CNH2

(3)

(4)

(B) 1 > 3 > 2 > 4 (D) 1 > 2 > 3 > 4

2

Q.10

Left to right sp , sp , sp, sp hybridization is present in : [JEE 2003] (A) H2C = CH – C  N (B) H2C = C = CH –CH3 (C) HC  C – C  CH (D) HC  C – CH = CH2

Q.11

Maximum dipole moment will be of: [JEE 2003] (A) CCl4 (B) CHCl3 (C) CH2Cl 2 (D) CH3Cl

OH with

[JEE 1998] OH

Amongst the following, the most basic compound is : [JEE 2000] (A) C6H5NH2 (B) p-NO2–C6H4NH2 (C) m-NO2–C6H4NH2 (D) C6H5CH2NH2

(A) 2 > 1 > 3 > 4 (C) 3 > 1 > 2 > 4

NO2

(D) CD2 = C – CD3

A solution of (+) 1-chloro-1-phenylethane in toluene racemizes slowly in the presence of small amount of SbCl5 due to formation of: [JEE 1999] (A) carbanion (B) carbene (C) free radical (D) carbocation

OH

(I) (II) (III) (IV) The order of acidity is : (A) III > IV > I > II (B) I > IV > III > II (C) II > I > III > IV (D) IV > III > I > II Q.4

(B) CH3 – C – CH3

CH3 – C = CH2

CH3 – C

[JEE 1996]

NO2

CH3

[JEE 1999]

The enol form of acetone, after Prolonged treatment with D2O, gives: [JEE 1999] O OD

(C) CH2 = C – CH2D

EXERCISE – IV (A)

OH

An aromatic molecule will: (A) have 4n  electrons (B) have (4n +2)  electrons (C) be planar (D) be cyclic

(A)

Total number of products obtained in this reaction Q.7 is?

Q.3

Br

OH

Cl2

(c)

Q.2

(D) CH3CH2CHBr

r i S . J . Q.6

CH3

Q.1

Br

+ HBr     C

(d)

(a)

(C) CH3CHBrCH2

H

CH3

Q.20

OH

OH

COOH

Q.12 H O 2N M

OH

when X is made to react with 2 eq. of NaNH2 the product formed will be : [JEE 2003]


51

Q.18  OOC



OH

(A)

O

OOC

(B) H



OH

O2N M

OH

O 2N M

 OOC

(D)

CH3

H

OH

O

O2N M

Ph – MgBr +CH3 –C – OH

A

r i S . J . CH3

y

(A)

[JEE 2004]

(B) CH3 – C – OPh CH3

COOH x

Correct order of acidic strength is: (A) x > y > z (B) z > y > x (C) y > z > x (D) x > z > y

OH

CH3

(C)

(D) CH3 – C – Ph

Order of rate of reaction of following compound with phenyl magnesium bromide is: [JEE 2004] Ph – C – Ph

Me – C – H

O (I)

O

O

(II)

(III)

(A) I > II > III (C) III > I > II Q.15

Q.19

CH3

NH3

H3N z M



O2N M





Q.14

O

HOOC

OH

(C)

Q.13

For 1-methoxy-1,3- butadiene, which of the following resonating structure is the least stable? [JEE 2005]   (A) H2 C – CH – CH  CH – O – CH3  (B) H C – CH  CH – CH   O – CH3 2   (C) H 2 C  CH – CH  CH – O – CH3  (D) H C  CH  CH – CH   O – CH3 2



Me – C – Me

(B) II > III > I (D) II > I > III

CH3

Q.20

When benzene sulfonic acid and p-nitrophenol are treated with NaHCO3, the gases released respectively are [JEE 2006] (A) SO2, NO2 (B) SO2, NO (C) SO2, CO2 (D) CO2, CO2

Q.21

(I) 1,2-dihydroxy benzene (II) 1,3-dihydroxy benzene (III) 1,4-dihydroxy benzene (IV) Hydroxy benzene [JEE 2006] The increasing order of boiling points of above mentioned alcohols is (A) I < II < III < IV (B) I < II < IV < III (C) IV < I < II < III (D) IV < II < I < III

Q.22

CH3NH2 + CHCl3 + KOH  Nitrogen containing compound + KCl + H2O. Nitrogen containing compound is [JEE 2006] (A) CH3 – C  N (B) CH3 – NH –CH3

1-Bromo-3-chloro cyclobutane on reaction with 2equivalent of sodium in ether gives [JEE 2005] Cl

Br

(A)

(B)

(C)

(D)

SO3H

N

CH3COONa(excess)

Q.16

(aq. solution)

Me

[JEE 2005]

+

+

–

–

(D) CH3N  C

(C) CH3–N  C

SO2COOCH3

(A)

(B)

Me SO3Na

(C)

Question No. 23 to 25 (3 questions) Comprehension I RCONH2 is converted into RNH2 by means of Hofmann bromamide degradation.

COONa

+ CH3COOH

(D)

O

+ H2SO4

O

Cl

Cl

NH2

Me

Q.17

Me

Conversion of cyclohexanol into cyclohexene is most effective in: [JEE 2005] (A) concentrated H3PO4 (B) concentrated HCl (C) concentrated HCl/ZnCl 2 (D) concentrated HBr

O

Cl (ii)

(i)

(iii)

– +

OM

O C

O H2N

– N–Br ••

NH–Br

Cl

Cl

N

N

Cl

H (vi)

(v)

(iv)


52

In this reaction, RCONHBr is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation reaction is an Intramolecular reaction.

Q.28

The structure of the intermediate I is: [JEE 2007] 









ONa

CH2Cl

CHCl2

(A)

(B) CH3

Q.23

Q.24

Q.25

How can the conversion of (i) to (ii) be brought about? [JEE 2006] (A) KBr (B) KBr + CH3ONa (C) KBr + KOH (D) Br2 + KOH

15







CH2OH 2

CCl3

(C)

(D) CH3

CH3

Q.29

Hyperconjugation involves overlap of the following orbitals: [JEE 2008] (A)  –  (B)  – p (C) p – p (D)  –

r i S . J .

The correct stability order for the following species is: [JEE 2008] 





O

D



(I)

(ii)

(i)



ONa

ONa

What are the constituent amines formed when the mixture of (i) and (ii) undergoes Hofmann Q.30 bromamide degradation? CONH2

CH3

 

Which is the rate determining step in Hofmann bromamide degradation? [JEE 2006] (A) Formation of (i) (B) Formation of (ii) (C) Formation of (iii) (D) Formation of (iv)

CONH2



ONa

(II)



NH2

NH2

,

D

,

NH2

D

NH2

NH2

,

D

(C) (D)

15

NH2

NH2

15

NHD

,



O

Q.31

15

,

NH2



(III) (IV) (A) (II) > (IV) > (I) > (III) (B) (I) > (II) > (III) > (IV) (C) (II) > (I) > (IV) > (III) (D) (I) > (III) > (II) > (IV)

15

(B)



15

15

(A)

In the following carbocation , H/CH3 that is most likely to migrate to the positively charged carbon is [JEE 2009]

[JEE 2006]

,

H

H

2

+

HO

H

1

4 5

H3C – C – 3C – C –CH3

Paragraph for Question Nos. 26 to 28 (3 questions) Riemer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehydes as depicted below.

N 





ONa

OH

CHO

(I) (Intermediate) CH3 (I)

Q.26

Q.27

(B) H at C-4 (D) H at C-2

EXERCISE – IV (B)

Q.1

OH

Complete the following, giving the structures of the principal organic products: [JEE 1997] (a)

CHO

Ph

H

+ KNH2

aq.HCl

CH3 (II)

CH3

(A) CH3 at C-4 (C) CH3 at C-2

Ph

(b)

CH3 (III)

A

Br

+ CHBr3 + t- BuOK

B

Q.2 Write the intermediate steps for each of the Which one of the following reagents is used in the following reaction. [JEE 1998] above reaction? [JEE 2007]  H O 3 (A) aq. NaOH + CH3Cl (B) aq. NaOH + CH2Cl2 (i) C6H5CH(OH) C  CH    C6H5CH = CH –CHO (C) aq. NaOH + CHCl 3 (D) aq. NaOH + CCl 4 The electrophile in this reaction is [JEE 2007] (A) : CHCl (B) +CHCl2 (C) :CCl2 (D) •CCl 3

(ii)

H

OH



O

CH3


53

Q.3

Out of anhydrous AlCl3 and hydrous AlCl3 which is more soluble in diethyl ether? Explain with reason. [JEE 2003]

Q.4

Match Ka values with suitable acid: [JEE 2003]

(i)

Ka –5 3.3 × 10

(ii)

4.2 × 10

–5

(iii)

6.3 × 10

–5

(iv)

6.4 × 10

–5

(v)

Acid (a)



NH3

NH3

COOH COOH

MeO

(e)

OH Which of the following is more acidic and why? [JEE 2004]

COOH

Cl

(d)

Give resonating structures of the following compound. [JEE 2003]



Me

(c)

30.6 × 10

Q.6 COOH

(b)

–5

Q.5

r i S . J . F

COOH

O2N

ANSWER KEY EXERCISE-1

Q.No. Ans. Q.No. Ans. Q.No. Ans. Q.No. Ans.

1 B 16 C 31 B 46 C

2 B 17 B 32 C 47 D

3 C 18 C 33 D 48 B

4 C 19 A 34 A 49 C

5 D 20 A 35 C 50 A

6 B 21 D 36 A

7 C 22 C 37 C

8 B 23 D 38 A

9 D 24 C 39 B

10 D 25 B 40 C

11 A 26 D 41 B

12 D 27 D 42 C

13 C 28 B 43 D

14 C 29 C 44 B

15 B 30 A 45 A

EXERCISE-II

Q.No. Ans. Q.No. Ans. Q.No. Ans. Q.No. Ans.

1 8 A,B,C 19 A,C,D 24

H

Q.1

A Q,R 9 A,B 20 B,C A Q,S

B P,S 10 A,B,C 21 A,C B Q,R

C Q 11 A,B,D 22 B C P,R

N H

H

P1 = CH3–C –CH2 –C –CH2 –C –CH3 Br

D

H

D

T

H


H

T

D P,S 12 A,B,D 23

2 A,B,C 13 A,B,D A S,T

D P

25

3 B,D 14 A,B,D B P,S,T A P,Q

4 B,C,D 15 A,B,D C U B Q,S

5 A,B,D 16 A,B,C D Q C P

6 B,D 17 A,B,D E T,U D R

7 B,C 18 C,D

EXERCISE-III Br

H

H

P2 = CH3–C –CH2 –C –CH2 –C –CH3 H Br

D D

T H

P4 = CH3–C –CH2 –C –CH2 –C –CH3 H

H

T


54

H

D

Br

T

H

Br

H

H

T

D

Q.2

H

H

H

H

H

T

P8 = CH3– C –CH2 –C –CH2 –C –CH3

P7 = CH3–C –CH2 –C –CH2 –C –CH3 H

T

P6 = CH3–C –CH2 –C –CH2 –C –CH3


D

D

Br

H

(i) a > b > c; (ii) a > b > c ; (iii) b > a > c Q.3

H

(a) Due to Resonance

Br

Q.4

(i)

OMe (ii)

& Br

• (iii) A : • CCl2

B: Cl

Q.5

(a) Free radical mechanism

(a)

CH3– CH2 – CH3

(b) Due to more electron density

Br2/h

Br2 CCl4

CH2 – CH – CH3

aq.KOH

CH3– CH – CH3 (B)

CH2 = CH – CH3 (C)

Br (D)

O

O

C – Cl

C – N3

(b)

OH

CH3– CH – CH3 (A)

Br

r i S . J .

Cl

Br

Q.7

COOEt

Ph

NaN3

OH

N=C=O

Ag2O 

NH2

NaNO2

H 2O

O

+HCl

CH3

CH2 – Br

CH2 Br

Q.8

,

and its enantiomer,

N

CH3

and its enantiomer,

Br

CH3

Br

and its enantiomer,

Q.11

1.15 times more reaction

Q.13

(a) A :

Br

CH3

Br

,

Q.12 A =

Br CH3 – C – CH2CH2CH3

(c) C: CH3 – CH –CH = CH2 Cl

CH3

D: CH3– CH = CH–CH2Cl

CH2Br

(b)B : S

(d) E:

Ph – CH –Et Cl


55

OH



Q.15

H

(b)



+

HOH/H

+

OH

Q.16

(a) II, (b) I

Q.19

(a)

Q.17 (a) II, (b) I, (c) II (b) Me2C(I) –Et,

Ph –CH – Et

(c) Br

Br

CH3

Q.20

(a)

Cl

H

r i S . J .

CH3

CH3

CH3 Br

Br

(b)

(diastereoisomers)

+ Br

Br

Me

Me

Me Et

Et

(diastereoisomers)

H

+ Cl

(c) H H

Cl

(d)

Me H H

Cl + Cl Cl Cl

(Enantiomers)

Et

EXERCISE –IV(A)

Q.No. Ans. Q.No. Ans. Q.No. Ans.

1 C 16 C 31 D

2 D 17 A

3 D 18 C

4 B 19 A

5 B,C,D 20 D

N Br

Q.1

7 D 22 D

8 D 23 D

9 B 24 D

10 A 25 B

11 D 26 C

12 C 27 C

13 D 28 B

14 B 29 B

15 C 30 D

EXERCISE –IV(B)

(A) PhC  CPh, (B)

C6H5–CH –C  CH

Q.2

6 D 21 C

OH

H

C6H5–CH –C  CH

–H2O

C6H5–CH –C  CH +



OH2

+

OH2

C6H5–CH =C = CH

+

H2O

C6H5–CH =C = CH

C6H5–CH =C = CH +

–H

tautomerism

C6H5–CH =CH – CHO

OH2


56

+

+ H

(ii)

•• OH ••

OH

–H + +

O|

CH3

O

CH3

H

Q.3

Anhydrous AlCl 3 is more stable then hydrous AlCl3 because it is having vacant 3p orbital of Al which can accept lone pair of electrons from oxygen of diethylether.

Q.4

(i)-(d), (ii)- (b), (iii) –(a), (iv) –(c), (v)-(e) Q.5



r i S . J . 

 OH



NH3

Q.6

is more acidic as overall effect of –F is electron withdrawing so loss of portion is easier from this

F

compound.

N


57

classnote-508568e7ede56.pdf

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