Conceptual Improvement of Isomerism Final DPP-372

N.J

. SIR

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1

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

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3

(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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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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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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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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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.

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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

N.J

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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

N.J

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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

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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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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

N.J

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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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22

(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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23



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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(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

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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

N.J

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26

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

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27

Q.12 For which of the following keq > 1?

(A)

(B)

(C)

(D)

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28

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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29

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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30

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

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31

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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32

(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

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33


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

N.J

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34

(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

N.J

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35



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.


(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)

N.J

. SIR




36

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

N.J

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37

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.

N.J

. SIR




38


Date: DPP NO- 13 Time: 15

minutes


(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

N.J

. SIR




39

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

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40

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

N.J

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41

(e)

&

(f)

&

Q.15 Calculate total no. of optical isomers, optically, active, meso & enantiomer pairs corresponding to

the following:–

(a)

N.J

. SIR




42


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.


(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




43

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

N.J

. SIR




44

(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




45


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


(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




46

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

N.J

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47

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

N.J

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48

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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49

(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)

N.J

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50



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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51

(a) (b)


(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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52




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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53


(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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54


(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)

N.J

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55





(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)

N.J

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56

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)

N.J

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57



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

N.J

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58



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)

N.J

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59



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)

N.J

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60


(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

N.J

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61





(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

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62





(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

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63





(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

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64



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

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65


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

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66





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

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67

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

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68




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

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69

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

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Conceptual Improvement of Isomerism Final DPP-372

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