Basic Concepts in Organic UNIT-7 Chemistry and Stereochemistry · Chemistry (Basic Concepts in...

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Chemistry (Basic Concepts in Organic Chemistry and Stereochemistry)

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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

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244

(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%





245

(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





246

(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





247

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





248

(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)





249

(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

....





250

(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





251

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.





252

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





253

(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





254

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.





255

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]


[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





256

H

H

H

H


[Aromatic]


non planer[Non-Aromatic]

H

HHH

HH


[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]





257

(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.





258

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





259

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





260

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.





261

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





262

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





263

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






264

(4)

H

OH

H

OH

= PresentSymmetric


(5)

H

H

OH

HO

i Present Present

C Present

2

SymmetricAchiral

optically inactive

180º

(6)

H

Cl

H

Cl

= PresentSymmetric


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





265

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





266

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.





267

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





268

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





269

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.





270

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





271

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.





272

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

*





273

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)





274

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)





275

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





276

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





277

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





278

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





279

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





280

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





281

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





282

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.





283

(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


CH3

(aa)Cis isomer

PresentAchiral

optically inactive

=

(ee)Cis isomer

(a, e)Trans isomer

= Absentchiraloptically active

= Absentchiraloptically active

CH3

CH3CH3 CH3

CH3


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



Basic Concepts in Organic UNIT-7 Chemistry and Stereochemistry · Chemistry (Basic Concepts in...

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