Isomerism

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ORGANIC LECTURE NOTES



 

(LECTURE No. 5 TO 14

TOPIC : ISOMERISM

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Lecture 5 & 6

Structural Isomerism

(a) Classification of Isomers (b) Definition of Chain isomers

(a) Alkanes, Alkenes, Alkynes

(b) R – OH, RNH₂ (c) With terminal F.G. (d) Aromatic side chains

Position isomers (a) C – C, C = C, C  C,(b) – OH, – NH₂, – SO₃H (c) Disubstituted, Trisubstituted Aromatic Compound Functional Isomers (a) G.F. = C_nH_2n+2 O (– OH, – O –) (b) G.F. = C_nH_2nO (R₂C = O, RCHO) (c) G.F. = C_nH_2nO₂ (RCOOH, O || OR C R  ) (d) G.F. = C_nH_2n+3N (1°, 2°, 3° Amines) (e) Amides 1°, 2°, 3° Metamers (a) R – O – R (b) R – NH – R, R | R N R  (c) O || OR C R  , O || R NH C R   , O O || || R C O C R   

Lecture 7

Lecture 8

Lecture 9

Lecture 10

1. Structure Indentification (a) Monochlorination (b) Catalytic Hydrogenation of C = C; C  C 1. Ozonolysis: 2. Applications: 1. Stereoisomers (Classification) 2. Configurational isomers 3. Conformational isomers 4. Geometrical Isomerism

Lecture 11

Lecture 12

Lecture 13

1. Number of geometrical isomers

2. E/Z Nomenclature : 3. Physical properties of Geometrical isomers (i) Dipole moment () (ii) Boiling Point (iii) Melting Point: (iv) Water Solubility

1. Optical Isomerism 2. Symmetry of elements 3. Chirality (Dissymmetry)

and Optical Isomers 4. Optically Active

Car-bon Compounds 5. Projection Formula of Chiral Molecules 1. Configuration nomenclature in Optical Isomers 2. R/S Configura-tion in Fisher Projection Formula 3. Compounds with 2 Asym-metric Carbons of Similar Nature 1. Meso isomers 2. Properties of Optical Diastere-omers 3. Chemical method of separation (resolution) by using optically active reagent

4. Optically active com-pounds without a chiral

carbon:-5. Optical activity without asymmetric carbon (I) Case of allene (II) Case of spiranes : (III)Case of cycloalkylidene (IV) Case of ortho-ortho-tetrasubstituted biphe-nyls

Phase Coordinator (F Batch)

CDS Sir

Lecture 14

1. Conformations/conformers 2. Conformational isomers 3. Conformational energy 4. Dihedral angle of rotation 5. Strains

6. Ethane, propane, n-Butane 7. X-CH₂–CH₂–X type

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Isomerism Lecture Notes

Isomers:

Compounds with same general formula or molecular formula but different physical and chemical property.

Ex:- CH₃ – CH₂ – OH and CH₃ – O – CH₃

Ex:- CH₃–COOH and HCOOCH₃

Homologs:

Compounds with same general formula differing by same structural unit – CH₂ – or molecular weight by 14 unit.

Ex:- CH₃ – OH and CH₃ – CH₂ – OH

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homologs:-Structural isomers:

When two or more number of organic compounds have same molecular formula but different structural

formula these are called structural isomers.

Stereo isomers:

When two or more compounds have same Molecular Formual (M.F.) and same Structural Formula (S.F.) but have different stereochemical formula (S.C.F.), these are called stereoisomers.

Stereo Chemical Formula (S.C.F) :

It indicates different arrangements of atoms or groups in space around a stereo centre or it indicates different spatial orientations of atoms or groups around a stereo centre.

M.F. C₄H₈

S.F. (i) CH₃ – CH₂ – CH = CH₂ (ii) CH₃ – CH = CH – CH₃ (iii)

CH3

CH – C = CH3 2

–

These are structural isomers. For (ii), S.C.F. are

H C3 CH3 H H C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C and H C3 CH3 H H Stereocentre C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C C = C

These are stereoisomers.

Structural isomers are also known

as:-(i) Skeletal isomers (ii) Linkage isomers (iii) Constitutional isomers

Chain isomers :

They have different size of main carbon chain or side – chain (of C – atoms)

(i) The two chain isomers should have same nature of F.G./multiple bonds/substituents (except –R group) (ii) The position of F.G./M.B./substituent (locants) is not. considered here.

Note : (i) In almost all functional group chain isomer starts from four carbon atoms except in alkene, alkanone,

ether, ester, 2º amine and 3º amine.

Ex:-

Alkanes:-(a) C₁ – C₃ chain isomers not possible

(b) C₄H₁₀ H₃C – CH₂ – CH₂ – CH₃ and CH₃ – – CH₃

(c) C₅H₁₂ CH₃ – CH₂ – CH₂ – CH₂ – CH₃ , CH₃ – CH₂ – – CH₃ , CH₃ – – CH₃

(d) C₆H₁₄ (i) CH₃ – CH₂ – CH₂ – CH₂ – CH₂ – CH₃, (ii) CH₃ – CH – CH₂ – – CH₃,

(iii) CH₃ – CH₂ – – CH₂ – CH₃, (iv) CH₃ – – – CH₃,

(v) CH₃ – CH₂ – – CH₃

(i) and (ii) – Chain isomers (i) and (iii) – Chain isomers (i), (iv) – Chain isomers (i), (v) – Chain isomers (ii), (v) – Chain isomers (ii), (iii) – Position isomers (iv), (v) – Position isomers

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Ex. (i) C – C – C – C C – C– – C – C (ii) C – C – C C– – C C– – C – C, (iii) C – C C– – C – C – C C– – C

(i), (ii) – Chain (ii), (iii) – Position (i), (iii) – Chain

Ex. C₆H₁₂ (Cycloalkanes):-(1) , (2) – , (3) – – , (4) – – , (5) – – – , (6) – – , (7) –C – C – C , (8) –C – C –C , (9) –C – C – , (10) –– –, (11) –C – C

Ans. (3) and (4) – Positional isomer (3) and (6) – Positional isomer (4) and (6) – Positional isomer (5) and (10) – Positional isomer (All are chain isomers of 1) Remaining are chain isomers

Ex. and shows which types of isomerism ?

Ans. Chain Isomers

Q. Write chain isomers of N-alkanamine (1º) containing 4 carbon atoms ?

Ans. C – C – C – C – NH₂, C – C C–

– C – NH₂

Positional isomers:

They have different position of locants Functional group (F.G.) or Multiple Bond (M.B.) or substituents in the same skeleton of C-atoms.

Nature of F.G. or M.B. should not change. The skeleton of C-atom should not change.

Ex. (1) C – C – C – C C– – C , C – C – C C– – C – C (2) C = C – C – C , C – C = C – C (3) C  C – C – C – C , C – C  C – C – C

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(4) C – C –

OH–

C – C , C – C – C – C – OH

Q. Write all Positional isomers of Dichlorobenzene ?

Ans. Cl Cl – _– , Cl Cl – – , Cl Cl – –

Q. Write all Positional isomers of Dichlorocyclopropane ?

Ans. – – Cl Cl , –– Cl Cl

Q. Write all Positional isomers of Dichlorocyclobutane ?

Ans. – – Cl Cl , – – Cl Cl , – – Cl Cl

Q. Write all Positional isomers of Dichlorocyclopentane ?

Ans. – – Cl Cl , – – Cl Cl , – – Cl Cl

Functional isomers:

They have different nature of functional group (F.G.). The chain and positional isomerism is ignored (not considered).

Compound Functional isomer

C – C – C – C Nil

C – C – C = C (Ring-chain isomers are Functional isomers)

Ex. Compound Isomer Functional Remarks

(1) C – C – C  C (a) C = C – C = C Alkadiene (a), (b) are not functional isomers

Alkyne (b) Cycloalkene among themselve

(c) Bicyclo (2) C – C – C – OH C – C – O – C Alcohol Ethers (3) C – C – CH = O C – C O – C Aldehydes Ketones (4) C – C – COOH C – C O – O – C Carboxylic acids Esters

(5) C – C – C – NH₂ (a) C – C – . .NH – C 1º, 2º, 3º amines are functional isomers

1º amine 2º amine

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(6) C – C – C  N C – C – NC

Cyanide Isocyanide

(Propanenitrile) (Ethane isocyanide)

O

O–Peroxy bond (unstable)

does not exist at room temperature. C = C – O – C – OH Hemiacetal R – O – CH₂ – OH Unstable

Conclusion:- Following compounds don’t exist at room

temperature:-(i) – – C – C – C – C – C – C – C – C – C – C – C – C – C – C – C = – C – C – C – C – C – C – C – C – C – C – C – C – C – C – C – OH (ii) – C  C – OH (iii) – OH– – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – OH (iv) – OR– – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – OH (v) – OH– – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – – C – – –

C – O – C = C (vi) Any peroxy compound

Q. Write acyclic isomers of C₅H₁₂O

Ans. (i) C – C – C – C – C – OH, C – C – C –

OH– C – C, C – C – OH– C – C – C, C – C – C C– – C – OH, C – C C– – C – C – OH, C – C C– C– – C – OH, C – C – C C– C– – OH (7 alcohols) C – C – C – C – O – C, C – C – C C– – O – C, C – C C– – C – O – C, C – C C– C– – O – C, C – C – C – O – C – C, C – C C– – O – C – C (6 ethers). Q. G.F. n M.F. C_nH_{2n + 3}N (1) 3 C₃H₉N

1º Amine (R – NH₂) 2º Amine (R – NH – R) 3º Amine R – N – R

–

R

Q G.F. n M.F.

C_nH_{2n + 1}N (1) 2 C₂H₅N

2

1º Amine 2º Amine 3º Amine

(1) C = C – NH₂ C – C = N C – N = C

Ethenamine Ethanimine N-Methylene methanamine

(Unstable at room temperature)

N H H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – H C – CH2 – 2 – – Azine Q. G.F. n M.F. C_nH_{2n – 1}N 2 C₂H₃N

Ans. Cyanides and isocyanides

H₃C – C  N CH₂ = C = NH HC  C – NH₂

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Metamers:

When two isomers have same functional group (containing a hetero atom –O, N, S) but have different nature of alkyl or aryl (aromatic radical) group attached to hetero atom, then these are called metamers.

(a) Same functional group

(b) Chain or position isomerism is not considered. Following functional groups show metamerism

(a) R – O – R’ Ethers (All isomeric ethers are metamers) (b) R’ – NH – R 2º Amines (c) R’ – N R'' ––––––––––––––– – R 3º Amines (d) R – S – R’ Thioethers (e) R – C O – O – R’ Esters (f) R – C O – O – C O – R’ Anhydrides (g) R – S O O – O – R’ Sulphonate esters

Q. How many esters are possible for C₃H₆O₂ ?

Ans. O || H C O C C    and O || C O C C   are metamers

Q. 3º Amines of M.F. C₅H₁₃N (All metamers)

Ans. C – C – N C–C–C–C–C–C–C–C–C–C–C–C–C–C–C– – C – C C – C – C – N C–C–C–C–C–C–C–C–C–C–C–C–C–C–C– – C C – C C– – N C–C–C–C–C–C–C–C–C–C–C–C–C–C–C– – C Q. – O – C – C and – O – C – C

Show which type of isomerism ?

Ans. Metamerism

Q. Acyclic compound (A) contains 18 1º H atoms, two types of ‘C’ atoms. All H are identical. Identify (A).

Ans. C – C C– C– – C C– C– – C  C₈H₁₈

Q. Cyclic compound (P) contains 18 1º H atoms, two types of ‘C’ atoms. All H are identical. Identify (P).

Ans. – – – – C C C or

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Q. Write structure of smallest cyclic ester (S.F.)

O

C O_{Cyclic esters are known as lactones.}

Q. Write structure of smallest cyclic amide ?

O C CH2 NH is amide. O C CH N is imide

Q. Compound X is an ether. It has 12 1º H atoms. It has 2 types of H-atoms and 3 types of C-atoms. Write it Structural formula ? Ans. H₃C – O – – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 – – C – CH3 CH3 CH₃

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STRUCTURE INDENTIFICATION

(1) Calculation of Degree of Unsaturation

(DU):-(a) It is the hydrogen deficiency index (HDI) or Double Bond Equivalence (DBE)

(b) ) O DU ( CH – C H – C H₃ ₂ ₃     –2H –2HCH₃ – C  CH or CH₂ = C = CH₂ or

That means Deficieny of 2H is equivalent to 1 DU

(c) (i) 1DU = Presence of 1 Double Bond or Presence of 1 Ring closure

(ii) 2DU = Presence of 2 Double bond or 1 Triple bond or two ring closure or 1 double bond + 1 ring.

(d) G.F. D.U. (i) C_xH_y (x + 1) –       2 y (ii) C_xH_yO_z (x + 1) –        2 o y (iii) C_xH_yX_s (x + 1) –        2 s y (iv) C_xH_yN_w (x + 1) –       2 w – y (v) C_xH_yO_zX_sN_w (x + 1) –        2 w – s y

Ex Calculate DU of following compounds

(a) C₆H₆O DU = 4 (b) C₆H₅Cl DU = 4 (c) C₆Br₆ DU = 4 (d) C₅H₁₁OCl DU = 0 (e) C₉H₁₂N₂ DU = 5 (f) C₆N₆ DU = 10 (g) C₁₀H₈SO₅ N₄Cl₂ DU = 8 E.g. M.F. DU S.F. 1. C₄H₆ 2 C–C=C=C H H H H H H C – C – C  C C – C  C – C

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C C CH2 2. C₂H₂Cl₂ =1 Total isomers = 3 3. C₆H₆ClNO (Aromatic) = 4 Cl NH2 OH NHOH Cl

Note : If the aromatic Compounds have minimum D.E. ‘4’. That means at least 1 Benzene ring is present.

Q. C₁₆H₁₆ is symmetrical aromatic alkene. Draw all possible structure.

ANS. DU = 9

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Structure Identification of Organic Compounds by Chemical Reactions:

Monochlorination:-(a) (i) CH₄Cl2/h_CH 3Cl + HCl (ii) CH₃ – CH₃Cl2/SunlightCH₃ – CH₂Cl + HCl (iii) Cl2/h Cl – + HCl (iv) C – C C– C– – CCl2/hC – C C– C– – C – Cl + HCl (v) Cl2/h Cl – + HCl

Remarks:- When an alkane or a cycloalkane is treated with halogen (Cl₂, Br₂, F₂, I₂), a photochemical reaction takes place and a C – H bond cleaves and a C – Cl bond is formed. So, on H-atom is substituted by one halogen atom. This is known as monohalogenation reaction.

Application:- If a molecule has more than one type of H-atom, then on monochlorination, it forms a mixture of

monochloroisomers. All these isomers are position isomers.

Conclusion:- Hence, it can be concluded that the total no. of position isomers (structural) of monochloro compounds

is equal to the number of different types of H-atoms present in the reactant. The different type of H-atoms are also known as non-identical Hydrogens or non-equivalent Hydrogens or chemically different Hydrogens.

Ex. (a) C – C – CMonochlorination 2

(b) C – C – C – C Monochlorination 2 (c) C – C – C – C – C Monochlorination 3 (d) C – C C– – C – C Monochlorination 4 (e) CH3 –          Monochlorination 5

Q. X(C₅H₁₂)Cl2/honly one monochloro isomer..

Ans. X = Neopentane

Q. P(C₆H₁₄)Cl2/h_{Two monochloro}

How many isomers of P will give two monochloro compounds ?

Ans. C – C

C– – C

C–

– C only one isomers

Remark : In aromatic hydrocarbons, the hydrogen atoms of the side-chain are chlorinated, but H-atoms of Benzene

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Ex. – CH3      Cl2/h – CH Cl2

Q. X(C₈H₁₀) (Aromatic)Cl2/h_{Two mohochloro}

Y(C₈H₁₀) (Aromatic)Cl2/hOne monochloro

Ans. (X) (Y)

Catalytic Hydrogenation of C = C; C  C

General reaction:-(a) R – CH = CH – R + H₂ Ni R – CH₂ – CH₂ – R (b) R – C  C – R + 2H₂Ni/Pt/Pd_{R – CH} 2 – CH2 – R H2 ) isolated Not ( R – CH CH – R   H2 _{R – CH} 2 – CH2 – R (c) CH₂ = CH – CH – CH₃2H2/NiCH₃ – CH₂ – CH₂ – CH₃ (d) 3H2/Ni (e) – CH = CH2 e temperatur room Ni / H₂          – CH – CH2 3 H /Ni (100 – 150ºC) 2 – CH – CH2 3

(f)  [Reaction cannot be stopped at any intermediate stage]

Remarks:-(a) Alkenes, Alkynes, polyenes or polyynes can be hydrogenated by using catalysts Ni/Pt/Pd at room temperature. (b) All C – C  bonds(C = C, C  C) are hydrogenated. The reaction can’t be stopped at any intermediate stage.

Exceptions:-Aromatic  bonds which are stable at room temperature but can be hydrogenated at high temperature.

 The no. of moles of H₂ consumed by 1 mole of compounds is equal to the no. of  bonds presents.

 All positional isomers of alkenes or alkynes (due to multiple bond) always give same product on hydrogena-tion.

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 During catalytic hydrogenation, all monochlorination and no rearrangement in carbon skeleton takes place. Ex:- C = C – C = C + Cl₂ X C – – C – C C – Cl Ex. (1) 2H2/Ni (X)   h / Cl₂ Y + Z Y = C – Cl Z = Cl (2) CH3     H2/Ni CH3      Cl2/h 5 Monochloro product (3) CH3     H2/Ni CH3 5 Monochloro product

Q. X(C₄H₆)H2/NiYCl2/hZ(only one monochloro product) Identify X, Y, Z

Ans. DU = 2

X , Y , Z

Cl –

Q. Identify the lowest molecular weight alkane which gives four structural isomeric monochloro products ?

Ans. C –

C

C– – C – C

C₅H₁₂ = 72g

Q. Identify the structure of hexane which gives 3 monochloro products ?

Ans. C –

C C–

C–

– C – C, C – C – C – C – C – C

Q. Find the no. of monochloro products of a fully saturated isomer of C₄H₆.

Ans. DU = 2

 2 monochloro product

Q. Find the structural isomers of fully saturated cycloalkane of M.F. C₆H₁₂ which gives two monochloro product? Ans. – – – Q. ) H C ( A 18 8      Cl2/h ) type one Only ( Cl H C₈ ₁₇ Identify A ?

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Ans. C – C C– C– – C C– C– – C

Q. Write all isomeric alkynes which produce an isomer of heptane which on further monochlorination gives

(a) three monochloro products.

Ans. (i) C  C – – C – C C – C – C (ii) C  C – C C– C– – C – C (b) two monochloro Ans. Nil

Q. Find the structure of lowest molecular weight hydrocarbon and maximum unsaturation which on hydrogena-tion produce such an alkane which gives two monochloro products ?

Ans. C = C = C or C – C  C

Q. Determine the M.W. of maximum unsaturated hydrocarbon which on hydrogenation gives C₆H₁₂ which on further chlorination gives two monochloro.

Ans. CH2 CH2 CH2 C₆H₆ = 78g

Ozonolysis

:

It tells about position of unsaturation.

Remarks:-(1) Alkene and polyalkene on ozonolysis undergo oxidative cleavage. (2) (a) The reagent of reductive ozonolysis is

(i) O₃ (ozone) (ii) Zn and H₂O or Zn and CH₃COOH or (CH₃)₂S (b) The reagent of oxidative ozonolysis is O₃ and H₂O₂.

(3) The products are carbonyl compounds (aldehydes or ketones). This type of ozonolysis is known as reductive ozonolysis.

(4) Ozonolysis does not interfere with other F.G.s.

General Reaction:- R – CH = CH – R(1) O (2) Zn/H O 3 2 R – CH = O + O = CH – R + ZnO + H₂O Ex:- (1) CH₂ = CH₂(1) O (2) Zn/H O 3 2 CH₂ = O + CH₂ = O (2) CH₃ – CH₂ – CH = CH₂(1) O (2) Zn/H O 3 2 CH₃ – CH₂ – CH = O + O = CH₂ (3) CH₂ = CH – CH₂ – CH = CH – CH₃ O3/ Zn CH₂ = O + O = CH – CH₂ – CH = O + O = CH – CH₃ (4)



O /



3



Zn





_{+ OHC – CH} 2 – CHO (Propandial)

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Applications:

The process is used to determine the position of C = C in a molecule.

If the products are rejoined, the position of C = C can be determined in the reactant molecule. All C = C (except aromatic ones) undergo oxidative cleavage under normal conditions.

At higher temperature, the aromatic double bonds can also undergo ozonolysis.

(1) C C Zn 3

O



 



O = C – C – C O – C – C = O + O = CH₂ (2) Zn

)

(

O

₃









(3) CH = CH – CH3 – Zn / O e temperatur low 3







– CH = O + O = CH – CH₃ (4) – CH = CH – Zn 3

O



 



C₆H₅ – CH = O Q. ) isomer no ( Compound Single H C ) P ( 2 – n 2 n     H2/Ni ) Q ( H C_n ₂_n_₂ Cl2/h 3 m ) products m ( Cl H C_n ₂_n ₁   H O2 2 O (oxidative)3 HCOOH + CH₃ – – C – CH3 – CH3 COOH Identify P ? Ans. P =

Q. An unsaturated hydrocarbon on ozonolysis produces 1 mole of , 1 mol CO₂, 1 mol

Find the structure of the hydrocarbon and the no. of monochloro products formed by hydrogenation.

Ans. , 5 monochloro product

Q. ) n hydrocarbo . Unsat ( X     H2/Ni – –      h Cl₂ ) 3 m ( products m  O (Zn/H O)3 2 O O identify structure of X ?

(18)

Q. .) C . H . Unsat ( X H2/Ni Cl2/h ) 7 m ( products m  O (Zn/H O)3 2

HCHO + 4(1-oxoethyl) Cyclohexan-1-one. Identify X ? Ans. X  C C = C – C– Q. Identify structure of X ? Ans. X is Q. X H2 C – C – C – C – C – C O (Zn)3

CH₃CHO + CHO – CHO Identify structure of X ? Ans. X is C – C = C – C = C – C Q. Q. CH3 – – CH3 O H , Zn O 2 3     2 + Q. CH3 – – CH3 Zn,H O O 2 3     O O CH3 C – C CH3 – – + 2

(19)

Stereoisomers (Classification)

Steriosomers :

The stereoisomers has different orientation of groups along a stereo centre. A stereocentre can be C = C (any double bond), a ring structure, asymmetric carbon atom (*C_abcd). These isomer has same general formula, structural formula and molecular formula but different stereochemical formula.

E.g. CH3 C = C H CH3 H (I) CH3 C = C H H CH3 (II) CH₃–CH₂–CH=CH₂ (III) I, III  Positional

II, III  Positional I, II  Stereoisomer

Configurational isomers :

Configurational isomerism arises due to different orientations along a stereocentre and these isomers can be seperated and these isomers do not convert into one-another at room temperature. Therefore, they are true isomers. They can separated by physical and chemical method.

E.g. Cis–2–Butene Trans-2-Butene

Conformational isomers :

When different optical orientations arise due to the free-rotation along a sigma covalent bond. Such isomers are called conformational isomers.

E.g. eclipsed ethane and staggered ethane

These change into each other at room temperature and can never be isolated. So these are not considered as true isomers.

Geometrical Isomerism

1. Cause of Geometrical isomerism :

Geometrical isomerism arises due to the presence of a double bond or a ring structure C = C, C = N, N = N, or Ring structure (Stereo centres)

Due to the rigidity of double bond or the ring structure to rotate at the room temperature the molecule exist in two or more orientations.

This rigidity to rotation is described as restricted rotation/hindered Rotation/No rotation.

E.g. C = C a b a b ( )I

The root form of geometrical isomers lie in restricted rotation.

a a

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Condition

(i) The two groups at each end of restricted bond must be different. C = C 1 2 3 4 C_aa C_aa X C_aa C_bd X C_ae C_b2 X C_ab C_ab



C_ab . C_bd



C_ab C_de



(ii) In two geometrical isomer the distance between two particular groups one the ends of the restricted bond must be changed.

Q. Which of the following compounds show G.I.

(1) Ethene (2) Propene (3) 2-Methylbut-2-ene 3 3 3 CH | CH CH C CH    (4) But-2-ene CH3 – CH = CH – CH3

(5) Penta-1, 3-diene C = C – C = C – C (6) 1, 2-Dideuteroethene

(7) Phenyle thene

H C = CH2

(8) Buta-1, 3-Diene C = C – C = C

Ans. 4, 5, 6

1. Geometrical isomerism across

– Nil

and By E / Z

2. Geometrical isomerism across

(a) Imine ( )

Imine compounds are produced from carbonyl compounds on reaction with ammonia.

(Syn and anti)

(21)

Q. Which of the following compounds show geometrical isomerism after reaction with NH₃. (a) CH₃ C CH₃ || O   (b) H C H || O   (c) H C D || O   (d) CH C H || O 3  (e) O || CH C Ph  ₃ (f) O || Ph C Ph  (g) (h) (i) Ans. c, d, e, g, i (b) Oximes _{C = N – OH} :

They are prepared by reacting carbonyl compuond with hydroxyl amine (NH₂ – OH)

C = O + H N–OH2 R H     H2O C = N–OH R H (Aldoxime) C = N R H OH ( )I (syn) and C = N R H OH ( )II (anti)

Syn and anti in aldehyde only not for ketones.

* Except formaldehyde (CH₂O) All other aldehyde form two oximes. * Unsymmetrical ketoes form two oximes.

Eg. Which of the following ketones will form two oximes. (1) Propanone C – C – C = O (2) Butanone _{C – C – C – C = O} H (3) 3-Pentanone C – C – C – C – C O (4) Acetophenone _{C H –C–CH} 6 5 3 O (5) Benzophenone _{C H –C–C H} 6 5 6 5 O (6) Cyclo hexane O (7) O Me

(22)

Q. The lowest molecular weight of acyclic ketone and its next homologue are mixed with excess of NH₂ – OH to react. How many oximes are formed after the reaction ?

Ans. 3 (c) Hydrazones C = N – NH2 : C = O + H N.NH2 2 R H hydrazine  H2O _{C = N–NH} 2 R H C = N R H NH2 ( )I C = N R H NH2 ( )II + (Geo. diastereomers)

Q. Two chain isomer of a cycloalkanone which are next higher homologue of lowest molecular weight, cycloalkanone reacted with hydrazine. Identify the structure and number of isomer of hydrazones prepared ?

Ans. O CH3 + O + H₂N – NH₂ N .. NH2₊ N CH3 .. NH2 + N CH3 .. NH2 Number of isomer = 3

(3) Geometrical isomerism across azo compounds (– N = N – )

(i) H – N = N – H (H₂N₂) (ii) Ph₂ N₂ (Azobenzene) N = N Ph Ph syn .. .. N = N Ph anti Ph .. ..

(4) Geometrical isomerism across ring structure

(i)

Restricted rotation

(ii)

(23)

(5) Geometrical isomerism in cycloalkenes across double bonds :

In cycloalkenes, G.I. exists across double bonds with ring size equal to or greater then 8 carbon atoms (due to ring strain)

Stereocentre

An atom or bond across which stereoisomerism exists (either G.I. or optical isomerism)

These compounds which are not mirror images of each other are diastereomers. These compounds which are non-superimponsible mirror-images of each other are enantiomers

are enantiomers

Number of geometrical isomers

Case I : Compounds having dissimilar ends

No. of G.I. = 2n

where n = number of stereocentre

Ex.

Ans. Stereocentre = 2

Geometrical isomerism = 4

(a) (b)

(c) (d)

Case II : Compounds with similar ends even no. of stereocentre.

No. of Geometrical isomerism = 2n–1₊ 2 1 n 2 

Ex.1 CH₃ – CH = CH – CH = CH – CH₃

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Ex.2 Cl – CH – CH CH – CH

Geometrical isomerism = 23_{+ 2}1_{= 10.}

Case III : Compounds with similar ends but odd number of stereocentre.

No. of Geometrical isomerism = 2n–1₊ 2 1 n 2  Ex.1 CH – CH = CH – CH = CH – CH = CH – CH₃ Ans. Stereocentre = 3 Geometrical isomerism = 6

E/Z Nomenclature :

Z (Zussamen = together) a > b and e > d

E (Entagegen = opposite) a > b and e > d

Rules :(i) The group with the first atom having higher atomic number is senior.

Thus – F > – CH > – NH₂ > – CH₃

(ii) If the first atom is identical, then second atom is observed for deciding the seniority of the group.

(a) (b) < (c) < (d) > Ex. Z Ex. E

(iii) If the first atom has same atomic number but different atomic mass, that is isotopes. Then heavier isotope has higher seniority.

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(iv) If the group has unsaturation, then a hypothetical hypothetical equivalent in drawn for it and it is com-pared with other group for seniority.

(1)

(2) – CH = CH₂ < – C  CH

Remark : Bond pair is always senior to lone pair.

Ex. (1) C == C CH = CH2 CH – CH = CH2 C(CH )3 3 HC C E (2) Z (3) H C = C = CH2 C == C C CH CH – C(CH )2 3 3 N C E (3) Z (4) E (5) E

(26)

Physical properties of Geometrical isomers :

The physical properties of organic compounds can be compared through the knowledge of their molecular formula, structural formula and stereochemical formula (S.C.F.)

(1) Dipole moment () : Polarity of an organic molecule.

The organic molecules are composed of covalent bonds, every covalent bond has its own dipole moment which is known as bond moment.

It is expressed by given formula :  = q × d

Where q is partial charge on individual atoms A and B whose value increases with increase in E.N. difference between A and B. and d is the internuclear separation.

If  = zero, E.N. difference is zero or very less and bond is non-polar.

Ex. (1) H – F > H – Cl > H – Br > H – I () (2) C – C < C – N > C – O < C – F ()

Dipole moment of the molecule :

It is the vector sum of the individual bond moments.

If the vector sum of individual bond moments becomes zero, then the molecule as a whole is said to be non-polar.

Ex. (i) O = C = O Non-Polar

(ii) S = C = S Non-Polar (iii) Polar (iv) Polar (v) Polar Note. CH₃ – Cl > CH₃ – F > CH₃ – Br > CH₃ – I because M = q × d F – CH Cl CH3 3 q q  _but _CH_–_Cl _CH _–_F 3 3 d d  Thus, CH₃ – Cl > CH₃ – F But, H – F > H – Cl.

Dipole moment of some Functional Groups : (order)

> – COX > – COOH > – OH > – NH₂ > – R – X > R – O – R

(2.8 – 3 D) (X = Cl, Br, I)

> R – C  C – H > – R – CH = CH₂ > R – CH₂ – CH₃ (non polar) (non polar)

(27)

Dipole moment for Geometrical isomerism :

(A)

(cis) (trans)

Thus Alkene C_ab = C_ab type,  = cis > trans (= O)

(B) Unsymmetrical Alkene (C_ab = C_ad )

(i) <

(ii) <

(iii) CH₃ – CH = CH – COOH (Trans > Cis)

(iv) >

Ortho > Meta > Para

If two isomers have different dipole moments then their all physical properties are different. (melting point, boiling point, solubility, refractive index, density)

All structural isomers have different , and thus different physical properties.

In case of stereoisomers, all diastereoisomers have different dipole moments and thus different physical properties.

All geometrical isomers are diastereoisomers and thus posses different diople moment and Physical prop-erties.

Separation of isomers : The isomers having different physical properties can be separated by normal

physical methods of separation. (2) Boiling Point :

Intermolecular forces of attraction :

H– bond > Dipole – Dipole interaction > Vander-wall force.

The boiling point of a compound depends on the intermolecular forces of attraction existing between two molecule. If a compound has strong forces of attraction, then the B.P. is high.

Among homologs, molecular weight increases, Vander Wall force increases, Boiling point also increases

Ex. (a) CH₄ < C₂H₆ < C₃H₈

(28)

Among structural isomers

(1) Functional isomers (H-bond > Dipole – Dipole > Vander Wall force)

–COOH > – OH > – NH₂ > > R – X > ROR > R – C  CH > R – CH = CH₂ > R – CH₂ – CH₃ (a) Among –COOH, –OH, –NH₂ –––––– H-bond

(b) – C– = O, R – X, R – OR, R – C  CH –––––– Dipole – Dipole (c) R – CH = CH₂, R – CH₂ – CH₃ –––––– Vander Wall force

In acids, there is dimer formation

O – H --- O O --- H – O

–

R – C C – R _dimer

In isomeric ketones and aldehydes, BP_Ketone > B.P.._Aldehyde

In chain isomers, as branching  , surface area of molecule  VDW forces  Boiling Point  Thus, BP  1º > 2º > 3º (–OH, –NH₂) Eg:- (i) C₃H₈O C – C – C – OH > C – OH– C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – C (ii) C₄H₁₀O C – C – C – C – OH > C – C C– – C – OH > C – C – OH– C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – C > C – C – OH– C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – C

Among stereoisomers:-In geometrical isomer, the dipole moment of geometrical isomer is always different.

More polar geometrical isomer has higher B.P. (a) C_ab = C_ab B.P._cis > B.P._trans

(b) R – CH = CH – X B.P._trans > B.P._cis

(c)

COOH COOH

C = C Maleic acid (Cis)

H H

>

COOH HOOC

C = C Fumaric acid (trans)

H

H C = C Fumaric acid (trans)

H

H H (d) CHO CH3 – + C = C H H > CHO CH3 C = C H H C = C H H C = C H H C = C H H C = C H H C = C H H C = C H H C = C H H C = C H H C = C H H C = C H H C = C H H C = C H H C = C H H C = C H H (e) NH2 CH3 C = C H H > NH2 CH3 C = C H H

Among Disubstituted Aromatic

Compounds:-(i) C₆H₄Cl₂

ortho B.P. > meta B.P. > para B.P. (ii) C₆H₄(OH)₂

ortho B.P. < meta B.P. < para B.P. (due to H-bonding)

O – H O – H Ortho 5-membered Intramolecular H-bonding OH OH Less Intermolecular H-bonding – – OH OH More Intermolecular H-bonding – –

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(3) Melting Point:

The melting point of an organic compound depends upon its packing in solid state. The packing depends on following

factors:-(i) Intermolecular forces of attraction:- (H-bond > Dipole – Dipole > Vander Wall force) in structural

isomer.

(ii) Symmetry of the molecule:- More symmetrical molecules are more closely packed.

In geometrical isomers the trans isomers are more symmetrical than cis isomers (C_ab = C_ab type of alkenes) So trans form have higher M.P. than cis isomers.

Eg:- (I) (a) CH₂ = CH₂ < CH₃ – CH = CH₂ < CH₃ – CH₂ – CH = CH₂

Homologues: Molecular weight increases vander wall force increases melting point increases.

(b) CH₂ = O < CH₃ – CH = O < CH₃ – CH₂ – CH = O (c) < CH3 – < Et – <

(II) Structural (a) Functional

isomers:-(i) R – COOH > H – C

O

– O – R

(H-bonding) (No H-bonding)

H-attached to ‘O’ H with ‘R’

(ii) CH₃ – CH₂ – CH₂ – CH₂ – OH > CH₃ – CH₂ – O – CH₂ – CH₃

(H-bond) (No H-bond)

(iii) CH₃ – CH₂ – CH₂ – CH₂ – NH₂ > CH₃ – CH₂ – CH₂ – NH – CH₃ > CH₃ – CH₂ – N(CH₃)₂ (1º) (2º) (3º) (b) Chain isomers:-(i) C – C – C – C > C – C C– – C **Exception :- (ii) CH₃ – 3 3 3 CH | CH C | CH  > CH₃ – CH₂ – CH₂ – CH₂ – CH₃ > CH₃ – 3 CH | CH – CH₂ – CH₃ (M.P.).)

Neopentane (spherical) (Close-packing)

For B.P.:- n > iso > neo For M.P.:- neo > n > iso (III) Geometrical

isomers:-C_ab = C_ab type b a

.

* C = C a b C = C a b C = C a b C = C a b C = C a b C = C a b C = C a b C = C a b C = C a b C = C a b C = C a b C = C a b C = C a b C = C a b C = C a b b a a C = C b C = C b C = C b C = C b C = C b C = C b C = C b C = C b C = C b C = C b C = C b C = C b C = C b C = C b C = C b

Elements of symmetry:- 2 Planes of Symmetry 1 Plane of Symmetry

It has a centre of symmetry (). No centre of symmetry. The symmetry of a molecule can be described by observing symmetry elements.

Plane of Symmetry (P.S.) :- It is defined as any plane in the molecule which bisect the molecule into 2

(30)

Centre of Symmetry (C.S.) :- If we start from one point of the molecule, reach to centre and then go to

equal and opposite distance (in same direction), then if similar point is obtained, then the molecule has a C.S.

(It should be applicable to all the points of the molecule).

The trans isomers has more no, of symmetry elements than cis isomer; so it is more closely packed in solid state and has higher M.P.

Melting point in geometrical isomers :

Ex.:- (1) H H C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl > H H C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl C = C Cl Cl (2) > Me H Cl C = C Cl C = C Cl C = C Cl C = C Cl C = C Cl C = C Cl C = C Cl C = C Cl C = C Cl C = C Cl C = C Cl C = C Cl C = C Cl C = C Cl C = C Cl

(3) Maleic acid < Fumaric acid

(4) Stilbene (Ph – CH = CH – Ph) E > Z

(IV) ortho, meta, para isomers

(1) _Cl– (p) Most symmetric Cl – – (m) less polar Cl – Cl – (o) more polar Cl – _Cl

para > ortho > meta (we see polarity now, if H-bonding doesn’t exist)

(2) – (p) Symmetrical H-bond (intermolecular) OH – OH – (m) Intermolecular H-bond OH – OH – (o) Intramolecular H-bond OH – _OH

Thus, para > meta > ortho (if H-bonding) (4) Water Solubility :

Criteria:- H-bond > Dipole – dipole > Vander Wall forces (insoluble) between solute and solvent

(I)

Homologs:- Hydrophilic –––––– water loving –––––– H-bond

 Hydrophobic –––––– water repelling –––––– Hydrocarbon

_{As M.W..  , solubility  .} Ex:- (i) CH₃OH > C₂H₅OH > C₃H₇OH > C₄H₉OH (ii) OH – > OH – Me >

(iii) Formic acid > Acetic acid > Propanoic Acid > Butanoic acid

(II) Structural

(1) Functional

isomers:-(a) –COOH > –NH > –OH H-bond

2 > > R – CH = CH > R – CH – CH

insoluble (VDW)

(31)

(b)

(c) R – COOH > HCOOR (d) R – CHO > R₂’C = O

(2) Chain isomers (Effect of

Branching):-As branching (in Hydrocarbon)  , Surface Area of hydrocarbon  so, water solubility  (as area repelling water decreases) Degree:- 3º > 2º > 1º (–OH) Ex:- (i) M.F. = C₃H₈O C – C – C – OH < C – OH– C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – C...Solubility C – C – C – OH > C – OH– C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – C...B.P.. (ii) C₄H₁₀O C – C – C – C – OH < C – C C– – C – OH < C – C – OH– C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – C < C – C – OH– C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – – C – C (iii) C₄H₈O₂ (acid) CH₃ – CH₂ – CH₂ – C O – O – H < CH₃ – CH – CH | CH₃ – C O – O – H

(III) Geometrical Isomers:- More Polar geometrical isomer is more soluble in water.

(i) > Cis > Trans

(ii) < Trans > Cis

(iii) H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH > H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH C = C H COOH Cis > Trans

Optical Isomerism

Some organic compounds can rotate the plane of plane-polarised light. Such compounds are called optically active compounds.

The optically active compound can show optical isomerism.

Measurement of optical activity

It is measured by an instrument called polarimeter.

(1)

=

(3) (5) (6)

(8) Recorder

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(1)Source of light

(2)Polychromatic Non-polarised light (3)Slit (Monochromator)

(4)Monochromatic Non-polarised light (5)Polariser (Prism-setting)

(6)Monochromatic plane-polarised light (7)Sample tube ( = path length)

(8)Recorder for measurement of optical rotation

Observations:- When plane polarised light is passed through sample tube, then following changes can be observed

in the

Angle of rotation () Inference

(1)  = 0º Compound is optically inactive.

(2)  = +xº (clockwise rotation) Optically active, dextrorotatory or d or (+). (3)  = –xº (anticlockwise rotation) Optically active, laevorotatory or l or (–).

Specific Rotation:- The specific rotation of a compound indicates the optical rotation of unit concentration (1 g/mL)

present in a sample tube of 1 dm of path-length at given temperature and given wavelength of light. It is represented as

follows:-









__

c

]

[

t 25₅₈₀ºC_nm

where  = observed angle of rotation

c = concentration in g/mL (or density)

 = Path length of Sample tube in dm

Q. Compound ‘X’ has  = +70º for 2 g/mL solution in sample tube of  = 1 dm. Calculate  ?

Ans.  =

2 70

= +35º

(1) Its concentration is made twice, the observed rotation() will be 140º but  = 35º

(2) If  = +70º, it will be a

(i) d compound or (ii)  compound of  = –(360º – 70º) = –290º

It can be decided by changing the concentration or by changing the length of the tube (c or ).

If concentration is reduced to half, d will have +35º and  will have –145º (not distinguish) still halfed, it will give +17.5º and –72.5º to distinguish.

( ) With symmetrical molecule [ ] = 0 I  ( ) With asymmetric molecule [ ] 0 II  Optically inactive

compounds:-These compounds have symmetrical molecule, so the optical rotation observed after interaction of light is zero.

But in case of optically active compounds. The molecules are asymmetric in nature and show non-zero optical rotation. Symmetry of elements :-(1) Centre of symmetry (C.S.) (2) Plane of symmetry (P.S.) Ex:-R C = C H R H C = C H R H C = C H R H C = C H R H C = C H R H C = C H R H C = C H R H C = C H R H C = C H R H C = C H R H C = C H R H C = C H R H C = C H R H C = C H R H C = C H R H

Centre of symmetry _{Centre of symmetry}Cl

Cl

(33)

Dissymmetry:- The molecules or objects in which minimum two elements of symmetry (Centre of symmetry and

Plane of symmetry) are absent are called dissymetric molecules/objects.

Asymmetry:- When all the elements of symmetry (23 including Centre of symmetry and Plane of symmetry are

absent) the molecule is said to be asymmetry.

So, all asymmetric molecules are always dissymmetry, but the reverse is not true.

Dissymmetry and chirality:- All dissymmetric molecules are chiral and are optically active.

Dissymmetry, Chirality and Optical Activity:- Dissymmetry/Chirality is the minimum and sufficient condition for a

molecule to be optically active. So, if in a molecule, a C.S. and P.S. are absent, the molecule will be optically active. All asymmetric molecules are also optically active.

Chirality (Hand-like property or Handedness):- The human hand does not have Centre of symmetry and Plane

of symmetry , so it is dissymmetry. It does not have any element of symmetry, so it can be called asymmetry. Any molecule which has this property is called chiral.

Note : All dissymmetric compounds are chiral.

Optical

Isomers:-Enantiomers:- The mirror-image stereoisomers are called enantiomers. Two

types:-(a) Dextrorotatory (b) Laevorotatory  Enantiomers are always mirror image isomers.

 They have same molecular formula, structural formula but have different orientation in space.  They have dissymmetry/chirality.

 Every enantiomer is optically active.

 The two enantiomers can rotate the plane-polarised light with equal magnitude and opposite signs.  They have similar physical properties except the sign of optical rotation.

 These are the isomers which have maximum resemblance with each other.  These can be distinguishedonly by polarimeter.

Chirality (Dissymmetry) and Optical

Isomers:-(a) The dissymmetric molecules have two orientations in space. These two orientations are called stereoisomers, optical isomers, enantiomers.

(b) These are mirror images (enantiomers).

(c) Non-superimposability of enantiomers:- The dissymmetric molecules are always non-superimposable

on their mirror-image orientations.

(d) The non-superimposable orientations are non-identical orientations, so are called isomers.

(e) Because of dissymmetry, these two isomers are capable of rotating plane-polarised light. So, these are called optical isomers.

Optically Active Carbon

Compounds:-If a carbon-atom is attached with four different group, then if does not have any element of symmetry. It is known as asymmetric carbon atom, which is represented as *C_abde.

a

b d

C* e

If a molecule contains only one asymmetric carbon atom, then the molecule as a whole becomes chiral and optically active and show optical isomers.

Centre of symmetric Plane of symmetric Optical active

(1) H H _H C H Absent Yes No H

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(3) H Br H C Cl Absent Yes No (4) H Br F C Cl Absent No Yes

Ex.1:- Mark the chiral

objects:-(i) Cup (ii) Plate (iii) Letter A (iv) Letter G (v) Fan (vi) Door (vii) Shoe (viii) Glove

Ans. IV, VII VIII

Projection Formula of Chiral

Molecules:-(i) Wedge-Dash Projection formulae

up down

Ex:- (i) Butan-2-ol

H CH3

OH C C H2 5

(ii) Fisher Projection formula

Ex:- (i) Butan-2-ol (CH₃ – CH₂ – CH–

OH – CH₃)

Rules of writing Fisher Projection formula

:-(i) It is represented by a cross (+).

(ii) Groups at Vertical line are away from observer. (iii) Groups at Horizontal line are towards the observer. (iv) Centrol ‘C’ atom of the cross is chiral.

(v) High priority group lies at the top of vertical line (Numbering starts from top).

OH H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 H CH CH2 3 CH3 // // // // // // // // // // ////

Ex:- Draw Fisher Projection formula of following

molecules:-(1) 2-chlorobutane

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(3) O OH– CH – CH – CH2 O OH– CH – CH – CH2 O OH– CH – CH – CH2 O OH– CH – CH – CH2 O OH– CH – CH – CH2 O OH– CH – CH – CH2 O OH– CH – CH – CH2 O OH– CH – CH – CH2 OH– * O OH– CH – CH – CH2 O OH– CH – CH – CH2 O OH– CH – CH – CH2 O OH– CH – CH – CH2 O OH– CH – CH – CH2 O OH– CH – CH – CH2 O OH– CH – CH – CH2 (4) OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– * * 1 2 3 4 5 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 OH– CH – CH – CH – CH – CH3 2 3 , , ,

(I, II), (III, IV)Enantiomers. All other diastereomers.

They are true isomers (not superimposable) (5) Glucose ––––––––CHO CHOH– – – – CHOH CHOH CHOH CH OH2 –––––––

Mark:- (i) No. of C*4

(ii) Draw one Fisher Projection Formula

H OH H OH H OH H OH CH = O

Configuration nomenclature in Optical Isomers

:-Relative configuration : The experimentally determined relationship between the configurations of two

molecules, even though we may not know the absolute configuration of either. Relative configuration is expressed by D-L system.

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) : Application on correct Fisher Projection Formula

This method is used to relate the configuration of sugars and amino acids to the enantiomers of glyceralde-hyde. 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.

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Examples :

Sugars have several asymmetric carbons. A sugar whose highest numbered chiral centre (the penultimate carbon) has the same configuration as (+)-glyceraldehyde (– OH group on right side) is designated as a D-sugar, one whose highest numbered chiral centre has the same configuration as L-glyceraldehyde is desig-nated as an L-sugar. e.g. (I) R/S Configuration : RRectus SSinister Examples : d a b _c (1) Seniority order a > b > c > d

(2) Put junior most group at dotted line

 _c

b

a _d

(3) Now go from a to b, b to c.

(i) If it follows clockwise route, it is (R).

(ii) If it follows anticlockwise route, its configuration is (S).

Q. _Cl Br I F Cl Br I F (R) (S)

R/S Configuration in Fisher Projection Formula

:-Ex:- Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d) Et(b) Pr (a) Me(c) H(d)  (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d) (b) (a) (c) (d)  wise Clock c b a          R

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 If the juniormost group is at horizontal line clockwiseS anticlockwiseR

 If juniormost at vertical line, then clockwiseR anticlockwiseS Q. (1) Ans. R (2) Ans. R (3) Ans. S Properties of enantiomers :-One chiral carbon :

(i) Number of optical isomer = 2 (dor )

(ii) Number of Racemic MixtureEquimolar mixture of d and . [] = 0 due to external compensation of optical rotation.

Properties:-(i) Dipole moment same

(ii) Boiling Point same

(iii) Melting Point same

(iv) Solubility same

(v) Specific rotation [] different

Molecules with more than one chiral carbons(I) Calculation of No. of optical isomer

:-Case I:- When all Chiral carbon atoms are differently substituted (all are dissimilar C*)

optical isomer = (2)n

Case II:- When all the Chiral carbon atoms are similar at ends.

(i) n = even  x = 2n – 1₊ 2–1 n

2 (ii) n = odd  x = 2n – 1

Molecules with two Asymmetric Carbon Atoms of Dissimilar Nature:-(i) Structural formula CH₃ – CH – CH

Cl– CH – CH Cl– CH – CH Cl– CH – CH Cl– Cl– CH – CH Cl– CH – CH Cl– CH – CH Cl– CH – CH Cl– CH – CH Cl– CH – CH Cl– CH – CH Cl– CH – CH Cl– CH – CH Cl– CH – CH Cl– CH – CH Cl– – CH₂ – CH₃

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(iii) Stereochemical Formula:-H CH3 Cl H Cl C H2 5 Cl CH3 H Cl H C H2 5 ; Cl CH3 H Cl H C H2 5 Cl CH3 H Cl H C H2 5 I II III IV [] = +xº –xº +yº –yº

Analysis:-(a) I, II – Enantiomers III, IV – Enantiomers I, III – Diasteromers I, IV – Diasteromers II, III – Diasteromers II, IV – Diasteromers

(b) No. of Racemic Mixture – Two (I + II, III + IV).

Q. CH₃ – Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– CH – CH – CH Cl– Cl– – CH₂ – CH₃ Calculate total number of optical isomer.

Ans. n = 3,

x = 23_{= 8 (4 d-}__pairs)

(i) (ii) (iii) (iv)

and their mirror images.

(ii) 4 Racemic Mixtures (4 different boiling point on fractional distillation)

Compounds with 2 Asymmetric Carbons of Similar

Nature:-Ex:- CH₃ – CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– CH – CH OH– OH– – CH₃ (i) n = 2 (ii) x = 2n – 1₊ 2–1 n 2 = 3 (Optical stereoisomers).

(iii) Stereo chemical formula

Me H OH H OH Me

(I) (II) (III) (IV)

In (I and II) Plane of symmetry present.

Superimposable on its mirror image.

Thus, I and II are identical.

[] = 0, optically inactive.

Meso isomer

In (III)Specific rotation [] = +xº In (IV)Specific rotation [] = –xº

No. of d –  pairs = 1 (III + IV) = No. of Racemic mixtures

In (I) and (II) : [] = 0 due to internal compensation and Non-Resolvable In (III) and (IV) : Resolvable [can be separated into two isomers (enantiomers)]

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Meso

isomers:-The optical stereoisomers which have more than one no. of asymmetric carbon atoms but have a plane of symmetry are called meso isomers.

They are achiral (optical rotation = 0).

They have [] = 0 due to internal compensation of optical rotation.

They are diastereomer of d –  pair. So, it has different physical properties than d – -pair..

Presence of more than one asymmetric ‘C’ atoms.

They are non resolvable.

Q. Mark meso isomers among following

(1) COOH H OH COOH H OH (2) Cl Me Me Cl (3) (4) H H OH OH * (5) Ans. (1) (2) (4) and (5)

Molecules with three similar chiral carbon:-Ex:- CH₃ – Cl– CH – CH – CH Cl– Cl–  – CH₃  M.F.. n = 3 x = 2n – 1_{= 4 stereoisomers.}

Stereo Chemical Formula

:-\\\\\\\\\ \\\ \\\ \\\ \\\ \\\ \\

(I) (II) (III) (IV)

2S 2S 2S 2R

3S 3R 3achiral 3achiral

4R 4R 4S 4R

Optically active (i) Achiral due to Plane of symmetric. (Optically inactive).

(ii) Achiral due to Plane of symmetric. (Optically inactive). Properties of Optical

Diastereomers:-The optical isomer which are neither mirror image nor superimpossible to each other are called optical diastereomers.

Conclusion:- Optical Diastereomers have different physical properties, so these can be separated by normal

physical methods of separation.

(i) By fractional distillation (different B.P.)

(ii) By fractional crystallization (different solubility) (iii) Chromatography (different solubility)

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Assertion:- All diastereomers have different polarities (Both Geometrical and Optical) – True

Assertion:- Geometrical isomer are always diastereomers – True

Resolution of d –  mixture (Racemic Mixture):- The enantiomers in a d –  pair have identical physical proper-ties, so these cannot be separated by using normal physical methods of separation. The special methods used for separation of d –  pair is known as resolution.

Chemical method of separation (resolution) by using optically active

reagent:-General scheme of

resolution:-d + + resolution:-d' Racemic Optically mixture active Reagent  Hydrolysis d + d' Hydrolysis

Example:- (I) Resolution (by using Esterification) of RCOOH and R'OH

General Reaction:- R – C – O – H + H – O – R '  O O H – ) SO H ( tion Esterifica 2 4 2         R* – C O – O – R' (d) H COOH Et Me ) ( H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me H COOH Et Me + OH H D Me (d') OH H D Me (d') OH H D Me (d') OH H D Me (d') OH H D Me (d') OH H D Me (d') OH H D Me (d') OH H D Me OH H D Me OH H D Me OH H D Me OH H D Me OH H D Me OH H D Me OH H D Me (d') O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me O H D Me (dd') H C – O Et Me + O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( O H D Me H C – O Et Me ) ' d ( Pair of Diastereomers Ester upon still hydrolysis will give back the carboxylic acid and alcohol.

(II) Separation of (d ) pair of alcohol by using optically active

+ (dd') O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me O H D Me H O – C Ph Me + O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d ( O H D Me H C – O Ph Me ) ' d (

(III) By salt formation (R_{COOH + R’}_NH 2) ) d ( COOH R    ₊ ) ' d ( NH ' R ₂ salt ) ' d ' dd ( NH ' R COO R – ₃     