Isomerism

ISOMERISM INTRODUCTION The name was given by Berzilius. Two or more than two organic compounds having the same molecular formula and molecular weight but different physical and chemical properties are called isomers and the phenomenon is called isomerism. 2. Position Isomerism (PI) : The compounds which have same molecular formula, same functional group, same parent carbon chain but different position of functional group or multiple bond or substituents, shows position isomerism. Ex. But–1–ene But–2–ene Ex. 1–Butanol 2–Butanol STRUCTRAL ISOMERISM 1. Chain Isomerism (CI) : The compounds which have same molecular formula, same functional group, same position of functional group or multiple bond or substituent but different arrangement of carbon chain (different parent name of compound) shows chain isomerism. Ex. 1–Chlorobutane 2–Chlorobutane EXAMPLE OF CI AND PI Ex. 1–Butane (4C) (i) C 4H10 have two isomers : Both butane and isobutane are chain isomers. Ex. Butane 2–Methyl propane (3C) Ex. Butene (4C) 2–Methyl-1-propene (3C) Isobutane (ii) C5H12 have three isomers : All of three structures are chain isomers because only carbon chain (parent) is different. Ex. 1–Butanol (4C) Pentane 2–Methyl–1–propanol (3C) 2–Methyl butane 2,2–Dimethylpropane (iii) C6H14 has 5 isomers (a) (d) CH3CH2CH2CH2CH2CH3 Hexane 2,3–Dimethyl butane (b) (e) 2–Methyl pentane (c) 3–Methyl pentane (iv) C7H16 has 9 isomers  a–b, b–d, a–c, c–d Chain Isomers  2,2–Dimethyl butane b–c, d–e Position Isomers (v) C3H6Cl2 has 4 isomers : Position of chlorine atom is different in all the structure, so these are position Ex.19 Structures CH3—CH2—CH CH2 and Isomers. are :– 1. Ans. Chain Isomers Molecular formula C4H10 C5H12 C6H14 C7H16 C8H18 C9H20 C10H22 2. 3. 3. No. of Isomers 2 3 5 9 18 35 75 Ring chain isomerism (RCI) : Same molecular formula but different mode of linking (open chain or closed chain) of carbon atoms. 4. [open chain] [closed chain or ring] They have same molecular formula so they are Ring chain isomers. (vi) C5H11Cl has 8 isomers (vii) C8H10 has 4 aromotic isomers Ex.20 Relate a,b and c:– (o,m,p) (viii) C6H4X2  3 Aromatic isomers C6H4XY  3 Aromatic isomers C6H3X3 3 Aromatic isomers  C6H3XYZ  Ans. a–b  a–c , b–c  Functional Isomers Ring-chain Isomers, Functional Isomers Special points :  Alkenes with cycloalkane and alkynes (Alkadienes) with cycloalkenes show Ring-chain Isomerism.  Ring-chain Isomers are also Functional Isomers but priority is given to Ring-chain Isomers. 10 Aromatic isomers Ex.21 CH3—CH CH—CH3 and (1) Ring-chain Isomers (3) Functional Isomers are :– and Ex.23 (2) Chain Isomers (4) Position Isomers are ? Ans. (1) OR (1) Functional Isomers (3) Chain Isomers (2) Position Isomers (4) Metamerism Ans. (1) OR (1) Functional Isomers (3) Ring-chain Isomers (2) Position Isomers (4) 1 and 3 Both Ans. (4) Ans. Molecular formula same, Functional group same, position of Functional group same but different parent carbon atom chain so both are Chain isomers Ex.24 How many minimum carbons required for Chain isomerism and Position isomerism in alkanes ? Ans. 4, 6 Ex.25 How many minimum carbons required for Chain isomerism and Position isomerism in alkenes ? Ans. 4, 4 Ex.22 Relate structures a,b,c and d. (a) CH3—CH2—C CH (b) CH2 C CH—CH3 Ex.26 How many minimum carbons required for Chain isomerism and Position isomerism in alkynes ? Ans. 5, 4 4. (c) Functional Isomerism :- Same molecular formula but different functional groups. Following compounds show Functional isomerism, as they have same molecular formula and different functional group. (d) (i) Alcohol and ether a, b  Functional Isomers a, c.  Ring-chain Isomers and Functional Isomers b, c  Ring-chain Isomers and Functional Isomers  CH3—CH2—OH and CH3—O—CH3 (ii) Aldehydes and ketones  and (iii) Acids and ester  and a, d.  Ring-chain Isomers and Functional Isomers c, d.  Chain Isomers (iv) b, d.  Ring-chain Isomers and Functional Isomers CH3—CH2—CH2—CN and Cyanide and isocyanide  CH3—CH2—CH2—NC (v) Nitro and Nitrite  Ex. CH3—O—CH2—CH2—CH3 ; and CH3—CH2—O—N O 1–Methoxypropane 2–Methoxypropane (vi) Keto and enol   Both are metamers and Position isomer but priority is given to metamerism.  Metamers may be Position isomer or Chain isomer but priority is given to metamerism. Ex. CH3—CH2—NH—CH2—CH3 CH3—NH—CH2—CH2—CH3 and (vii) Amide and Oxime  and CH3—CH NOH (viii) 1°, 2°, 3° amines (i) CH3—CH2—CH2—NH2 N–Ethyl ethanamine N–Methyl propanamine (ii) CH3—NH—CH2—CH3 They are only metamers not CI Ex. (iii) (ix) CH3—NH—CH2—CH2—CH2—CH3 Alcoholic and phenolic compounds : Both are metamers and Chain isomer and Ex. (x) Alkyl halides do not show Functional isomerism. 2–Hexanone 5. Metamerism: Same molecular formula, same polyvalent Functional group but different alkyl groups attached to polyvalent Functional group. Polyvalent Functional group [More than one valency] are : —O— , —S—, , , , 3–Hexanone Both are metamers and Position isomer Ex. 2–Hexanone —NH—, Ex. , , CH3—O—CH2—CH2—CH3; 2–Methyl–3–pentanone Both are Metamers, Chain and Position isomers Both are metamers. Condition for Tautomerism : Ex.27 Structures and (a)For carbonyl compounds :- Carbonyl compounds having atleast one –H show tautomerism (i) 3 H, show tautomerism. (ii) 6 H, show tautomerism are Ans. 6. Both are metamers. Tautomerism or Desmotropism : Tautomerism was introduced by "Laar". It's also called desmotropism. (iii) 1 H, show tautomerism (iv) No H, No tautomerism (v) No H, No Tautomerism (vi) 3 H,  Desmotroism means bond turning. [Desmos = Bond ; Tropos = Turn]  Tautomers have same molecular formula but different structural formula due to wandering nature of active hydrogen between two atoms. Ex. shows tautomerism (Aceto phenone) (vii) No H, No tautomerism (Benzophenone) Ex. (viii) 2 H, shows tautomerism keto ene + ol = enol (ix) 4 H, shows tautomerism (x) H, attached sp2 carbon Note : (1) Tautomers are also F. I. and exist in dynamic equilibrium [ is used to show tautomerism] (2) By shifting of H–atom bond also change its position. does not take part in tautomerism (c) H—C N and H—N C are tautomers [also Functional isomers] while R—C N and R—N C are only Functional isomers. Ex.28 Which of the following show keto enol tautomerism. H—C N C N—H Active H (1) (2) and (d) H—O—N O are tautomers. ENOL CONTENT (3) Ans. 1. (4) "keto" (  99%) (1), (2) and (4) "enol" (  1%) 2. (b)For nitro compounds : Nitro compounds having atleast one – H show tautomerism Less stable or unstable  1%  99% 3. Nitro form Aci nitro form "keto" (  1%) (acidic form so soluble in base) Note : Nitro compounds with at least one –H are soluble in NaOH. "enol" (stable by resonance and aromatic nature) (  99%) 4. Aceto acetic ester (AAE) Ex. 29 Which of the following show tautomerism. (1) CH3CH2—NO2 (3) (CH3)3C—NO2 Ans. (2) (CH3)2CH—NO2 (4) Ph—CH2—NO2 1, 2 and 4 – In aqueous solution Ex.30 Which of the following do not soluble in NaOH. (1) CH3CH2—NO2 (3) (CH3)3C—NO2 (2) (CH3)2CH—NO2 (4) Ph—CH2—NO2 Ans. Only 3 – In liquid state  Keto from is stablised by intermolecular H–Bonding. Other Examples  Enol form is stabilised by intramolecular H–Bonding. KEY POINT (1) Enol content number of >C O group. (2) If number of >C O groups are equal then proportional to number of –H. (3) Group —OH attached to sp2 carbon or double bond is less stable or unstable. (4) More active H, more take part in tautomerism. (5) Stability of enol form depnds on (i) Resonance and (ii) H – Bond. Ex. 31 Arrange the following in correct order of enol content. 1. (1) C4H10O Isomers 7 [4 alcohol and 3 ethers] total 7 isomers are possible. Alcohol : CH3—CH2—CH2—CH2—OH and Ethers : CH3—O—CH2—CH2—CH3, C2H5—O—C 2H5, (2) C5H12O Isomers [8 alcohols and 6 ethers] total 14 isomers are possible. (i) CH3—O—CH2—CH2—CH2—CH3 (ii) 2. 3. (iii) 4. Ans. 1 < 2 < 3 < 4 (iv) C2H5—O—C3H7 Ex.32 Which have maximum stable enol form (v) (1) (vi) (2) (3) C4H10O (2 aldehyde and 1 ketone) total 3 isomers are possible. (i) (3) (ii) (4) Ans. 4. Stable by Resonance and is Aromatic (iii) (4) C4H8O2 (2 acids and 4 esters) total 6 isomers are possible. (i) and Ex.33 (ii) are (iii) (a) Position isomers (c) Geometrical isomers (b) Functional isomers (d) Chain isomers Ans. (b) Functional group different so Functional isomers. (iv) Note : If Functional group is changed then they will be never Chain, Position isomer and metamers. (v) Ex.34 CH3—S—CH2—CH3 and CH3—CH2—S—CH3 are – (vi) Ans. Identical (5) Aromatic isomers of C7H8O. and Ex.35 (a) (b) are – Ans. (c) (d) Identical STEREO ISOMERISM Two or more than two compounds having same molecular formula, same structural formula but different arrangements of atoms or groups in space. (e) (A) CONFIGURATIONAL ISOMERISM :  a, b  b, c – Functional isomers – Position isomers  c, d  a, d – Position isomers – Functional isomers  a, e – Functional isomers Note : Alcoholic and phenolic groups are Functional isomers. Stereo isomers which have following characteristics. 1. Stereo isomer which cannot interconvert in each other at room temperature due to restricted rotation known as Geometrical isomerism. 2. Stereo isomer which cannot super impose on each other due to chirality know as optical isomerism. GEOMETRICAL ISOMERISM (G. I) Alkenes ( >C C<), oximes (>C N—) and azo compounds [—N N—], show G. I. due to restricted rotation about double bond while cycloalkanes show G. I. due to restricted rotation about single bond. Not Geometrical isomers Not Geometrical isomers G. I. IN ALKENES Reason :- Restricted rotation about double bond : It is due to overlapping of p–orbital. Restricted Rotation Free Rotation EX.36 Which of the following show Geometrical isomerism – (1) 1,1–diphenyl–1–butene (2) 1,1–diphenyl–2–butene (3) 2,3–dimethyl–2–butene (4) 3-phenyl–1–butene Ans. (2) NOMENCLATURE SYSTEMS OF GEOMETRICAL ISOMERS (a) Cis–Trans System : If same groups at same side then cis and if same groups at different side then trans. [Same groups, same side] [Same groups different side] Ex. cis trans cis cis Ex. cis–2–butene Note : cis break. trans–2–butene trans is possible only when  bond Ex. Condition for Geometrical isomerism 1. Only those alkenes show G. I. in which "Each sp2 carbon have different atoms or groups" trans-2–pentene It does not show Geometrical isomers So no cis–trans Geometrical isomers Geometrical isomers Priority Rules : Properties of Cis–Trans Isomers : (i) Stability — cis < trans Rule I : Priority is proportional to atomic number of atom which is directly attached to sp2 carbon. [less repulsion between similar groups in trans] (ii) Dipole moment — cis > trans (iii) Polarity — cis > trans [more , more polarity) 'Z' (iv) Solubility — cis > trans [because cis > trans ] (v) Boiling point — cis > trans [ionic character] 'E' (vi) Melting point — cis < trans [more packing capacity of trans.] Rule II : If rule-I is failed then consider next atom Dipole moment [] : Ex.1  cis  0 trans  = Zero 'Z' decreasing order of atomic number Ex.2 trans  0 cis  0 Rule III :– If multiple bond is present then consider them as :- Ex.3 trans  0 Ex.37 If dipole moment of chlorobenzene is , then dipole moment of Ans. is – Zero (b) E – Z System: E (Entgegen) : When high priority groups are opposite side. Z (Zussaman) : When high priority groups are same side. 'E' 'Z' HP – High priority and LP – Low priority 'Z' Rule IV : If isotopes are present then consider atomic weight. 'Z' Ex.38 Which of the following show Geometrical isomerism – (1)CH3—CH2—CH N—OH (2)H2C N—OH (3) Ex.1 'Z' (4) Ex.2 'E' Ans. (1), (4) GEOMETRICAL ISOMERS IN Ex.3 AZO COMPOUNDS ( ) 'E' Ex. 4 'E' GEOMETRICAL ISOMERS IN OXIMES [>C = N]  Oximes show G. I. due to restricted rotation about double bond.  Only those oximes show Geometrical isomerism in which sp2 carbon have two different groups. [CH3—CH O + H2N—OH] CH3—CH N—OH (oxime) Ex. Acetaldoximes has two Geometrical isomers – syn anti When H and OH are on the same side When H and OH are on the opposite side Ex. Ph—CH N—OH Benzaldoxime Ph—N N—Ph (Azo benzene) GEOMETRICAL ISOMERS IN CYCLOALKANES Cycloalkanes show Geometrical isomers due to restricted rotation about single bond. Only those cyclo alkanes show Geometrical isomers in which atleast two different carbons have two different groups. Optical Isomers Ex.39 Which of the following show Geometrical isomerism – (1) (2) (3) (4) Ans. (1), (2) and (4) NUMBER OF GEOMETRICAL ISOMERS IN POLYENES R1—CH CH—CH CH ........ CH CH—R2 (a) If R1 R2 then number of Geometrical isomers = 2n [n = number of double bonds.] Ex. CH 3 —CH C H— CH C H— CH CH— CH2CH3 As n = 3 number of Geometrical isomers 3 =2 =8 (b) If R1 = R2 then number of Geometrical isomers = 2n – 1 + 2p – 1 where  and  (when n is even)  (n is odd) Ex. CH3—CH CH—CH CH—CH CH—CH3 [n = 3] Number of Geometrical isomers = 2 2 + 21 Introduction and Definitions (I) Optical activity & plane-polarised light : Ordinary light is an electromagnetic wave, which has oscillation in all the directions perpendicular to the path of propagation. When ordinary light is passed through Nicol prism it has all its oscillations in the same plane and is called plane-polarised light. Certain compounds rotate the plane of polarised light in a characteristic way when it is passed through their solutions. These compounds are referred to as optically active compounds. The angle of rotation can be measured by an instrument called polarimeter. (II) dextrorotatory compounds : If the substance rotates plane-polarised light to the right i.e. in clockwise direction it is called dextrorotatory & indicated by 'd' or (+). (III) laevorotatory compounds : If light is rotated towards left i.e. in anticlockwise direction the substance is said to be laevorotatory and indicated by ' ' or (–) 2n – 1 + 2p – 1 [ p =  ] =4+2=6 Special Point : CH2 CH—CH CH2 [1,3–butadiene] show Geometrical isomerism due to resonance. CH2 CH—CH CH2  . (IV) Specific rotation []. Specific rotation is the number of degrees of rotation observed if a 1-dm (10-cm) tube is used and the compound has concentration 1 gm/mL. Thus specific rotation [] is  [] t = C [] = Specific rotation  = observed angle of rotation (degree)  = Pathlength (dm) c = concentration (gm/ml)  = wavelength (nm) t = temperature (25°C) CAUSE OF OPTICAL ACTIVITY The foundation of modern theory of stereochemistry was laid by Louis Pasteur when he observed two different kind of crystals, which were mirror images of each other. In aqueous solutions of both types of crystals showed optical rotation that was equal in magnitude but opposite in direction. Pasteur believed that this difference in optical activity was associated with the three dimensional arrangement of atoms in the two types of crystals. Later van't Hoff and LeBel proposed that all the four valencies of carbon are directed towards the four corners of regular tetrahedron, and if all the four substituent attached to such a carbon are different the resulting molecule lack symmetry and such a molecule is referred to as asymmetric molecule and asymmetry of the molecule is responsible for optical activity in such organic compounds. (III) Asymmetric and dissymmetric compounds : A molecule which does not possess any element of symmetry (there are all 23 elements of symmetry) is called asymmetric. A molecule which does not possess plane of symmetry and centre of symmetry is called dissymmetric. (IV) Condition for optical activity : The minimum condition for a compound to show optical activity is molecular dissymmetry i.e. absence of plane of symmetry and centre of symmetry. (V) Chirality : OF MOLECULAR DISSYMMETRY/ A compound which is non-superimposable to its mirror image is called chiral while a compound which is superimposable to its mirror image is called achiral. All dissymmetric (or asymmetric) compounds are chiral & vice-versa. ASYMMETRY AND CHIRALITY. (VI) Chiral centre : ELEMENT OF SYMMETRY AND CONCEPT (I) Plane of symmetry : It is an imaginary plane which bisects the molecule in two equal halves in such a way that each half of the molecule is the mirror image of the other half. A compound in which a carbon is atttached with four different groups lacks symmetry and is called chiral carbon or asymmetric carbon. It is represented by C*. e.g. (a) CH3  CH  CH3 (2-chloropropane) : | Cl (II) Centre of symmetry : The centre of symmetry is defined as the point in a molecule through which if a straight line is drawn from any part of the molecule this line encounters identical groups at equal distances in opposite direction. I has no chiral centre since two groups (a & b) are identical. Its is superimposable on its mirror image II ( III). * (b) CH3  C H  C2H5 (2-chlorobutane) : Ans. H H | | * (A) CH3  C  C *  COOH | | OH OH | Cl H H | (B) HO H H H H | | | | * * |  C  C  C  C  C  C*  O | | | | | H * H * OH OH OH OH PROJECTION FORMULA IN OPTICAL ISOMERISM (I) Wedge-dash projection formula : I has one chiral centre it is asymmetric & it is not superimposable to its mirror image II ( III). The necessary condition for chirality is not just the presence of asymmetric carbon atoms but the asymmetry of the molecule as a whole. It is a convenient way of depicting three dimensional structure in two dimension. In this projection four bonds of a tetrahedral molecule is shown by two lines (in the plane) one wedge (up the plane) one dash (down the plane) Ex.40 Which of the following compounds has chiral carbon ? (A) CH3 – CH = CH – CH3 (B) CH3 – CHOH – CH2 – CH3 (C) (D) All of these Sol. The starred carbon is chiral carbon because all the four substituents are different CH3 – C* HOH – CH2 – CH3 Ex.41 Mark the asymmetric carbon atoms in the following compounds (A) CH3 – (CHOH)2 – COOH (B) HOCH2 – (CHOH)4 – CHO (II) Fischer projection formula : It is also a convenient way of depicting three dimensional structure in two dimension. Rules for writing Fischer projection formula : (i) The molecule is drawn in the form of cross (+) with the chiral carbon at the intersection of horizontal & vertical lines. (ii) On vertical line, main chain is taken with first carbon at the top. (iii) The horizontal lines represent the bonds directed towards the viewer and vertical lines away from the viewer e.g. CONFIGURATIONAL NOMENCLATURE * CH2  C H  CHO | | OH OH IN OPTICAL ISOMERS Relative configuration : the experimentally determined relationship between the configurations of two molecules, even though we may not know can be represented in two different Fischer projection the absolute configuration of either. Relative as configuration is expressed by D-L system. (a) (glyceraldehyde) Absolute configuration : The detailed stereochemical picture of a molecule, including how the atoms are arranged in space. Alternatively the (R) or (S) configuration at each chirality centre. (I) D - L System (Relative configuration) : * (b) CH3  C H  COOH (Alanine) | NH2 can be represented in two different Fisher projections as Ex.42 Draw the wedge–dash projection formula of (A) Ans. (B) (B) This method is used to relate the configuration of sugars and amino acids by the help of enantiomers of glyceraldehyde. The configuration of (+)glyceraldenyde has been assigned as D and the compounds with the same relative configuration are also assigned as D, & those with (-) glyceraldehyde are assigned as L. Examples : Sugars have several asymmetric carbons. A sugar whose highest numbered chiral centre (the penultimase carbon) has the same configuration as D - (+) - glyceraldehyde (– OH group on right side) is designated as a D-sugar, one whose highest numbered chiral centre has the same configuration Ex.43 (i) as L-glyceraldehyde is designated as an L-sugar. (ii)  e.g. (iii)  (iv) (II) R and S configurations in Fischer projection : (absolute configuration) Rule I : The priorities of groups which are attached with the asymmetric C-atom are assigned by sequence rule.  Ex.44 Assign Ror S configuration to each of the following compounds (A) (B) Rule II : The lowest priority group is brought to the bottom of Fischer projection by two or even simultaneous exchanges. Rule III : Then an arrow is drawn from first priority (C) group to second priority group to third priority group. If the arrow is clockwise the configuration assigned to the projection is R & If it is anticlockwise Ans. (A) R the configuration assigned is S. (B) R (C) R CONVERTING A WEDGE-DASH FORMULA INTO FISCHER Ex.45 2-Chlorobutane : PROJECTION FORMULA Step 1 : Decide the priority of groups by sequence rule. Step 2 : Bring the lowest prior group to dash by even simultaneous exchanges. Step 3 : Draw an arrow from first prior group to second prior group till third prior group. Ex.46 Chloroiodomethanesulfonic acid : Step 4 : If the direction of arrow is clockwise the configuration is R and if anticlockwise it is S. Step 5 : Draw the Fischer projection formula having equivalent configuration to the wedge-dash formula. Examples : Ex.47 Lactic acid : (a)  Here the lowest prior group is already on dash, there is no need for exchanges. (b) Ex.48 Alanine :   ENANTIOMERS Stereoisomers which are non-superimposable mirror Ex.49 images of each other are called enantiomers. e.g. in all the following molecules there is no centre of symmetry or plane of symmetry. Therefore their two mirror image isomers are non-superimposable and hence enantiomers. That means 72% is pure (+) 2-Butanol and 28% is (± mixture) PROPERTIES OF ENANTIOMERS Properties Remarks (1) Molecular formula Same (2) Structural formula Same (3) Stereochemical formula (str. formula with orientation) (4) Dipole moment (5) (6) Different Same Physical properties (m.p., b.p., density, solubility, refractive index etc.) Chemical properties (a) with optically inactive compound (b) with optically active compound Same Total (+) isomer = 72 + 14 = 86%, (–) isomer = 14% ENANTIOMERIC EXCESS To compute the enantiomeric excess of a mixture we calculate the excess of predominant enantiomer as a percentage of the entire mixture. The calculation of enantiomeric excess gives the same result as the calculation of optical purity.  optical purity = enantiomeric excess = |d  | × 100 d  = excess of one enantiomer over other × 100 entire mixture Same Different Thus for above example RACEMIC MIXTURE optical purity = enantiomeric excess = d –  = 72 % A mixture of equal amounts of enantiomers is called & d +  = 100 % a racemic mixture or racemic modification. A racemic so 2d = 172  d = 86% &  = 14 % modification is always optically inactive when (composition of mixture) enantiomers are mixed together, the rotation caused by a molecule of one enantiomer is exactly cancelled by an equal and opposite rotation caused by OPTICAL DIASTEREOMERS The optical isomers which are neither mirror image a molecule of its enantiomer. nor superimposable to each other are called The prefix (±) is used to specify the racemic nature diastereomers. Diastereomers have different physical of the particular sample. e.g. (±) Lactic acid, (±) and chemical properties and they can be easily Alanine. separated by physical methods. OPTICAL PURITY e.g Let us consider the stereoisomers of Sometimes we deal with mixtures that are neither 3-chlorobutan-2-ol optically pure nor racemic mixture. In these cases we specify the optical purity of the mixture. It is defined as the ratio of its rotation to the rotation of pure enantiomer.  Optical purity = observed optical rotation × 100 optical rotation of pure enantiomer e.g. If we have some 2-butanol with observed rotation of + 9.72, we compare this rotation with + 13.5 rotation of the pure (+) enantiomer. 9 . 72 optical purity = 13 . 5 × 100 = 72%. There are 4 stereoisomers of 3-chlorobutan-2-ol. NUMBER OF OPTICAL ISOMERS In which (I & II) & (III & IV) are enantiomeric (STEREOISOMERS) pairs. (I & III) or (I & IV) or (II & III) or (II & IV) all the isomers in each pair are neither mirror image nor superimposable to each other. Therefore these pairs are optical diastereomers. MESO COMPOUND A meso compound is one whose molecules are superimposable on their mirror images even though they contain chiral centres. A meso compound is optically inactive since it has a plane of symmetry. A meso compound is nonresolvable. e.g. Let us consider the stereoisomers of 2, 3-Butanediol e.g. Let us draw the total isomers of n = 3 (odd chiral centres with similar ends.) Total isomers = 23–1 = 22 = 4 In all the possible isomers I & II are enantiomers. But III & IV are not enantiomers since they have plane of symmetry they are superimposable to each other (all symmetrical compounds are superimposable to their mirror images). Thus III & IV are identical & meso compounds. Thus total stereoisomers of 2, 3-butanediol is 3. Two enantiomers and one meso isomer. REACTION OF CHIRAL MOLECULES WITH OPTICALLY ACTIVE REAGENT (OPTICAL RESOLUTION) Resolution refers to the method of separating a racemic mixture into its enantiomeric constituents. A racemic mixture is allowed to react with another optically pure compound. This changes a racemic mixture into a mixture of diastereomers which have different melting and boiling point and solubilities. These can be separated from one another by conventional method of separation of compounds. The separated diastereomers is then broken down to give pure enantiomers. Suppose a racemic mixture (±) A is to be separated. It is reacted with an optically OPTICALACTIVITY WITHOUT pure compound (+) B. Thus the schematic diagram ASYMMETRIC CARBON for resolution will be. (I) Case of allene : (a) Allenes with even  bonds : e.g. the orbital diagram of this structure will be Example : + + Diastereomers (Separable) Since the groups at the end of allene are in perpendicular plane, it will not show geometrical isomerism. The molecule lacks centre of symmetry as well as plane of symmetry. Overall the structure has molecular dissymmetry which is the sufficient condition for optical activity. The molecule will exist in two enantiomeric forms. (b) Allenes with odd  bonds : (IV) Case of ortho-ortho-tetrasubstituted biphenyls : e.g. the orbital diagram of this structure will be The groups at the end of allene structure lie in same plane (ZX plane). Therefore it will have a plane of symmetry (ZX plane). The molecules lacks molecular dissymmetry & it will not show optical activity hence optical isomerism. But the compound will exist in two geometrical diastereomeric forms. becomes non-planar at room temperature in order to have minimum electronic repulsion among the substituent. In this orientation (phenyl planes perpendicular to each other) the free rotation of C – C single bond is restricted and molecule shows optical activity due to molecular disymmetry. e.g. COMPOUNDS HAVING CHIRAL (II) Case of spiranes : A similar case like allenes is CENTRES OTHER THAN CARBON observed in spiranes. The spiranes with even rings The tetrahedral atom with four different groups and different groups at terminal carbons show optical attached to it is a stereocentre (chiral centres). activity & optical isomerism, while the spiranes with Thus when atom like nitrogen, phosphorous, sulphur, odd rings shows geometrical isomerism. silicon, germanium has four different groups attached (a) spiranes with even rings : with them they are chiral and show optical activity & hence optical isomerism. e.g. shows optical isomerism. (b) spiranes with odd rings : shows geometrical isomerism. (III) Case of cycloalkylidene : ASYMMETRIC NITROGEN AMINE INVERSION Amines with all different groups attached to N-atom have chiral centres (R1R2R3 ) as nitrogen atom. Since the geometry of the molecule is tetrahedral it has molecular dissymmetry. It will exist as two enantiomers but the two enantiomers of amines cannot be separated because they rapidly interconvert to each other. Therefore they always exist as racemic mixture. .. e.g. Ethylmethyl amines CH3 NHC 2H5 . Interconvertible ethylmethylamine enantiomers (non resolvable) CONFORMATIONAL ISOMERISM Newman projection : For conformational analysis, a special type of structural formula is convenient to use which is called newman projection formula and another type is a sawhorse formula. To write newman projection formula we imagine ourselves taking a view from one carbon atom directly along the selected bond axis to the next atom. The front carbon and its other bonds are represented Free rotation : as For defining free rotation let us consider the bonding in ethane. Dihedral angle : and those of the back carbon as The angle between C – X and C – Y in X – C – C – Y when it is visualised along C – C bond. We find that the -bond joining the carbon atoms is cylinderically symmetrical about the line joining the two carbon nuclei ; if the energy does not differ much in different arrangements the molecule can rotate about this carbon-carbon  bond, we describe this freedom to change by saying that there is free rotation about the carbon-carbon single bond. Conformations : Different arrangements of atoms that can be converted into one another by rotation about single bonds are called conformations. Conformational isomers : There are infinite arrangement (conformations) which arise due free rotation around carbon-carbon  bond, different conformations corresponding to energy minima are called conformational isomers. The conformational isomerism arises due to free rotation along a bond. Staggered, eclipsed and skew conformations : (I) The staggered conformation of a molecule is that conformation where the dihedral angle between the bonds at each atom of carbon-carbon bond is 180°. (II) In then eclipsed conformation the atoms bonded to carbons at each end of carbon-carbon bond are directly opposite to one another. The dihedral angle between them is 0°. e.g. (III) Skew conformation : All conformations other than staggered or eclipsed are skew conformations. Factors affecting stability of conformations : (I) Angle strain : Any atom tends to have bond angles that match those of its bonding orbitals. For example 109°.28 for sp3-hybridized carbon. Any deviation from "normal" bond angles are accompanied by angle strain. Energy profile of Eclipsed and Staggered forms of ethane (II) Torsional strain : Any pair of tetrahedral carbons attached to each other tend to have their bonds staggered for minimum repulsion between their bonds. Any deviation from the staggered conformation are accompanied by torsional strain. (III) van der Waals strain : Non bonded atoms or groups that just touch each other i.e. they are about as far apart as the sum of their van der Waals raddii, attract each other. If brought any closer together they repel each other. Such crowding together is accompanied by van der Waals strain. Conformational analysis of ethane : Ethane molecule contains a carbon-carbon  bond and each carbon is further attached to 3 H-atoms. It exists in two extreme conformations i.e. (i) eclipsed conformation (ii) staggered conformation The potential energy barrier between the two conformations of ethane is about 12.5 kJ/mol. The potential energy of ethane molecule is at a minimum for the staggered conformation, increase with rotation and reaches a maximum at the eclipsed conformation most ethane molecules, naturally exist in the most stable. staggered conformation. There are only three energy minima, that is ethane has only three conformers. Since they are indistinguishable they are degenerate. Conformational analysis of butane : If we consider rotation about the C2 - C3 bond of butane, we find that there are six important conformations shown as I - VI below : Energy profile of conforminations of butane : Strains and stability in n-butane dihedral angle () Torsional strain vander Waal strain I : Anti 180° Absent Absent II : Partially eclipsed 120° Maximum Present (between – CH3 & – H grps) Intermediate - 1 III : Gauche 60° Absent Present (between two – CH3 grps) Intermediate - 2 (> Intermediate1) 0° Maximum Maximum (between two – CH3 grps) Minimum Conformation of n-butane IV : Fully eclipsed The stability order will be : Anti > Gauche > Partially eclipsed > Fully eclipsed. n-Butane exists as three conformational isomers one anti (I) and two gauche (III & V). The gauche conformers III and V are mirror images of each other and hence are (conformation) enantiomers. Gauche conformations (III & V) and anti conformation ( ) are not mirror images of each other and hence are conformational diastereomer n-butane spends the greater part of its time as the anti conformer, and divides the smaller part equally between the two gauche conformers. As a result of the rapid inter conversion these isomers can't be separated. Stability Maximum Case of intramolecular hydrogen bonding : In case of G – CH2 – CH2 – OH, where G = – OH, – NH2 , – F, – NR2, – NO2, – COOH, – CHO the Gauche form is more stable than the anti form due to intramolecular hydrogen bonding i.e. stability : Gauche form > anti form.