IITIANS SPECTRUM EDUTECH

ASSIGNMENT - 1
Identify E/Z configuration:-
Q.1

HOCH2 CH2Cl
(1) C=C Me (2) C=C
N C CH3

CH3
F Cl
(3) H (4) C=C
C I
Me Br
O O
HO—CH2
(5) O (6) O
C=C C=C
O N C O
O O
O

HO—C CH2O—Cl HOCH2 CN

(7) C=C (8) C=C
HOH2C COCl OHC CH2 — OH
CH3 CH3

Me CH2 = CH C C—H
(9) C=C (10) C=C CH3 CH3
CH2 — CH C CH
Me Me
Me
CH C C—H F H
(11) Me C=C (12) C=C
OH — CH2 C N Cl D

H
I CH2 CH2—C—I Cl — CH 2 — CH2 CH2 — CH3
(13) C=C H (14) C=C
Cl2 CH CH2—Cl HO — CH 2 — CH2 CH — CD3
D
HS OH
(15) C=C (16)
CH3 — CH2 CH3 H
Br Cl

(17) (18)
H H
Me Et
CH 3 CH3 CH3 H
(19) C=C (20) C=C
H H H CH3

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 H H H

(21) (22) H (23) H

H

O
(24) (25)

H CH2—CH2—Br
(26) (27)
Br CH2—CH2—Cl

H O Br OCH3
(28) (29)
Cl H CH3
O

Q.2
1. Which of the following will show G.I. in acidic medium

(A) O (B) O (C) O

H
(D) C=O (E) O (F) O
H

2. Which of the following will show G.I.

O O
|| ||
(A) CH3 – CH2 – C – NH2 (B) CH3 – C – NH – CH3
O O
|| Me
(C) C – NH – CH3 (D) C–N –
Me

3. Write cis/trans in the following of it show G.I.

(A) CH3 (B) (C)

CH3 Cl Cl
C Cl Cl

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 (D) (E) (F) (G)

Cl Cl Cl

(H) (I)

Cl

(J) Cl (K) Cl (L) Cl

Cl Cl Cl

4. Following will show G.I.

(A) (B) (C) (D)

C C C C
H CH3 H H H H H CH3

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 IITIANS SPECTRUM EDUTECH
ASSIGNMENT - 2
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
n 1
If n = odd ; P=
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)

H
H
(11) Me Me (12)

H H
H H
Me Me
(13) (14)
H H Me Me
Me
(15) (16)
Me

(17) (18)

(19) CH3–CH=CH–CH=CH–CH=CH–CH3 (20) CH2=HC CH=CH–Me

(21) Me–CH=C=CH–CH2–CH=C=C=CH–Me (22)

(23) Br–CH=C=C=CH–CH=CH–Br (24)

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 CH3 H
(25) (26)
H CH=C=CH2

(27) CH3—CH=CH CH=CH—CH3 (28) CH3–CH=C=C=CH–CH=CH–CH3

(29) (30)

CH3
(31) (32) CH3 – CH = CH – CH = N – OH
H
O
(33)
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)

CH3
Q.5 C=N Correct name of above compound is
Et OH
(A) syn – methylethyl ketoxime
(B) anti – methylethyl ketoxime
(C) syn – ethylmethyl ketoxime
(D) syn – methylethyl ketoxime

Q.6 is
C
H Cl
(A) E (B) Z (C) none (D) can’t predict

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 IITIANS SPECTRUM EDUTECH
ASSIGNMENT - 3

CH3 CH3 CH3 H

Q.1 C=C=C C=C=C
H l1 H l1 CH3
H
(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

CH3 l1 CH3 CH3 l2 H

Q.2 C=C=C=C C=C=C=C
H H H CH3
(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)

CH=CH–CH3
(D) CH3 — (CH = CH)3 — Ph (E) (F)CH3—CC—CH=CH—CH3
H

Q.5 Which of the following compound can show geometrical isomerism.

Br Cl CH3
(A) C=C (B) C
I Cl CH3
F Et CH3 CH3
(C) C=C (D) C
CH3 CH3
Cl Et

Q.6 The geometrical isomerism is shown by

CH2 CH2 CHCl
(A) (B) (C) (D)

CHCl

Q.7 Which of the following double bond will not exhibit geometrical isomerism.
Me Ph Me
(A) C = C (B) C=O
CH3 C 6H 5 Ph
Me
(C) C = N — Me (D) Me — N = N — Me
Ph

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Q.8 Which of the compound show geometrical isomerism.

CH — Br H H
N N CH3 N
(A) (B) (C) (D)
Br CH3 O
O
Cl H Cl Me C–OEt
N N
(E) Me (F) Me (G) N-H
EtO–C
N
O Me
COOMe OMe
Me
NC COOMe N
(H) N N (I) (J) H2C = CH – C OCH2Ph
N
O H Me
N
H
CH=CH–COOH
(CH3)3C (CH3)3C
C=C=C=C
(K) (L)
Cl Br

Cl Cl
O
(M) (N)
O

Cl

Q.9

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:–

H H H Cl
C=C C=C
CH3 Cl 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 CH3 H CH3
C=C C=C
H H Et H
1 2
(A) 1 > 2 (B) 1 < 2 (C) 1 = 2 (D) none

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

(A)

(B)

(C)

(D)

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 IITIANS SPECTRUM EDUTECH
ASSIGNMENT - 4
Q.1 Compound P.O.S. C.O.S. Optically active

1.
CH3 CH3

H H

2.
H CH3

CH3 H

3. H
H
CH3
CH3

H H
4.

CH3 CH3

5. H H

Cl Cl

6. Cl H

H Cl

7. Br
Cl H
Cl
H Cl
H
8. H
Cl
H
Cl
Cl

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9. H H
C=C=C
Cl Cl

10. H Cl
C=C=C
Cl Br

11. H H
C=C=C=C
Cl Cl

12. H H
C C
Cl Cl

13. Cl H
C C
H Cl

14. H H
Cl Cl

15. H
CH 3
CH3

16. Cl H
H Cl

17.

C
H Cl
CH3 CH3
18.

C
H H

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 CH3 CH3
19.

C
H Cl

20.

21.

22. Cl I

I Cl

23.

24. O = C = O

25. H
O=C=O

H H
26. O=C=O

27.
N=N=N
H H

H
28. C No. of POS _ _ _ _ _ _ _ _
H H
H
H
29. C No. of POS _ _ _ _ _ _ _ _
Cl H
H

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 H H
C=C
30. No. of POS _ _ _ _ _
CH 3 CH 3

CH 3
H
C=C
31. No. of POS _ _ _ _
H CH 3

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

CH=CH–CH 3
(B) CH3 – CH = C (Q) 6
CH=CH–CH 3

(C) CH3 – CH = CH CH = CH – CH3 (R) 4

(D) Cl – CH = CH – CH = CH – CH = CH – CH3 (S) 2

H5 C2 Cl H5 C2 Cl
Q.3 Statement 1: C=C and C=C are structural
CH3 Br Br CH3
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? H

CH3 Cl Cl
N O
N–H CH3
(A) (B) C=C=C (C) (D)
H3 C
O N
H H
H

Q.5 Find out the correct option(s) ?

NH
CH3 CH3 CH3
(A) C=C Orientation is E (B) Orientation is Z
NH H H
H

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 D
C=C
CH3
(C) H Orientation is Z (D) geometrical isomers are not possible
CH3

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

Me
(iv) (v)
D
(vi) Br – CH = C = C = CH – CH = CH – Br

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