Reaction Mechanism.pdf
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IIT-JEE ChEmistry by N.J. sir
ORGANIC ChemIstRy
S . J .
REACTION MECHANISM
N
r i
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1
IIT-JEE ChEmistry by N.J. sir
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
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2
IIT-JEE ChEmistry by N.J. sir
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
170C
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)
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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
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4
IIT-JEE ChEmistry by N.J. sir
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
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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
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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
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7
IIT-JEE ChEmistry by N.J. sir
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 - - - - - - - - - - - - - -
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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 Intermediate3 (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
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9
IIT-JEE ChEmistry by N.J. sir
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)
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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
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11
IIT-JEE ChEmistry by N.J. sir
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
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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)
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13
IIT-JEE ChEmistry by N.J. sir
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
CC
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
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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
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15
IIT-JEE ChEmistry by N.J. sir
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)
HBr CCl4
or
(e)
MeOH
(f)
HBr CCl4
(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
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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
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18
IIT-JEE ChEmistry by N.J. sir
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 ZnH2O 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
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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
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20
IIT-JEE ChEmistry by N.J. sir
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
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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.
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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
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23
IIT-JEE ChEmistry by N.J. sir
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
AlHg C2H5OH
(2)
Na DE
14
(7)
I I I+
Cu 120260C
I
Cl
(4)
(6)
Br
O
(3)
S . J . Cl
Br
r i
Na DE
(5)
Br
(8)
N
Na Dry ether
Na Dry ether
(9)
Na Dry ether
Cl
Na DE
Br
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24
IIT-JEE ChEmistry by N.J. sir
ORGANIC chemistry
DPP NO- 16 Q.1
Time: 15 minutes Cl
Identify the major product:CH2 — I
(12)
Na + Dry ether
(1)
Na Dry ether
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
Na Dry ether
(9)
Cl
(1) (2) (3) (4)
Cl
Na Dry ether
(10) Cl
Na Dry ether
(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 .
Na Dry ether
(5)
r i
Which of following is correct matching of energy profile diagram? CH3 Cl CH3 — Cl
Na Dry ether
(3)
Na Dry ether
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°?
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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
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26
IIT-JEE ChEmistry by N.J. sir
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
?
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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–CC–CH2–Br BrMg–CH2–CC–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
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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
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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
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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
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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
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32
IIT-JEE ChEmistry by N.J. sir
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
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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
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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
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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 - - -- -
–
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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
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H
37
IIT-JEE ChEmistry by N.J. sir
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/ _ _ _ _ _
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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
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39
IIT-JEE ChEmistry by N.J. sir
ORGANIC chemistry
r i S . J .
Reaction
Intermediates
N
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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
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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
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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
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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
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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
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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
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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
CC
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
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47
D
KNH
2 ?
Q.20
Cl
Q.21
NH3 ,33C
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,
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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
HBr
CH3
CH3 – CH = CH – CH2 – CH – CH2
HBr
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
HBr
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
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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 65 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
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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]
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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)
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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
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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
P3 = CH3–C –CH2 –C –CH2 –C –CH3 Br
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
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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
P5 = CH3–C –CH2 –C –CH2 –C –CH3 Br
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
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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
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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
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57
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