Isomerism

ORGANIC LECTURE NOTES

   (LECTURE No. 5 TO 14

TOPIC : ISOMERISM

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Contents

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

Metamers (a) R – O – R

(b) R – NH – R, R  N  R | R

(c) R  C  OR , R  C  NH  R , R  C  O  C  R || || || || O O O O

(c) Disubstituted, Trisubstituted Aromatic Compound Functional Isomers (a) G.F. = CnH2n+2 O

(– OH, – O –)

(b) G.F. = CnH2nO

(R2C = O, RCHO)

(c) G.F. = CnH2nO2

(RCOOH, R  C  OR ) || O

(d) G.F. = CnH2n+3 N

(1°, 2°, 3° Amines)

(e) Amides

1°, 2°, 3°

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Contents Lecture 7

Lecture 8

1. Ozonolysis: 1. Structure 2. Applications: Indentification (a) Monochlorination (b) Catalytic Hydrogenation of C = C; C  C

Lecture 9

Lecture 10

1. Stereoisomers (Classification) 2. Configurational isomers 3. Conformational isomers 4. Geometrical Isomerism

Lecture 11

Lecture 12

1. Configuration 1. Optical Isomerism nomenclature 2. Symmetry of elements in Optical 3. Chirality (Dissymmetry) Isomers and Optical Isomers 2. R/S Configura4. Optically Active Cartion in Fisher bon Compounds Projection 5. Projection Formula of Formula Chiral Molecules 3. Compounds with 2 Asymmetric Carbons of Similar Nature

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

Lecture 13 1. Meso isomers 2. Properties of Optical Diastereomers 3. Chemical method of separation (resolution) by using optically active reagent

4. Optically active compounds 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-orthotetrasubstituted biphenyls

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 (Saw horse Newman projection formula, stability, P.E. curves)

Phase Coordinator (F Batch) CDS Sir

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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 Difference between isomers and homologs:-

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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. C4H8 CH3 (iii) CH 3 – C = CH 2 –

M.F.

S.F. (i) CH3 – CH2 – CH = CH2 These are structural isomers. For (ii), S.C.F. are H3C H

CH3 C=C

H

H3C

and

(ii) CH3 – CH = CH – CH3

C=C

H CH3 Stereocentre These are stereoisomers. H

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 –

(c) C5H12

CH3 – CH2 – CH2 – CH2 – CH3 , CH3 – CH2 –

(d) C6H14

(i) CH3 – CH2 – CH2 – CH2 – CH2 – CH3, (ii) CH3 – CH – CH2 – (iii) CH3 – CH2 –

(v) CH3 – CH2 –

– CH3

– CH2 – CH3, (iv) CH3 –

– CH3 ,

–

CH3 –

– CH3

– CH3,

– CH3,

– 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 Page-5

–

–

(iii) C – C – C – C – C – C C C

–

(ii) C – C – C – C – C – C, C C –

(i) C – C – C – C – C – C C–C –

Ex.

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

C6H12 (Cycloalkanes):-

– –

– –

,

(10)

–

–

–

Ans.

(4)

C C–C

C–C–C

,

(7)

, –

(11)

,

,

(8)

,

– C–C

(9)

– ,

– –

(6)

(3)

–

,

,

–

(2)

–

(5)

,

–

(1)

–

–

Ex.

C–C –

(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 ? –

C C – C – C – NH2

Ans.

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. 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

–

(1) C – C – C – C – C , C –

Ex.

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–

(4) C – C – C – C , C – C – C – C – OH OH Write all Positional isomers of Dichlorobenzene ?

–

–

–

–

Cl

Ans.

Q.

Cl

Cl

Cl ,

–

, –

Q.

Cl

Cl

Write all Positional isomers of Dichlorocyclopropane ? Cl

Cl– Cl

–

–

Ans.

, –

Cl

Write all Positional isomers of Dichlorocyclobutane ? Cl Cl –

–

–

Q.

Cl Cl– ,

–

Cl

Ans.

–

, Cl

Write all Positional isomers of Dichlorocyclopentane ? Cl

–

Cl –

Ans.

Cl Cl

–

Cl

––

Q.

,

, –

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 Ex. (1)

(Ring-chain isomers are Functional isomers)

Compound C–C–CC

Isomer Functional (a) C = C – C = C Alkadiene

Remarks (a), (b) are not functional isomers

Alkyne

(b)

Cycloalkene

among themselve

(c)

Bicyclo

(2)

C – C – C – OH Alcohol

C–C–O–C Ethers

(3)

C – C – CH = O

C– C –C O Ketones

(4)

C – C – COOH Carboxylic acids

(5)

C – C – C – NH2 1º amine

C– C –O–C O Esters .. (a) C – C – NH – C 2º amine .. (b) C – N – C C 3º amine

1º, 2º, 3º amines are functional isomers

–

Aldehydes

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(6)

C–C–CN Cyanide (Propanenitrile)

C – C – NC Isocyanide (Ethane isocyanide)

O

–Peroxy bond (unstable) O does not exist at room temperature. C = C – O – C – OH R – O – CH2 – OH

Hemiacetal Unstable

–

Conclusion:- Following compounds don’t exist at room temperature:(ii) – C  C – OH

(iii) – C – OH –

–

–

(i) – C = C – OH

–

–

OH

(v) – C – O – C = C

OR

Q.

(vi) Any peroxy compound

–

–

(iv) – C – OH

OH

Write acyclic isomers of C5H12O –

C C – C – C – C – OH,

C – C – C – C – C, OH C C – C – C – OH C – –

(7 alcohols)

– –

C C – C – O – C, C

C – C – C – O – C, C

C – C – C – O – C – C,

C C– C –O–C–C

(6 ethers).

G.F. CnH2n + 3N

n (1) 3

M.F. C3H9N

2º Amine (R – NH – R)

3º Amine R – N – R R

G.F. CnH2n + 1N 2

n (1) 2

M.F. C2H5N

1º Amine (1) C = C – NH2 Ethenamine

2º Amine 3º Amine C–C=N C–N=C Ethanimine N-Methylene methanamine (Unstable at room temperature)

–

C – C – C – O – C, C –

C – C – C – C – O – C,

– –

–

C C – C – C – C – OH,

C – C – C – C – C, OH C C – C – C – OH, C

–

(i) C – C – C – C – C – OH,

–

Ans.

1º Amine (R – NH2) Q

–

–

Q.

–

H2C– – CH2 Azine – N H

Q.

G.F. CnH2n – 1N

n 2

M.F. C2H3N

Ans.

Cyanides and isocyanides H3C – C  N

CH2 = C = NH

HC  C – NH2

H3C – N



C Page-8

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 Ethers (All isomeric ethers are metamers)

(b) R’ – NH – R

2º Amines

(c) R’ – N – R R'' (d) R – S – R’

3º Amines

(e) R – C – O – R’ O

Esters

–

(a) R – O – R’

Thioethers

(f) R – C – O – C – R’ Anhydrides O

O

O (g) R – S – O – R’

Sulphonate esters

O Q.

How many esters are possible for C3H6O2 ?

Ans.

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

Q.

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

C

Ans.

–

C–C– N–C–C C C–C–C– N–C –

O–C–C

–

Q.

–

–

C C C– C – N–C

and

–

O–C

Show which type of isomerism ? C

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  C8H18 C C

Q.

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

– –

Metamerism

– –

Ans.

–

–

C – –

C

–

Ans.

–

C

C

or

C

C

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Q.

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

C

Cyclic esters are known as lactones.

O Q.

Write structure of smallest cyclic amide ?

NH C

Ans.

CH2 is amide. C

CH is imide

O O 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 ? CH3 H3C – O – C – CH3 – –

Q.

N

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) –2H (b) H3 C – H2C – CH3   (DU  O)

–2H   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) CxHy

y (x + 1) –   2

(ii) CxHyOz

yo  (x + 1) –   2 

(iii) CxHyXs

ys  (x + 1) –   2 

(iv) CxHyNw

y–w  (x + 1) –   2 

(v) CxHyOzXsNw

ys– w  (x + 1) –  2  

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 E.g. M.F.

DU = 8 DU

S.F. H

1.

C4H6

2

H

C–C=C=C

H H

H H

C–C–CC C–CC–C C=C–C=C

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C

C CH2

2.

C2H2Cl2

=1

Total isomers = 3

Cl 3. C6H6ClNO (Aromatic)

NHOH OH

=4

Cl

NH2 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 or

or

–C–C=C–C–

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Structure Identification of Organic Compounds by Chemical Reactions:

Monochlorination:(a)

Cl / h  CH3Cl + HCl (i) CH4  2  Cl / Sunlight (ii) CH3 – CH3 2    CH3 – CH2Cl + HCl –

Cl

Cl / h 2  

(iii)

+ HCl – –

– –

C C Cl / h  C – C – C – Cl + HCl (iv) C – C – C 2  C C Cl / h 2  

(v)

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.

Monochlori nation (a) C – C – C        2 Monochlori nation (b) C – C – C – C        2

(c) C – C – C – C – C

Monochlori nation        3

–

–

Monochlori nation (d) C – C – C – C        4 C CH3 Monochlori nation        5 (e)

Q.

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

Ans.

X = Neopentane

Q.

Cl / h  Two monochloro P(C6H14) 2 

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 ring are stable. In pure benzene, no monochlorination

Page-13

CH3 Cl / h 2  

Ex. Q.

–

–

CH2Cl

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

Ans.

(X)

(Y)

Catalytic Hydrogenation of C = C; C  C General reaction:Ni  R – CH2 – CH2 – R (a) R – CH = CH – R + H2  Ni / Pt / Pd  R – CH2 – CH2 – R (b) R – C  C – R + 2H2     H2

H2  R – CH2 – CH2 – R R – CH  CH – R  (Not isolated) 2H2 / Ni   CH3 – CH2 – CH2 – CH3 (c) CH2 = CH – CH – CH3   3H2 / Ni    

H2 / Ni       room temperature

H 2/Ni (100 – 150ºC)

(f)

 

CH2 – CH3

–

–

CH = CH2

(e)

CH2 – CH3

–

(d)

[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 hydrogenation. Page-14

During catalytic hydrogenation, all monochlorination and no rearrangement in carbon skeleton takes place.

Ex:- C = C – C = C + Cl2 X

Ex.

C– C–C C – Cl –



2H2 / Ni    

(1)

C – Cl

Y =

Cl / h 2   Y + Z

(X)

Z = Cl

CH3

CH3 H / Ni 2

(2)

CH3

CH3 H / Ni 2

(3)

Q.

Cl / h 2   5 Monochloro product

5 Monochloro product

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

Identify X, Y, Z Ans.

DU = 2 –

Cl

X

Y

,

Z

Identify the lowest molecular weight alkane which gives four structural isomeric monochloro products ? C– C –C–C –

Q. Ans.

,

C

C5H12

= 72g

Identify the structure of hexane which gives 3 monochloro products ?

Ans.

C C – C – C – C, C

Q.

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

Ans.

DU = 2

– –

Q.

C–C–C–C–C–C

 2 monochloro product Find the structural isomers of product?

fully saturated cycloalkane of M.F. C6H12 which gives two monochloro

–

Q.

Q.

–

–

Ans.

Cl / h 2   A C8H17 Cl (C8H18 ) (Only one type)

Identify A ?

Page-15

– –

Q.

– –

Ans.

C C C– C – C –C C C

Write all isomeric alkynes which produce an isomer of heptane which on further monochlorination gives (a) three monochloro products. C–C (i) C  C – C – C – C

(b) Ans.

two monochloro Nil

Q.

Find the structure of lowest molecular weight hydrocarbon and maximum unsaturation which on hydrogenation 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.

C (ii) C  C – C – C – C C – –

–

Ans.

CH2

Ans. 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)2 S (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.

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

Ex:-

(1) CH2 = CH2

(1) O3 CH2 = O + CH2 = O (2) Zn/H2O

(2) CH3 – CH2 – CH = CH2

(1) O3 CH3 – CH2 – CH = O + O = CH2 (2) Zn/H2O

(3) CH2 = CH – CH2 – CH = CH – CH3

O3 / Zn

CH2 = O + O = CH – CH2 – CH = O + O = CH – CH3 (4)

O / Zn 3  + OHC – CH2 – CHO (Propandial) Page-16

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.

C

O3   O = C – C – C – C – C = O + O = CH2

(1)

Zn

O

C

O () 3

(2)

(4)

Q.

CH = CH – CH3 –

low temperatur e

      

–

(3)

Zn

CH = O

O3 / Zn

– CH = CH –

+ O = CH – CH3

O3   C6H5 – CH = O Zn

H / Ni Cl / h 2 CnH2n 2  2   CnH2n1Cl (P) CnH2n – 2 ( Q) (m products ) Single Compound m3 (no isomer )

H2O2 O3(oxidative)

– –

CH3 HCOOH + CH3 – C – COOH CH3

Identify P ?

Ans.

P=

Q.

An unsaturated hydrocarbon on ozonolysis produces 1 mole of

Ans.

Find the structure of the hydrocarbon and the no. of monochloro products formed by hydrogenation. , 5 monochloro product

–

, 1 mol CO2, 1 mol

X H / Ni 2 (Unsat. hydrocarbon) –

Q.

Cl2   m products h (m  3 )

O 3 (Zn/H2O)

O

identify structure of X ? O

Ans.

X = Page-17

Q.

H / Ni 2 X (Unsat. H.C.)

Cl / h 2   m products (m  7 )

O 3 (Zn/H2O)

HCHO + 4(1-oxoethyl) Cyclohexan-1-one. Identify X ?

C X

–

Ans.

C=C–C

Q.

Identify structure of X ?

Ans.

X is

Q.

H2 C–C–C–C–C–C X 

O3 (Zn)

CH3CHO + CHO – CHO Identify structure of X ? Ans.

X is C – C = C – C = C – C

Q.

–

CH3

CH – 3

Q.

–

CH3

Q.

A

O3   Zn, H2O

O3   Zn, H2O

2

CH3

+

CH3 C – C– +2

–

CH – 3

Q.

O O

Methyl glyoxal + Formaldehyde Page-18

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 (*Cabcd). These isomer has same general formula, structural formula and molecular formula but different stereochemical formula. E.g.

CH3

C=C

H CH3

(I)

CH3 H H

(II)

C=C H CH3 CH3–CH2–CH=CH2 I, III II, III I, II

(III)

Positional Positional 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.

a

a

b

C=C (I) b

The root form of geometrical isomers lie in restricted rotation.

a b

(I)

a

a

b

b

a (II)

b Page-19

Condition (i) The two groups at each end of restricted bond must be different. 1 3 C=C 2 4 Caa Caa X Caa Cbd X Cae Cb2 X Cab Cab  Cab . Cbd Cab Cde

 

(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 CH 3  C  CH  CH3 | CH3

(4) But-2-ene CH3 – CH = CH – CH3

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

(6) 1, 2-Dideuteroethene

H2C = CH

(7) Phenyle thene Ans.

(8) Buta-1, 3-Diene C = C – C = C

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) Imines prepared from unsymmetrical addehydes and ketones, always show geometrical isomerism. Page-20

Q.

Which of the following compounds show geometrical isomerism after reaction with NH3. O || (a) CH3  C  CH3

O || (d) CH3  C  H

(g)

Ans.

O || (b) H  C  H

O || (c) H  C  D

(e) Ph  C  CH3 || O

(f) Ph  C  Ph || O

(h)

(i)

c, d, e, g, i (b) Oximes

C = N – OH :

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

R

C = N–OH H (Aldoxime) OH

R

C=N H

R

H2O + H2 N–OH   

C=O

(I) (syn)

OH

C=N

and

H

(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 6H 5–C–CH 3 O

(5) Benzophenone

C 6H 5–C–C 6H 5 O

O (6) Cyclo hexane O Me (7) Methyl Cyclohexanone Ans.

1, 2, 4, 7

Page-21

Q. Ans.

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 ? 3 (c) Hydrazones

C = N – NH2 :

R

R

H2O C = O + H2N.NH2    H hydrazine

C = N–NH2 H R

NH2

R +

C=N H

(I)

NH2

C=N H

(II)

(Geo. diastereomers)

Q.

Ans.

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 ? .. .. N NH2 NH2 N N .. O + + H2N – NH2 + + NH2 CH 3 CH3 O CH 3

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

syn

Ph

..

Ph ..

N=N anti

Ph

(4) Geometrical isomerism across ring structure

(i)

Restricted rotation

(ii)

(iii) Page-22

(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. n

Ex.1 Ans.

No. of Geometrical isomerism = 2n–1 + 2 2 CH3 – CH = CH – CH = CH – CH3 Stereocentre = 2 Geometrical isomerism = 21 + 20 = 3

1

Page-23

Ex.2

Cl – CH – CH

CH – CH

Ans.

Stereocentre = 4 Geometrical isomerism = 23 + 21 = 10.

Case III : Compounds with similar ends but odd number of stereocentre. No. of Geometrical isomerism = 2n–1 + 2 Ex.1 Ans.

n 1 2

CH – CH = CH – CH = CH – CH = CH – CH3 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. Page-24

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

(1) (2) – CH = CH2 < – C  CH

Remark : Bond pair is always senior to lone pair. CH 2 = CH

Ex.

(1)

HC

C

C == C

C(CH3)3

E CH – CH = CH2

(2)

Z

N (3)

CH2 – C(CH3)3

C

H2C = C = CH

C == C

E

C

CH

(3)

Z

(4)

E

(5)

E

Page-25

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 nonpolar. 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 qCH Cl  qCH 3

3 –F

but

Thus,

CH3 – Cl > CH3 – F

But,

H – F > H – Cl.

dCH3 – Cl  dCH3 – F

Dipole moment of some Functional Groups : (order) > – COX > – COOH > – OH > – NH2 > – R – X (2.8 – 3 D)

> R–O–R

(X = Cl, Br, I)

> R – C  C – H > – R – CH = CH2 > R – CH2 – CH3 (non polar) (non polar) Page-26

Dipole moment for Geometrical isomerism :

(A)

(cis)

(trans)

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

Unsymmetrical Alkene (Cab = Cad )

(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 properties. 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 (b) CH3OH < CH3 – CH2 – OH < C – C – C – OH (c) H – CHO < CH3 – CHO < C2H5CHO

Page-27

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

C–R

–

R–C

dimer

O – H ------ O

 In isomeric ketones and aldehydes, BPKetone > B.P..Aldehyde  In chain isomers, as branching  , surface area of molecule  VDW forces  Boiling Point  Thus, BP  1º > 2º > 3º (–OH, –NH2)

(i) C3H8O

C – C – C – OH

> C– C– C –

Eg:-

OH

–

C

–

–

–

C – C – C – C – OH > C – C – C – OH > C – C – C – C > C – C – C OH C OH

(ii) C4H10O

Among stereoisomers:-In geometrical isomer, the dipole moment of geometrical isomer is always different. More polar geometrical isomer has higher B.P. (a) Cab = Cab B.P.cis > B.P.trans (b) R – CH = CH – X

B.P.trans > B.P.cis

COOH H H > C=C C=C H COOH HOOC COOH Fumaric acid (trans) Maleic acid (Cis) H

(c)

CH3

CH3

H C=C

(d)

H

– +CHO

CH3

(e) H

H

H

NH2

CH3

H

C=C

>

C=C

CHO C=C

>

H

NH2

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 –

OH

–

OH Less Intermolecular H-bonding

–

Ortho 5-membered Intramolecular H-bonding

–

OH

O–H

OH More Intermolecular H-bonding Page-28

(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 (Cab = Cab 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 –

<

Et

–

(c)

CH3 <

<

(II) Structural isomers:(a) Functional isomers:(i) R – COOH

>

H– C –O–R O (No H-bonding)

(H-bonding) H-attached to ‘O’ (ii) CH3 – CH2 – CH2 – CH2 – OH (H-bond)

H with ‘R’ CH3 – CH2 – O – CH2 – CH3 (No H-bond)

>

(iii) CH3 – CH2 – CH2 – CH2 – NH2 > (1º)

CH3 – CH2 – CH2 – NH – CH3 (2º)

> CH3 – CH2 – N(CH3)2 (3º)

(b) Chain isomers:–

C (i) C – C – C – C > C – C – C CH3 | **Exception :- (ii) CH3 – C  CH3 > CH3 – CH2 – CH2 – CH2 – CH3 > CH3 – CH – CH2 – CH3 (M.P.) .) | | CH3 CH3

Neopentane (spherical) (Close-packing) For B.P.:- n > iso > neo For M.P.:- neo > n > iso (III) Geometrical isomers:Cab = Cab type a b

b

.C C= * a

a

a C=C b

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 equal halves, which should be mirror images of each other. A molecule can have more than one P.S.

Page-29

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

Me (2)

>

C=C Cl

Cl

(3) Maleic acid < Fumaric acid (4) Stilbene (Ph – CH = CH – Ph) E > Z (IV) ortho, meta, para isomers

Cl

Cl

–

–

–

Cl

Cl

–

–

–

Cl Cl (m) (o) (p) less polar more polar Most symmetric para > ortho > meta (we see polarity now, if H-bonding doesn’t exist) (1)

–

OH –

OH

–

OH

OH –

–

–

OH (p) (2) Symmetrical H-bond (intermolecular)

OH (m) Intermolecular H-bond

(o) Intramolecular H-bond

Thus, para > meta > ortho (if H-bonding)

(4) Water Solubility : Criteria:- H-bond > Dipole – dipole > Vander Wall forces (insoluble) between solute and solvent Homologs: Hydrophilic –––––– water loving –––––– H-bond  Hydrophobic –––––– water repelling –––––– Hydrocarbon  As M.W..  , solubility  .

Ex:-

(i) CH3OH > C2H5OH > C3H7OH > C4H9OH OH OH (ii)

–

–

(I)

>

Me >

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

Structural isomers:(1) Functional isomers:(a) –COOH > –NH2 > –OH > H-bond > R – CH = CH2 > R – CH2 – CH3 insoluble (VDW) Page-30

(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) C – C – C – OH < C – C – C.........Solubility OH C – C – C – OH > C – C – C.........B.P.. OH –

(i) M.F. = C3H8O

–

Ex:-

–

(ii) C4H10O

–

–

–

C C C – C – C – C – OH < C – C – C – OH < C – C – C – C < C – C – C OH OH

(iii) C4H8O2 (acid)

CH3 – CH2 – CH2 – C – O – H O (III)

CH3 | CH3 – CH – CH – C – O – H O

<

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

>

(ii)

Cis > Trans

<

COOH

COOH C=C

(iii) H

Trans > Cis

H

COOH

H C=C

> COOH

Cis > Trans H

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)

Recorder (8)

Page-31

(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 recorder:Angle of rotation () (1)  = 0º (2)  = +xº (clockwise rotation) (3)  = –xº (anticlockwise rotation)

Inference Compound is optically inactive. Optically active, dextrorotatory or d or (+). 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:-

t  25 º C

[ ] 580 nm  where

 c

 = 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.

=

(1) (2)

70 = +35º 2 Its concentration is made twice, the observed rotation() will be 140º but  = 35º 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.

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

Cl H

R C=C

Ex:-

H R Centre of symmetry

H

H

Cl Centre of symmetry

Page-32

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 distinguished only by polarimeter.

Chirality (Dissymmetry) and Optical Isomers:(a) (b) (c) (d) (e)

The dissymmetric molecules have two orientations in space. These two orientations are called stereoisomers, optical isomers, enantiomers. These are mirror images (enantiomers). Non-superimposability of enantiomers:- The dissymmetric molecules are always non-superimposable on their mirror-image orientations. The non-superimposable orientations are non-identical orientations, so are called isomers. 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 *Cabde. a C* e d If a molecule contains only one asymmetric carbon atom, then the molecule as a whole becomes chiral and optically active and show optical isomers. b

Centre of symmetric

Plane of symmetric

Optical active

H C

(1) H

H

H

Absent

Yes

No

Absent

Yes

No

H C

(2) H

H

Cl

Page-33

H C

(3) Br

H

Cl

Absent

Yes

No

Absent

No

Yes

H C

(4) Br

F

Cl

Ex.1:- Mark the chiral objects:(i) Cup (ii) Plate (vii) Shoe (viii) Glove Ans.

(iii) Letter A

(iv) Letter G

(v) Fan

(vi) Door

IV, VII VIII

Projection Formula of Chiral Molecules:(i) Wedge-Dash Projection formulae down up

Ex:-

(i) Butan-2-ol CH3 C H

OH C2H5

(ii) Fisher Projection formula (i) Butan-2-ol

(CH3 – CH2 – CH – CH3) –

Ex:-

OH

CH3 H

OH CH2CH3

Ex:-

///////////////////////

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).

Draw Fisher Projection formula of following molecules:-

(1) 2-chlorobutane

(2) Pentan-2-ol

Page-34

–

–

* (3) CH2 – CH – CH OH OH O

1

2

3

4

5

–

–

* * (4) CH3 – CH – CH – CH2 – CH3 OH OH

,

,

,

(I, II), (III, IV)  Enantiomers. All other diastereomers. They are true isomers (not superimposable)

CHO CHOH CHOH CHOH CHOH CH2 OH

– – – – –

(5) Glucose

Mark:- (i) No. of C*  4 (ii) Draw one Fisher Projection Formula CH = O H

OH

H

OH

H

OH

H

OH

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 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.

Page-35

Examples :

Sugars have several asymmetric carbons. A sugar whose highest numbered chiral centre (the penultimate carbon) has the same configuration as D-(+)-glyceraldehyde (– OH group on right side) is designated as a Dsugar, one whose highest numbered chiral centre has the same configuration as L-glyceraldehyde is designated as an L-sugar.

e.g.

(I) R/S Configuration : R  Rectus

S  Sinister

Examples : a b

c

d

(1) Seniority order a > b > c > d (2) Put junior most group at dotted line b



(3)

Q.

c a d 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). Br Br

Cl

I

F

Cl F (S)

(R)

I

R/S Configuration in Fisher Projection Formula :(c)

Me(c)

Ex:-

H(d)

Et(b)  (d) Pr (a)

(b) 

abc     Clock wise

R

(a)

Page-36

 

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

Properties:(i) Dipole moment (ii) Boiling Point (iii) Melting Point (iv) Solubility (v) Specific rotation []

same same same same 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. n

–1

–

–

(i) n = even  x = 2n – 1 + 2 2 (ii) n = odd  x = 2n – 1 Molecules with two Asymmetric Carbon Atoms of Dissimilar Nature:(i) Structural formula CH3 – CH – CH – CH2 – CH3 Cl Cl n 2 (ii) Optical isomer = 2 = 2 = 4

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(iii) Stereochemical Formula:-

CH3

CH3

CH3

H

Cl

Cl

H H

H

Cl

Cl

H; Cl

C2H5

C2H5

I [] = +xº

CH3 Cl

Cl

H

H

C2H5

II –xº

Cl C2H5

III +yº

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

H

IV –yº

III, IV – Enantiomers I, IV – Diasteromers II, IV – Diasteromers

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

n = 3, x = 23 = 8 (4 d-  pairs)

–

Ans.

–

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

Q.

(i)

(ii)

(ii)

(iii)

(iv)

and their mirror images. 4 Racemic Mixtures (4 different boiling point on fractional distillation)

–

–

Compounds with 2 Asymmetric Carbons of Similar Nature:Ex:CH3 – CH – CH – CH3 OH OH (i) n = 2 n

–1

(ii) x = 2n – 1 + 2 2 = 3 (Optical stereoisomers). (iii) Stereo chemical formula :Me H

OH

H

OH Me (I) (II)  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

(III)

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

(IV)

Page-38

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 H

H

COOH OH

(1)

(2) H

Ans.

OH COOH

Cl Me

Cl

(3)

(4)

Me

OH H

(5)

* OH

(1) (2) (4) and (5)

–

–

–

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

\\\\\\\\\\\\\\\\\\\\

(I) 2S 3S 4R

(II) 2S 3R 4R

(III) 2S 3  achiral 4S

(IV) 2R 3  achiral 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) (iv) Differential Melting (M.P. diff.) Page-39

 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 properties, 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+ + d' Racemic Optically mixture active Reagent

Hydrolysis

Hydrolysis

d + d'   (I) Resolution (by using Esterification) of RCOOH and R'OH

Example:-

 Esterifica tion   General Reaction:- R – C – O – H + H – O – R '      R* – C – O – R' (H2 SO 4 ) O O – H2O Me H

COOH (d)

Me

Et Et

+ OH

H

H D (d')

COOH ( )

H

H + H (dd')

C–O O

(d')

O D ( d ')

(dd')

Me

H ( d' )

C–O Me

D

Et

Me

Et

Me

Me

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 acid:D

Ph +

H

H (dd')

O–C O Me

Me

Ph

Me

+

H

H

C–O O ( d' )

D

Me

(III) By salt formation (RCOOH + R’NH2)

R COOH + R' NH2 (d   ) (d' )

R COO – R' NH3 ( dd' d' ) salt



Page-40

Optical Purity or Enantiomeric Excess (O.P. or E.E.):Ex:Compound X has [] =  70º We have mixture of enantiomers in different percentage. %E.E. 

(obs)  100% []specific

Optically active compounds without a chiral carbon:Rapid R R Inversion –  Asymmetric N:R' – N  N – R' R'' R''  A nitrogen atom attached with 3 different groups (sp3) is chiral, asymmetric.  In case of acyclic compounds, the Nitrogen, having trigonal pyramidal shape undergoes rapid inversion of shape at room temperature. So at room temperature, everytime a racemic mixture of ‘d’ and ‘  ’ forms exists. This racemic mixture can never be resolved at room temperature.  Nitrogen converts into its enantiomer..

–

– –



So, acyclic molecules with chiral nitrogen are chiral but optically inactive ([] = 0) and are non-resolvable. If asymmetric Nitrogen atom is present in ring structure, then inversion does not take place and such molecules are optically active []  0  . (otherwise ring will break)

Ex:-

 Me N

[ ]  0

Optical activity without asymmetric carbon : CCC–C–C (Cumulated double bond)  One after another

CC–C C–C (Conjugated double bond)  Alternate

C C–C–C C (Isolated double bond)  Separated

(I) Case of allene : (a) Allenes with even  bonds :

e.g.

the orbital diagram of this structure will be 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.

Page-41

(b) Allenes with odd  bonds :

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.

(II) Case of spiranes : A similar case like allenes is observed in spiranes. The spiranes with even rings and different groups at terminal carbons show optical activity & optical isomerism, while the spiranes with odd rings shows geometrical isomerism. (a) spiranes with even rings : shows optical isomerism.

(b) spiranes with odd rings : shows geometrical isomerism. (III) Case of cycloalkylidene :

(IV) Case of ortho-ortho-tetrasubstituted biphenyls : 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.

Page-42

Lecture # 14 CONFORMATIONAL ISOMERISM : 1.

Free Rotation : A sigma covalent bond undergoes free rotation at room temperature. (C–C, C –O, C–N, N–N, O–O)

2.

Conformers / Rotamers or conformations : The infinite number of spatial oreintation of molecule is arise due to free rotation along a sigma covalent bond or rotamers.

3.

Conformational Isomers : The conformations of extreme energy different are called conformational isomers but these are not true isomers and can never be isolated.

4.

Conformational energy : The potential energy different b/w 2 conformational isomer is called conformational energy. It has very low value i.e. (10–20 kJ/ mole) which is available at room temperature. Projection formula (a) Saw horse Projection formula Ex.

CH3 – CH3

rotation   

Eclipsed

Staggered

(b) Newmann Projection formula

      1 12.5 kJ / mol

Energy - Level Diagram

Page-43

Ex.

CH3–CH2–CH3

Saw horse Projection formula

Eclipsed

Staggered

Newman Projection Formula CH3 H

CH3 H

  60 º

H

H

H H II Staggered

Eclipsed

º 120  

HH

H III Eclipsed

CH3 H H H



H

H

H H IV Staggered

Propane :

Butane : Ex.

Ethyl-hydrogen repulsion is less than methyl-methyl repulsion. CH3

CH3 CH3

60 º  

H H

HH (I)

CH3 CH3

H

º 120  

H

H (II)

H

º 180  

H

CH3 H H 240 º  

H

CH3

(IV)

H

H CH3 H H (V) Page-44

300º   

360º   

I/VII = Fully eclipsed II/IV = Gauche form or skew form III/V = Partially eclipsed IV = Anti form Stability Order : IV > II > III > I P.E. Order : IV < II < III < I Potential Energy Diagram :

Conformational Isomers : Due to free rotation from 0º to 360º those conformations which are most stable and have minimum P.E. are defined as conformational isomers. They maay have same energy and such isomers are called degenerate isomers or equienergic isomers. The conformational isomers with different energies are called non-degenerate isomers. The conformational isomes lie at the P.E. minima in the P.E. diagram with respect to rotational angle. Conformational Energy (C.E.) The rotational energy barrier is known as conformational energy. It is the P.E. difference between conformational at potential energy minima and maxima. Angle of rotation/Dihedral Angle (D.H.A.)/Torsion Angle : The interfacial angle between the groups attach at two -bonded atoms is defined as dihedral angle.

Strains : (i) Torsional Stran (eclipsing strain) : It is defined as the electronic repulsion between the bond-pairs of two adjacent eclipsed bonds. It is active at torsional angles 0º, 120º, 240º which the molecule has eclipsed conformation. It is considered almost zero in the staggered conformation. (D.H.A. = 60º, 180º, 300º) (ii) Vander Wall strain : It is the repulsion between the group attached at adjacent bonds. The vander wall strain is maximum in eclipsed conformation, and minimum in anti conformation, while intermediate in gauche (skew) conformations. It is almost zero for H-atom since the sum of vander wall radii is less than internuclear distance between two H atom. (iii) Angle Strain : It arises due to distortion in normal bond-angles. In acyclic compounds, there is no distortion of bond-angle due to free rotation. So, all conformational have angle strain zero. Page-45

Page-46

Ques. Draw conformation isomers of following compound (a) CH3 – CH – CH3

(b) CH3 – (CH2)3 – CH3

CH3 Ques. Draw conformation isomers of following compound CH3 – CH –CH – CH3 with respect to C2 and C3 carbon CH3 CH3

atoms. H H

H 60 º

180 º



CH CH3 CH3 3 Eclipsed

CH3

CH3



H3CCH

CH3

CH3

CH3

120 º

 

H3C

H3C

H

CH3

H

H

3

H

H3C

CH3

H Staggered

CH3

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. Example 1.

CH2–OH (Glycol). CH2OH OH OH

H







H H (II) Gauche

H

Hydrogen Bonding : (II) O–H H

O–H

H H Gauche form most stable due to hydrogen bonding. Stability Order : II > IV > III > I H

Example 2.

O2N – CH2 – CH2 – OH

Page-47

Example 3.

Tartaric Acid : COOH – CHOH – CHOH – COOH

Sterioisomers : 3 1. Meso

COOH H

OH

H

OH COOH

2. d/

COOH

COOH H

Meso



COOH H

OH

H

OH COOH



OH

OH



H H

COOH OH

COOH

COOH COOH

H H

COOH OH

H

OH

COOH

 

OH

H

OH

H

OH

d/  COOH

COOH COOH OH H

OH

COOH

 H OH

 H

H

OH

COOH

COOH OH

HH

COOH  H OH

COOH COOH

OH

H

OH COOH

COOH

H

H

 

OH H COOH OH OH Conformation isomers of cyclohexane : Superficial idea of chair & boat form (This is discuss in orientation class) Page-48 H

OH