geometrical isomerism by s.k.sinha

A visual Studyin

GEOMETRICAL ISOMERISM

By

S.K.SinhaResonance Kota

Checkout the unique 3D visual presentation for the 1st time at

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STEREOISOMERISM1. Isomers with different orientation

or distribution of atoms or groupof atoms in space are calledstereoisomers.

2. Stereo-isomers have identicalconnectivity of atoms and groups( bond pattern)( bond pattern).

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Checkout the unique 3D visual presentation for the 1st time at https://play.google.com/store/apps/details?id=air.geo.isomer

1. Example: cis and trans 2-

STEREOISOMERISM

pbutene has identical bondpattern.

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1. Geometrical Isomerism is a

STEREOISOMERISM

form of configurational stereoisomers.

2. Any two configurationalisomers are stereoisomers, butcan not be interconvertedcan not be interconverted

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1. These isomers differ in the

STEREOISOMERISM

configuration (The spatialarrangement) or Arrangementof atoms in space.p

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1. Configurational isomerism

CONFIGURATIONAL ISOMERISM

garises due to non-interconvertibility of bonds atroom temperature.p

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1. They are non interconvertible ,

CONFIGURATIONAL ISOMERISM

y ,they can be separated byphysical and chemicalmethods.

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1. Isomers with different


1. Isomers with differentgeometrical arrangement ofgroup of atoms or atoms inspace are known asspace are known asgeometrical isomers

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1. Example


1. Example

O

C C

C O HH

HCO

O H

OO H

O HFumaric Acid

C O HCO

C C

HH HH

Maleic Acid

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2. Example2. Example

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Conditions for G.I.


Conditions for G.I.

Cis-trans isomerism

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Conditions For GI :


1. For GI to occur, presence of a functional group or condition of restricted rotation isof restricted rotation is required . It is represented by GRR (i.e. -C=C- , -C=N- , -N=N-or ring str )or ring str.)

2. GRR = Group with restricted rotationS.K

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Conditions for G.I :


Co d t o s o G2. Non-convertability

aa ba

Xbb ab

X

a ba a

Xb ab b

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3. Different groups should beattached at each doublybonded atom (or on the sidesbonded atom (or on the sidesof GRR GROUP)

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aa baa

b

a

a

b

a

a

a

and are identical but not G.I.

aa ba

bb ab

andaa

can show G.I.

cbS.K.S

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4. Different groups should be in one plane (Planarity)

1. example

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1 following types of compound


1. following types of compound can show geometrical isomers :

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A. Compounds with C=C(Alkene )


p ( )

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trans-2-Butene has lesser


dipole moment than cis -2-butene.

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Cis has high B.P. than trans.


gBut trans has high mp than cis.

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A Compounds with C=C


p

O

C

O

OHH

C C

HCO

OOH

HCO

OH

C

O

OHC

OH

O

Fumaric Acid

C C

HH HH

Maleic Acid

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Maleic acid form Anhydride


y

CCO

OH O

OH C CO

OO

C C

HH

CC

H H

Maleic Acid Maleicanhydride

Fumaric acid: No AnhydrideMaleic Acid y

HCO

OH

C C

HCO

CH OH

No anhydride

O

Fumaric Acid

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cis and trans 1,2-Dichloroethene


,

Z: 1.28 g/cm³Boiling Z: 60.3 °CDensity

g/

E: 1.26 g/cm³

o g

point

60 3 C

E: 47.5 °C

Melting

point

Z: ‐81 °C

E: 81 °C

Dipole

t

Z (cis): 1.9

D

E (t ) 0point E: ‐81 C moment E (trans): 0

D

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cis and trans StilbeneGEOMETRICAL ISOMERISM

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cis and trans Stilbene


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cis-Stilbene has a melting point GEOMETRICAL ISOMERISM

g pof 5-6 °C, while trans-stilbenemelts around 125 °C

CC H

C HCH

C HC H

H

C C

CC

C C H

CH C H

H

CHCHCH

CH C H

C H C H

CH

CH

CHCH

CHCHCH

CHCH

CHC

CH

C C

CCH

H H

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B. Compound with C=N bond


p(imine, oxime )

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Compounds with C=N: Hydrazones


p y

NNH2

CH3

CH

E-AcetaldehydehydrazoneNH2

NNH2

CH3

CH3

Z-Acetaldehydehydrazone

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Compounds with C=N: Phenyl


p yhydrazones.

HNHC

CHCH

CH

Ph

CNCHCH

Ph

E-Benzaldehydephenylhydrazone

H

C

H

NC

CHCH

CHPhNHC

CHCH

CH

Z-Benzaldehydephenylhydrazone

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Compounds with C=N:


pSemicarbazone

C

O

NHCH3

C N

NH NH2

H

O

C

O

N H 2H

C N

N H N H 2

CH 3

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C. Compounds with N=N (Azo)


p ( )

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Compounds with N=N :


pAzobenzene

CHCH CHCHCH

CH

CH

CH CH

CHCH

CH

CCH CHC

N N

CHCH

CH NC

CHCH

CCH

CHN

NC

CHCH

CH

CH

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D. Compounds with Cycles


p y

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D Compounds with Cycles



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D. Compounds with Cycles


p y

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p y

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E. Cyclooctene or higher alkene


E. Cyclooctene or higher alkene

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F. Cumulene with odd no


F. Cumulene with odd no

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G. Spirane with odd rings


p g

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spirane with odd no of rings


p g

trans Spiranetrans‐Spirane

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H. spiroallene with odd no of


p( + rings)

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I. Decaline


I. Decaline

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CONCEPT TESTING


CONCEPT TESTING

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1 Is G I possible in given compound ?


1. Is G.I. possible in given compound ?

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2 Is G I possible in given compound ?2. Is G.I. possible in given compound ?

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A. 1,2-dichloropropene


B. 2,3- dichloro -1-butene

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A. 1,1-dichlorocyclopropaneB 1 2 –dichlorocyclopropane


B. 1,2 –dichlorocyclopropaneC. 1,3-dichlorocyclohexene

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Solution

1. Yes2 No

Solution

2. No3. No4. Yes5. No6. A Yes B No7. A No B Yes C No

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Formation of Oxime.Formation of Oxime.Symmetrical aldehyde /ketone

form one oxime.

C

CH2

CH2

CH2

CH2

CH2 O N OHH

HC

CH2

CH2

CH2

CH2

CH2N

OH

22 2

sym ald/ketone only one oximeS.K.S

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Formation of OximeFormation of OximeSymmetrical aldehyde /ketone

form one oxime.Unsymmetrical aldehyde

/ketone form two oxime.

C

CHCH

CH O N OH

H

C

CHCH

CH2 N

OH

CH2CH2 OH

C

CH2CH2

CH2 O N OH

H

C

CH2CH2

CH2 N

CH2CH2 H

C

C 2

CH

C 2

CH

CH2 N

OH

unsym ald/ketone form two oxime

C

CHCH

CH2 O N OH

H

y S.K.S

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Formation of OximeFormation of OximeUnsymmetrical aldehyde

/ketone form two oxime.

C

CH3

O N OH

H

C

CH3

N

OH

C

CH2CH3

O N OH

H

C

CH2CH3

N

CHCH

C

CH2

CH3

CH3

N

OH

C

CH2

CH3

CH3

O N OH

H

H

unsym ald/ketone form two oximeS.K.S

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Formation of HydrazoneFormation of HydrazoneUnsymmetrical aldehyde

/ketone form two

H H H NH2

Hydrazones.

C O

CH3

N NH2

H

C N

CH3

C N

NH2CH3

C

CH3

H

O N NH2

H

H

H

unsym ald/ketone form two hydra

H H

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Formation of HydrazoneFormation of HydrazoneSymmetrical aldehyde /ketone

form one Hydrazone.

H H H NH2

C O

H

N NH2

H

C N

H

sym ald/ketoneform one hydra

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Formation of OximeFormation of OximeSymmetrical aldehyde /ketone

form one oxime.

C

CH3

O N

H

H

OH C

CH3

N

OH

CHCH3 H CH3

sym ald/ketoneform one oximS.K

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Nomenclature in G I

Cis / trans: similarity of groups1 Similar groups on the same

Nomenclature in G.I

1. Similar groups on the same side or nearest : cis

2. Similar groups on the opposite side or farthest : transS.K

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CONCEPT TESTING

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Q1 Are these two cis trans GEOMETRICAL ISOMERISM

isomers?

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GEOMETRICAL ISOMERISMQ2 Are these two cis trans isomers ?

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Q8 Relation between the given pair

Cl

Q8 Relation between the given pair of compound is …………………………..

C CH2

Cl

HHH

C C

CC

H

Cl HH H

Cl

CC

H

Cl

H

H

C CH2

Cl

HHCl

C C

CC

H

H HH H

CC

H

H

H

H

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Cl


C

C CH2

C

Cl

ClHH

C C

CCH HH H

H HCl

C CH2H

HH

C C

CC

H

Cl HH H

CC

H

Cl

H

H

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Q10 Relation between the given pair Q10 Relation between the given pair of compound is …………………………..

CH2

CH3

C CH2CH3 C CH3

ClCl

CH

CH3

CH3CH2

CC

CH2CH3

Cl

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CH3CH2

CH2CH

CHCH2

CH2CH3

CH3CH

CHCH2

CH2CH2

CH2CH3CH CH2 CH2 CH3

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CH3CH2

CH2CH

CHCH3

CH3 CH CH2 CH3CH3CH

CHCH2

CH2CH2

CH3

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H


C H 2 C

H

CH 2

C H 2 C

C C H 2

C H 3 C H 3H

H

CH2 C CH2

H

CH3 C CH2 CH3

HH

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H


CHCH2

OCH2

CC

CHCH3 O C CH3

H

H

CH2 C

CH

O CH2 C H

CHCH3 CH3

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CH2 CH2


CH2CH CH

CH2

CH2C

CHC

H

H

H H

CH2 C CH CH

H

CHC

CCH2

CHCH2

H

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ClB


C CCl

ClBr

CH2

C

C

C

H HCH2 H

ClCl

C CH

ClCl

CH 2

CH

CH H

CH 2 Br

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C H 3


C HCH

CH C HC H

3

C H 2

C H 2

CH 2

CH 2

C H

C H 2C H 2C H 2

CH 2 C H 2C H 2

CH2

CH2CH2 CH

CH2 CH3

CHCH

CH2 CH2CHCH2 CH

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Answer

1. No2. No3. No4. No4. No5. Yes6. No7. No7. No8. Geometrical8. Positional10 Geometrical10. Geometrical11. Positional12. Homologous13 Identical13. Identical14. Geometrical15. Geometrical16 Geometrical16. Geometrical17. Geometrical

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Nomenclature in G I

Cis / trans: similarity of groups1 Similar groups on the same

Nomenclature in G.I

1. Similar groups on the same side or nearest : cis

2. Similar groups on the opposite side or farthest : transS.K

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Nomenclature in G I

E and Z : Based on seniority Seniors /high priority groups on the

Nomenclature in G.I

Seniors /high priority groups on thesame side or nearest : Z

(Z = zusammen = together)

Seniors /high priority groups on the opposite side or farthest :E

(E = entgegen = opposite)S.K.S

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Sequence rules : (CIP rules)

• Rules for seniority


Rule I : Greater the atomic number of 1st atom , greater the senior group.

I > Br > Cl > S > F > O > N > C

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Sequence rules : (CIP rules)GEOMETRICAL ISOMERISM


Rule II : If atomic number is same then isotopes with the greater mass no has greater priority .

Ex. (a) – T > – D > – H

(b) –C14H3 > – C12H3

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Sequence rules : (CIP rules)Sequence rules : (CIP rules)

Rule III : If the first atom is identical than second atom is observed for seniority.

Ex. (a)– CH2Cl > – CH2OH > –CH2NH2 > –CH2CH3 > – CH3

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Rule IV : Groups containing double or triple bonds are assigned seniority as if the bonds are connected with

t tseparate atoms..

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Ex. for deciding seniority among – C ≡ C and – CH= CH2 , their h pothetical eq i alents arehypothetical equivalents are compared.

CH CH CCH3

CH3 CH CH2 CH3 C CH3H

CH3 C CH CH3 CCH3

CH33CH3S.K

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Rule V : Bond pair is senior to lone pair. H

NHHH

H

H N+

H

H

Greater Priority

H N+ H

Greater PriorityH

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CONCEPT TESTING

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Q1 Check if the given compound is Q1 Check , if the given compound is Z or E.

CH3Cl

C C

CH3Cl

C C

CHB CHCH2Br CH3

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Q2 Check if the given compound is Q2: Check , if the given compound is Z or E.

OH

CHNH2 CH3

OH

C C

2 3

CH2Br CH3

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CHCH3OH

CCH2

CH2 CH2

C

HCH2 CH2 HBr

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CH2

CHOH CH3

CH3

CCH2

CH2 CH2

C

CH3Br 3Br

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C CH

H

CH3

C

C

CH3

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CH CH

CH3

C C

CH2 C

CH CH

CH

CH2 C CHCl

CH CH

CH CHS.K.S

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O HCH 3

C NC N

H

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CH

CH

C

CH

OH

C

CH

CH

C OH

Cl

C C

C CHCH2

CH CH

CH CH

I

CH CH

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Q9 Check if the given compound is

CH

Q9: Check , if the given compound is Z or E.

CCH CH

CH3

C C

CCH3 CH2

C C

CCH2CH3 CCH2

CH

CH3

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CCH CH

CH3

C

CH3

C C

CCH CH2C

CH3

CCH2

CH

CH2

CHCH CHCHCH2

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Answer

1. Z2. E3. E4 E4. E5. E6. Z7. Z8. Z8 Z8. Z10. E

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Number of G I:


Number of G I:

Numbers of geometricalNumbers of geometrical isomers can be found by calculating the number of gstereocenters in the compound.

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Number of G I:

Step-1: Count total no of

Number of G I:

Step-1: Count total no. of possible stereocenterwhere GI is possible.

CH3CH

CHCH

CHCH

CHCH2

CHCH2

G.I not possibleG.I possibleS.K.S

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Number of G I:

Step-2: See if both ends are

Number of G I:

Step-2: See if both ends are identical or not.

If both ends are not identical then total no of possible GI = 2n ,

where n= no of stereo-centers.S.K

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If both ends are not identical then total no of possible GI = 2n , where n= no of stereocentersstereocenters.

CH3CH

CHCH

CHCH

CHCH2

CHCH2C 2 2

Here n=3

G.I not possibleG.I possible

Total no of GI = 23 =8

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Step-i: if both ends are identical .Then see if the no of possibleno of possible stereocenters are odd/ even.

ii. If n = no of stereocenter is even then total no of GI = 2(n-1)+2 ((n/2)-1)

CH C H C H C H C HCH 3C H

C HC H

C HC H

C HC H

C HC H 3

Here n=4

G.I possible

Total no of GI = 10

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Step-i: if both ends are identical .Then see if the no of possibleno of possible stereocenters are odd/ even.

ii. If n = no of stereocenter is odd then total no of GI = 2(n-1)+2 ((n-1)/2)

CH3CH2

CHCH

CHCH

CHCH

CH2CH3

G I ibl

Here n=3 Total no of GI = 6

G.I possible

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CONCEPT TESTING

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Q:1 Find total no of geometrical isomer for

CHCH3

CCH

CHCH

CH

CH

CH2

CHCH

CCH2

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H

C HC H 2

CH 2

C HC

CH 2 C l

H

C H 2C

C HC H 2

C H

C H 2

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C

HCH3

CHCH

CH2

CHC

CH3

CH2C

2

CCH3

3

CH3

CCH3

CH22S.K.S

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C H 3

C H 2

C HC H

CH 32

CC

CCH 3

CH 3 C H 3

CCH 3

N H

C H 3

N HS.K.S

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CH3

CH2 CH C C CH CH CH CH3

CH3

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Q:6 How many products are formed in given reaction

O

C

O

+NH2

OH

CH3 CH3

Q 7 H d t f dQ:7 How many products are formed in given reaction

O

+NH2

OHCH2

CCH2 +

CH2CH2

CH22

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Q:8 How many products are formed in given reaction

O

C

O

CH2

C

CH2

CH2+

NH2 NH2

CH22

O

Q:9 How many products are formed inQ:9 How many products are formed in given reaction

C

O

CH3 C

CH2

+ NH2 OH

CH3

CCH3 CH3

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10. How many products are formed in given reaction

C

CH

CH2CH2

C

O

+

NH2 OH

CH2C

C

O

O

O

11. How many products are formed in given reaction

CCH2

COO

NH2 OH

CH2C

CH2

CH2

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12 Write all Isomers of C H12.Write all Isomers of C4H8

13.Write all Isomers of C5H10

14.Write all GI of C2HClBr2

15 Write all GI of C HClBrI15.Write all GI of C2HClBrI

16.Write G.I of 1,3,5,7‐Octatetraene

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geometrical isomerism by s.k.sinha

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