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Isomerism and Stereo-Isomerism


  • Existence of one molecule in different forms, molecular formula remaining same.

Chain Isomerism

  • It arises due to variation in chain length, total no. Of carbons remaining the same. It is shown by only acyclic compounds. Methane, Ethane and Propane do not show Chain isomerism. Butane shows two chain isomers, normal Butane and iso-butane (IUPAC Name is 2-methylpropane.      


Effects of chain isomerism

  • The branched isomer has lesser boiling point than unbranched isomer. We can say that the more the branching, lesser is the boiling point. Because surface area decreases with branching and hence the Van der Waal’s force of attraction.

  • But in case of melting point, the case is just opposite. More is the branching, more is the melting point. Because symmetry increases with branching and hence the melting point.

  • More the branching, lesser the dipole moment.

Problem: Four organic compounds having dipole moment in debye are A: 0.03, B: 0.7, C: 1.1, D: .001. Arrange them in the decreasing order of melting points.

Ans: D > A > B > C

Melting point is directly related to branching and dipole moment is lowest for the most symmetrical structure. So, the one with lowest dipole moment will have the highest melting point and so on.

Note: Order will be opposite in case of boiling point.

Position Isomerism

  • Due to change in position of functional group along the same carbon chain.

  • Chain and position isomerism cannot take place together. If there is confusion of an isomer being a chain or position isomer, assign it as Chain. e.g. Below given isomers are chain isomers. First one will be called Butane nitrile and second one will be called 2-propane nitrile.

  • Principal functional groups like –CHO, -COOH, -CN, -NC, -COCl, -COOR, -CONH2,

-SO3H will always be at first position while considering isomerism.

  • Mono-substituted cyclic compounds do not show position isomerism. E.g.

Problem: Two compounds both containing –OH group and having molecular formula C7H8O are given. One gives violet-red coloration with Ferric Chloride solution but the other does not. Identify both of them.

Ans: A phenol gives red violet colour with ferric chloride. One which does not give positive Ferric Chloride test is and the one which gives is ortho/meta/para .

Functional group isomerism

  • In chain & position isomerism, only physical properties of molecules change whereas chemical properties usually remain same.

  • Chain in Functional group- Functional group isomerism.

        e.g. Alcohols & ethers

        Mol. Formula C2H6O can be CH3CH2OH or CH3OCH3

Problem: Two isomers (mol. Formula C2H6O), one has a boiling point of 78oC, gives positive Ceric ammonium nitrate test and gives hydrogen gas on reaction with sodium metal while the other has boiling point 11oC and does not react with Ceric ammonium nitrate or sodium  metal.

Ans: The first isomer is ethyl alcohol (ethanol) CH3CH2OH and the second is dimethyl ether (ethoxy ethane) CH3OCH3.

  • Ring-chain isomerism:- special case of functional group isomerism.

e.g. CH3CH=CH2   and are also functional group isomers.


  • Change in no. Of carbon atoms along divalent O, trivalent N and oxo group.

e.g. CH3CH2OCH2CH3 and CH3OCH2CH2CH3

  • Metamerism is special case of position isomerism.


  • Arises due to migration of acidic hydrogen from less electronegative carbon to more electronegative oxygen or nitrogen. E.g. keto-enol tautomerism

  • Tautomerism is a dynamic chemical equilibrium. We can apply Le Chatelier Principle.

  • Both different forms are called tautomers of each other. They exist as equilibrium mixture whose composition is fixed.

  • It is not a physical isomerism like other structural isomers.

  • As tautomerism is a chemical process, laws of chemical kinetics can be applied to it and rate constant can be calculated for tautomeric change.

  • Tautomerism is catalysed by acid or base.

  • Rate of tautomerism is higher in solvent which acts as both acid and base

  • We can say that tautomerism is special case of functional group isomerism.

Keto-enol tautomerism



Stereoisomers (stereomers) have the same bonding order of atoms but differ in the way these atoms are arranged in space. They are classified by their symmetry properties in terms of certain symmetry elements. The two most important elements are as follows:

1. A symmetry plane divides a molecule into equivalent halves. It is like a mirror placed so that half the molecule is a mirror image of the other half.

2. A center (point) of symmetry is a point in the center of a molecule to which a line can be drawn from any atom such that, when extended an equal distance past the center, the line meets another atom of the same kind.

A chiral stereoisomer is not superimposable on its mirror image. It does not possess a plane or center of symmetry. The nonsuperimposable mirror images are called enantiomers. A mixture of equal numbers of molecules of each enantiomer is a racemic form (racemate). The conversion of an enantiomer into a racemic form is called racemization. Resolution is the separation of a racemic form into individual enantiomers.

Chirality:- sp3 hybridised carbon having 4 different groups attached. A chiral carbon has the ability to rotate plane polarised light. Plane ploarised light is light vibrating in one plane.

e.g. The blue carbon is chiral carbon.

The molecule which rotates plane-polarised light to the right is called dextrorotatory. Similarly, the one which rotates to the left is called leavorotatory.

Stereomers which are not mirror images are called diastereomers. Molecules with a plane or center of symmetry have superimposable mirror images; they are achiral.

The specific rotation [α]Tλ is an inherent physical property of an enantiomer which, varies with the solvent used, temperature (T in °C), and wavelength of light used (λ).

It is defined as the observed rotation per unit length of light path, per unit concentration (for a solution) or density (for a pure liquid) of the enantiomer;

θ -observed rotation, in degrees

l - length of path, in decimeters (dm)

c -concentration or density, in g/mL_kg/dm3

and where, by convention, the unit deg·dm2/kg of [α]Tλ  is simply written degrees (°).

α is an additive/extensive property.

Problem: How could it be decided whether an observed dextrorotation of +60° is not actually a levorotation of -300°?

Ans: By halving the concentration or the tube length. That would halve the number of optically active molecules, and the new rotation would be +30° if the substance was dextrorotatory or -150° if it is levorotatory.

If one chiral centre is present then the molecule is always optically active.

Two optical isomers

If more than one chiral centre is present then the molecule may or may not be optically active. The compounds which are optically inactive in such cases are called meso compounds.

Case I: Compound containing ‘n’ chiral carbons in which first and last carbons are asymmetrically substituted.

Total no. Of optically active forms = 2n

No. Of enantiomeric pairs= 2n/2

No. Of meso compounds = 0

Case II: Compound containing ‘n’ chiral carbons in which first and last carbons are asymmetrically substituted.

Case II a: n=even

Total no. Of optically active forms = 2n-1

No. Of enantiomeric pairs= 2n-1/2

No. Of meso compounds = 2(n/2)-1

Case II b: n=odd

Total no. Of optically active forms = 2n-1- 2(n-1/2)

Total no. Of optically active forms = [ 2n-1- 2(n-1/2)] /2

No. Of enantiomeric pairs= 2n-1/2

No. Of meso compounds = 2(n-1/2)

Problem: optical activity disappears when (R)-2-chlorobutane is allowed to stand in aqueous H2SO4 and when (S)-2-iodooctane is treated with aqueous KI solution.

Ans: Optically active compounds become inactive if they lose their chirality because the chiral center no longer has four different groups, or if they undergo racemization. In the two reactions given, C remains chiral, and it must be concluded that in both reactions, racemization has occured.

DIASTEREOMERS: They are configurational isomers not related as mirror images.

Their properties are different.

For all practical purposes, they are two different compounds and they can be separated by any conventional method of separation. E.g.

Geometrical isomerism/Cis-Trans isomerism

Due to restricted rotation around C=C, C=N and cyclic compounds.

Cis-same group along the same plane

Trans- same group across the diagonal

The C atom around which restricted rotation is present, must have asymmetric substitution.


Molecules like will not show geometrical isomerism.

Syn-Anti isomerism

In case of oximes, lone pair of electrons is also counted in isomerism. A lone pair is given the least priority in Cann-Ingold-Prelog system of nomenclature. E.g.

Properties of Cis and trans isomer

  1. Boiling point of cis isomer is higher than that of trans while melting point of trans isomer is higher than that of cis.

  1. Trans isomer is stabler than cis because of lesser repulsion.

(3.)  Dipole moment of cis isomer is always higher than trans isomer. Dipole moment of     trans isomer is zero if the compound is of the type abC=Cba because it gets cancelled out.

        E.g. a) 1,1-dichloroethylene, (b) cis- and trans-1,2-dichloroethylene.

  1. Both cis and trans isomers are two different compounds for all practical purposes.


Due to rotation around C-C

Any compound having more than one carbon atom may be represented as conformations.

  • Dihederal angle: Angle between two planes that contain atoms Ha and Hb is called dihederal/torsional angle (θ).

  • The values of θ for the closest pairs of C-H bonds in the eclipsed and staggered conformations are 0º and 60º, respectively. There are an infinite number of conformations with energies between staggered and eclipsed conformations. All intermediate conformations are called skew; their θ values lie between 0º and 60º. For simplicity, we are concerned only with conformations at minimum and maximum energies.

Newmann Projection of conformations of ethane

Newmann Projection of conformations of n-butane

  • Staggered form is always more stable than eclipsed form in any case. So, its percentage will be more in any mixture.

Problem: 1,2-Dibromoethane has a zero dipole moment, whereas ethylene glycol, CH2OHCH2OH, has a measurable dipole moment.

Ans: 1,2-Dibromoethane exists in the anti form, so that the C-Br dipoles cancel and the net dipole moment is zero. When the glycol exists in the gauche form, intramolecular H-bonding occurs. Intramolecular H-bonding is a stabilizing effect which does not occur in the anti conformer.