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. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
CONCEPTUAL IMPROVEMENT
OF ISOMERISM
N.J
. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry DPP-1 Time: 20 Minutes
Q.1 Relationship between molecules:–
(a)
&
(b)
&
(c)
&
(d) CH2 — OH & O — CH3
(e)
&
(f)
&
(g)
&
(h)
&
(i)
&
(j) CH3 — CH2 — CH2 — NH2 & CH —CH — N3 2
CH3
H
(k) O — CH3 & CH2 — OH
COOH COOH
OH
OH
N.J
. SIR
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(l)
OH
CH3
&
CH —OH2
(m) & CH3 — C C — CH3
(n) CH3 — CH2 — CH2 — CH2 — NH2 &
(o) CH —C — OH2
O
& C — O — CH3
O
(p) &
(q)
&
(r)
&
Q.2 Fill in the blanks:–
(a) — C — O —
O
& — C — O —
O
are _ _ _ _ _ _ _
(b) — O —
& — O — are _ _ _ _ _ _ _
(c)
HCH3
C = C
H
O
&
HCH3
C = C
H
O
are _ _ _ _ _ _ _
(d) CH3 — CH2 — CH2 — N|H
— CH3 & CH3 — CH2 — N|H
— CH2 — CH3 are _ _ _ _ _ _ _
(e) CH3 — CH2 — CH2 — C N
& CH3 — CH2 — CH2 — N C are _ _ _ _ _ _ _
CH — CH — N 3 2
CH 3
CH3
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(f) CH2 — SH
& S — CH3 are _ _ _ _ _ _ _
(g) CH — CH —CH —C3 2 2
H
O
& CH3 — CH2 —
O||C
— CH3 are _ _ _ _ _ _ _
(h) CH — CH —CH —C3 2 2
O
CH3
& CH3 — CH2 —
O||C
— CH2 — CH3
are _ _ _ _ _ _ _
Q.3 Column matching
Column I Column II
(A) — N
CH3
H & CH2 — NH2 (P) metamers
(B) H —
O||C
— OCH3 & CH3 —
O||C
— OH (Q) functional isomers
(C)
& (R) position
(D)
& PhO—C—O
HH
C = C
H
O
(S) identical
H H
C = C
H
O
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 2 Time: 15 minutes
Compounds Relationship
1. &
2. O
&
O
3. &
4.
&
5.
Br
Cl
&
Cl
Br
6. CH2 = CH – CH2 OH & CH2 = CH – OCH3
7. O
&
O
8. CH — CH — CH OH2 2
O
& CH — C — OCH3 3
O
9.
&
10.
&
11.
&
12. CH3OCH3 & CH3OCH2CH3
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. SIR
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Compounds Relationship
13. &
14.
O
&
CHO
15. OH
SH
& OHHS
16.
CH3
H
H
Et
&
CH3
HH
Et
17. OH
&
OH
18. OCH3
&
OCH3
19.
CH3
CH CH2 3
H
H
Br
Cl &
CH3
CH CH2 3
H
H
Br
Cl
20.
CH3
Me
H Cl
&
CH3
Me
HCl
21.
&
22. N
H
& N
N.J
. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 3 Time: 15 minutes
Q.1 Which of following is 2° alcohol ?
(A)
OH
(B) OH (C) OH (D) OH
Q.2 Which of following is 3° alcohol ?
(A)
OH
(B)
OH
(C)
OH
(D)
OH
Q.3 Which of following is 2° amine ?
(A)
NH2
(B)
NH–CH3
(C) N
CH3
(D)
NH2
Q.4 Which of following compound has presence of 1°, 2°, 3°, 4° carbon ?
(A)
(B)
(C) (D)
Q.5 How many ether is/are possible for molecular formula C4H10O.
(A) 1 (B) 2 (C) 3 (D) 4
Q.6 How many ketone is/are possible for molecular formula C4H8O
(A) 1 (B) 2 (C) 3 (D) 4
Q.7 How many alcohol is/are possible for Molecular farmula C4H10O (only structural)
(A) 2 (B) 3 (C) 4 (D) 5
Q.8 O–H
O
O
Number of Functional group in above compound is
(A) 3 (B) 4 (C) 5 (D) 6
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. SIR
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Q.9 The functional groups in Cortisone are:
O
O
HOH C2
CH3 OCH3
Cortisoe (A) Ether, alkene, alcohol (B) Alcohol, ketone, alkene, ether
(C) Alcohol, ketone, amine (D) Ether, amine ketone
Q.10
OO
O – H
SH
OHOH
H
How mnay types of functional groups are presention given compound.
(A) 6 (B) 5 (C) 4 (D) 7
Q.11 The hybridization of carbon atom in the given compound
(A) sp2, sp (B) sp3, sp2 (C) sp3, sp (D) only sp2
Q.12 Bond X is made by the overlap of which type of hybridized orbitals?
X (A) sp and sp3 (B) sp and sp2 (C) sp2 and sp3 (D) none of these
Q.13 The number of sp2 – sp2 sigma bonds in the compound given below is:
(A) 1 (B) 3 (C) 4 (D) 5
Q.14 Present functional group is
O
OCH3
CH3 (A) ketone (B) ester (C) ether (D) alcohol
Q.15 Present functional group is/are
O
O
OCCH3
O
(A) ketone (B) ester (C) ether (D) A and B both
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Q.16 Find out the degree of carbon in the following compounds
(a)
CH — CH — CH — CH CH — CH3 2 2 2 2 3
CH — CH — CH — CH — CH3 2 3
4-ethyl-3-methyloctane
(b) CH3
CH3CH3
1,2,7-trimethylcyclopentadecane
(c)
CH3
CH3
CH3
CH3
1,1,2,5-tetramethylcyclopentane
(d)
CH3
Q.17 Calculate DBE value of C4H8 and draw possible structural isomer
Q.18 Calculate C3H6O, DBE value & draw all possible structural isomer. Q.19 Calculate DBE value of C5H10 & draw all possible structural isomer. Q.20 Calculate DBE value of C4H6 and draw all the possible structural isomers. Q.21 Calculate DBE value of C2H4O2 and draw all the possible structural isomers.
N.J
. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 4 Time: 15 minutes
Q.1 Identify relationship between following molecules.
(a) &
(b) &
(c) &
(d) &
(e) CH3 — CH2 — CH2 — CN & CH3 — CH — CH3 | CN
(f) &
(g) &
(h) CH3 — CH — CH2 — CH3 & CH3 — CH2 — CH2 — CH2 — OH | OH
(i)
NH 2
&
NH 2
(j)
&
(k)
&
Q.2 Calculate I.H.D. in the following molecules:–
(a) (b)
O
CH 3
(c)
CH—NH 2 2
(d)
O
O
Cl
CH 3
(e)
O
OH
O
H N 2
Q.3 Calculate No. of & / bonds in the following molecules:–
(a) (b) (c)
(d)
CN
O
O = C—OH
Cl (e) N C — C N
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O O || ||
(f) C —H (g) C — OH | | C —H C — OH || || O O
Q.4 Which of the following is the staggered conformation for rotation about the C1 C2 bond in the
following structure?
CH CHCH CH3 2 3
1 2 3 4
CH3
(I)
H
CH CH2 3
H
H
CH3
H (II)
CH CH2 3
H
CH3
H
CH3
CH3
(III)
H
HH
CH3
H
CH3
(IV) HH
C H2 5
H
H
H
(V)
HH
H
CH3 CH3
H
(A) I (B) II (C) III (D) IV and V Q.5 Which of the Newman projections shown below represents the most stable conformation about the
C1–C2 bond of 1-iodo-2-methyl propane?
(A) HH
H
I
CH3
CH3
(B) HH
HI
CH3
CH3
(C)
H
H
HI
CH3
CH3
(D)
H
H H
I CH3
CH3
Q.6 Among the butane conformers, which occur at energy minima on a graph of potential energy versus
dihedral angle? (A) gauche only (B) eclipsed and totally eclipsed (C) gauche and anti (D) eclipsed only (E) anti only Q.7 Which of the following best explains the reason for the relative stabilities of the conformers shown?
(I)
H
H H
CH3
CH3
H
(II)
H
H H
CH3
CH3
H
(A) I has more torsional strain (B) I has more steric strain (C) II has more torsional strain (D) II has more steric strain Q.8 Draw the Newman projection that represents the least stable conformation of 3,3-dimethylhexane
viewed along the C3-C4 bond.
N.J
. SIR
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Q.9 Draw the Newman structure for the most stable conformation of 1-bromopropane considering rotation about the C1–C2 bond.
Q.10 Draw a Newman projection of the most stable conformation of 2-methylpropane. Q.11 (a) Write Newman projections for the gauche and anti conformations of 1,2-dichloroethane
(ClCH2CH2Cl) (b) The measured dipole moment of ClCH2CH2Cl is 1.12 D. Which among the following statements
about 1,2-dichloroethane is/are false? (1) It may exist entirely in the anti conformation. (2) It may exist entirely in the gauche conformation. (3) It may exist as a mixture of anti and gauche conformations.
Q.12
H
H
CH3
CH3
H
H
; If front carbon is rotated by 180° the conformation formed is:–
(A) gauche (B) anti (C) partially eclipsed (D) perfectly eclipsed
Q.13
H
H
CH3
CH3
H
H
In the given energy diagram for butane the above conformation is represented at
which point
A
B
C
D
E
angle of rotation (A) A (B) B (C) C (D) D
Q.14 If for 1,2-dibromoethane is 0.75 D and Xanti = 0.7 Calculate for gauche. Q.15 Compare relative stabilities of given conformers.
(a)
H
H H
H
CH3
CH3
(b)
H
H
HH
CH3
CH3 (c)
H
H
HH
CH3
CH3
Q.16 Draw most stable conformer of (a) 3-methyl pentane
(b) 3-methyl hexane Q.17 Draw most stable forms of Z — CH2 — CH2 — OH
Z = OH, F, Cl, NH2, OMe, —
O||C — OH, —
O||C — H
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Q.18 (Me3)N
— CH2 — CH2 — O —
O||C — CH3; Most stable form of the compound.
Q.19 Draw most stable conformers of 3 2 2CH — CH — NH
& CH3 —
3
CH|CH
— O — H
Q.20 What is the effect on dipole moment of 1,2 – dichloroethane when the temperature is increased?
Q.21 Most stable conformer of 2
2
CH — COOH|CH — COOH
(succinic acid)
(a) at very low pH (b) at very high pH. Q.22 Draw more stable conformation for the following–
(a)CH3 — 2
3
CH — CH|CH
— CH3 (b) CH3 —
3 3
CH — CH||
CH CH
— CH3
(c)
(d)
(e) — C—— C — CH3
CH3
CH3 H
H
Q.22 Compare Rotational barrier.
(a) 3 2 2 3CH — CH — CH — CH
3 2 3CH — CH — CH
3 3CH — CH
(i) (ii) (iii)
(b) — CH — CH —2 2
— CH — CH —2 2
(i) (ii)
(iii)
(c)
CH — CH 2 2
Cl Cl
CH — CH 2 2
Br Br
CH — CH 2 2
I I
CH — CH 2 2
F F
(1) (2) (3) (4)
(d)
(i) (ii) (iii)
(e)
(i) (ii)
— CH — CH —
CH 3
CH 3
— CH— CH — 2 2
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 5 Time: 15 minutes
Q.1 For the conformation of lowest energy estimate the atomic angles in the cation and the neutral molecule drawn below. Provide one number only for each question.
C – O:
:
+
O:
:
(A) C – O – C bond angle in cation. ………………………………………..
(B) C – O – C bond angle in neutral molecule ……………………………. (C) C – O – C – C dihedral angle in the cation …………………………… (D) C – O – C – C dihedral angle in the neutral molecule………………..
Q.2 Which of the following is the most stable conformation of bromocyclohexane?
(I) H
Br
(II)
HBr
(III) Br
H
(IV)
Br
H
(V) HBr
(A) I (B) II (C) III (D) IV (E) V
Q.3 Which of the following correctly lists the conformations of cyclohaxane in order of increasing energy?
(A) chair < boat < twist-boat < half-chair (B) half-chair < boat < twist-boat < chair (C) half-chair < twist-boat < boat < chair (D) chair < twist-boat <boat < half-chair Q.4 In the boat conformation of cyclohexane, the ―flagpole‖ hydrogens are located: (A) on the same carbon (B) on adjacent carbons (C) on C-1 and C-3 (D) on C-1 and C-4 (E) none of the above Q.5 The Keq for the interconversion for the two chair forms of methylcyclohexane at 25°C is 18. What %
of the chair conformers feature an axial methyl group? (A) 95 (B) 75 (C) 50 (D) 25 (E) 5 Q.6 Which of the following describes the most stable conformation of trans-1-tert-butyl-3-
methylcyclohexane (A) Both groups are equatorial (B) Both groups are axial (C) The tert-butyl group is equitorial and the methyl group is axial (D) The tert-butyl group is axial and the methyl group is equitorial (E) None of the above Q.7 Name the compound shown below.
ClCl
(A) trans-1, 2-dichlorocyclohexane (B) cis-1, 2-dichlorocyclohexane (C) trans-1, 3-dichlorocyclohexane (D) cis-1, 3-dichlorocyclohexane (E) trans-1, 4-dichlorocyclohexane
N.J
. SIR
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Q.8 What can be said about the magnitude of the equilibrium constant K for the following process?
H H
H H
CH3
CH3CH(CH )3 2
CH(CH )3 2 (A) K = 1 (B) K > 1 (C) K < 1 (D) No estimate of K can be made
Paragraph for Question Nos. 9 to 11
Groups bonded by only a sigma () bond (i.e., by a single bond) can undergo rotation about that bond with respect to each other. The temporary molecular shapes that result from rotation of groups about single bonds are called conformations of a molecule. Each possible structure is called a conformer. An analysis of the energy changes associated with a molecular undergoing rotation about single bonds is called conformational analysis.
Q.9 Most stable conformer of given compound is
HO—CH —CH —F2 2
(A)
F
H
OH
H
H
H
(B)
F
H
OHH
H
H
(C)
FH
OH
H H
H
(D)
HH HH
FHO
Q.10 Most Stable conformer of
(A) (B) (C) (D)
Q.11 Which of the following pairs of structures represent conformational isomers?
(A) (B) H HH H
CH3 CH3
H HH
HCH3
CH3
(C) Cl Br
Cl
Br
(D)
H H
H H H
H
H H
H
H H
H
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Q.12 Which of the following pairs of structures represent conformational isomers?
(A) and (B) and
(C) and (D) C and C
Q.13 Draw the Newmann projection formula of the most stable conformation of 3-hydroxy propanal
across C2 and C3. Q.14 (a) Below are six conformations for a specific compound. With respect to the biggest groups,
determine which structure are eclipsed, anti, gauche, highest in energy and lowest in energy.
(A)
(D)
(B)
(E)
(C)
(F)
HPr Me
MeEt
HEt
H
H
+1 +1 +1 +1 +1
Anti All Gauche All eclipsed Highest in energy
Lowest in energy
H H
H
H
Pr
Pr
Pr
Pr
PrMe
Me
Me
Me
Me
Me
Me
Me
Me
Me
EtEt
Et
EtH
H
H
H
(Pr = propyl)
Q.15 The equilibrium constant for the ring-flip of fluorocyclohexane is 1.5 at 25°C. Calculate the
percentage of the axial conformer at the temperature. Q.16 Following eclipsed form of propane is repeated after rotation of
HHH
H
CH3
H
(A) 45° (B) 90° (C) 120° (D) 180°
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 6 Time: 15 minutes
Q.1 Column I ColumnII
(A) H
CH3
CH3
(P) cis-form
(B) CH3
CH3 (Q) trans-form
(C) CH3
CH3
(R) Keq is greater than one or equal to one when compound
undergo flip.
(D)
CH3
CH3
(S) Keq is less than one when compound undergo flip.
Q.2 Write correct order of stability of different form of following compound X with suitable reason.
CMe3
Me
(I)
CMe3
Me
(II)
CMe3
Me
(III) Me C3
Me
(IV)
Me C3
Me
Q.3 Identify most stable form of given compound
Et
CH3
(A)
CH3
Et (B)
CH3
Et (C)
CH3
Et
(D)
CH3
Et
Q.4 Dipole moment of a compound W – CH2 – CH2 – W is 1.5 D. If dipole moment of its gauche form is
6.0 D. what will be mol fraction of its anti form.
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Q.5 Compare stability:
(a)
OH
HO
and
OH HO
(b) OHHO
and
(c) and
(d) and
(e) and
(f) and
(g)
and
(h)
and
Q.6 Draw 1,2,3,4,5,6-hexamethyle cyclohexane in which all (a) methyl at axial position (b) methyl at equitorial position. Q.7 Draw most stable form of methyl-cyclohexane.
OH
HO
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Q.8 Compare stability of the following:–
(a) (b)
(c) (d)
Q.9 Draw structure & compare stabilities of following (a) cis & trans – 1, 2 – dimethyl cyclohexane. (b) cis & trans – 1, 3 – dimethyl cycohexane. Q.10 Explain:–
H
R
Keq
R Keq
– H 1
– CH3 18
– CH2 — CH3 23
– CHCH3
CH3
38
3800
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 7 Time: 15 minutes
Identify E/Z configuration:- Q.1
(1) C = C Me (2) C = C
CH Cl2
CH3
HOCH2
N C
(3) C
H
Me
CH3
(4) C = C Cl
Br
F
I
(5) C = C
O
O
O
O
(6) C = C O
O
O
O
HO—CH2
N C
(7) C = C
CH O—Cl2
COCl
HO—C
HOH C2
O
(8) C = C
CN
CH — OH2
HOCH2
OHC
(9) C = C Me
(10) C = C
C C — H
C CH
CH = CH2
CH — CH2
CH3 CH3
CH3 CH3
Me Me
(11) C = C
C C — H
C N
CH
OH — CH2
Me
Me
(12) C = C
H
D
F
Cl
(13) C = C
CH —C—2 I
CH —Cl2
I CH2
Cl CH2
H
H (14) C = C
CH — CH2 3
CH — CD3
Cl — CH — CH2 2
HO — CH — CH2 2
D
(15) C = C
CH3
OH
CH — CH3 2
HS
(16)
H
(17)
HH
(18)
Cl
EtMe
Br
(19) C = C
CH3
H
CH3
H
(20) C = C
CH3
HCH3
H
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(21)
HH
(22) H
H
(23)
H
H
(24)
O
O
(25)
(26)
(27) H
Br CH —CH —Cl2 2
CH —CH —Br2 2
(28)
H
Cl
O
O
(29) H
Br
CH3
OCH3
Q.2 1. Which of the following will show G.I. in acidic medium
(A) O (B) O (C) O
(D) C = O
H
H (E) O (F) O
2. Which of the following will show G.I.
(A) CH3 – CH2 –
O||C – NH2 (B) CH3 –
O||C – NH – CH3
(C)
O||C – NH – CH3 (D) C–N
Me
MeO
–
3. Write cis/trans in the following of it show G.I.
(A) CH3
CH3
C
(B) Cl
Cl
(C)
ClCl
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. SIR
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(D) (E) (F)
(G)
(H)
Cl
Cl
(I)
ClCl
(J)
Cl
Cl
(K) ClCl
(L) Cl
Cl
4. Following will show G.I.
(A)
CCH3H
(B)
CH H
(C)
CH H
(D)
CCH3
H
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. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 8 Time: 15 minutes
Calculate of geometrical isomer:
Case I If both the ends are different
2n when n is number of stereogenic area or bond which can show G.I.
Case II If both the ends are same 2n–1 + 2p–1 If n = even ; P = n/2
If n = odd ; P = n 1
2
Hint Rules for cyclic system:- (i) 3 member to 7 member cyclo alkene exist in only cis form. (ii) 8 to 11 member can form cis & trans but cis is more stable. (iii) from 12 member trans is more stable. Q.1 Calculate total number of only geometrical isomers in following compounds (Theoritical).
(1) (2)
(3) (4)
(5)
(6)
(7)
(8)
(9) (10)
(11)
H
MeMe
H
(12)
H
H
(13)
H
MeMe
H
(14)
H
MeMe
H
(15)
Me
Me
(16)
(17)
(18)
(19) CH3–CH=CH–CH=CH–CH=CH–CH3
(20) CH=CH–MeCH =HC2
(21) Me–CH=C=CH–CH2–CH=C=C=CH–Me (22)
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. SIR
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(23) Br–CH=C=C=CH–CH=CH–Br (24)
(25)
(26) H
H
CH3
CH=C=CH2
(27) CH3—CH=CH CH=CH—CH3 (28) CH3–CH=C=C=CH–CH=CH–CH3
(29)
(30)
(31) H
CH3
(32) CH3 – CH = CH – CH = N – OH
(33) O
O
Q.2
&
are -------------------------
Q.3
&
are ---------------------
Q.4 Calculate total no. of geometrical isomers for the following:–
(A) (B) CH3 – CH = CH – CH = C= CH – CH3 (C) CH3 – CH = C = C = CH – CH3 (D) CH3–CH=CH–CH=C=CH–CH=C=C=CH–CH3
(E)
Q.5 C = N
Et OH
CH3
Correct name of above compound is
(A) syn – methylethyl ketoxime (B) anti – methylethyl ketoxime (C) syn – ethylmethyl ketoxime (D) syn – methylethyl ketoxime
Q.6 C
ClH
is
(A) E (B) Z (C) none (D) can‘t predict
N.J
. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 9 Time: 15 minutes
Q.1 C = C = C
CH3
H
CH3
H l1
C = C = C
CH3
HCH3
H l1 (I) (II) I and II are not geometrical isomers of each other because (A) I1 = I2 (B) I1 > I2 (C) I2 > I1 (D) they are geometrical isomers
Q.2 C = C = C = CCH3
H
CH3
H
l1
C = C = C = CCH3
HCH3
H
l2
(I) (II) I and II are not geometrical isomers of each other because (A) I1 = I2 (B) I1 > I2 (C) I2 > I1 (D) they are geometrical isomers Q.3 MeCH = CH — CH = C = CH — CH = CH2 Total number of geometrical isomers possible for above compounds are: (A) 16 (B) 8 (C) 4 (D) 2 Q.4 Find total number of Geometrical isomerism of following compounds.
(A) CH3 — CH = CH — CH = N — OH (B) (C)
(D) CH3 — (CH = CH)3 — Ph (E)
CH=CH–CH3
H
(F)CH3—CC—CH=CH—CH3
Q.5 Which of the following compound can show geometrical isomerism.
(A) C = C
Cl
Cl
Br
I (B) C
CH3
CH3
(C) C = C
Et
Et
F
Cl
(D) C
CH3
CH3CH3
CH3
Q.6 The geometrical isomerism is shown by
(A)
CH2
(B)
CH2
(C)
CHCl
(D)
CHCl
Q.7 Which of the following double bond will not exhibit geometrical isomerism.
(A) C = C
Ph
C H6 5
Me
CH3
(B) C = O
Me
Ph
(C) C = N — Me
Me
Ph
(D) Me — N = N — Me
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. SIR
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Q.8 Which of the compound show geometrical isomerism.
(A)
CH — Br
Br
(B)
N
H
CH3
(C)
N
H
O
(D)
NCH3
(E) N
Cl
Me
H
(F) N
Cl
Me
Me
(G) N-H
C–OEt
O
N
Me
EtO–C
O
(H) N
N
H
N (I)
ON
H
COOMe
COOMe
NC (J) H C = CH – C2
Me
MeN
OMe
OCH Ph2
(K)
C = C = C = C
Cl Cl
(CH ) C3 3(CH ) C3 3
(L)
BrCl
CH=CH–COOH
(M) O
O
(N)
Q.9
Cl
Cl
The above conformation is (A) cis (B) trans (C) will not show G.I. (D) can‘t predict
Q.10 for the boiling point of given compounds:–
Cl
H
C = C
CH3
H
Cl
HC = C
CH3
H
1 2
(A) 1 > 2 (B) 1 < 2 (C) 1 = 2 (D) none Q.11 For the melting point of given compounds
Et
HC = C
CH3
H Et HC = C
CH3H
1 2 (A) 1 > 2 (B) 1 < 2 (C) 1 = 2 (D) none
N.J
. SIR
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Q.12 For which of the following keq > 1?
(A)
(B)
(C)
(D)
N.J
. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry DPP-10 Time: 20 Minutes
Q.1 Compound P.O.S. C.O.S. Optically active
1.
CH 3 CH 3
H H
2.
CH 3
CH 3
H
H
3.
CH 3
CH 3
H
H
4. H
CH 3
H
CH3
5. H
Cl
H
Cl
6. H
H
Cl
Cl
7.
H
H
Cl
Cl
Cl
Br
8.
Cl
H Cl
H
Cl
H
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. SIR
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9. C = C = C
H H
Cl Cl
10. C = C = C
H Cl
Br Cl
11. C = C = C = C
H H
Cl Cl
12.
C H
Cl
H
Cl C
13.
C H
Cl
Cl
H C
14. H
Cl Cl
H
15.
3 CH CH 3
H
H
16.
H
Cl H
Cl
17.
C H Cl
18.
H C
CH 3 CH 3
H
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. SIR
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19.
H C
CH 3 CH 3
Cl
20.
21.
22. Cl
Cl I
I
23.
24. O = C = O
25. O = C = O
H
26.
O = C = O H H
27. N = N = N
H H
28.
C
H H
H
H
No. of POS _ _ _ _ _ _ _ _
29.
C
H H
H
Cl
No. of POS _ _ _ _ _ _ _ _
N.J
. SIR
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30.
No. of POS _ _ _ _ _
31.
No. of POS _ _ _ _
Q.2 Match the following structural formulae with their possible geometrical isomers? Column-I Column-II
(Structural formula (Total geometrical isomers excluding mirror image)
(A) CH3 – CH = CH CH2 – CH3 (P) 8
(B) CH3 – CH =
CH=CH–CH 3
CH=CH–CH 3
C (Q) 6
(C) CH3 – CH = CH CH = CH – CH3 (R) 4
(D) Cl – CH = CH – CH = CH – CH = CH – CH3 (S) 2
Q.3 Statement 1: C = C
CH3
H C5 2
Br
Cl
and C = C
CH3
H C5 2
Br
Cl
are structural
Statement 2: The above mentioned compounds can show geometrical isomerism. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B Statement-1 is true, statement-2 is true and statement-2 is NOT correct explanation for
statement-1. (C) Statement-1 is false, statement-2 is true. (D) Statement-1 is true, statement-2 is false.
Q.4 Which compound(s) will show Geometrical isomerism?
(A)
CH3
N–H (B) C = C = C
H
Cl
H
Cl
(C) H C3
CH3
(D)
O
O
H
H
N
N
Q.5 Find out the correct option(s) ?
(A) C = C
H
CH3
NH
NH
Orientation is E (B)
H
CH3CH3
HOrientation is Z
C = C
H H
CH 3
CH 3
C = C
H
H CH 3
CH 3
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. SIR
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(C)
C = C
H
D
Orientation is Z (D) CH3
CH3
geometrical isomers are not possible
Q.6 Calculate total number of geometrical isomers in following compounds. (Excluding mirror image) (i) CH3 – CH = CH – CH = CH – CH = CH – CH3
(ii) CH2 = HC CH = CH – Me (iii) Me – CH = C = CH – CH2 – CH = C = C = CH – Me
(iv) Me
D
(v)
(vi) Br – CH = C = C = CH – CH = CH – Br
N.J
. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
Date: DPP- 11 Time: 15
minutes
Q.1 How many chiral carbon atoms are present in the following compounds?
(i)
H
N
O
H
H
CH3
(ii)
O
O
S PhOCH2 – C – NH
N
COOH
Coccinellin Penicillin V
(iii)
O
(iv)
COOH
HO
Estrone Betulinic acid
Q.2 Assign priority number to the following groups as per Cahn, Ingold, Prelog sequence rule
(a) – CH2OH, –CH3, –CH2CH2OH, –H
(b) – Cl, –Br, –CH = CH2, –CH3
(c) O
– C – H , –OH, – CH3, – CH2OH
(d) – CH(CH3)2, –CH2CH2Br, –Cl, –CH2CH2CH2Br
(e) – CH = CH2, – CH2CH3,
, – CH3
(f) – CH = CH2, – C CH, – CH – CH2
CH3 CH3
– C – CH
CH3 CH3
CH3 CH3
(g) – CH2CH2CH2I – CH – CH – CH3
Br
– CH – CH2CH3
Cl
–F
Q.3 Indicate whether each of the following structure has the R configuration or the S-configuration.
(a)
C
CH3
CH2CH3
CH2 – Br
CH(CH3)2
(b)
C
CH3CH2
CH2CH2Cl
CH2 – Br
OH
(c)
C
Cl H
F
Br
(d)
C
H
OH
CH2 – Br
CH3
(e)
C
CH3CH2 H
D
CH3
(f)
C
CH3CH2
H
Br
CH3
(g)
CH2CH3
Cl
CH3
H
(h) CH3CH2
Cl
F
H
(i) CH =CH2
CH3
F OH (j)
SH
Me
H NH2
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. SIR
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(k)
CH3
CH2 – CH3
H Cl (l)
CH2
CH2 – Br
H NH2
(m)
CH3
CH2CH3
H
CH2I
(n)
CHO
CH2 – OH
H OH
(o)
CH3
CH2 – CH3
H COOH
(p)
CH3
CH2 – OH
HO
CH2CH2CH2OH
(q)
CH3
CH2CH3
HO H
H Cl (r)
CH3
CH2CH3
Br H
H Br
(s)
CH3
CH2CH3
Br H
H Br
(t)
CH3
CH2CH3
H
Br
CH2CH3
CH3
(u)
OH
H
COOH
OH
H
COOH
Q.4 Indicate whether each of the following structure has the R configuration or the S-configuration.
(a)
C
H
CH3
COOH
CH = CH2
(b)
C
CH3
CH2CH2Cl
SH
CH2 – Cl
(c)
C
OH
CH3
NH2
F
(d)
C
D
OCH3
T
CH2 – OH
(e)
C
CH3O
OCH3
H
CH3
18
(f)
C
HS
D
NH2
I
(g)
DCH2
SO3H
CH3
SCH3 (h)
CH2
CH3
CH2F
CH2OH
NH2
(i)
CH3
CH2CH3
(j)
H
SR
Et
NR2 (k)
CH3
CH2CH3
CH2CH2CH3
I (l)
CH3
CH2OH
OPh
(m)
CH3
H
D CH2Cl (n)
CH3
COOH
H OH (o)
Me
Et
D CHO
(p)
CH2NH2
CH2CH2CH2NH2
H2N CH2CH3 (q)
Me
CH2CH2CH3 2
H2N D
D Br
(r)
CH3
CH2CH2CH3 2
I Et CH3 I
(s)
CH2Cl
CH2CH3 2
H OMe
H Br
(t)
CH2CH3
CH2CH3 2
CH3 H
I H
(u)
SO3H
H Cl
SO3H
Cl H (v)
CH2CH2CH2COOH
H
CH3 H
CH3 (w)
Et
D
CH2CH3
HO
Me
D (x)
CH2CH2CH3
H
Br
Cl
CH2CH2CH3
H
(y)
COOH
OH H
CH2OH
OH
OH
H
H
(z)
CH3 H
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. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 12 Time: 15 minutes
Q.1 Column-I Column-II
(A)
O
ClMe
Me
Cl
(P) Total number of stereo isomers are odd.
(B)
Br
Br (Q) Total number of stereocenter are even or have
centre
(C)C
C
H
H Cl
Cl
(R) Compounds having plane of symmetry or axis of
symmetry
(D) HO
H OH
COOH
COOH
H (S) Compounds have zero dipole in given form.
Q.2 Column-I Column-II
(A) Plane of symmetry (P) H
C
Me
Me H
C
(B) Centre of symmetry (Q)
MeMe
(C) Meso Compound (R) C
H
Me
Me
(D) Chiral atom is / are present (S)
H H
HH Ph
Ph
COOHCOOH
(T) C = C = C = C
H
Cl
Cl
H
Comprehension (Q.3 to Q.5)
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On the basis of the following set of compounds, answer the following questions.
(I)
HOH
HOH
CHO
H
CH OH2
HOOHH
(II) H
OHHOH C2
HO
CHO
HO
H
HO H
H (III) HHOH
CHO
H
CH OH2
HOH
OH
OH
(IV) HOH
HOH
CHO
H CH OH2
HO
OH
H (V)
HHOH
CHO
H
CH OH2
HO H
HOHO
Q.3 Which of the following represent enantiomeric pair? (A) I & II (B) II & V
(C) I & V (D) II & IV (E) III & V
Q.4 Which of the following does not represent active diastereomeric pair? (A) III & IV (B) I & II
(C) I & III (D) I & IV (E) None of these
Q.5 Which of the following represent ‗D‘ sugar. (A) I (B) II
(C) III (D) IV Q.6 Select the pair of enantiomer and diastereomers out of the following:
CH3CH3
CH3
HH
HA
CH3CH3
H C3
HH
HB
CH3
CH3
CH3
H
H
HC
CH3
H C3
H
H
HD
H C3
s Q.7 Which of the following compounds should have the larger energy barrier to rotation about the
indicated bond ? (a) Me3C
CMe (b) Me3Si
SiMe3
Q.8 How many compounds are theoretically possible for formula C3H6O (excluding stereoisomers)?
Q.9 How many acyclic isomers of C5H10 are possible which are incapable of showing Optical
Isomerism? Q.10 How many stereoisomers are possible for the following?
CH=C=C=CH–Me
Me CH=CH–CH=CH 2
Me (A) 16 (B) 4
(C) 6 (D) 8
Paragraph for question nos. 11 to 13
Answer the following questions based on given reaction
Cl2hv
(monochlorination)
Products.
Q.11 The number of theoretically possible products (including stereo) are
(A) 6
(B) 8
(C) 10 (D) 12 Q.12 How many products are resolvable. (A) 4 (B) 6 (C) 8 (D) 10
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Q.13 How many factions are present on fractional distillation? (A) 4 (B) 5 (C) 6 (D) 8 Q.14 (a) How many plane of symmetry are present in prismane (C6H6)?
(b) How many chiral centres are present in the following compound?
O
S
O
Ph
Br (c) Minimum carbon atoms required for an alkane to show optical isomerism.
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
Date: DPP NO- 13 Time: 15
minutes
Q.1 Column-I Column-II
(A) OH
OH
H
HCO H2
CO H2
(P) chiral
(B) OH
OH
H
CO H2
CO H2
OH
(Q) achiral
(C)
H
H
CO CH CH OH2 2 2
CO H2
(R) meso
(D) CO CH CH O C2 2 2 2
CO CH CH O C2 2 2 2
H
H H
H
(S) compounds containing even number of chiral
center Q.2 Column-I Column-II
(A)
MeH
Cl Cl
and
Me H
Cl Cl
are (P) Structural isomers
(B)
CH3
H
CH3
H
and
CH3
H CH3
H
are (Q) Compounds are optical
isomers and enantiomers
(C) C
C
CH –CH –COOH2 2
Me H
H
and Et
C
C
Me H
O CH3C
O
are (R) Compounds which aregeometrical isomer and
diastereo isomers
(D) OHH
COOH
H
COOEt
OH
and HHO
COOH
HO
COOEt
Hare (S) Compounds are
geometrical isomers and enantiomers (T) Not isomers
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Q.3 Statement 1:
F
F
F
F
(I) (II)
(I) and (II) are optically inactive molecules.
Statement 2: Molecules containing plane of symmetry or centre of symmetry are optically inactive.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B Statement-1 is true, statement-2 is true and statement-2 is NOT correct explanation for statement-1.
(C) Statement-1 is false, statement-2 is true.
(D) Statement-1 is true, statement-2 is false.
Q.4 Minimum C atoms required for a compound to show geometrical isomerism:
(A) 2 (B) 3 (C) 4 (D) None of these
Q.5 The correct stability order of the following species is
Me
Me
H
Me
H
HH
MeH
H
Me
Me
(a) (b) (c)
(A) c < a < b (B) c = b < a
(C) c < a = b (D) a = b = c
Q.6
Et
Me
This compound shows:
(A) geometrical isomerism (B) optical isomerism
(C) both (D) none
Q.7 (+)-Tartaric acid has a specific rotation of +12.0°. Calculate the specific rotation of a mixture of 68% (+)-trataric acid and 32% (–)- tartaric acid.
(A) 4.32°
(B) – 4.32° (C) – 12° (D) 12°
Paragraph for question nos. 8 to 10
24 gm of optically pure tartaric acid is dissolved in water to make 240 ml solution. It is kept in 20 cm polarimeter tube & plane polarized light is passed through it to product rotation of –2.4°.
Q.8 If mixture of d and l tartaric acid has the specific rotation – 4.0°, calculate the % of optical purity of this mixture?
(A) 50% (B) 66.67% (C) 33.33% (D) None
Q.9 Calculate the % of d tartaric acid in a mixture of d and l tartaric acid which has the observed specific rotation + 6.0°.
(A) 25% (B)75% (C) 50% (D) 66.67%
Q.10 If original solution is diluted by 2 times and length of polarimeter is increased four times of previous length. What will be the specific rotation.
(A) – 4.8 (B) + 4.8 (C) – 6 (D) – 12
N.J
. SIR
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Q.11 Select resolvable compounds.
(A)
HO
O
H
N
CH3
(B) Cl
OH
O
(C)
H
(D)
CH3
H
N–H
H
O
Me
O
O
Q.12 Calculate total number of stereocentre, prochiral carbon and theoretical stereoisomer in the
following compound.
SN
O
Number of stereocentre = v ; Number of prochiral carbon = x and number of stereoisomer =
yz. Represent your answer as vxyz. For example v = 4, x = 4 and yz = 34 so represent it as 4434.
Q.13 How many geometrical isomers are possible for the following structure.
H Q.14 Relationship between molecules:–
(a)
CHH
CH3CH3
&
CHH
CH3CH3
(b) C = C = C = CH H
ClCl
& C = C = C = CH
H
Cl
Cl
(c)
CH3 CH3
Cl
&
CH3 CH3
Cl
(d) CH3
CH3
& CH3
CH3
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. SIR
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(e)
&
(f)
&
Q.15 Calculate total no. of optical isomers, optically, active, meso & enantiomer pairs corresponding to
the following:–
(a)
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP-14 Time: 15 minutes
Q.1 Match the following : Column I Column II
(A)
Me
O
Me C – Me
O
Progesteron
(P) Meso compound
(B)
O
Me OH
C C – Me
Me2N
(RO–486/mifepristone)
(Q) Compound having even no. of chiral carbon
(C)
H
O
O
H
N H
O
(R) Optically active compound
(D)
H
OH
COOH
H
COOH
OH
(S) Compound having odd no. of chiral carbon
(T) Compound having odd no. of prochiral carbon.
Q.2 Match the following : Column I Column II
(A) Compound having only plane symmetry and (P)
Et
N
N N
Et
Et
axis of symmetry (Consider the given chair form only)
(B) Compound having center of symmetry, plane of (Q)
Me Me
symmetry and axis of symmetry
(C) Compound having axis of symmetry (C3) (R)
:N
(D) Compound having C2 axis of symmetry but absence (S) 2,2,3,3-Tetramethyl butane in of plane of symmetry staggered conformer
N.J
. SIR
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Question No.3 to 4 (2 questions) Questions below consist of an "Assertion" in column I and "Reason" in column II. Use the
following key to choose the appropriate answer (A) If both " Assertion" and "Reason" are correct and "Reason" is the correct explanation of the
"Assertion". (B) If "Assertion" and "Reason" are correct, but "Reason" is not the correct explanation of the
"Assertion". (C) If "Assertion" is correct, but "Reason" is incorrect. (D) If "Assertion" is incorrect, but "Reason" is correct.
Q.3 Assertion : Cyclopropane is planar while cyclobutane is non-planar. Reason : Angle strain in cyclopropane is more than that in cyclobutane.
Q.4 Assertion :
Me Et
Me
is most stable of conformer of
Me
Me
Et
Reason : Torsional strain and flag pole interactions cause the boat conformation to have considerably higher energy than the chair conformation.
Q.5 Correct statement about D-mannitol (in given form) :
OH
HO
OH OH
OH
OH
(A) C3 axis of symmetry (B) C2 axis of symmetry (C) Centre of symmetry is present (D) 3-chiral centre are present
Q.6 Which of the following pair of compounds can be separated by fractional crystallisation.
(A)
D O
H
Me
O – C
H
OH
Me and
Me O
D
H
O – C
Me
H
OH (B)
Cl
Cl
O
and
Cl
O
Cl
(C)
H
OH
COOH
HO
H
COOH
and
H
HO
COOH
HO
H
COOH
(D)
Me Me
and
Me
Me
Q.7 An unknow compound weighing 4.5 gm is dissolved in enough carbon tetrachloride to make a total
volume of 250 c.c. The observed rotation of this solution is +357.75° in a 25 cm cell using the sodium D line. But if 4.5 gm is dissolved in 125 cc we observed rotation is +355.50°. Calculate specific rotation for this compound. (assuming length of polarimeter tube is 1 dm)
Q.8 Identify chiral and achiral compounds from the list given below :
(a)
Br
F
Br F
(b) I
C = C = C Cl
(c)
Cl
Cl
(d)
OH
CO2H
H
H
CHO
OH
(e)
NO2 SO3H
NO2 SO3H
(f)
H
CH3 CH3
H
CH3
H
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. SIR
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(g)
Me
Me
Et
Cl
H
H
H H
(h)
C = C
CH3
H H
CH3
(i)
CH3
CH3
N – N
CH3
CH3
CO2H
HO2C
(j)
HO2C CO2H
CO2H HO2C
H5C6 C6H5
N N
(k)
N C6H5
H
H5C2O2C
H
(l)
C
H3C
C
CH3
C
C
C
O
H H
CH3 H
(m)
C = C = C = C
NO2
NO2
Q.9 Calculate the specific rotations of the following samples taken at 25° using the sodium D line. (a) 1.00 g of sample is dissolved in 20.0 mL of ethanol. Then 5.00 mL of this solution is placed in a
20.0 cm polarimeter tube. The observed rotation is 1.25° counterclockwise. (b) 0.050 g of sample is dissolved in 2.0 mL of ethanol, and this solution is placed in a 2.0 cm
polarimeter tube. The observed rotation is clockwise 0.043°. (c) Indicate the stereo centres in the following molecule and total number of stereomers in the
following molecule. Also draw the structures of pair of distereomers.
COOH
H N H
N O
O
Q.10 Select chiral molecule out of the following list compound.
(i)
CH3
O
(ii)
H
H
H
OH
COOH
HO
HO
HOOC
(iii)
C = C = C (iv)
N N
C C6H5
O
H
HO2C
(v)
HO2C
N = N
CO2H
(Azodiformic acid)
(vi)
CH3
CH3
(vii) Cl
CH3
CH3
(viii)
CH3 CH3
(ix)
CH3 CH3
CH3 CH3
N.J
. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP-15 Time: 15 minutes
Q.1 Match the following :
Column I Column II
(A)
H C H
Cl (P) It can show geometrical isomerism
(B) C
H
CH3 CH3
H (Q) Optically active compound
(C)
O
NH
O
NH (R) Presence of odd, number of chiral carbon
(D)
C = C — C – C = C
Br
H H
Br
H H Br
H
(S) Resolvable compound
(T) Presence of Pseudo chiral centre
Q.2 Match the following : Column I Column II
(A)
OH O H
H
CH3
(P) Total number of stereoisomers is odd for the structure
(B)
H
CH3 – C
CH2
H
CH3
O
(Q) Total number of stereoisomers is even for the structure
(C)
OH H
CO2H
OH H
CO2H
(R) Odd number of chiral centre
(D)
HO
H
CH2NH2
HO
OH
(S) Even number of chiral centre
(T) Optically active diastreomers possible (U)
Question No.3 to 4 (2 questions) Questions below consist of an "Assertion" in column I and "Reason" in column II. Use the
following key to choose the appropriate answer (A) If both " Assertion" and "Reason" are correct and "Reason" is the correct explanation of the
"Assertion". (B) If "Assertion" and "Reason" are correct, but "Reason" is not the correct explanation of the
"Assertion". (C) If "Assertion" is correct, but "Reason" is incorrect. (D) If "Assertion" is incorrect, but "Reason" is correct.
N.J
. SIR
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Q.3 Assertion :
CH3
Cl
O
Cl CH3
is an optically active compound.
Reason : No symmetry element is present in above compound.
Q.4 Assertion : Cis form of 1, 3-dimethyl cyclohexane is more stable than its trans form. Reason : Heat of combustion of trans form of 1,3-dimethyl cyclohexane is more compared to its cis
form.
Q.5 Which of the following Fischer projection is the enantiomer of the following molecule ?
Cl
Br F
HO
H3C CH3
H
F
(A)
OH
Br
Cl CH3
H3C
F
H
F (B)
OH
Br
Cl
H3C
F
H
F
H3C
(C)
OH
Br
Cl
CH3
F
H
F
H3C
(D)
OH
Br
Cl
F
H
F
H3C
H3C
Q.6 How many plane of symmetry are present in following compound.
(A) 2 (B) 1 (C) 0 (D) 3
Q.7 (i) The specific rotation of (S)-2-iodobutane is +15.90°. (a) Draw the structure of (S)-2-iodobutane (b) Predict the specific rotation of (R)-2-iodobutane (c) Determine the percentage composition of a mixture of (R) and (S)-2iodoobutane with a
specific rotation of 7.95°.
(ii) Dextrorotatory -pinene has a specific rotation 20D = + 50°. A sample of -pinene containing
both the enantiomers was found to have a specific rotation value 20D = 30°. The percentage
of the (+) and (–) enantiomers present in the sample are, respectively.
Q.8 Consider the following six structures :
(I)
OH
HO
HN
O
CH3 CH3
(II)
OH
HO
HN
O
CH3 CH3
(III)
OH
HO
HN
O
CH3 CH3
(IV)
OH
HO
HN
O
CH3 CH3
(V)
OH
HO
HN
O
CH3 CH3
(VI)
OH
HO
HN
O
CH3 CH3
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. SIR
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Establish the stereochemical relationship between : (a) I and II, (b) III and IV, (c) II and III, (d) I and V, (e) IV and (VI)
Q.9 Select resolvable compounds
(i)
Br I
Br I
(ii)
NO2 SO3H
NO2 SO3H
(iii) Ph – S = O
CH3
(iv)
N
N Me
Me
(v) MeCHBrCH2Me (vi)
Me
Br N
..
Me
(vii)
H Cl
N
..
Me
(viii)
Me
O Me
N
(ix) MeN DH2Br
(x) MeCH2CHCH2Me
OH
(xi)
NO2 SO3H
NO2 SO3H
(xii)
C = C = CH2
H
Cl
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry DPP NO- 16 Time: 15 minutes
Q.1 Find relationship between given pairs Identical Enantiomer Diastereomer Constitutional Other
isomer
(1)
(a) (b)
(2)
OH
OH
OH
OH
OH
OH
(a) (b) (c)
(3)
CO H2 CO H2 (a) (b)
(4)
H
H
Me
Me
H H
MeMe
(a) (b)
(5)
H
H
(a) (b)
(6)
OH
OHH
H
Me
Et
H
HHO
HO
Me
Et
(a) (b)
(7)
BrBr
Cl Cl
BrBr
Cl Cl
(a) (b)
(8)
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(a) (b) Identical Enantiomer Diastereomer Constitutional Other
isomer
(9)
OH
OH
(a) (b)
(10) OH
O
OH
O
(a) (b)
(11) H
NH2
COOHHO
HO
(a)
H NH2
HOOC
OH
OH
(b)
(12)
(a) (b)
(13)
Cl
Cl
(a) (b)
(14)
O
ClH
O
ClH
(a) (b)
(a) (b)
(15)
O
S—O
O
Me
(a)
O
O—S
O
Me
(b)
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 17 Time: 15 minutes
Q.1 Find relationship between given pairs
Identical Enantiomer Diastereomer Constitutional Other isomer
(1)
Cl
Cl
(a) (b)
(2) N
N
(a) (b)
(3)
H
H
Me
OHO
HMe
MeO
O
(a) (b)
(4) OH
OH
COOH
COOEt
H
H
H
HHO
HO
COOH
COOEt
(a) (b)
(5)
H Me
ClCl
HMe
ClCl
(a) (b)
(6)
OH
H
OH
H
(a) (b)
(7)
H H
HMe
Br
H
HH
Br
(a) (b)
(8) SH
OH
Me
Et
H
H
H
H
Me
Et
HS
HO
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(a) (b)
Identical Enantiomer Diastereomer Constitutional Other isomer
(9)
Cl
Cl
Cl
Cl
Cl
Cl
(a) (b)
(10)
O
NH
OH
N
(a) (b)
(11)
(a) (b)
(12)
H
HMe
H
HMe
(a) (b)
(13)
Cl
Br
Cl
Br
(a) (b)
(14) HHO
OH HHO
OH
(a) (b)
(15) Cl
Et
H
H
Cl
Cl
Et
H
HCl
(a) (b)
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 18 Time: 15 minutes
Q.1 Find relationship between given pairs
Identical Enantiomer Diastereomer Constitutional Other
isomer
(1)
H C = N2CH = NH
O
O
(a)
H C = N2 CH = NH
O
O
(b)
(2) C
O
H Me
OH
(a)
C
O
H
Me
OH
(b)
(3) CN
Br
OH
COOH
H
H
CN
Br
HO
COOH
H
H
(a) (b)
(4)
O
Br
O
Br
(a) (b)
(5)
Br
Br
H
H
Br
Br
Me
HH
(a) (b)
(6)
O
OH
(a) (b)
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Identical Enantiomer Diastereomer Constitutional Other isomer
(7)
CH3
H
OH
CH3
OH
H
CH3
H
H
CH3
OH
HO
(a) (b)
CH3
HO
OH
CH3
H
H
(c)
(8)
(a) (b)
(9)
CO H2
H
OH
CO H2
OH
H
CO H2
H
CO H2
HO
HHO
(a) (b)
(10) Me
Et
Me
Et
(a)
Et
Et
Me
Me
(b)
(11)
O
O
(a)
O
O
(b)
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Identical Enantiomer Diastereomer Constitutional Other isomer
(12) N
N
(a) (b)
(13) N N
(a) (b)
(14)
I
Cl
Br
I
Cl
Br
(a) (b)
(15)
O
O O
O
(a)
O
OO
O
(b)
O
OO
O
(c)
(16)
NH Me
NH2
(a) (b)
NH2
NH
(c) (d)
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 19 Time: 15 minutes
Q.1 Find relationship between given pairs
Identical Enantiomer Diastereomer Constitutional Other isomer
(1)
OH
O
(a) (b)
OH
(c)
(2)
N
H
O
N
OH
(a) (b)
(3) OH O
(a) (b)
OH OH (c) (d)
(4)
(a) (b) (c)
(5)
N
Me
O
N
OH
(a) (b)
(6) Me — C N Me — N —— C
(a) (b)
(7) OH OH
(a) (b)
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Identical Enantiomer Diastereomer Constitutional Other
isomer
(8) OH
O
OH O
H
(a) (b)
O
HO
(c)
(9) O
O
(a) (b)
(10) CH3OH OH
(a) (b)
(11)
OH
CH3
OH
(a) (b)
(12)
Cl
Cl
Cl
Cl (a) (b)
(13) CN
CN
(a) (b)
(14) CO H2
CO H2
(a) (b)
(15)
(a) (b)
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 20 Time: 15 minutes
If there is presence of given symmetry then writes () otherwise (x)
Compound P.O.S(s) C2 C3 S4 S2
(1) N
(2)
Cl
(3)
C
C
CHH
HH
(4)
(5) C
Br
CH3CH3
CH3
(6)
CH3
CH3
(7)
Br
Br
Br
Br
Br
Br
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 21 Time: 15 minutes
If there is presence of given symmetry then writes () otherwise (x)
Compound P.O.S(s) C2 C3 S4 S2
(1)
(2)
(3)
O
(4)
(5) C = C
ClCl
HH
(6) C = C
Cl
Cl H
H
(7)
H
HBr
Br
(8)
CO H2
H
CO H2
HO
OHH
(9)
(10)
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 22 Time: 15 minutes
1. Find (m) meso compounds & optically active (a) and total stereoisomer (a + m) of following compounds
Compound Meso (m) Active isomer (a) (a + m)
(1)
CH3
CH3H C3
H C3
(2)
BrCl
Cl
(3)
(4)
NH
ClCl
(5) CH3
H C3
Cl
Cl Br
Br
(6)
O
O
Cl
Cl
ClCl
OH
HO
(7)
Cl
CH3H C3
(8) H C3
CH3
Cl
H C3O
O
(9)
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Compound Meso (m) Active isomer (a) (a + m)
(10)
(11)
Br
Br
(12)
Br
Cl
(13)
C
CH3H C3
HH
(14)
Cl
ClCl
ClClCl
(15)
CH3
CH3
H C3
H C3
(16)
H C3
HN
CH3
Br
CH 3
CH 3
H3C
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 24 Time: 15 minutes
1. Find (m) meso compounds & optically active (a) and total stereoisomer (a + m) of following compounds
Compound Meso (m) Active isomer (a) (a + m)
(1)
(2)
CH3
Cl
Cl
(3)
(4)
Cl
CH3
Br
CH3Br
(5)
Br Br
Br
(6) H C3
Cl
CH3
Cl
CH3 SH
Br
(7) CH3
H C3
(8)
CH3
H C3
Cl
Cl
CH3
CH3
Cl
N.J
. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 24 Time: 15 minutes
1. Find (m) meso compounds & optically active (a) and total stereoisomer (a + m) of following compounds
Compound Meso (m) Active isomer (a) (a + m)
(1)
(2)
Cl
ClCl
Br
(3)
Cl
CH3
CH3
(4)
(5)
OH
CH3
H C3
HO
(6)
ClBr
(7)
CH3CH3
Cl Cl
(8)
CH3CH3
Cl Br
(9)
OH
Cl
O
O
ClCl
Cl Cl ClCl
Cl
HO
OH
CH 3 HO3
N.J
. SIR
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63
IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 25 Time: 15 minutes
1. Find (m) meso compounds & optically active (a) and total stereoisomer (a + m) of following compounds
Compound Meso (m) Active isomer (a) (a + m)
(1) Cl Cl
(2) H C3
CH3
Cl
Br
(3)
CH3
CH3
(4)
Cl
CH
CH
(5)
(6)
CH2
CH2
(7) CH3
H C3
(8) CH3
H C3CH3
H C3CH3
H C3
(9)
BrH C2 CH3
(10)
HO
H C3 CH3
(11)
OH
Cl
O
O
ClCl
Cl Cl ClCl
Cl
HO
N.J
. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 26 Time: 15 minutes
1. Identify correct matching and find out the total number of chiral centre.
S.No. Compound Optically active Chiral Achiral Optically No. of
molecule molecule inactive Chiral Center
(1) N
Et
(2)
C
Me
Me
(3)
COOH
H
HO
(4) S
Me
O
(5) OH
OH
(6) H
OH
OH
H
(7) Cl
Cl
(8)
O OH
HO OH
OH
HO
N.J
. SIR
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S.No. Compound Optically active Chiral Achiral Optically No. of
molecule molecule inactive Chiral center
(9) HMe
HCl
Cl
Me
(10)
H
OO
HMe
MeO
O
(11) Cl Cl
H H
CH OH2
CH OH2
(12)
OMeMe
(13)
Br
Cl
F
(14)
(15)
Me
H
H
Cl
Cl
Me
(16)
Me
Me
H
H
Br
Br
N.J
. SIR
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 27 Time: 15 minutes
1. Identify correct matching and find out the total number of chiral centre.
S.No. Compound Optically active Chiral Achiral Optically No. of
molecule molecule inactive Chiral Center
(1)
Me
Me
(2) MeMe
H
H
(3)
NO2
NO2 F
COOH
(4)
ClCl
F F
(5)
Me
Me
Cl
Cl
ClH
H
H
(6) O = C = O
(7) C C
Br
Br
H
H
(8) C = C = C = CNO2
MeCl
F
N.J
. SIR
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S.No. Compound Optically active Chiral Achiral Optically No. of molecule molecule inactive Chiral Center
(9)
C = C = C = C
Me
Me H
H
(10) CO CH CH O C2 2 2 2
CO CH CH O C2 2 2 2
H
H H
H
(11)
HH
OHHO CO H2
CO H2
(12) CO CH CH OH2 2 2
CO H2
H
H
(13)
H
OH
CO H2
CO H2H
OH
(14) NH
HN O
O
H
H
Me
Me
(15) NH
HN O
O
H
H
Me
Me
(16)
ClClH H
N.J
. SIR
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F-106, Road No. 2, Indraprastha Industrial Area, End of Evergreen Motors (Mahindra Showroom), BSNL Office Lane,
Jhalawar Road, Kota, Rajasthan (324005)
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IIT-JEE Chemistry by N.J. Sir ORGANIC chemistry
DPP NO- 28 Time: 15 minutes
1. Identify correct matching and find out the total number of chiral centre.
S.No. Compound Optically active Chiral Achiral Optically No. of molecule molecule inactive Chiral Center
(1)
COOH
COOH H
HH H
Ph
Ph
(2) HH
COOHMe
(3)
H
H
COOHMe
(4) Br
FBr
F
(5) C = C = C
Cl
I
(6)
(7)
Cl
Me
Me
(8)
(9)
Me
Me Me
Me
N.J
. SIR
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S.No. Compound Optically Chiral Achiral Optically No. of active molecule molecule inactive Chiral center
(10) H
H
H
MeMe
(11)
Cl
Cl
(12)
ClMe
MeMe
Me
COOHBr
(13)
Me
MeMe
Br Me
MeMe
Br
BrH HOH
Br OH
(14)
H
D
Me
COOH
(15)
O
(16)
HN
N
O
HO C2
C H6 5
(17)
CH3
CH3
Cl
Cl H
H