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1. ELECTRONIC EFFECTS
1. Inductive Effects : Permanent displacement of sigma electrons along the carbon chain whenan atom or group of atoms having different electronegativity is attached with this chain is calledinductive effect.
When a covalent bond is present between two atoms which have same electronegativity, theshared electron pair lie exactly between the two atoms.
C C C C:or
Covalent bond Electron pair occupied central position
Electronegativity of both atom same. Non polar covalent bond
Equal distributions of electrons
In a covalent bond between unlike atoms, the electron pair forming the sigma bond is nevershared equality. It tends to be attached a little more towards the more electronegative atom. As a resultof this polarity develops in the molecule.
C X or CS+ : XS+ or CS+ XS�
Electronegativity X > C Polar Covalent bond
The characteristics of the inductive effect are as follows :
(1) It is a permanent effect.
(2) Inductive effect is distance dependent electronic factor. The effect weakens steadily withincreasing distance from the substituent.
(3) The inductive effect depends on the electronegativity of the substituent.
(4) The inductive effect is transmitted through sigma bond.
3 2
+ -
CH CH Cl I effect will draw electrons.
A slightly positive charge on the carbon atom
Types of Inductive Effect
(A) + IE (+I effect)
Groups or atoms which are less electronegative than hydrogen atom are said to have +IE
Decreasing Order of +IE :
� O > � COO > (CH3)3C � > (CH3)2CH � > CH3CH2 � > CT3 > CD3 > CH3 > T > D > H
All alkyl groups have +IE
Havier the isotope, more will be the +IE
Basic Concepts in OrganicChemistry and Stereochemistry
UNIT-7
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(B) �IE (�I effect)
Groups or atoms which are more electronegative than hydrogen atom are said to have �IE
Decreasing Order of �IE :
3 32 2 3 3 2OF OR NF NCl NR NH NO
2CN SO R CHO C C OH F Cl Br I|| ||O O
6 5 2OR OH C CH C H CH CH H
Greater the number of carbon atoms of the alkyl group greater is the +IE
2 5 3C H CH IE
Applications of Inductive Effect
(1) Stability of Intermediates
(2) Acidic nature of compounds
(3) Basic nature of Amines
(1) Stability of Intermediates
(A) Stability of Carbocations :
IEStability of Carbocations
IE
Examples :
(1) CH3 C+ >
CH3
CH3
CH3 C
H
CH3
> CH3 C
H
H
+ IE of methyl groups decreases;Stability decrease
+ +
(2) CH CH > CH CH Cl > CH CH Cl > CH CCl3 2 2 2 2 2 2 3
+ + + +
no�s of Cl increase;
� IE increase;
Stability of Carbocations decrease.
(3) CH CH > CH = CH > CH C3 2 2 + + +
% s characters increases;� IE increase;
Stability of Carbocation decreases.
sp3 sp2 sp%s = 25% 33% 50%
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(B) Stability of Carbanions :
IEStability of Carbanions
IE
Examples :
(1)
CH CH CH >3 2 2 CH3 C
H
CH3
> CH3 C
CH3
CH3
+ IE of methyl groups increases;Stability of carbanions decrease.
CH2 CH2
(2)% s Character increases;
� IE increase;
Stability of Carbanions increase.
CH CH < CH = CH < HC C3 2 2 2
(2) Acidic nature of Compounds :
IEAcidity
IE
Examples :
(1) CH � C � OH > CH CH � C � OH > CH CH CH � C � OH3 3 2 3 2 2
O O O
+ IE increase;Acidity decrease.
(2) FCH COOH > Cl � CH COOH > Br CH COOH > ICH COOH2 2 2 2
� IE decrease;
Acidity decrease.
(3) C Cl COOH > CH Cl COOH > CH Cl COOH > CH COOH3 2 2 3
� IE of Cl decrease;
Acidity decrease.
(4) CH CH CH COOH < CH � CH � CH COOH < CH CH CH COOH2 2 2 3 2 3 2
Distance of Cl form COOH decreases;� IE on COOH increase;
Acidity increase.
Cl Cl Cl
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(3) Basic Nature of Amines
Basic nature of amines depends upon lone pair availability on N atom which it can donate.
When one of more hydrogen atom(s) of ammonia are replaced by the alkyl groups, then the
electron pair availability on N atom increases due to + IE of alkyl group.
But in case of aromatic amines due to - I effect of phenyl group, basicity of aromatic amines
increase in comparison to ammonia.
CH � NH3 2
Methyl aminep = 10.64ka
> NH3
ammoniap = 9.25ka
> C H NH6 5 2
anilinep = 4.62ka
In aqueous solution; the basicity of amines are
(CH3)2NH > CH3NH2 > (CH3)3N > NH3
Therefore, the order of basicity can be easily remembered by the formula
2 1 3 NH3
but in the gas phase, the basicity of amines are
(CH3)3 N > (CH3)2 NH > CH3 NH2 > NH3
Basicity of Aromatic Amines
Case-I : When alkyl group attached to benzene ring, the order or basicity is as follows :
NH2
..
CH3
>
NH2
..
>
NH2
..
>
NH2
..
CH3
CH3
Case-II : When methoxy group is attached to the benzene ring, the order of basicity is :
NH2
>
NH2
>
NH2
>
NH2
OCH3
OCH3
OCH3
Case-III : When a nitro group is attached to the benzene ring, the order of basicity is :
NH2
>
NH2
>
NH2
>
NH2
NO2
NO2
NO2
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Case-IV : In the case of N-alkyl substituted aryl amines, the order of basicity is :
N
>
NHCH3
>
NH2
CH (+ IE)3
CH (+ IE)3
Resonance or Mesomeric Effect
The resonance effect (Mesomeric effect) involves delocalisation of electrons through resonancevia the -system.
Resonance : Two or more structures of the same molecule each differing in the distribution ofelectrons.
The properties of these structures will not be those to be expected of any of the resonatingstructures, but they will be those to be expected of a combination called a resonance hybrid of them.The phenomenon is called resonance and represented by the symbol
The resonance hybrid is the real structure of the chemical species.
For example,
We can not explain all the properties of Benzene if we consider a single structure of benzenei.e
However, if we consider following two kekule structures for benzene we can explain all theproperties of benzene.
Resonance hybrid
Some Examples of Resonance are -
(i) Carbonate Ion
O C O
O�
O C O
O�
�
O C
O�
O��
(ii) Aniline
NH2
..NH2
..NH2
�
+NH2
�
+NH2
�
+NH2
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(iii) Nitrobenzene
N
OO+
�
N
O+
�
N
OO+
�
N
O+
� O�
N
OO+
�
+
+
O O� � �
N+
+
O
(iv) Phenol
OH OH OH OH OH OH++ +
�
�
�
Rules for Resonance Contribution of a Resonating Structures
(1) All the canonical structures (Resonating structures) must be written according to the lewismethod showing bonds, lone pair of electrons, and formal charge.
e.g. -
H
C
H
H O H+
(2) The position of the nuclei in each of the canonical structures must be in the same relativeposition. Only electrons movement is allowed.
e.g.- O N O+ �
I
O N O+�
II
Thus I and II are canonical structures
C
H
H
CH3
CH3
C
H
CH3
CH3
H
C
III IV
C
III and IV are not resonating structures because H atom has been moved.
(3) All the resonating structures must have the same number of unpaired electrons.
R CH CH2 R CH CH R CH CH2
��+ +
R CH CH2
(not resonating structure)
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(4) All the atoms taking part in resonance must lie in same plane or nearly in the same plane
so that maximum overlapping of p-orbital can be possible.
(5) The energy of actual molecule is lower than the energy of any contributing structure.
(6) All the resonating structures do not contribute equally to the resonance hybrid. The more
stable structure is the great contribution.
The stabilities of resonating structures :
(1) Resonating structures with maximum numbers of covalent bonds are normally more
stable e.g. :
CH2 CH CH CH2 CH2 CH CH CH2 CH2 CH CH CH2
+ � +�
(more stable)
(2) If different resonating structures have the same number of covalent bond, them uncharged
structures are more stable. e.g. :
CH2 CH NH2
..CH2 CH NH2
+�
Uncharge(more stable)
Charge(less stable)
(3) Structures in which all of the atoms have a complete octate are preferred. e.g. :
CH2 O CH3
Carbon atom has6 electrons in itsvalence shell
here octate is notcompleted
(less stable)
..
..+ +
CH2 O CH3
Carbon atom has8 electrons in itsvalence shell
here octate iscompleted
(more stable)
..
(4) Structures in which there is maximum distance between like charges or minimum distance
between unlike charges, are more stable. e.g. :
N N� �
+
+
(more stable) (less stable)
(5) Structures with negative formal charge is present on a more electronegative atom and
positive charge on the least electronegative atom is more stable. e.g. :
CH2 CH C H CH2 CH C
(more stable)
H CH2 CH C H
O....
�
� +
+
O.... ..O
....
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(6) Equivalent resonating structures are more stable than non equivalent resonating structures.e.g. :
R CO
O H
.. ..
..
..
R CO
O H
.. ..
..
..
+
�
non-equivalentresonating structures(less stable)
R C RO
.. ..
..
�
O
.. ..
C O
.. ..
..
.. ..�O equivalentresonating structures(more stable)
Resonance Energy
The resonance energy is a measured of the extra stability of the resonance hybrid. The differencein energy between the actual structure (resonance hybrid) and the most stable resonating structure iscalled resonance energy. e.g. :
Resonance hybridResonating Structures
� 36 k cal/mol
most stableresonating structure
resonatinghybrid
+ 3H� 49.8 k cal/mol
2+ 3H� 85.8 k cal/mol
2
Cyclohexane
P.E.
Thus benzene has 36 k cal/mol less energy than that of the most stable resonating structureexpected for a typical compound with three double bonds. Hence resonance energy of benzene is 36k cal/mol.
Types of Mesomeric Effect :
There are two types
(i) + M Effect : The capacity of a group to donate electrons into a molecular -system isdescribed as a +M effect.
Groups which shows '+M' effect are :
2 2O NH NHR NR OH OR NH C R||O
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2R Ph CH CH F Cl Br I N OO C||O
(ii) � M Effect : The capacity of a group to withdraws electrons from a molecular -system isdescribed as a �M effect.
Groups which shows '�M� effect' are :
Me3N+ > � NO2 > � CN > � SO3H > � CHO > � COMe > � COX > � CO2H > � COOR
Applications of Resonance
(1) Stability of Intermediate
Resonance has been used to explain stability of the reaction intermediate like carbocations,carbanions, carbene free radicals etc.
(2) Bond Order
Total no's of covalent bond between two
bonded atom is all resonating structuresBond order
Total resonating structures
For example,
(1) CO32� C
O
O O� �
C
O
O O
�
�C
O
O O�
�
Bond order = 2 1 1 4
1.333 3
(2)ClO4 C
O
O O
�
C
O
O O
C
O
O O�
�
O�
O O
C
O
O O�O
Bond order = 1 2 2 2 7
1.754 4
(3) NO3� N
O
O� �
N
O
O O
�
�N
O
O O�
�
+ + +
O
Bond order = 2 1 1 4
1.333 3
(3) Acidity and Basicity
(1) Acidity of Carboxilic Acids : Carboxilic acid is acidic due to the resonance stabilization ofthe carboxylate ion.
Carboxylate ions exist as two equivalent resonating structures and moreover the negative chargeis shared.
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CRH
O
O....
H+
CRO
O....
..
.. ..
�CR
O
O
..
..
..
.. ..
�
Carboxilate ionsequivalent resonating structuresmore stable
(2) Acidity of Phenols : Phenol itself is resonance stabilized and formed resonance stabilizedanion namely phenoxide anion.
OH O�H+
�
..O�H
+ ..O�H
+ ..O�H
..
�..
[Contributing resonating structures of phenol]
O�H O
+ H +
� ..
Phenol
�
O O
�..
� ..
Contributing resonating structures of phenoxide ion
O O
�..
O
Phenoxideion
� �
If electron withdrawing group (EWG) is attached to the benzene ring, it increase the acidic natureof phenol.
If electron releasing group (ERG) is attached to the benzene ring, it decrease the acidic natureof phenol.
Examples :
(1)
OH OH
CH3
(ERG)OH
CH3
OH
CH3
P-cresol10.19(iv)
m-cresol10.08
(iii)
o-cresol10.28
(ii)
Phenol9.9(i)
pka
Acidic strength order = I > III > IV > II
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(2)
OH OH
NO2
(EWG)OH
NO2
OH
NO2
P-nitrophenol7.14(iv)
m-nitrophenol8.35(iii)
o-nitrophenol7.2(ii)
Phenol9.9(i)
pka
Acidic strength order = II > IV > III > I
(3)
OH OH OH
10.19(iii)
9.9(ii)
7.14(i)
pkaNO2 CH3
(ERG)(EWG)
Acidic strength order = I > II > III
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2. STERIC EFFECTS
(1) Ortho Effect : Ortho effect is observe in benzoic acids and Anilines
COOH NH2
Benzoic acid Aniline
Ortho substituted benzoic acids are highly acidic than benzoic acid.
While orthosubstituted Anilines are weaker base than anilines.
COOH COOH NH2 NH2
G G
(Acidity) (Basicity)
Acidity Order :
COOH COOH NH2 NH2
G G
(2) Steric inhibition of Resonance (SIR Effect) : The most important condition for resonanceto occurs is that the resonating structures must be coplanar. There are many examples in whichresonance is inhibited because orbitals are sterically forced out of planarity. This phenomenon is calledsteric inhibition of resonance.
N
CH3
NO2 CH3
These two orbital are not in same plane
due to steric hinderance
Aromaticity
The term 'Aromaticity' may be defined as the ability of an organic compound like benzene toundergo substitution a rather than addition reaction.
A Compound is said to be aromatic if it fulfils the following conditions -
(1) The structure must be cyclic having conjugated -bonds.
(2) The ring must be planar.
(3) It must obey huckel's (4n + 2) electrons rule. It states, "if in a conjugated planar cyclicpolyene, the number of delocalized electrons is (4n + 2) where n is an integer.
(4) Each C atom of the ring must be sp2 hybridized (or occasionally sp hybridized)
(5) It must undergo substitution reaction.
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Aromatic, Anti-aromatic and Non-Aromatic CompoundCyclic conjugated polyenes which contain (4n + 2) electrons (n = 0, 1, 2, 3, �) are aromatic
e.g. -
PlanarCyclicConjugated6 electrons, follow huckel�s ruleAromatic
In contrast a conjugated cyclic system with 4 electrons is called antiaromatic.
PlanarCyclicConjugated4 electronsAntiaromatic
A cyclic compounds in which continuous overlapping of p-orbitals disturbed cannot be aromaticor antiaromatic then it is termed as 'non aromatic�.
e.g. -
PlanarCyclic4 electronsbut no conjugation
Non aromaticExamples :
(1) Annulenes : The monocyclic conjugated system of the general formula CnHn have beencalled annulenes.
The ring size of an annulene is indicated by a number in brackets.
[4] Annulene4 electrons
[Antiaromatic]
[6] Annulene6 electrons
[Aromatic]
[8] Annulene cyclooctatetraene (COT)
Tubshape non planer
[non Aromatic]
H H X
[10] Annulene[Non Aromatic]due to strainCaused by theinside hydrogens
[10] Annulene10 electronsBridge cycodecapentaene[Aromatic]
(X = CH , O, NH)2
HHHH
[12] Annulene[Non Aromatic]
due to strain caused by the inside hydogens
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H
H
H
H
[14] Annulene14 electrons
[Aromatic]
[16] Annulene16 electrons
non planer[Non-Aromatic]
H
HHH
HH
[18] Annulene18 electrons
[Aromatic]
(2) Aromatic Ions :
(1)
Cl
+ Ag BF4
HBF4
+ Ag Cl [Aromatic ion]
(2)
O O
[Aromatic ion]
Ph PhPh Ph
(3)
C
Ph PhPh Ph
+
C NNCC C
C CNNNC NC
[Aromatic ion]
(4)
Ph
Ph
Br
Br
Ph
Ph
Sn F5
SO� 60ºC
2
Ph
Ph
Ph
Ph
2+
[Aromatic ion]
(5)
H
K/C H6 6 �
k[Aromatic ion]
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(3) Heterocyclic Compounds :
N
H
..O S.. .... .. N
Pyrrol Furan Thiophene Pyridine(4) Metallocenes :
Fe Cr
CO CO CO
(Ferrocene)Aromatic
BenzenetricarbonylchromiumAromatic
(5) Homoaromatic Compounds : Compounds that contain one or more sp3 hybrid carbon areknown as homoaromatic compounds and aromaticity of these compounds are known as homoaromaticity.
Conc H SO2 4
H
H
HH+
Cyclooctatetraene (COT)Tub-Shapenon planer
(Non Aromatic)
Homo aromaticcompound
When COT is dissolve in concentrated sulphuric acid, a proton adds to one of the double bondto form homotropylium ion.STEREOCHEMISTRY
The branch of chemistry which deals with three-dimensional structures of molecules and theireffect on physical and chemical properties is known as stereochemistry.Introduction
Compounds that have the same molecular formula, same molecular weight but different physicaland chemical properties are called 'Isomers' and phenomenon is known as 'Isomerism'.
There are two types of isomerism1. Structural isomerism2. Stereoisomerism
Isomerism
Structural Isomerismor
Constitution Isomerism
Stereoisomerism
This isomerism arries dueto different sequence ofbonding of atoms withinthe molecule.
This isomerism arries due to 3-D arrangement of atoms around the sterocentre within the molecule.
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3. STEREOISOMERISM
Conditions
1. Stereoisomerism differ only in the way that atoms are oriented in space.
2. Stereoisomer have identical IUPAC names (except prefix like cis and trans)
3. Stereoisomers have the same functional group.
Ex.1 C6H12 C6H12 Ex.2 C6H12 C6H12
CH3
CH3
CH3
and
CH3
Cis-1, 2-dimethylCyclobutane
Trans-1, 2-dimethylCyclobutane
Same molecular formulasame name except prefix
Stereo isomers
CH3
CH3
CH3
andCH3
Cis-1, 2-dimethylCyclobutane
1, 1-dimethylCyclobutane
Same molecular formuladifferent names
Constitutional isomers
It has been further classified in two categories.
1. Geometrical isomerism
2. Optical isomerism
Optical Isomerism :
Any substances which are capable of rotating the plane of polarized light are called opticalisomers and the phenomenon is called optical isomerism.
Optical isomers are two types -
1. Enantiomers
2. Diastereomers
Optical Isomers
Enantiomers Diasteromers
Enantiomers are mirrorimages of each other
These cannot be super-imposed upon their mirrorimage.
Diastereomers are not mirror image of each others.
These have atleast two chiralcentres having same configurationat one chiral centre but there is aoppose configuration at anotherchiral centre.
Enantiomers are also calledenantiomorphs.
Optical antipodes
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Ex.(i) (Enantiomers)
Mirror-image isomers.
CH3
COOH
HO
H Mirror HC3 H
OHC
COOH
C
(2R)-lacticacidI
(2S)-lacticacidII
I and II are non-super impossible mirror images isomers.
Ex.(ii) (Diasteromers)
HO
H
COOH
H
OH
COOH
HO
HO
COOH
H
H
COOH
*
*
*
*
(2S, 3S)-Tartaric AcidThreoisomer
(Similar groups lieson oppose side)
(2R, 3S)-Tartaric AcidErythroisomer
(Similar groups lieson same side)
Cis-trans isomers, Anomers, Epimers are Diasteromers.
Relationship between the Molecules -
Molecule with the samemolecular formula
Homomers(identical)
isomers
Yes No
Superimposable
Steroisomers ConstitutionalIsomers
Yes No
Same Constitutation
Enantiomers Diastereomers
Yes No
Non Suerimposable
Optical Activity : The rotation of plane polarized light is known as optical activity and substancespossessing this properties are called optical active. The optical activity is observed and measured by aninstrument called polarimeter.
Plane Polarized Light (PPL) : In ordinary light vibration occurs in all the planes perpendicularto the direction in which light travel. When ordinary light passed through Nicol prism. Then vibration takesplace in only one direction.
It is called plane polarized light (PPL)
Ordinary Light Plane-Polarized Light
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OrdinaryLight
Polarizer(Nicolprism)
PPL Sample Tube(Chiral Compound)
Analyzer
Eye
LightSource
Figure : Polarimeter
Solid line : before rotation
Broken line : after rotation
the angle of rotation
Angle of Rotation and Specific Rotation
Optically active compounds rotate the plane of the polarized light through an angle of .
The angle , measured in degree (°) is called the angle of rotation ()
A compound that rotate the plane of polarized light is said to be optically active.
Optically active compound can exist in four forms :
1. Dextrorotatory (d) or () form : If the rotation of the plane is to the right (clockwise) thenthe substance is called dextrorotatory (Latin : Dexter = right).
2. Laevorotatory () or (�) form : If the rotation of the plane is to the left (Anticlockwise)then the substance is called laevorotatory. (Latin : laevus = left)
3. Mesoform : Optical isomers which have atleast two chiral centre but posses plane ofsymmetry is called Mesoform.
It is optically inactive due to internal compensation i.e. the rotation caused by the upperhalf of the molecule is nautralized by the lower half of the molecule.
Examples :
COOH
COOH
OH
OH
H
H
Plane ofSymmetry
HH
OHOH
H
OH
H
OH
4. Racemic mixture (dl) or (±) : Equimolar mixture of Pairs of enantiomers is called racemicmixture. It is optically inactive due to external compensation i.e. the rotation caused by thed form is neutralized by the from.
Angle of rotation depends upon
1. The nature of substrate
2. Concentration of the sample (c)
3. Length of tube ()
4. Wavelength of the light
5. Temperature (t)
The common polarimeter use the sodium D-line or Mercury green line. Now new modern spectra-polarimeter use photocells because they give more accurate value.
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Specific Rotation : Specific rotation is the observe angle of rotation if 1-dm (10 cm) tude is usedand concentration of sample is 1g/ml.
t
SP D c
Specific rotation = observed angle of rotaion ( )
length (dm) g / ml
If pure sample of optically active substance is used then
t
SP D d
here d represents density of a pure sample
That means specific rotation is independent on
1. Length of tube ()
2. Concentration of the sample (c)
Enantiomeric Excess : Some times we deals with a mixture that is neither optical pure enantiomernor a racemic mixture. One enantiomer is present in excess of the other.
The enantiomeric excess (ee) also called the optical purity, tells how much more there is of oneenantiomer.
Enantiomeric excess (ee) = % of one enantiomer � % of the other enatiomer
Ex.1 If a mixture contain 75% of one enantiomer and 25% of the other. Then the enantiomeric excessis ?
Sol. ee = 75% � 25% = 50%
There is a 50% excess of one enantiomer over the racemic mixture.
Ex.2 If the enantiomeric excess is 75% then how much of each enantiomer is present?
Sol. 75% ee means that the solution contains an excess of 75% of an enantiomer and 25% of theracemic mixture of both enantiomers.
Because racemic mixture contains an equal amount of both enantiomers, it has 12.5 % of eachenantiomer.
Total amount of one enantiomer = 75 + 12.5 = 87.5%
Total amount of other enantiomer = 12.5%
100 ee% major enantiomer
2100 ee
% minor enantiomer 2
The enantiomeric excess can also be calculated as -
mixture
pure enantiomer
[ ]ee 100%
[ ]
[] = specific rotation
(moles of major enantiomer) (moles of minor enantiomer)ee 100
moles of both enantiomer
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Conditions for Optical Activity : A compound is said to be optical active if it is non- superimposableon its mirror image due to asymmetry.
Chirality is the necessary and sufficient condition for a molecule to be optically active.
1. All Asymmetric and dissymmetric molecules are chiral and optical active.
An Asymmetric molecule has no element of symmetry.
Elements of Symmetry1. Centre of Symmetry (i)
2. Plane of Symmetry ( )
AsymmetricAbsent
Absent
DissymetricAbsent
Absent
Symetrici-Present
or-Present
3. Axis of Symmetry (C )n Absent Presentor
Both Present
Chiral
Optically Active
Achiral
Optically Inactive
Elements of Symmetry
1. Plane of Symmetry : The plane which cuts through the middle of a molecule in such a waythat one half of the molecule is a mirror image of the other half.
Example
MeMe
OH CN
MeMe
OH
CNPlane of Symmetry run throughcentral carbon, OH and CN
HH
COOH COOH
HH
COOH
COOHPlane of Symmetry run throughcentral carbon, and both COOH
CH3CO H2
H OH
No plane of Symmetry becauseboth are non-idential atom
2. Centre of Symmetry(i) : A point in the molecule through if a line is drawn in one direction, thenextended to equal distance in opposite direction. It meets another similar group or atom.
HH
ClBr
BrCl
HH
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3. Axis of Symmetry (Cn) : An imaginary axis through which the object is rotated by an angle,leave the orientation of the object indistinguishable from its original orientation.
Cl
H
H
Cl
= 180º
n = 2Cl
H
H
Cl
180º 180º
IdenticalStructures
Ex. Which of the following molecules are chiral ?
(1)H
OH
OH
H
H
H
H
H
(2)H
Cl
H
OH
H
Cl
H
OH
(3)
Cl
HH
Cl
(4)
H
OH
H
OH
(5)
H
H
OH
HO
(6)H
Cl
H
Cl
Ans. (1)
Sol.
(1)
H
OH
OH
H
180º
= Absenti = AbsentC = Presentn
Dissymmetric moleculechiral optically active
(2)
H
Cl
H
OH
H
Cl
H
OH
= PresentSymmetric
Achiraloptically inactive
(3)
Cl
HH
Cl
PresentSymmetric
Achiraloptically inactive
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(4)
H
OH
H
OH
= PresentSymmetric
Achiraloptically inactive
(5)
H
H
OH
HO
i Present Present
C Present
2
SymmetricAchiral
optically inactive
180º
(6)
H
Cl
H
Cl
= PresentSymmetric
Achiraloptically inactive
2. Compounds with Asymmetric Carbon Atom
A carbon atom is asymmetric if four different groups or atoms are attached to it
a
c
c
b b* Chiral carbon atom(Asymmetric Carbon atom)
Example :
COOH
HOCH3
Lactic Acid 2-Bromo-2-Chlorobutane
H
*
CH3
C H2 5Br
Cl
*
2-Butanol
CH3
C H2 5OH
H
*
C
C
C
C
HO
H
H
H
OH
OH
*
O
CH OH2
*
*
CH OH2
3-Chiral Centre
COOH
(CHOH)4
COOH4-Chiral Centre
*
Br1-Chiral Centre
C C NHCH3
H H
OH CH3
**
2-Chiral Centre
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3. Optical Activity Even in The Absence of Chiral Carbon Atom
A. Optical Activity Due to a Chiral Axis :
1. Allene :
ca
bc
b
a
(a b)
If no�s of duble bonds
are evenIf no�s of double bonds
are odd
c c cb
ab
a
InPlane
Perpendicularsto plane
The groups at the end of allenesare perpendicular to each other
Here Axis of symmetry Present
C2
c = c = c
c c c ca
b
a
b
InPlane
InPlane
The groups at the end of allenesare in same plane
Here Plane of symmetry Present
= Presenti = Absent
C = Absent2
Symmetric, Achiraloptical Inactive
Gives geometricalisomerism
c
c
c
c
c
c
a
b
b
a
cis
Trans
c
c
a
a
b
b
C = Present2
i = Absent = Absent
Dissymmetric moleculeChiraloptically active
Here,
c ca
bc
a
b
a
b
a
bc cc
Thus Allanes of type abc = c = cab and abc = c = ccd are chiral, when it has even no's of doublebond.
Cl
H
Me
CMe3
C C CH
HOOC
COOH
HC C C
1-chloro-3, 4, 4-trimethyl-1-2-pentadiene
(R)�(�)�gutinic acid
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2. Sprains :
A
B
A
BA B
Since the replacement of a double bond in allene by a ring does not alter the basic geometry ofthe molecule thus the conditions for Chirality in spirans are quite analogous to those in allenes.
here, C2 = Present
i = Absent
= Absent
Dissymmetric molecule
Chiral
Optically active.
A
B
A
BC C C
Allene
A
B
A
B
A
B
A
B
Example :
H
NH2
H
H N2
H
COOH
H
HOOC
Diaminospiro-heptane Spiro-heptanedi carboxilic acid
3. Hemispirans :
A
B
A
BAsymmetric ChiralOptically active
(A B)
Example :
H
COOH
Me
H
CNH2
H
H N2
H4. Biphenyls : Biphenyls show enantiomerism when the following two conditions are satisfied.
(i) Neither of the rings should possess plane of symmetry ()
A
A
B
B
A
B
A
BBoth rings have planeof symmetry achiral
optically inactive
Chiral AxisChiral
optically active
(ii) The ortho-substituents should be larger in size so that the two rings must not be complainer,they should be orthogonal to each other
Here A and B = COOH, NO2, I, Br etc.
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Example :
SO H3
H
H
SO H3
COOH
NO2
NO2
COOH
COOH
OH
COOH
OH OH
OHHO
OH
5. Bridged Biphenyls :
a
(CH )2 n
bhere a = b
ora b
non planaroptically active
Example :
HOOC
CH2
CH3 Me
CH2
Me
CH2
CH2
CH2
COOEtEtOOC
-
6. Bipyrroles :
N N
XX
YY
7. Binepthols :
OHOH
8. Compounds having chiral centres other than carbon.
Example :
CH3
Si
C H2 5
C H3 7
H
CH3
Ge
C H2 5
C H3 7
H
CH3
Sn
C H2 5
C H3 7
I
C H6 5
N
CH3
CH C H2 6 5
C H2 5
X
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X
R1
R2
R3
X = N, P, As, Sb
Here pyramidal inversion takes place at room temperature.
X
R1
R2
R3
X
R3
R2
R1
X
R1
R2
R3
Due to pyramidal inversion both forms are interconvertable therefore they always exist as racemicmixture it is chiral but optically inactive.
B. Optical Activity Due to Chiral Plane
1. Ansa compounds :
(CH )2 n
2. Para cyclophanes :
(CH )2 n(CH )2 n
3. Trans cyclooctane :
Calculation of optical isomers :
Optical isomer can be calculated by using following flow-chart
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Calculate the total no�s of
chiral centre (n)
Chiral molecule( = Absent
i = Absent)
Achiral molecule( = Present ori = Present orboth present)
1. Total enantiomers (Total optically active forms) (E) = 2n
2. Total optically inactive forms (mesoforms) m = 0
3. Total stereisomers T = E + m 2 + 0 = 2n n
4. Total pair of enantionmersnE 2
P2 2
5. Total racemic mixture or Total racemic modification
n2R M P
2
n = even n = odd
(n 1)E 2 (n 1) (n 1)
E 2 22
n 2m 2
2
n 1
m 22
T = E + m T = E + m(n 1)E 2
P2 2
P = E
R = m + P R = m + P
Example :
(1)
COOH
(CHOH)2
COOH-Present
Achiraln = 2 (even)
E = 2 = 22�1
2 2m 2 1
2
T = 2 + 1 = 32
P 12
R = 1 + 1 = 2
(2)
COOH
(CHOH)3
COOH-Present
Achiraln = 3 (odd)
E = 2
m = 2
T = 4
P = 1
R = 3
(3)
CHO
(CHOH)4
CH OH2
-PresentChiraln = 4
E = 16
m = 0
T = 16
P = 8
R = 8
Que. Calculate the value of total optically active isomers ?
(1) CN
Br
(2)
Cl ClBr Cl (3)
CHO
(CHOH)2
CH OH2
(4)
C
CH OH2
(CHOH)3
CH OH2
O
Ans. (1) 9, (2) 2, (3) 4, (4) 9
Configuration : The arrangements of atoms that characterizes a particular stereoisomer iscalled its configuration.
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A. Representation of a configuration
1. Wedge-dash Representation :
4
21
3
Chiral Centre
Lies on plane of paperrepresent principal axis
wedge
Lies towardthe viewer
dash
Lies awayfrom theviewer
2. Fischer Projection Formula :
4
2
31
Principle chain(away from the viewer)
Towards the viewerChiral Centre
4
2
31or
Characterstics
1. Fischer projection formula is always eclipsed conformation.
2. One inter change or odd number interchange of ligands at one chiral centre, givesenantiomers
Two interchange or even number interchange of ligands at one chiral centre, gives identicalconfiguration.
1
3I
24 I Changesst
2
3II
14 II Changesnd
2
4III
13
Ist and IInd = Enantiomers
IInd and IIIrd = Enantiomers
Ist and IIIrd = Idential
3. When fischer projection formula is rotated 180º on the plane of paper then it gives idential
configurations.
1
3
24180º
3
1
42on the plane
of paper
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4. When fischer projection formula is rotated 180º out of the plane of paper, then it gives
enantiomers.
1
3
24 180º
1
3
42out of the plane
of paper
5. When fischer projection formula is rotated 90º on the plane of paper then it gives
enantiomers.
1
3
24 90º
4
2
13on the plane
of paper
6.
1
3
24
1
2
43
Constant
(identical)
Inter Conversion of Wedge-dash and Fischer Projection Formula :
The group which is below the plane of the paper (dash) is kept at the bottom and remaining threegroups are attached in the same sequence (i.e., clockwise or anticlockwise) as they appear in the flyingWedge formula.
(1)
COOH
OH
CH3
H
COOH
CH3
OHH
(R) (R)
(2)
OH
H
Ph COOH
OH
H
COOHPh
R
(3)
NH2
H
Me COOH
H
NH2
COOHMe(4)
NH3
OH
CH3Ph
CH3OH
NH3
Ph
Assignment of Configuration :
1. D/L System (Relative configuration)
2. R/S Configuration (Absolute configuration)
1. D/L System (Relative configuration) : In relative configuration we deals only with theconfiguration of one chiral Centre.
This method is used to assign the configuration of sugars and amino acids.
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In the case of sugars the reference substances is glyceraldehyde.
CHO
CH OH2
(OH groupright side)
D(+)-glyceraldehyde
OHH
CHO
CH OH2
(OH groupleft side)
L(�)-glyceraldehyde
HHO
While in the case of amino acids the reference substances is alanine.
COOH
CH3
D(+)-Alanine
NH2H
COOH
CH3
L(�)-Alanine
HNH2
Rules for Determining Relative Configuration
Rule 1 : First the main chain determined and its atoms are numbered.
Example :5CH OH2
CHO
H
H
H
HO
HO
HO
1
2
3
4
5CH3
COOH
OH
NH2
OH
H
H
H N2
1
2
3
4
I IIRule 2 : Numbering should move from up to down. If it moves from down to up then we
rotate the projection formula about 180º on the plane of paper.
5CH OH2
CHO
H
H
H
HO
HO
HO
1
2
3
4
I
180º
1CHO
5CH OH2
OH
OH
OH
H
H
H 4
3
2
5CH3
1COOH
OH
NH2
OH
H
H
H N22
3
4
II
180º
1COOH
5CH3
NH2
H
H
HO
H N2
HO 4
3
2
Rule 3 : In the case of sugars; we will see the highest numbered chiral centres while in the caseof Amino Acids, we will see the configuration of lowest no. of chiral centres
1CHO
CH OH2
OH
OH
OH
H
H
H 4
3
2
D-form
1COOH
5CH3
NH2
H
H
H
H N2
HO 4
3
2
D-form
*
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Examples :CHO
CH OH2
CH OH2
OH
OH
CHOH
OH
OH
H
HHO
H
H
[D]
[D]
COOH
NH2
H
H C3
[D]
H
OH
[L]
HOH C2
CHO
H
OH
CH OH2
H
OH
CHO
HO
HO
H
H
[L]
COOHH
H N2
NH2
HCH3
[D]2. R/S Configuration : To assign an enantiomer as R or S, first assign a priority (1, 2, 3 or 4)
to each groupPriorities are assigned to groups by a series of rules known as the Cohn-Ingold-Prelog (CIP)
sequence rule, named after the three chemists who developed them.Rule 1 : The atom of highest atomic number gets the highest priority.
4
2
13 F Br
H
Cl
2
1
34 H SO H3
Cl
I
Rule 2 : If two atoms are isotopes of the same element, the atom of higher mass numbered hasthe highest priority.
4
1
32 H C3 D
H
I
1
3
24 H Cl35
Cl37
CH3
Rule 3 : When first atom is same then the next sets of atoms are examined.The process is continued.
3
4
21 Cl CH CH2 3
CH3
H
(C C H H ) C C H
H H
H H(C H H H)
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Rule 4 : Groups containing double or triple bond are assigned as.
C O equals C O
O CConsider thiscarbon bonded to 2 O�s
Consider this oxygenbonded to 2 C�s
C N equals C N
N CConsider thiscarbon bonded to 3 N�s
Consider this nitrogenbonded to 3 C�s
N C
equalsC
CHHCC
C
Example :
C
COOH
OH
H CH OH2
1
3
2
4Consider this carbonbonded to 1 oxygen
C
O
O( H
Consider this carbonbonded to 3 O�s
Assignment of R or S to a Chiral Centre
(A) R/S configuration in Wedge-Dash formula :
Example :
C H2 5
COOH
CH3
H
Enantiomer-A
COOH
Enantiomer-B
CH3
HH C5 2
Step 1 : Assign a sequence of priorities 1 to 4 to the four atoms or group of atoms bonded tothe chiral centre.
COOH > C H > CH > H2 5 3
1 2 3 4
Step 2 : We visualize the molecule oriented so that the ligand/atoms of lowest priority is directedaway from the viewer (on the dash)
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C H2 5
COOH
H
Lowest Prioritytowards the
viewerEnantiomer-A
H C3
1
24
3
CH3
COOH
C H2 5
H
1
32
4
Lowest Priorityaway fromthe viewer
rotate
CH3
COOH
H
Lowest Prioritytowards the viewer
H C5 2
1
3
4
2
H
COOH
C H2 5
Lowest Priorityaway fromthe viewer
H C3
1
4
2
3rotate
Step 3 : Trace a circle from priority group 1 2 3. If the arrow point of the circle goes in theclockwise direction, the configuration is specified R (Latin : Ractus = right).
If Anticlockwise the configuration is specified S (Latin : Sinister = left).
CH3
COOH
C H2 5
H
1
3
2
CH3
COOH
C H2 5
H
1
3 2
AnticlockwiseS-isomer
ClockwiseR-isomer
Simple method for Assignment of R or S
4
4
4
4
Answeras such
Answerreverse
even change
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Examples :
OHH
COOHPh
23
1
4 H
COOHme
23
4
OH1 Cl
NH2me
23
1
H4
Lowest Priorityaway from theviewerAnswer as suchR-Configuration
Lowest Priorityaway from theviewerAnswer as suchR-Configuration
Lowest Prioritytowards the viewerAnswer reverseS-Configuration
H
Cl C
C H2 5
CF3
4
3
2
1 EvenChange
CF3
Cl C
HC H2 5
2
4
3
1
R-Configuration
R and S configuration in fischer projection
Example :
Cl
Br
CH3H
H
Cl
CH3Br
1-bromo-1-chloro-ethane
Step 1 : Assign a sequence of priority 1 to 4 and trace a circle from priority group 1234.
Cl
Br
CH3H
H
Cl
CH3Br
2
1
34
4
2
1 3
Step 2 : Case-I if the group of lowest priority is on the horizontal line then clockwise sequencegives S and Anticlockwise sequence R configuration.
Cl
Br
CH3H
H
Cl
CH3Br
2
1
34
4
2
1 3
Lowest Priority left sideclockwise
[S]-Configuration
Lowest Priority isabove the plane
Anticlockwise
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Step 3 : Case-II If the group of lowest priority is above or below the plane then clockwisesequence gives R and anticlockwise sequence S configuration.
H
Cl
CH3Br
4
2
1 3
S-ConfigurationExamples :
CHOH2
Br
ClH
1
24
3
[R]
COOme CH3 NH2
CHBr2 CH25 COOEt
H NH2 COOHHO Cl H
2 3 2
4 2 31 1 4
3 4 1
[R] [R] [R]
R R R R
Resolution :
The Separation of a racemic modification into its constituent enantiomers is known as resolution.
Enantiomers have identical, physical and chemical properties so these cannot be separated bynormal physical method.
Separation of pair of enantiomer by using Chemical method (Via Diastereomers)
This is the best method of resolution. In this method a racemic modification is converted into amixture of diastereomers by using a pure enantiomer of another compound.
R/SR/SRacemicmodification
Optically Pure Compound (R� or S�)
Mixture of diastereomersRR'+SR'
RR'
R + R�
SR'
S + R'
Fractional Crystallisation
Hydrolysis Hydrolysis
R�
The diastereomers have different physicalproperties, so can be separated byany physical method
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Properties of Enantiomers
Enantiomers have identical physical and chemical properties except :
1. Odour
Ex. (+) � Limonene has smell of oranges and
(�) � Limonene has smell of lemons
2. Optical rotation (equal magnitude but opposite direction)
3. Physical activity
Ex. (�) � Nicotine is poisonous than (+) � Nicotine;
(+) � histidine is sweet whereas (�) � histidine is tasteless.
4. Reactivity towards optically active compound because it causes them to behaves asDiastereomers.
R S
RR� SR�
Diastereomers
+ R� + R�
Diastereomers have different (optically active) physical properties
Diastereomers react with O.A (optically active) as well as optically inactive (OI) compounds withdifferent rate.
Conformational Analysis
Special orientation of molecule which differ only in the dihedral angle and easily inter convertibleby rotation about single bond are called conformers. They are also known as rotational isomers orrotamers a molecule may have a number of conformers each different conformation will have a differentenergy.
Conformational analysis is the analysis of physical and chemical properties of a compound interm of various conformations in ground state, transition state and also excited state.
Dihedral Angle
The angle that separates a bond on atom from a bond on an adjacent atom is called a dihedralangle. There are three conformers are possible.
}-dihedral
angle
Eclipsed = 0º
dihedralangle
Staggered = 60º
Skew or gauche
= 60º > > 0º
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4. CONFORMATIONS OF ACYCLIC SYSTEM
Conformations of Ethane
When ethane molecule rotates about carbon-carbon single bond.
Two extreme conformations can result
1. Staggered conformation 2. Eclipsed conformation
Eclipsed = 0º
Staggered = 60º
HH
HHH
H
rotate 60º
H
H
H
HH
H
(Newman Projection of Ethane)
The staggered and eclipsed conformations of ethane interconvert at room temperature. Thestaggered conformations are more stable (lower in energy) than then eclipsed conformations.
Reason : In eclipsed conformation carbon-hydrogen bonds are closest so electron-electronrepulsion between the bonds in the eclipsed conformation increase its energy compare to the staggeredconformation.
The potential energy barrier in ethane is 2.9 k cal/mol. The extra energy of the eclipsed conformationis called torsional strain.
Energy profile diagram of conformations of ethane :
600 120 180 240 300
E E E
S S S StaggeredP.E
Eclipsed
2.9 K cal/mol
Dihedral angle ( )
2.9 K cal/mol
Conformations of Butane :
CH3
1
CH2
2
CH2
3
CH3
4
Consider rotation at C � C2 3
CH3 C C CH3
H H
H Heach C has 2H�s and 1 CH group3
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In the case of n-butane rotation about C2�C3 bond gives rise to six distinct conformations asshown below.
The conformations are raised out of successive rotations by a dihedral angle of 60º each time.
Fully eclipsed = 0º
[i]
Gauche = + 60º
[ii]
rotate 60º
[60º]
meme
HHH
H
me
H
H
H
H
merotate 60º
[120º]
Eclipsed = +120º
[iii]
meH
HmeH
H
rotate 60º
[300º]
rotate 60º
[240º]
rotate 60º[180º]
Anti conformation = +/� 180º
[iv]
me
H
H
me
H
H
Eclipsed = � 120º
[v]
me
HHH
me
Gauche = � 60º
[vi]
me
me
H
H
H
H
rotate 60º[360º]
H
The staggered conformation (II, IV, VI) are lower in energy (more stable) than the eclipsedconformations (I, III, V)
The �CH3 groups are further apart in the anti conformation (IV), than the gaucheconformation (II, VI) so among the staggered conformations IV is lower in energy (morestable) than II and VI
Conformation I is higher in energy than III or V because two longer CH3 groups are forceclose to each other.
Over all potential energy = I > III ~ V > VI ~ II > IV
Stability = IV > II ~ VI > V ~ III > I
Energy Profile Diagram of Conformations of Butane
600 120 180 240 300
P.E
I
II
III
IV
V
VI
19 KJ/mol
3.8 KJ/mol 16 KJ/mol
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5. CONFORMATIONS OF CYCLOHEXANES
For Cyclohexanes there are two extreme conformations
(A) Chair Conformation : A planar cyclohexane would experience angle strain (bond angle isdifferent from desired tetrahedral bond angle of 109.5º) and torsional strain (caused by repulsion between
bonding electrons)
120º
Angle > 109.5º
Angle strain
Torsional strain
H
H
H
H
So in order to minimize above repulsion, cyclohexane adopt�s a puckered conformation called the
chain form which is more stable than any other possible conformation.
H
H
H
H
H
H
HH
axial H�s( to the molecular plane)
equatorial H�s(|| to the molecular plane)H
HH
H
Fig. : Chair Conformation of Cyclohexane
Cyclohexanes has six axial H�s and six equitorial H�s
It is a staggered conformationH
H
H
H
H
H
H
H
(Newman Projection of the Chair Conformation)(B) Boat Conformation : It is a eclipsed conformation of cyclohexane. It is less stable than the
chain conformation due to torsional strain furthermore in the boat form there is steric repulsion betweenthe two hydrogens (flagpole hydrogens)
H H
HHH
HH
H
H
HH
H
Flagpole hydrogens
Boat Conformation of Cyclohexane.
H H
HH H
H
Newman Projection of the Boat ConformationH H
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Ring Flipping
At room temperature cyclohexane rapidly interconvert (flip) to mirror image chair conformations.
ChairForm
InvertedChair Form
Axial and equatorial H atoms are interconverted during a ring flip. Axial H atoms become equatorialand equatorial H atoms becomes axial H atoms
Conformations of Monosubstituted Cyclohexanes
When one hydrogen in cyclohexane in replaced by a larger atom or group crowding occurs.
Ex. methyl cyclohexane
The two chair conformers of a monosubstituted cyclohexane such as methyl cyclohexane are notequivalent
The methyl substituent is in an equatorial position in one conformer and is an axial position in theother.
Equatorial-methylmore stable
Chair conformer
axial-methylless stable
Chair conformer
CH3
H
ring flip
CH3
H
The Chair conformer with the methyl substituent in an equatorial position is the more stablebecause a substituent has more room and therefore steric interaction. These steric interactions arecalled 1,3-diaxial interaction.
H C
H H
H
H
H
H
H
HH
H
(1, 3-diaxial interaction)
Conformations of disubstituted Cyclohexanes :
Rotation around the C-C bonds in the ring of a cyclohexane is restricted. So disubstitutedCyclohexanes gives geometrical isomerism
One has both methyl substituent on the same side of the cyclohexane ring (up-up or down-down)it is called cis isomer.
The other has the two methyl substituent on opposite side of the ring (up-down), It is called thetrans isomer.
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(1) 1, 2-dimethyl Cyclohexane :
CH (up)3
CH (up)3
(ae or ea)Cis isomer
chiralbut optically
inactive due toconformationalenantiomerism
CH (up)3
(aa)Trans isomer
CH (down)3
CH (up)3
(ee)Trans isomer
CH (down)3
= Absenti = Absentchiraloptically active
= Absenti = Absentchiraloptically active
(2) 1, 3-dimethyl Cyclohexane :
CH3
(aa)Cis isomer
PresentAchiral
optically inactive
=
(ee)Cis isomer
(a, e)Trans isomer
= Absentchiraloptically active
= Absentchiraloptically active
CH3
CH3CH3 CH3
CH3
(3) 1, 4-dimethyl Cyclohexane :
CH3
a, e or e, aCis isomer
Plane = PresentAchiral
optically inactive
( )
a aTrans isomer
Plane ( ) = PresentAchiral
optically inactive
e e
Plane ( ) = PresentAchiral
optically inactive
Trans isomer
CH3
CH3
CH3
CH3CH3