IlsomerismDISCLAIMER “The content provided herein are created and owned by various authors and licensed to Sorting Hat Technologies Private Limited (“Company”). The Company disclaims all rights and liabilities in relation to the content. The author of the content shall be solely responsible towards, without limitation, any claims, liabilities, damages or suits which may arise with respect to the same.”INTRODUCTION Degree of Unsaturation (DU) The main objective of an organic chemist is the determination of the structure of a new organic compound which has been obtained in pure state either from a natural source or synthesised in the laboratory. In order to establish the correct structure of an organic compound, necessary to detect skeleton of compound, elements and functional groups present in the organic compound, CHa, CH saturated open _ unsaturated chain alkane (ou=1 (OU = 0) Deficiency of hydrogen generates unsaturation in the compound to compute this unsaturation is called DU. How many bond cleavage give open chain saturated compound is DU of that compound. Open chain saturated compound having DU = 0. In compound unsaturation is observed in the form of n-bonds & rings. Gase-1: if molecular formuta is given. u=(C+ [ES (Where C = carbon, H = hydrogen, N = nitrogen) ii) iv) CH, unsaturated (OU = 2) (Tips and Trick ® DU = 1— Ring’ or Double bond’ DU = 2-+ =" bond + ‘=' bond bond — '=' bond + ring — Ring + Ring CH, 6 (4+ -3=5-3=2 C\H,0 (6+)-5=7-3=4vy) CHe 8 8+1)--= (B41)=5 vil) C,H,8rFClI 4+4 8 +1)- =9-4= ( )-(424) 9-4=5 ix) CHyN (o-9-(2) =4-4=0 Case-II : If molecular formula is given. vi) viii) x DU = Number of x -bond + number of rings i) Toulene (4) p-xylene (4) v) Resorcinol (4) OH iv) vi) C,H,O 8 8+1)-229-4=5 (8+1)-5 CHyN +1)-( 9-1 2S) 6-42 C,H,8°FNO 342-1 (ery-(4) =9-2=7 o-xylene (4) o-cresol (4) p-toluidine (4) OH ; CH; NH) CH| c=o vii) o-salicyaldehyde (5) viii) Anisol (4) ° con ix) ° Aspirin (6) ») O Piperidine (1) O | H Tropolone (5) xii) Azulene (7) xiii) ade, Acetone (1) xiv) Ph-C-Ph Benzo phenone (9) xv) Mesitylene (4) xvi) O Triphenyl methane (12) cL xvii) MmCase-II: If molecular structure is given in 3-D OU = number of faces - number of box Ad CyHy dD DU=6-1=5 Cubane a (a = CH, ” O ; ° D ° Note : in case of salt DU is fractional. Example: 9 I CH,-C-0 DU=(C+1) =@+n-( wie JessIntroduction of Isomerism and Structural Isomers *# Two or more different compounds having same CH,CH,OH &CH,OCH, molecular formula are known sores as isomers. Iso + same Definitions mers — unit Example : C,H,O CHARACTERISTIC OF ISOMERS * They have same molecular formula, * They have same molecular weight. * They have same emperical formula. + Emperical formula : Formula in simplest ratio or in condense form. Eg. MF => EF CH, => CH CiH,,0, = — CH,O * They have came emperical weight. * They have came vapour density. 2 « vapour density = Molecular weight * They have same DU. * Atleast one physical or chemical property must be different. ‘soMeEnisM Structural womertem seereoisomeriem ott 1.Chain isomers 2, Position isomers conformers configurational 3. Furetional isomers L 4. ing chain isomers +. Geometiclisemerism 5. Metameris 2. optel tsomerism 6, Tautemerism ‘STRUCTURAL ISOMERS Two or more different compound having same molecular formula but have difference in connectivity of atoms or groups.STEREOISOMERS aS Example: Definitions H CIN eH IN et INCL emg two or more different cl SH Ho SH HH Set compound having same ) (e) () molecular formula as well as (A, 8) = Structural, (A, C) = Structural, structural formula but have (B.C) = Stereoisomers difference in orientation of Boiling point « Dipole moment co &igroups:n shacesor Dipole moment of B= 1 40 " Dipole moment of C> n=0 Dipole moment and boiling point are physical property. So here 8 & C have different physical properties. ‘STRUCTURAL ISOMERS S| 1. Chain Isomers: Definitions Example : ¢ Two or more different i) A compound having same —_ molecular formula but different in parent carbon chain or side chain. "OA iii) “No : ox CN I iv) CH, - CH, - CH, -CN, CH,—CH—CH, c ¢ | vy) C—c—c—c, c—E—e-c, ete Points to remember c * Chain isomers must have same functional group. vi) “\A™., assy2. Positional Isomers SS) © For positional isomers, parent carbon Definitions chain must be same. © For positional isomers, functional group remain same. * Two or more different compound having same. molecular formula but Examples (Solved) : difference in position of IAA ~WY substituent, multiple bond and functional group are known as positional isomers. Chain isomers ¥ They are chain isomers because PCC is not same wae AZ Positional isomers ¥ SH ww Yn AN Positional isomers ¥ SH vy) “_«<~ AK Positional isomers ~ cH, a bo 2 a CH,-CH-CH,-C=N i 230 4 ional is ¥ YW) oy, 2u-du,du, Positional isomers 3. Functional isomers Examples (Solved ae eS © Alcohol - ether CH, -CH, -OH & CH, -O-CH, + Two or more different compound having same molecular formula but different functional groups. © Aldehyde - ketone 1 g CH,-CH,-C-H & CH,-C-CH,© Alkyne — diene CH=C~CH, © 1°, 2°, 3° amine are different functional group. CH, ~ CH, ~ CH, CH, ~ NH, CH, ~ CH, ~ NH= CH, - CH, CHs— N ~ GH: ~ GHs CHa © Nitrile & isonitrile CH3 - C= N or CH, -| Points to remember »® © 1°, 2°, 3° amide are different functional group Example ° H,C-CH,— mom ts, H, OCH, EN CH, H. ol So z “cH,Structural lsomerism RING CHAIN ISOMERS. SS) Examples : Definitions Ln + Two or more different 1 “_~ compound having same alkane alkene molecular formula but one is Ring chain isomers or functional isomers. chain & another one is close chain with same DU. —o —~" [| ~ 8 Ether _ aldehyde Ring chain isomers or functional isomers 3. Find relations in (a & b), (a & d), (a &f), (b &c), (b &e), (68g), (c&e), (c &g), &F) and ea: O Ll XP~ » >> y AY vi) “Pw ‘Ans. (a) & (b) - chain isomer (@) & (d) - ring chain (a) &(f) - ring chain & functional isomer (b) & (©) -ring chain & functional isomer (b) & (@) - ring chai (b) & (@) - ring chain (© & (@) - functional isomer (©) & @ - functional isomer (d) & (f) - functional isomer (0) & (g) - chain isomerMETAMERISM =) Monovalent functional groups Definitions a OH 2, NH If alkyl or aryl group changes 3. -C—NH, 4. -C-OH around polyvalent functional i I group then these isomers are known as metamers. 5. —C-H 6. -c-x I I 0 0 7 -N= ! Polyvalent functional groups Paints to remember. x» 1. -0- 2 -s- © Those group which have 3. -C- 4. -¢-0- more than one free valency ! are known as polyvalent functional groups. 2 5. -$-0- 6. -NH- No 7% -C-N- -¢-0-C {ot I Il ° 0 Examples (Solved) : Find relation between given isomers : i) JO\ CH,-O-\ Metamers Metamers NK Fe Identical Identical iy FRO yO Positional isomers a 9 Oro oro netames 3 0 10.CH,-C-CH, ‘C-CH,-CH, y) u Metamers , c~<] vi) (a) i (b) Metamers Points to remember © Functional group remain same for metamerism. TAUTOMERISM SS) Protonotropy / cationotropy / desmotropism / _ Definitions keto enol tautomerism Examples : * Those functional isomers which are present in dynamic “i 9H equilibrium are called i) (CHLC-CH, => CH,=C-cH, tautomers. This process is | known as Tautomerism. H Keto Enol ° OH Intrdution ic ii) & Tautomers & © AS in tautomerism, proton functional isomers migrates, therefore it is also known as protonotropy. ¢ In tautomerism bonds also migrate, therefore it is also iii) & Functional isomer only known as desmotropism. » For tautomerism atleast one a-hydrogen is required. Tips and Trick & Priority order : R>C>T>F>M>C>P Radio city top FM near Carrer point n.Examples (Solved) : Among the following compounds how many will show tautomerism ? Yes ii) o; No Yes iv) . No v) 2-pentanone Yes vi) Acetone Yes vii) Yes . Examples (Solved) : Define relation between the following: 1, Or, a Chain isomers 2 NO Different compound 3. Different compound fw BS i p a An AA Functional isomer BAN NS, Identical Jie 6. ane Identical H AX ore Positional isomer ‘i 2.9 0. yo Functional isomer 8 1. A a Chain isomers N i 2 AG CN Functional isomer ° 1. Nu eyes Functional isomer 14, Identical ° 1.7 I runctional isomer ° ° 16. oS AA Metamers & positional isomer 17. Dy DS Functional isomer ° J eaintomer o 19. CHs ~ CH: - NZ ° CH,- CH,-O-N=0 Functional isomer QO OF ° 20. i oO ° I S-0 I Metamers 1B.Projection Formulas in Organic Chemistry WEDGE-DASH FORMULA Introduction { Group behind 11 o the plane \ sy roup in plane The configuration of organic CE molecules can be visualised Pi i x by three-dimensional (3D) Y structures, which may be Group in front of plane depicted by any of the following representations : The thick solid (solid wedge) line indicates the bonds lying above the plane of paper (projecting towards the viewer). Thin lines indicate the bonds lying within the plane of paper. ‘The dotted lines indicate the bonds lying below the plane of paper. Such a representation is called Wedge-Dash formula FISCHER PROJECTION FORMULA The horizontal lines represent the bonds directed towards the viewer and the vertical lines away from the viewer. For example, (~CHO) group in glyceraldehyde, (COOH) group in lactic acid etc, The C atoms are then numbered according to the IUPAC nomenclature. CHO Groups on the verteal te ae below 4 on the plane of paper faway rem thevewe) CH,OH Groups.on the horzonial line aro above the plane of paper (towards the viewar) INTERCONVERSION OF WEDGE-DASH FORMULA TO FISCHER FORMULA For Example: GeO sate SHO CrP iguaie he structure such J — ay PRCHOH siarmanc-conevener en Lnse20H Sans HOH \Wiedap-2ash formule Fischer projection ot dyeeratehya. forma 14.a a i) dmg d 5 Monks) of *e b (conRHS.) (conLH.s.) and (donR.HS.) a a iti) a ote Oe b b (conRHS.) a ok 4 (onRHs) Rc (cool HS) INTERCONVERSION OF FISCHER FORMULA TO WEDGE-DASH FORMULA ii) CHO CHO HO H HO» HOH n@ SCH,OH Fischer formula Wedge-Dash formula THE FOLLOWING RULES MUST BE OBSERVED 1. Afischer projection should not taken out of the plane of paper and turned over. It would lead to the Fischer projection of the enantiomer, since it changes the configuration at the chiral centre. a a Forbidden d ‘© ~Tumover 7 oD b Enantiomer 2. Even number of interchanges clockwise or anti-clockwise leads to the same stereoisomer, while odd number of interchanges results in an enantiomer. 15.CHO CHO CHO wow —=— no—L-on,oH #—foox,on H,C-OH «/ mierhange H inerchange == OH 0 © a (1) and (3) are equivalent Fischer projections. 3. Rotation of Fischer projection by 180 on the plane of paper results in the equivalent Fischer projection. cHo CH,OH Rotate by 180° HO H ‘on the plane: H ‘OH H,C-OH of paper CHO 0 wy (1) and (2) are equivalent Fischer projections. SAWHORSE AND NEWMAN PROJECTION FORMULAE. Sawhorse projection formula : i) ev ae Newman Projection Formula : These projection formulae are obtained by viewing the molecule along the bond joining two carbon atoms. (back C atom) 16.Newman Projection : i) Front carbon as dot (.) Points to remember ® ii) Back carbon as circle ‘O° Examples : * Most of the compound could be represented in H, newman projection formula if particular atoms about which projection has to be made is mentioned. st 1. — oR iy ok H glen Bog 4 Front 3. Nee — cHe HM CONVERSION OF SAWHORSE AND NEWMAN PROJECTION FORMULAE IN FISCHER AND VICE VERSA: Hoon HOH _ 00H COOH HO = COOH CH. H = HO: H 1. HO H te HO, - bee ie roaton about oleae I (Newman) CC bond HO NH CC bond uILLUSTRATIONS ©) raw Fischer projection formulae for the following molecules : i CHO aCe a 5 —cH,0H (8) HOH,C’ HCHC coon HCH, cjne*\"H (0) HOmC—C 0H Ha HC ‘CH,OH CH,OH CHO Sol ont cry @) won CH,CH, CH,OH cHO COOH ils © waton Onc on CHa HOH Convert wedge formula to Fischer formula : CHO 7 Wey £00 @ Hey . HNT H Ho” CHOH COOH CHO Sol (yk H @) oat Hy H,0H COOH wonton SoHs - Convert Fischer formula to wedge formula : Br ® ne—Low 18.COOH Br Br oral ra tee enti (on Hy Wm Hof = “CoHs nH” “CH, onc? *H “ @ Vertical line Sol Convert Sawhorse formula to Fischer formula : Br COOH ome MOSH (e) me B77 “Me COOH Br Br ‘COOH J Me Rotate the rotecule to 180, H—]-Me ceificiek he mobail e180, | 44 “or theplane of paper io Kee Aye | H the (COOH) group on tha top Me—>—H COOH br (Fiscer projection) Hooc roar Meo COOH ©) Me A, ” i Me——H coon = | Br Br, Br (Fischer projection)Conformer and Conformation ‘CONFORMER S| There are &% conformers across given s bond. Definitions Energy needed for C-C bond rotation is 3 Kcal per mole in ethane. @ Isomers formed by rotation across o bond are called Conformers are freely interconvertable into each other and not separable, they are not true isomers. conformers. Conformers are present in dynamic equilibrium. During conformational isomerism bond angle, bond length and configuration of molecule are not changed. || Which of the following shows conformational isomerism : H I 0. A) C, (C) H~ be AN ©) HH H iit (0) yO ON, ) Ht tean HHH AL @, (ee) BOND ANGLE (0) It is an angle between two valency of same atom. Points to remember x » Methane molecule (CH,) { an has bond angle between 2 C-H bonds is 109°28. H H DIHEDRAL ANGLE (4) =) Dihedral angle is not fixed, it keep on changing at _—_—Definitions room temperature, that's why itis called rotational angle of conformer during conformation. * The angle between two valency at different atom obtained in sigma bond rotation, is called dihedral ring. 20.TG I ; H. Br % Ht a a F ZF-C-C-1=180° 2F-C-C-Cl=60° %e © ECLIPSED FORM ver if Introduction — ny H When value of ¢ = 0°, 120°, 240°, o=0 360° then, that form is known H as eclipsed form in eclipsed ee form, valencies of one central Eclipsed conformation atom completely overlapped of ethane by valencies of adjacent atoms. STAGGERED FORM When value of = 60°, 180°, 300° then that form is known as staggered form. In staggered form, two biggest sized valencies attached at central atom must be opposite to each other. H H H $= 180° H H staggered conformation of ethane a.‘SKEW FORM Analysis of conformation : Let's understand @ © Interelectronic Repulsion EC> st Infinite forms which are © Interelectronic Distance ST> EC present in between eclipsed & © Stability ST>EC staggered are known as skew O Energy EC > sT forms. © Mole Fraction ST> EC 0° < 6 < 60" MAJOR FACTORS TO BE CONSIDERED WHEN DEALING WITH CONFORMATIONAL ISOMERISM Torsional energy (strain) : © Bond Pair-Bond Pair electronic repulsion © Itis only considered in eclipsed conformer. Van der waal strain : © Steric strain or strain due to their size is Van der Waals repulsion. © tis maximum in eclipsed conformer. Rotation Barri © Energy difference between most stable conformer & least stable conformer is known as rotation barrier. H £0" rotation, H: H 25 kiimot” H. H H Eclipsed form Staggered form ILLUSTRATIONS : Draw most stable structure. Q1 Propane (2 n-hexane (C, - ¢,) Sol, | oy Sol H, H H H H H CH, GH, 22.QS 2-3 dimethyl butane (C, - C,) OT 2-methyl pentane (C, - ¢,) Sol Ww Sol én, - Guu, - oH, - oH, cu,-chc- cH, be CH,CH, i H CH, H chy cH, H GH, chy cH, CH, H 4 H H CH,— OH Q4 | H é CH,— F on” (H-bonding) Zé N uv lu Sol stability: Gauche > Anti U Sol 4 QS s-hydroxy propane (C, - ¢,) CH. CH, Sol 9° GHP Sn on ¢ b oH @ H O92 Nme,- cn,P en, -coo® H H Qc Sol 4 He H Ht 4 il a) * ih H ‘NMe, 8. coo" Sol (H-bonding) Gauche > Anti Gauche > Anti 23.CH,— COOH Qi0f CH,- COOH Sol @ Low pk (most stable) Case-(b) (b) High pH (stable) High pH= Solution (Basic) Case-(a) COOH + H*® T+ coo* Low pH= Solution (Acidic) cof H. H H. H H COOH H H coor cof Among the given compound/s which having zero dipole moment in any of ‘coformation? RU un’ HO WK «c) (D) None of these H A H H Sol w ; wh 4 221 Sor Bah, — @) ,, H ot Hoa 4 (C) Always same y= 0 -. Only (C) 24,BAEYER ANGLE STRAIN THEORY Anglestrain« «Reactivity 1 Stabilbity Stability order of ring (cycloalkane) 6>52=7>4>3 Conformational Analysis in Cyclic Compound Introduction t ir According to Baeyer cyclic compounds are less stable then corresponding open chain compunds because in cyclic compound angle strain is present. Z\ 109°28" 109°28" 109°28" 109°28" Ob U Cyclopropane © Present in forced eclipsed confirmation. © Highly reactive due to its high angle strain. 60° g0° 108° 109° © Highly reactive, therefore gives test of unsaturation, 49°28 19°28" 1928 20 x x 25.cyclobutane Cyclopentane [7/=\\ Q-28 It undergo jelly flipping. Envelope flipping cyclohexane Racaoeten ifine= © In real sense, cyclohexane is non-planar structure with negligible angle strain so that it is less reactive in nature & highly stable in nature. © It's negligible angle strain also releases by various conformation. CONFORMER OF CYCLOHEXANE al[b elif clld a axial e + equatorial 26.a &e-> axial & equatorial Possible conformer of cyclohexane : res troments * Half chair form is very unstable due to eclipsing of hydrogen. + Boat form is unstable due to flag pole repulsion. ») flag pole «© Twist boat is relatively less repulsion stable than chair form due to strain. XD © Chair form has negligible angle strain. Therefore, most stable Twist boat conformer of cyclohexane © Rigid form does not exist because of eclipsing of HOH twelve hydrogen. H Ht d) ®) H H HOHFlipping x chair <—— chair i} Stability ] Cet = Of twist twist boat boat During flipping, all equitorial bond will convert into axial & all axial bond will convert into equitorial bonds. Hc. HC HC Pets B TB: 7 c Ic TB B c Newman Projection of Cyclohexane Conformer (chair) H H ® 4 ‘CHa H 2 ja H GH: H H H Newmann projection of the pair conformer 28.Q1 Which conformer of methyl cyclohexane is more stable? fe ood ¢ Equitorial is more stable than axial. ¢ Bulky group must occupy equitorial position. @ (A) is unstable because in (A), 1,3 diaxial repulsion is present Compare relative stability : (©) (D) Sol * Chair * Twist boat * Boat Stability * Half chair * Rigid 29.© Find the direction of equilibrium and also identify K,.. Hy , wide orf be fe of" = oLf27 =——T Sol @) forward, K,>1 (B) forward, K_,>1 (C) backward, K,, <1 (0) backward, K,, <1 : : oO H R “RTRRT Explain why substitution at equatorial position changes ,, value drastically up. Sol -r K., -R K, -H 1 -cH, 18 < 35 << 4000 Bulky group is more stable at equatorial as bulkiness increases stability at equatorial position increases. ; ; Rte rt , a -R K, 1s 24 2.2 2.2 Explain? Sol in case of F and Cl, size factor dominates therefore more bulky group at equatorial is more stable. But in case of Cl, Br and | bond length factor dominates. 30.Geometrical Isomerism GEOMETRICAL ISOMERISM ic) Those stereoisomers which are not intercon- Definitions vertible in each other at normal temperature are called configrational isomers. Configurational isomers which are not interconvertible in each other at normal temperature due to restricted rotation are called geometrical isomers, @ Those stereaisomers which have different geometry around restricted rotation unit. Three Condition for Geometrical Isomers. © Restricted rotation © Different axial distance © Different unit at terminal (terminal valency should be different) Free Rotation System WY oft ci (free rotation) HF a | if For example : Butane molecule about C, - C, sigma bond shows rotation. wow Interconvertible at room temperature 1. Restricted Rotatory System : > Prssive aig enperatre CH CH, CH ntercamersenis notpassbe, cal Koc! HMA Wo Non, + wotinerennvenane at roemtenpe aie Ina molecule restricted rotation arises due to _—Peints to remember a) Presence of double and triple bond Gi ipls bend neler show el —— / Ne: 0 QQ. pb only this Example ; CH,-CaC-CH, % AA stow —— (Gt) C=C \Z 3b) Around single bond in cyclic compounds (« bond pure) DuU=1 Example : Ln LP CO act like double bond i te 2 RR RR ©) Polyphenyl having Steric Hindrance Example : H H BG BG H H BG BG Free rotation BG = Bulky group + Small group * Bulky group + Steric hindrance decreases + Steric hindrance increases + Rotation (free) * Rotation (restricted) (This is becouse of steric hindrance rings become perpendicular so, that they don't come back.] d) Cyclic atkene Wox €) Allene System a a a a i. “ Ty a poore=cQ, | Dewa= cK f) Spirane System x, C 7 x x, C C 7 x y y y y We will study this d, ¢, f cases in next lecture. 32.2. Different aerial distance on a a a) eee ‘|s an ® , DoF “es ON ‘SPECIAL POINT TO LEARN Example : Dipole moment and Boiling Point : oot a HH + rex (A) BP x (Dipole moment) mS cme ao ae red (8) Here [¢, #@ © * Geometrical isomerism have different physical & chemical properties Points to remember 33.42 Efficient packaging * MP,< MP, Efficient packaging v (mp) (MP) Geometrical isomerism in cyclic alkane compound cr Examples : Select among the given compounds, _mtreduetion {( which is/are capable to show geometrical Pin veycloalkanes it two isomerism. different sp? hybridised x 7 atom contain different groups then molecule show + No = iss Geometrical isomerism. x a x x x a 3. yes 4. No H x x ‘ ‘en ws CL ws x x ap 7 Yes 8. No A, x x a 3s. of No 10. Yes x 34,n. y Yes 12, Yes z— D HY N 2. [ ) Yes Ts ILLUSTRATION Qi ix} A2 Which of the following double bond will not exhibit geometrical isomerism. Me. Ph Me. (a) Noa’ (B) Se=0 HC Sc gHs, Ph Me © ‘C=N—Me (D) Me - N=N- Me pnt (A, 8) Which of following will not show geometrical isomerism ? a QB (a) c=c () c=He / y / \ Br Br 1 HAC cH, Bt © c=c () =< H H Ph ci (D) 35.ao o A3 04 Aa AS Hy Cr " cf ® aw Which among these are isomers ? (A) land II (C) and 11 A) oH Hy Hy ay OH, (B) | and Il! (0) all of these Which of the following compound can show geometrical isomerism. Bgl @) j=. ' cl fog ©) p= cl ‘et () The geometrical isomerism is shown by si a CHC “ (0) ol >< CH 36.Poly Alkene CUMULATED ALKENE CH, =C =C=C =C=Ch, CONJUGATED ALKENE CH, = CH-CH =C-CH=CH, ISOLATED ALKENE CH, = CH - CH, - CH = CH-CH, - CH = CH, Stereochemistry in even p-bond allene: central carbons sp hybridized nd carbons ate sp! hybridized | and eabane ates hybraitad \e ee S GC, EK ‘of the overlap of the p ortitals ‘rust be at right angles t9 each other ot onty ae the two = bends perpenatcular, but the two methylene erours are 100 Stereochemistry in Odd x-bond allene : ‘Both terminal sides valencies are in same plane Condition for cumulated alkene : © Even cumulated alkene here show geometrical isomerism © Odd cumulated alkene can show geometricalisomerism if terminal valencies are different. / ) p=c=c, i =C iii) 37.GI IN CYCLO ALKENE Points to remember | oO © They can not show geomet- A rical isomerism about alkene due to angle strain in trans form. Baeyer angle strain theory : A Oo yose28' = 109°28'— 109°28" =0° 60° 90° 108° 49°28 128" ss H H H cis-cyclooctene __ trans cyclooctene stability 8 to 11! ——» cis > trans 11 ——+ trans > cis Example: H 7 O ? CO H cis-decene trans-decene 38.7) Minimum number of carbon required for cycloalkene to show geometrical isomerism Al 6 Sol Q2 Minimum number of carbon required to show geometrical isomerism. A2 oa Qa _ 2 ol "NSN H Si GEOMETRICAL ISOMERISM IN SPIRANE 3 Minimum number of carbon required for alkene to show geometrical isomerism A3 4 Hy CHy Sol cad rN HO O4 Minimum number of carbon required for cycloalkane to show geometrical isomerism Q4s5 CHy Sol A CH, By replacing ring from n-bond or n-bond from ring basic geometry remain same. dol O00 HO =O <= Above all have same geometry. odd number of rings —+ Always show G.l. If terminal : : even number of rings never show G.I. from terminal position xx y ¥ valencies are different can show G.I. from non-terminal position a. Db SX yGEOMETRICAL ISOMERISM IN POLYPHENYL SS) Atropisomers : Definitions: /AsMes GH» * Stereoisomers arises due to me sion are called atropisomers. Seed HOH H H H H ef H D D H, H H 3. 4 cf cl H a ch H H p x 5. 6. H ci ai a cl H % a H ( S cl GG, Restricted Free rotation rotation G,-G, = bulky groups at ortho position 40.Points to remember “s © Small groups : -H,-D,-T,-OH, -NH,,-F, Cl, -CH-, -C = C-,-C =Nete. Me ots © Bulky groups : ~Br, -l, -ON2, -COOH, -SO,H, — N + CHs, SH. : Nile CHS Hs — cen, --AsHs ete Nc Restricted rotation unit ¥ Terminal valency different ¥ geometrical isomerism ¥ Note: Even phenyl system never show geometrical isomerism Odd phenyl system can show geometrical isomerism If terminal valencies are different at ‘ortho position. a.Gometrical Isomerism TYPES OF CONFIGURATION IN GEOMETRICAL ISOMERISM Cis trans E-2 Syn - Anti 1. CIS-TRANS Example : HY . vu cae 0“ Sb Trans Cis H D Nee e7 a er, H » °c We" o "Nes cc 0g (Cis) Dp” 4 H 4, ¢ 4 (#=Ph group) rn) H (Cis) (Trans) H H A ia Fe PN ag7! co ™ ci NB Introduction © Same group oriented to same side at terminal valencies- cis © Same group oriented to different. side at terminal valencies- TRANS (Cis) (Trans) fire= 42.2. E-Z SYSTEM E = Entgegen & Z = Zusamman Points to remember @ There is no relationship between E-Z & cis-trans means a molecule having £-Z configuration, may be cis or trans & vice versa. * Same priority group are on opposite side = € © Same priority group are on same side > Z Priority order or rule : (Cahn, Ingold & Prelog ,CIP rule) Rule-1 : Assign priority on the basis of atomic number, higher is the atomic number of connecting atom from Geometrical Isomer unit higher will be its priority. @ Priority order : -| > -Br > -Cl > -F > -OH > -NH > -CH > 2 Concept Building Exercise

Example : UF HH. OK’ a) -CH, - Cl> -CH,- CH, -1 b) —CHQ_ <-OH ¢) “mn, SC = C: ) CH, 27 OH, 1 8) CHE Gray 43.Rule-3 : If connecting valencies having multiple bonds then duplicate the double bond & triplicate the triple bond then check priority. Duplicate -A=B —A-8 BA Triplicate Oo oc a, MACK EN H.C=HC“ = NCH=NH Kone Sol. C=C. He=He~ NcH= NH @ Rule-4 : If cyclic groups are present then priority will be decided by breaking single bond to obtain open chain compound Examples : ma 4 Sol. O-CH,-C 0 -CH,-CH,-C CH, - CH,-O-C Rule-5 : If real group is exactly identical to hypothetical (imaginary) group then priority will be given to real group. Real > Imaginary -CH, - CH, “CH cH > CH, CH, -CH, - CH, ‘Real lod Real @ os Imaginary Example : a4.Rule. If lone pair is present then it will 9° considered as least priority group as lone W cae pair atomic number is zero. 3 YQ nM Example : 7 ron,c% “coct Rule-7 : in case of isotope, priorities will be decided according to atomic mass, higher will be atomic mass, higher is the priority. H D a Yet (trans) / (2) ~ ™T F. H 4 Noc% @) a” ob ‘ S, F @ : _“7N Hoe! Li Hs cN No ac @ @) Cl-CH,-CH, H.-CH, onc’ Ncu,-oH i. Neo crs on HO-CH,-cH”~ No cD, D (CH=CH C—-G—H Zi Br. L 7. mC, © CH, CH. cH ei” Noten” 15. the te Mi Et («e) (2) (E) (E) (Ee) (2) (e) (E) 45.CH CH, Ht CH, CH, Br a 2. €) 16. ‘ = < (2) oc cnc ie) H H ° HS. OH n Now L 22. «) @ a ch, sou Seu, o 18. ° 3) @ oO HOH 19. CO (@ (EEE) 20. (e) 3. SYN- ANTI “c= i Introdueti ime -~ =N-OH (Oxime compound) Be aeeaeenl © Applicable only for nitrogen SS : containing compound where - =N-H (Imine compound) geometrical isomerism is possible. - N= N - (Azo compound) Examples : a) ". e (2) \ P @) "\ P Non Me“ “OH H~ Now Me. H. Nes oO \-2 N (4) (s) C=N, © c ane or HOM Me NY 46.“0 a Ny wn 2 or Now Me 2) 14) C=N. Hc oH 6)Calculation of Geometrical Isomer CALCULATION OF GEOMETRICAL ISOMER Case | = If both the ends are different 2° when n is number of stereogenic area or Pi bond which can show G.I. Case I> If both the ends are same Qe ae ifn = even; P= 0/2 net ifn = odd; P= 22? n= odd Fi Case-1 : Unsymmetrical compounds : 1. CH, - CH = CH — CH = CH ~ CH, ~ CH, Points to remember “— Rules for cyclic system : hs © 3member to7 member cyclo alkene exist in only cis form. Neu, ¥ » 8 to 11 member can form cis 2. CH,-CH=CH-CH=CH-CH’ fost & trans but cis is more stable Case-2 : symmetrical compounds : : © from 12 member trans is 3. CH,~ CH, - CH= CH ~ CH=CH ~ CH, ~ Me more stable. Ans. 3 4 H AOS (Axis of symmetry) ©) L_,, 480s (alternative axis of symmetry) cc) PLANE OF SYMMETRY (POS) (c) SS) In compound POS may or may not be present. Definitions In amolecule more then one POS may be present. While imaging POS thickness of molecule is also considered. + It ie an imaginary plane which bisect the molecule in exact ‘two equal half and they are Those molecule which are planar, they may cut sie ikea SF eich Cobre. along their thickness which is known as molecular plane (slice cut). Example: Bread Following alphabetical letters may be used for elementry understanding of P.O.S. : TA/ A 1S/$ $ tee one, 56.Identify number of following letters which have plane of symmetry. AB C DE

. - Definitions ¢ Aimaginary point in the plane from if we draw two lines in ‘opposite direction meets like atom/group at equal dis- tance, called COS. Points to remember &® * COS is absent in eclipsed conformer. * COS is absent in fischer pro- jection. * COS may present in stag- gered conformer. cos x cos” cos * cos v cos x 58.AXIS OF SYMMETRY (AOS) (C,) n= 360° At 360° molecule repeat itself and that axis is called natural axis and it is not considerable. Linear molecule has infinite AOS. Example : * If a compound is rotated by 360° by n, about a i axis and if we get in- ee guishable compound (config- e uration repeat itself) then +HE=€-H) -O-¢20-) tt is called axis of symmetry. Pe a H H Definitions Axis where value of n is maximum are known as Principle axis or Main axis. 20" iH HBC WH n=3 Ifq = 180° then n=2 CG, If q = 120° then n=3 Cc If q = 90° then n=4 C Weq= 72° then on=5 Ifq =60° then n=6 c . Find out A.0.S. in following molecules : H 1 o=c=0 2 H-BO NH cl cl a. or a Soecc" ore cl H H H. H Ht cl 8 Sescd 6 cect cle Set ow “H ALTERNATE AXIS OF SYMMETRY (AAQS) (C,) Example : Clo HEH cl - C=C c= H Nc EB ci SH 59.Example: Q. Comment about POS, COS and AOS in following molecules : 1. CH, a z Ss. i Ans. POS(s) > ae \ Gs +s c, + Ai Ans. POS(s) > ¢, > * Cc, > ‘ G > 2 Co G > a H \ f Ans. POS(s) > v 6. f=e Cs > oY a v & > Ans. POS (3) + c + ¥ ; c, + ° cy + Cc, > 2 Ans. POS(s) > Ma ¢, = Ans. > Cc > x oy c + ; 3 4 Y 8. Ho Ans. POS (s) > v c, _ ov cS > Ans. POS (3) > c + c 3 2 cy + Cc, > RR RS SRN ersDPP-23 Answer about following molecules in below pattern : Chiral Carbon >I) Pos alls cos >) Geometrical isomerism > ___ (¥/*) 4 y cos 4 c=c=c Geometrical isomerism x \ Ans. <<” Chiral Carbon > * 5 POs > Vv Hc 4 cos > * Ans. Geometrical isomerism >» = * Chiral Carbon Pos Hac fis cos 2 YSc=e, Geometrical isomerism H \ Ans. i Chiral Carbon > * © Pos Sy * 4 cos > * Hy Geometrical isomerism > * Ans. HG Chiral Carbon 3. ‘cx: Pos Hu cos ane Geometrical isomerism Chiral Carbon > * POs >» ¥ Hs cos > * Geometrical isomerism > ¥ Hy 4 Ans. ane Chiral Carbon Chiral Carbon > * FOS: Pos > coe Geometrical isomerism ve vydd tise thsd xxKK KR RK CR KROO Ans. Chiral Carbon Pos cos Geometrical isomerism Ans. Chiral Carbon Pos cos Geometrical isomerism bee bres xe K® Og ON Ans. Chiral Carbon Pos cos Geometrical isomerism e t 14. © Sinister (Left) R-S nomenclature is assigned as follow : Step-1:- By the set of sequence rule, we give the priority order of atom or group connected through the chiral carbon. Step-II :- If atom/group of minimum priority present on the vertical line, then Movement of eyes in clockwise direction = R Movement of eyes in anticlockwise = $ Movement of eyes taken from 1-> 2 > 3 through low molecular weight group (if needed) Step Ill :~ If minimum proirity group present on the horizontal line, then clockwise rotation > S anticlockwise rotation => R 1 1 2 4 4 3 R ACW R ACW 3 2 Example: COOH COOH 2 — Acw 4 4 4H ton 2 Syne Rls R CHs 63.Q1 Indicate whether each of the following structure has the R-configuration or the S-configuration : - 2 > 2R-3S 28-3 2R-3R 2s-3S 12, 13. H—+—D 14. nob, 15. yr 7. oH, cl oY F 1 19. HS: D 3S-3R 2s-3S 28-3R 1R-3R 18-28 64,DIASTEREOMER Example : cis-trans /—=\ AS CH CHy Erythro-threo {JOH = HO——H HOH, H—{—OH CH CHy Two chiral centre + one interchange SS — > Enantiomer RR < RS —> Diastereomer SR — Diastereomer HO Different Relationships of Stereoisomers + Those stereoisomers which are not mirror image of each other. Definitions Three chiral centre + two interchange SSS —> Enantiomer RRS —> Diastereomer RSR —> Diastereomer RSS — Diastereomer RR + Diastereomerhaving different physical properties melting, boiling point, solubility. * Diastereomer are having differentchemicalproperties. + If only one chiral centre then diastereomer is not possible but if there is chiral centre with x bond then diasteromer possible.Example: * 180° cc ——* Identical “Br Br Br B 180° ° — — 7 Enantiomer ci cl Br Br > Enantiomer a “el Br etl ‘ans, diastereomer “el Br Definitions = Ly» postion ¢ Those compounds which superimposed on ___ their cf mirror images are known as identical. IDENTICAL (i) COS / centre of symmetry / C, (ii) POS / plane of symmetry / s Points to remember (iii) AOS / axis of symmetry / C, . (iv) AAOS/ Alternate axis of symmetry /S, # ‘These compounds whichiare ( having chiral centre even Meso = two or more C.C. + POS / COS / AAOS 5 1 . though optically inactive. Br Br ‘. ra » Meso compound are optical- or —f—|-- pos ly inactive because half com- wil pound rotates PPL in clock- ‘Br Br Note: Meso compound are optically inactive but consider in optical isomer because it rotates PPL light but the net result is zero. wise direction and remaining half part in anticlockwise di- rection. 66.Identify meso : COOH COOH H: ‘OH ol H * H—}-0H 2 H—-t—0H ‘COOH COOH Br Br 5. 6. =< cl Cl cl Solutions : 1. Meso 3. Meso 5. Optically active 7. Optically active CH; Br, Br H ‘OH a x aH oH Bre ‘Br H OH H. OH CH, 0 Cl, 7 Cl 0 2. Optically active . Meso 6. Geometrical Isomer, Optically inactive 5 Note : Meso compounds are identical or diastereoisomer but not enantiomer. || Identify following meso or diastereoisomer sa Br 1 and Br ‘Br COOH COOH H OH OH—7—H aH oH EOH—W—H COOH GOOH Answer : 1, Identical 2. diastereoisomers 3. Identical 4. Identical Br Br Bry per 2, and Br r Me Me HHENANTIOMER aS Superimpose ALA eae © Those stereoisomers which Non superimpose BIG are mirror image of each other but do not superimpose COOH COOH ‘on each other are known as enantiomer. Anti Clockwise Hye FROH BR HORS cHy Enantiomers are having same physical properties like melting point, boiling point. solubility except rotation towards plane polarised light. Enantiomers are having same chemical properties except reaction with enzyme. © Equimolar mixture of enant omer is known as Racemic mixture = 50% d + 50% ¢ Optically inactive 68.Chiral-Achiral Molecule POLARIMETER CHIRAL-ACHIRAL MOLECULE ELECTROMAGNETIC WAVE OF LIGHT : OPTICAL ACTIVITY Introduction { ite » Substances which can rotate the plane-polarised light are called optically active substances and the property of a substance to rotate the plane-polarised light is called optical activity. ey Normal light Basic understanding of PPL using slit * The instrument used for measuring the optical rotation called a polarimeter. It consists of two Nicol prisms, one called the polariser (near the light source) and the other called the analyser, a glass tube containing the solution of an optically active compound is placed. Definitions 69.As the arrows indicate, the optically active substance in solution in the cell is causing the plane of the polarized light to rotate. of the emerging light is at a different angle than that of the entering polarized light. OPTICAL ISOMERISM a @ A substance which rotates the plane-polarised light towards right (in the clockwise direction) is called dextrorotatory (Latin : dexter = right) and is usually represented by the prefix d or (+). On the other hand, a substance which rotates the plane- polarised light towards the left (in the anticlockwise direction) is called laevorotatory (Latin : leaves = left) and is usually represented by the prefix ‘’ or (-). Behaviour of Optical Active and Inactive Molecules towards PPL [the plane of polarization nas not been rotated _tirection of ight propagation WE y iG Wl Cy): Nt A oral eee sample tube plane-polarized 70._dicection of light eropagation _ Chirality of molecules and objects : An inspection of the molecular structures of optically active compounds indicates that all these compounds have Chiral* (or dissymmetric) molecules. A molecule (or an object) is said to be chiral or dissymmetric, if it is not superimposable on its mirror image, and the property of non-superimposability is called chirality. On the other hand, a molecule (or an object) which is superimposable on its mirror image is called achiral (non-dissymmetric or symmetric). Specific rotation : 0 G Where Wis the optical rotation, / is the length of the tube in decimetre, C is the concentration of the solu! n in g mt", D is the wavelength of sodium light, and t is the temperature. Racemic forms and Enantiomeric Excess : A sample of an optically active substance that consists of a single enantiomer is said to be enantiomerically pure or to have an enantiomeric excess of 100%. An enantiomerically pure sample of (S)-(+)-2-butanol shows a specific rotation of +13.52°. On the other hand, a sample of (S)-(+)-2-butanol that contains less than an equimolar amount of (R)-(-)-2-butanol will show a specific rotation that is less than +13.52° but greater than 0°. Such a sample is said to have an enantiomeric excess less than 100%. The enantiomeric excess (ee) is defined as follows : % Enatiomeric excess _ moles of one enantiomer —moles of other enantiomer 155 . total moles of both enantiomers % Enantiomeric excess = observed specific rotation __ 99 specific rotation of the enantiomer2 A mixture of the 2-butanol enantiomers showed a specific rotation of +6.76°. We would then say that the enantiomeric excess of the (S)-(+)-2-butanol is 50%. +6.76" 413.52° Sol enantiomeric excess = Conditions for optical activity : For example : CI—G © It must have non-superimposable mirror image. © Itshould net contain any element of symmetry, ie, i) Plane of symmetry or (mirror plane or s-plane). ii) Centre of symmetry or centre of inversion. i) Alternating axis of symmetry. Example: 4 2 -CO,CH,CH, 0, C\Z Cones E™CO,CH,CH, 0, C A A Optically active > * Chiral molecule > x Achiral molecule > = ¥ Optically inactive +3 Vv Reason : POS ¥ COS ¥ *100 = 50% Introduction Points to remember It must have an assymmetric C atom or the chiral C atom. C atom which is attached to four different atoms or groups is called an asymmetric C atom or the chiral centre or stereogenic or stereocentre. Chiral, optically active then POS as well as COS should be absent. Achiral, optically inactive then POS and COS both ab- sent. Compound to be achiral it should have POS or COS or both. There is no relationship be- tween AOS and optically ac- tivity of molecule. (We @ 72.CO,CH,CH,OH 2 CO,H Optically active = > Chiral molecule > Achiral molecule > Optically inactive > a H ch 3, Hin Optically active > Chiral molecule > Achiral molecule > Optically inactive > Reason: POs ¥ HH 4, ‘COOH o% Optically active > Chiral molecule > Achiral molecule > Optically inactive > nF 5. & ‘Br ony Optically active > ih Sse we aK Chiral molecule Achiral molecule Optically inactive Reason : POS * cos Me ‘Me Optically active Chiral molecule Achiral molecule Optically inactive Reason : POS ¥ cas ¥ % % Optically active Chiral molecule Achiral molecule Optically inactive Reason : Me “H Optically active Chiral molecule Achiral molecule Optically inactive tue vise bude bod aoe x wR RK wRReason : POS * cos « Optically active Chiral molecule Achiral molecule Optically inactive Reason : POS ¥ 10. Hockey shaped molecule Optically active Chiral motecule Achiral molecule Optically inactive Reason : pos v n Hin, Ho! ‘COOH Optically active Chiral molecule Achiral molecule Optically inactive Reason : Optically active Chiral molecule ved * = = > > > =. eR RK Cin x wR OK Achiral molecule > wv Optically inactive + Reason : Pos v 07% von 13. Ho ‘OH OH Optically active > v Chiral molecule > Achiral molecule > * Optically inactive > — * Reason : POS x COS * Me a 14. > Ke d "4 Optically active + x Chiral molecule > i Achiral molecule > ¥ Optically inactive > v Reason : cos v Optically active > 0 Chiral molecule > 0 Achiral molecule > id Optically inactive >» Reason: Pos ¥ 74,D/L, Epimer, Anomer / Erythro-Threo / Amine flipping D/L NOMENCLATURE : Rule: (This nomenclature is only applicable for fischer projection of carbohydrate and amino acid. (i) Carbohydrate means optically active polyhydroxy carbonyl compound. (iil) Amino acid is the monomer of protein. (iv) For D-L nomenclature, fischer must be correct. Condition : (a) Carbon containing main functional group at top (b) Maximum carbon on vertical line. (©) For one chiral centre if OH/NH, group on right side then D and if on left side then L. HO CHO H UNH 8 NH2/HO- HOH HOH D L (@) If two or more chiral centre are present then in amino acid consider first chiral centre and in carbohydrate consider last chiral centre for D, L configuration. ©. Identify the D/L configuration COOH H——NH, 1. HOH L-Carbohydrate CHy COOH HTH) Sol. HO: H Right; D-Amino CH, CHO H—}—0H D-Carbohydrate 2 HO—}—H 75.CHO H: OH 4 HO H H ‘OH H ‘OH CH,OH CHO H—,;—OH ‘Sol. HO——H D-Glucose ERYTHRO AND THREO DESIGNATION The stereoisomers in which the same groups are present on the same side of the Fischer projection are called erythro isomers while the stereoisomers in which these groups are present on the opposite sides are known as threo isomers. Example-1: CHO. t CHO HORN | | er] t—H ntron) | I wor H CH,OH ' CH,OH D-Erytwose =’ = Enantiomor (OH) on the same side D-Carbohydrate L-Carbohydrate + Compound with two adjacent chiral (stereogenic) C atoms are sometimes designated by prefix ‘erythro or threo! Definitions Example-2: CHO { CHO (oy | 0) H | # 68) cHOH | CH20H D-ttreese Eramtiorer (OH) or diferent sides 76.Example-3 = COOH COOH H—-7B8r Br->—H n—tter) 4 41 a, CH. one met Let's understand @ 2.3-Dibromo butanoic acid Such an amine and carbanion with a lone pair of electrons (being considered a fourth different group) are chiral OPTICAL ACTIVITY OF TERT-AMINES AND TERT-CARBANION OF THE TYPE (R,R,R,N:) AND (RRR CY because of their pyramidal The energy required for this inversion is very low at geometry, but they are optically room temperature, since this inversion does not inactive. A rapid ‘umbrella’ type involve bond breaking and bond formation. Thus inversion converts either of racemisation occurs and enantiomers cannot be the enardomers-to a facemic: isolated. mixture. Ry oN: " Ry C font Natragen \, Cinversion \ RS inversion 4 RS 4 But amine salts are optically active. Nitrogen inversion in not possible, e.g. [R.R,R,N*H]CI. AMINE FLIPPING (UMBRELLA INVERSION / AMINE INVERSION) 1G, = 6 kcal/mol 1.“Why compound (8) has rapid inversion tendency compare to compound (A) Q =g noNST ome} Set Do et (A) (B) g g N Sol abe < Me] Set Bulky group D et (A) (8) Bulky valencies at sp* hybridised nitrogen atom tends more flipping. AG (A) > (B) ‘SOME MORE EXAMPLES OF OPTICALLY ACTIVE N-CONTAINING COMPOUNDS : Me. #8 |. Me 0 vi i MeN Me’ Me (+) form I (©) form 1,2.2-Trimethyl aziridine ) 1-Chloro-2-methyl aziridine shows four stereoisomers. RACEMIC MIXTURE Introduction { The optical activity of a racemic mixture is due to external compensation, i.e. the rotation » When equal amounts of caused by molecules of one enantiomer is exactly two enantiomers are mixed cancelled by an equal and opposite rotation together, it gives an optically caused by the same number of molecules of the inactive form called racemic Gther enantiomer mixture or modification. + It is denoted by using the prefix + or di before the same of compound. 78.Points to remember chemical reagents. Example: Lactic acid and Tartaric acid Resolution of Racemic Mixture: © Enantiomers have identical physical and chemical properties except towards optically active reagents, they cannot be separated by the usual techniques such as fractional, crystallisation, fractional _—_ distillation, chromatography, etc. Therefore, special methods are used to achieve their separation. Some of these are listed below i) Mechanical method ii) Biological method ili) Chemical methods (by making diastereomers) iv) Chromatographic special adsorbents. their separation by using Diastereomeric method These can be separated from one another by conventional methods of separation of compounds. The separated diastereoisomer is then broken down to give pure enantiomers. Chemically we can seperate enantiomer via diastereomerism i) d+t + dorl ——+ (Racemic mixture) M+) + (+4) or (-) ————+ @e © Aracemic mixture can also be obtained chemically. The process by which one enantiomer (+ or -) of an optically active compound is converted into a racemic mixture is called racemisation. Racemisation can be carried out by heat, light or Definitions © The process of separation of racemic mixture into its (+) and (-) enantiomers is called resolution. Introduction * One of the most common method is to allow a racemic mixture to react with an enantiomer of some other compound. This changes a racemic form into a mixture of diastereomers which have different solubilities and melting and boiling points. d-d+t-t (Diastereomer] G+, -) (Diastereomer) iit 79.Example-1: ‘RESOLUTION OF 1-PHENYLETHYLAMINE orm at 0) phendattamine (R,R}-(+)-tartaric acid £16/ka remains in solution Example-2 : oO 9° I 18 | a Il 18 R—C—OH + R—OH R—C—O—R + R—C—O—R + ++ + | (Racemic Mixture, + -) 80.The importance of stereochemistry : Chiral molecules can show their handiness in many ways, including the way they affects human being. One enantiomeric form of a compound called limonene is primarily responsible for the odor of oranges and the other enantiomer for the odor of lemons. o All naturally occurring sugars, including the sugars that occur in DNA, are of D-configuration. the yeast enzymes can specifically ferment D-glucose and not its L-enantiomer. © Stereochemistry also plays an important role in deciding the physiological properties of compounds. {-)-Nicotine is much more toxic than (+)-Nicotine, (+)-adrenaline is very active in construction of blood vessels than (-)-adrenaline. © Chirality is crucial for the effect of drugs as well. In majority of cases, only one enantiomer is found to have the desired effect while the other isomer may be totally inactive or has. an opposite effect. Example : It is the (8)-enantiomer of ibuprofen that has the pain-relieving action. (-)-Thyroxine, an amino acid of thyroid gland, speeds up metabolic processes and causes nervousness and loss of weight. Its enantiomer, (+)-thyroxine, has none of these effects but is used to lower the cholesterol levels. Ring Flipping fap od Non resolvable 81.Calculation of Optical Isomer (O.1.) CASE- For calcul: OR UNSYMMETRICAL MOLECULES ion of of isomers, in unsymmetrical molecule always apply Total Number of Optical Isomers = 2° where, n = Number of Chiral Centre / Number of even allene system / spiro / biphenyl (>) Calculate number of optical isomer (if possible) : * CHa 4. H,C—CH—CH,CH, 5. H——OH Sol. n = 1 chiral centre Hoe re2 HO, Hy HO Sol. n = 2 chiral centre rR is Bea 2. CH, ~ CH = C = CH- CH, Sol. n = 1 chiral centre 6 222 cl cH, "Nee M BN ce Sol. n = 2 chiral centre Hcy Nu 7 a4 HOH HOH 2% CHO CH,OH 2 chiral centre ‘oct Sol. n = 2 chiral centre Pea 82.Sol. n = 2 chiral centre Cl Pa4 ‘ai ° Br Sol. n = 2 chiral centre 10. Bea a) cl Sol. n = 0 chiral centre No optical isomerism possible ks “Br 6 rc . —a —— 1 If optical isomer and geometrical isomer nt. same centre then consider optical isomer here geometrical isomer included. Sol. n = 2 chiral centre CHs Bes 13. H H Br CH Sol. n = 1 chiral centre 222 cHs : cH, H H H H er 8r H CHa CHa (Number of possible optical isomers for following molecules : 2. 2,4-dichloropentane Sol. n = 2 chiral centre Optical isomer = : On Sol. n = 8 chiral centre Optical isomer = 256 Sol. n = 2 chiral centre Optical isomer = 4 83.cl 4. KX rorrcreon- 6 ot 8 Sol. n =1, spirane system Sol. n = 1 chiral centre Optical isomer - 2 Optical isomer = 2 Sol. n = 2 chiral centre Optical isomer = 4 CASE-II: Calculation of optical isomers in Symmetrical molecule : Condition : POS or / and COS present n= Even meso + optically active = optical isomer ax Total Optical Isomers. n= Odd We suggest to draw optical isomer / stereoisomer in case of n = odd number D5 catcutate all possible optical isomers for following molecules : Br ony 1 3. ‘Br ae Sol.n=2 ' Optical isomer = 3 Sol. n = 2 CH, Optical isomer = 3 2! ct ‘ cl H CH. cl cl Sol. n = 2 Sol. n = 2 chiral centre Optical isomer = 3 Optical isomer = 3 84,| Tips and Trick & symmetrical molecule + POS / COS Meso Optically active Optically isomer 2: 1 2 3 3 2 2 4 4 2 3 10 5 4 12 16 9 CHs n—ton H—}—0H Sol.n CH; Optical isomer = 3 Sol.n=4 > 42" > 8+2=10 ‘24 raw all pessible optical isomers (consider pseudo chiral centre) : ca Br, BrH—-—OH 2. H—7—OH H——0H CHs Sol. n = 3 (one pseudo) Optical isomer = 4 Br 3. Br Br Br Br “QQ ef ‘er Bt ‘ar S ar br Optical isomer = 4 86.How to calculate stereo isomer (S.1.) In unsymmetrical, Stereoisomer = 2" Points to remember Ol -» Number of chiral (O1 catcutate stereo isomer : ee > Even allene system / 1. CH,-CH = CH - Br " spiro / biphenyl Sol. 2 GI + x-bond cis /—\,, Trans \S > Ring 5 > Odd system allene / spiro / phenyl a piro / pheny 2. a Sol. 2 =2 q Cis Trans a a CH =C=C=CH-CH, Sol. 2° = 2 H H H cr \ pa Trans = S=c=e=c ne! \ 4 HyC—CH—CH,— CH a Sol. 2’ = ets oy 87.Br ark Br er Bp ci R cl “aRot cH “a Br Sol. 2? = 4 CH=cH—CHy O er Sol. 2°= 8 Sqlealenion Cis “ " Cis Or oer Ore Trans Tess we Tame Trans 8. HaC CH CH=CH CH=CH, CHS Sol. 2'= 2cl 4. HxC—CH=CH—CH—CH=CH—CH, Sol. Stereo |somer = 4 a ql a oxfice vane—[ tans cf tans vare—[ 4 H. H Pos Pos a ‘enanitomere pair 10. ‘ | GL=8 SLO. 7 meso + 2 enantiomerSUBJECTIVE 5. Find: ENS s > 3 Mom 4 S + 2 OA > 2 Mo o+ 0 7. 2,4-dichloropentane OA > 0 ce * Mos 4 on > 2 Ss os «4 8. pdidichloropientans M > 0 mn 2 3 OA > 4 oma = ~ “ me 2 CH=CH 9. 1,2-dichloropentane Mow. @ Ss > 2 Mo> 0 OA > 2 OA > 2 os aa? 1 Ss > 2 6 om» F Mo o> 0 w Sf OA > 2 OA 3 (2 CH, #+—c * — " Hy Ss > 4 S + 4 Mo o> 0 Mo-+ 0 OA > 4 OA > 4 GH wa c—f—-H cH, Total number of stereoisomers Number of meso compound Number of optically active isomers 90.