Stereochemistry PDF

Stereochemistry 3D Chemistry Chemistry in Three Dimensions Dr. Madlen B. Labib Professor of Pharmaceutical Organic Chemistry 3/25/2024 Dr. Madlen B. Labib 1 ...

Stereochemistry 3D Chemistry Chemistry in Three Dimensions Dr. Madlen B. Labib Professor of Pharmaceutical Organic Chemistry 3/25/2024 Dr. Madlen B. Labib 1  Isomerism a phenomenon exhibited only by organic compounds two or more different compounds (not identical) that: have the same molecular formula (M.F.) but differ in physical and/or chemical properties. Example: Ethanol & Dimethyl ether O OH - the same M.F (C2H6O) but -different physical and chemical properties, 3/25/2024 -→ they areB. isomers Dr. Madlen Labib 2  Isomerism Constitutional Stereo Isomerism (structural) (3D) They have the same M.F. but differ in have the same bonding connectivity (arrangement) of atoms in sequence, the molecule differ in spatial arrangement of [the way that atoms are connected to their atoms (i.e. atom orientation each other] in space). 1) Chain (Skeletal) Isomers: 2) Positional Isomers: 3) Functional group Isomers: 3/25/2024 Dr. Madlen B. Labib 3  Constitutional (Structural) Isomers 1) Chain (Skeletal) 2) Positional 3) Functional group differ in the sequence of have the same carbon differ in type of C atom attachment which skeleton functional groups. constitutes (forms) their differ in position of the Examples: skeleton substituent group. Examples Ethanol & Dimethyl 1) n-Butane & iso-Butane Examples ether [M.F. = C4H10] 1) 1-Pentanol & 2- O Pentanol OH OH OH Acetone (propanone) 2- n-pentane & iso- 2) o-nitrophenol & m- & Propanal pentane & neo-pentane nitrophenol [M.F. = C5H12] 3/25/2024 Dr. Madlen B. Labib 4  QUIZ Q1: Give the Relationship between the following Isomers? 1) Acetic acid and Methyl formate. 2) 2-Pentanone and 3-Pentanone. 3) n-Pentane and 2,2-dimethyl propane. 5) m-Aminobenzoic acid & p-Aminobenzoic acid. 3/25/2024 Dr. Madlen B. Labib 5 3/25/2024 Dr. Madlen B. Labib 6  isomerism Constitutional (structural) Stereo Isomerism (3D) have the same bonding sequence, differ in spatial arrangement of their atoms (i.e. atom orientation in space). 1) Geometrical 2) Enantiomers 3) Conformers Isomers (Optical Isomers) 3/25/2024 Dr. Madlen B. Labib 7  Enantiomers (Optical Isomers) stereoisomers which are non-superimposable mirror image molecules just like right and left hands. have at least One Chiral carbon (C atom attached to four different groups, C*) identical physical properties (e.g. b.p., m.p, solubility.....etc.) and same chemical properties. they rotate plane-polarized light, so, they are called optically active isomers. But they differ in: i) Their effect on plane polarized light i.e. One Enantiomer rotates the plane- polarized light to Right and called dextrorotatory, (d), (+), the Other Enantiomer rotates the plane polarized light to Left and is called levorotatory, (l), (-). ii) Their interaction with optically active molecules such as enzymes and cell receptors. →their rate of reaction with these molecules and consequently their physiological effect 3/25/2024 Dr. Madlen B. Labib 8  stereoisomers which are non-superimposable mirror image molecules just like right and left hands. 3/25/2024 Dr. Madlen B. Labib 9  have at least One Chiral carbon (C atom attached to four different groups, C*) 3/25/2024 Dr. Madlen B. Labib 10  they rotate plane-polarized light to right or to left , so, they are called optically active isomers. they have identical physical properties (e.g. b.p., m.p, 3/25/2024 solubility.....etc.) and the same Dr. Madlen B. chemical Labib properties 11  enantiomers differ in: Their effect on plane polarized light One Enantiomer rotates the plane polarized light to Right “dextrorotatory, (d), (+)” the Other Enantiomer rotates the plane polarized light to Left “levorotatory, (l), (-)”. 3/25/2024 Dr. Madlen B. Labib 12  If the comp. NOT rotate the polarized light,... comp. optically inactive. If the comp. rotate the polarized light to right,... comp. optically active, d(+) If the comp. rotate the polarized light to left,... comp. optically active, l(-) **The compound to be optically active it must not be symmetric (must be chiral).... If the comp. has any element of symmetry (achiral compound) it will optically inactive. Chiral compounds are optically active compounds. 3/25/2024 Dr. Madlen B. Labib 13  enabtiomers differ in: 2. Their rate of reaction with optically active molecules (enzymes and cell receptors) 3. Their physiological effect 3/25/2024 Dr. Madlen B. Labib 14  CHIRALITY “Molecular Asymmetry” An object is called Chiral if it is non-superposable on its mirror image. (i.e. the compound is present in two forms: 2 non-superposable mirror images = 2 enantiomers) 3/25/2024 Chiral compounds are optically active Dr. Madlen compounds. B. Labib 15 3/25/2024 Dr. Madlen B. Labib 16  CHIRALITY “Molecular Asymmetry” An object is called Chiral if it is non-superposable on its mirror image. (i.e. the compound is present in two forms: 2 non superposable mirror images = 2 enantiomers) Chiral compounds are optically active compounds. Conditions of Chirality 1) Presence of one or more Chiral (asymmetric center): atom that bonded to different groups 2) Absence of any element of symmetry. * 3/25/2024 Chiral Dr. Madlen B. Labib center Achiral compound; 17 with no chiral center  CHIRALITY “Molecular Asymmetry” I II I & II are 2 mirror image pair, have different spatial arrangement They are non-superimposable. They are enantiomers 3/25/2024 Dr. Madlen B. Labib 18  CHIRALITY “Molecular Asymmetry” 2,3,5-Trimethylhexane H3C CH 3 H3C H CH3 H C C CH 3 H3C * * CH2 H2C H H CH CH H3C CH3 H3C CH3 (A) (B) A & B are 2 mirror image pair, have different spatial arrangement They are non-superimposable. They are enantiomers 3/25/2024 Dr. Madlen B. Labib 19  QUIZ Q4: Which of the following is optically active (i.e. contain C*) a) CH3CH2CH2OH. b) CH3CH2CH(Br)CH3. CH3 H3C CH CH C2H5 CH CH2 CH3 H3C CH CH2 CH3 CH3 CH3 CH2CH3 H3C CH2CH3 CH3 (H3C) 2HC H CH3 CH CH2CH3 3/25/2024 Dr. Madlen B. Labib 20  CHIRALITY Elements of Symmetry 1) Plane of Symmetry 2) Centre of Symmetry imaginary line divides the molecule into two halves each is mirror image to the other. Compounds containing planes of symmetry are called Meso-compounds They are optically inactive. 3/25/2024 Dr. Madlen B. Labib 21  CHIRALITY Elements of Symmetry 1) Plane of Symmetry 2) Centre of Symmetry imaginary point in the center of a molecule if a line is drawn from any element and extended an equal distance after this point, an identical element will be present. Compounds containing centers of symmetry are optically inactive. 3/25/2024 Dr. Madlen B. Labib 22  enantiomers: mirror-image isomers; pairs of compounds that are non superimposable mirror images. chiral : different from its mirror image; have an enantiomer. achiral (not chiral): identical with its mirror image Compounds have NO chiral atom are optically inactive (except atropisomers). Compounds have ONE chiral atom are optically active. While Compounds have More Than One chiral atom may be optically active. 3/25/2024 Dr. Madlen B. Labib 23  Meso compounds compounds divided into 2 equal parts each part is a mirror image to the other optically inactive (internal compensation) e.g. meso tartaric acid COOH H OH H OH plane of symmetry COOH 3/25/2024 Dr. Madlen B. Labib 24  Racemic Mixture Equimolar mixture of two enantiomers (R&S) (50% of each) optically inactive due to external compensation 3/25/2024 Dr. Madlen B. Labib 25  ATROPISOMERS Chiral compounds i.e. they are non-superpossible on their mirror images although They contain No chiral carbons. So, they are optically active compounds. They are classified into: 1) Biphenyls 2) Allenes 3) Spiranes 3/25/2024 Dr. Madlen B. Labib 26  ATROPISOMERS 1) Biphenyls 2) Allenes 3) Spiranes Two conditions are necessary for a biphenyl compound to be non-superposable on its mirror image (chiral): a) Neither ring contains plane of symmetry. b) Large substituent in ortho positions of one or the two rings, so the biphenyl lose its planar configuration by twisting its two rings and the angle between them approaches 90o. 3/25/2024 27 Dr. Madlen B. Labib  ATROPISOMERS 1) Biphenyls 2) Allenes 3) Spiranes 1 3 1 3 2 2 (No plane of symmetry, Optically active) The center carbon of Allenes [C-2] forms two π bonds, perpendicular to each other. Consequently, substituents at terminal carbons [C-1, C-3] lie in two perpendicular planes The whole compound is chiral. 3/25/2024 Dr. Madlen B. Labib 28  ATROPISOMERS 1) Biphenyls 2) Allenes 3) Spiranes They are compounds containing two rings at two perpendicular planes due to restriction of rotation. 3/25/2024 Dr. Madlen B. Labib 29  Molecular Representation 1- Compounds containing ONE chiral carbon vertical Below plane In plane horizontal Above plane In plane Line bond → in plane dotted bond → below plane wedged bond → above plane. 3/25/2024 Dr. Madlen B. Labib 30  Molecular Representation 1- Compounds containing ONE chiral carbon In Fisher representation: odd (1, 3,...etc.) interchanges OR rotation by 90o gives the mirror image (Enantiomers) I II III I, II → enantiomers II, III → enantiomers 3/25/2024 →Madlen I, III Dr. identical B. Labib 31 3/25/2024 Dr. Madlen B. Labib 32  Molecular Representation 1- Compounds containing ONE chiral carbon But even (2, 4,...etc.) interchanges OR rotation by 180° gives the same compound (Identical) I II III I, II → identical, the same II, III → identical, the same 3/25/2024 I, III → Dr.identical, Madlen B. Labib the same 33 3/25/2024 Dr. Madlen B. Labib 34  Molecular Representation 2- Compounds containing TWO chiral carbons R2 T L2 R2 L2 T L1 R1 R1 B B L1 3/25/2024 Dr.R T: Top, B: bottom, R1: first right, Madlen B. Labib 35 2: second right, L1: first left, L2: second left.  Molecular Representation 2- Compounds containing TWO chiral carbons Notes: 1) In Fisher projection, odd (1, 3,...etc.) interchanges gives the mirror image (Enantiomers). 2) In Fisher projection, even (2, 4,...etc.) interchanges or rotation in plane 180oC gives the same compound (Identical, the same). 3/25/2024 Dr. Madlen B. Labib 36 3/25/2024 Dr. Madlen B. Labib 37  Molecular Representation 2- Compounds containing TWO chiral carbons 3) In Sawhorse projection, In Sawhorse projection, rotation around the asymmetric carbon with any angle and at any direction does not affect the configuration of the compound. A, B and C are identical and so, their Fisher and Newmann projections are 3/25/2024 also the same. Dr. Madlen B. Labib 38 3/25/2024 Dr. Madlen B. Labib 39  Nomenclature of stereoisomers 1- RELATIVE CONFIGURATIONS 1) D & L Configuration 2) Erythro & Threo Cofiguration H OH HO H CH2OH CH2OH In all sugars D series L series 3/25/2024 Dr. Madlen B. Labib 40  RELATIVE CONFIGURATIONS 1) D & L Configuration 2) Erythro & Threo Cofiguration same groups in the same side called erythro same groups on opposite sides called threo 3/25/2024 Dr. Madlen B. Labib 41  2- ABSOLUTE CONFIGURATION How to determine absolute configurations (R & S) 1) Atoms around the chiral carbon are arranged in priority from 1 to 2 to 3 (1 → 2 → 3) according to their atomic number as follows: I > Br > Cl > S > F > O > N > C > D > H. S R Anticlockwise Clockwise 2) the smallest atom, [ # 4] is 3) the smallest atom, [ # 4] is vertical and the order from 1 to 3 is vertical and the order from 1 to 3 3/25/2024 Anticlockwise ( ) Dr. Madlen B. Labib is Clockwise ( ) 42  How to determine absolute configurations (R & S) 2 1 3 (S) 3/25/2024 Dr. Madlen B. Labib 43  How to determine absolute configurations (R & S) 4) If atom 4 (smallest atom) is horizontal and the order of arrangement from (1 → 2 → 3) is anticlockwise , so configuration is R. 5) If atom 4 (smallest atom) is horizontal and the order of arrangement from (1 → 2 → 3) is clockwise , so configuration is S. 2 2 4 1 1 4 Lowest group is horizontal 3 3 Lowest group is horizontal anticlockwise clockwise R-configuration S-configuration 3/25/2024 Dr. Madlen B. Labib 44  QUIZ Q5: Find the Relationship (Enantiomers, Identical) between the following: R R R S identical Enantiomers l 3/25/2024 Dr. Madlen B. Labib 45  (C) (D) (A) (B) Show relation between: A&B A&C A&D B&D B&C C&D 3/25/2024 Dr. Madlen B. Labib 46  How to determine absolute configurations (R & S) If two atoms attached to the chiral carbon are the same, we compare the atoms attached to these first atoms. (R) 2-chlorobutane Clockwise the smallest atom, [ # 4] is vertical and the order from 1 to 3 is 3/25/2024 Dr. Madlen B.Clockwise Labib 47  ABSOLUTE CONFIGURATIONS (R & S) The double or triple bond is considered to be equivalent to two or three atoms S Anticlockwise the smallest atom, [ # 4] is vertical and the order from 1 to 3 is Anticlockwise 3/25/2024 Dr. Madlen B. Labib 48  ❑ In case of two chiral carbons, we should consider each chiral center separately. the first chiral is simplifed as drawn CH3 & the groups are ordered. 3 3 CH3 H OH CH3 H OH H OH H OH 4 1 1 R clockwise but 4 in the horizontal C2H5 R 2 2... S A the second chiral is simplifed as drawn & the groups are ordered. 2 2 R R 4 1 1 H OH H OH C2H5 C2H5 Counterclockwise but 4 in the horizontal 3 3... R CH3 S H OH... R... 2S,3R-2,3-pentanediol H OH C2H5 3/25/2024 Dr. Madlen B. Labib 49 2S, 3R  H H H C3 H H CH3 2C H4 (S) C H3CH2C H 1 C C C H3CH2CH2C CH(CH3)2 C H (R) CH3 H2C H H H H C1 H C C CH3 3 2 C H CH3 4 3/25/2024 Dr. Madlen B. Labib 50  CH3 CH3 CH3 CH3 S R S R H OH H OH HO H HO H R S S R H OH HO H H OH HO H C2H5 C2H5 C2H5 C2H5 a b c d 2S,3S-2,3-Pentanediol 2R,3R-2,3- 2S,3R-2,3-Pentanediol 2R,3S-2,3-Pentanediol Pentanediol ** The relationship between a & c or d is diasteromers. also, the relationship between b & c or d is diasteromers Diasteromers: stereoisomers that have same molecular formula, structure but different spatial arrangement & different physical & chemical properties & not mirror images. 3/25/2024 Dr. Madlen B. Labib 51  Enatiomers Diastereomers 1- Mirror images & non superpossible. 1- not mirror images. 2- Identical physical and chemical properties. 2- different physical and chemical properties. 3/25/2024 Dr. Madlen B. Labib 52  CH3 CH3 (S) (R) H3CH2C H H CH2CH3 H3CH2CH2C CH(CH3)2 (H3C) 2HC CH2CH2CH3 (R) (S) H H (A) (B) CH3 CH3 (S) (R) H3CH2C H H CH2CH3 (H3C) 2HC CH2CH2CH3 H3CH2CH2C CH(CH3)2 (S) (R) H H (C) (D) A and B are enantiomers, C and D are enantiomers, A and C are diastereomers, A and D are diastereomers, B and C are diastereomers, B and D are diastereomers 3/25/2024 Dr. Madlen B. Labib 53  2) In 3D structure: Rules: 1) Rank the groups (atoms) bonded to the asymmetric center in order of priority, (according to their atomic number). 2) Orient the molecule so that the group or atom with the lowest priority (4) is directed away from you. (bonded by a hatched wedge). Then draw an imaginary arrow from the group or atom with the highest priority (1) to the group or atom with the next highest priority (2) then (3). If the arrow points clockwise, the asymmetric center has R configuration. If the arrow points counterclockwise, the asymmetric center has S configuration. [N.B. make sure the arrow is drawn from 1 to 2 to 3 whether there is another atom in between.] 3/25/2024 Dr. Madlen B. Labib 54  3) If the group with the lowest priority (4) is not bonded by a hatched wedge, interchange group 4 with the group bonded by a hatched wedge. Then complete as before. Draw an arrow from the group or atom with the highest priority (1), to the atom or group with the second highest priority (2). Because you have interchanged two groups, you are now determining the configuration of the enantiomer of the original molecule. So, if the arrow points clockweise, the means of the original molecule has the S configuration. On the other hand, if the arrow points counterclockwise, the enantiomer (with the interchanged groups) has the S configuration, which means the original molecule has the R configuration. 3/25/2024 Dr. Madlen B. Labib 55  Example: 1 the group with the lowest priority CH(CH3)2 is bonded by a hatched wedge. 4 2 H H3CH2C CH3 3 (S)-2,3-dimethylpentane 2 2 CH2CH3 CH2CH3 1 3 switch CH3 and H CH3 1 4 (H3C) 2HC H H (H3C)2HC CH3 4 3 S-configuration R -configuration 3/25/2024 Dr. Madlen B. Labib 56  QUIZ Q7: Determine the absolute configuration of the following? S R S 3/25/2024 Dr. Madlen B. Labib 57  QUIZ Q8: Draw the following? 1) (R) 2-Butanol. 2) (S) HOOCCH2CH(OH)COOH. 3/25/2024 Dr. Madlen B. Labib 58  NUMBER OF STEREOISIOMERS 1) Number of stereoisomers = 2n, where n is number of chiral carbons. Compounds containing One chiral carbon give two isomers (R&S). Compounds containing Two chiral carbons give four isomers. Example: CH2(OH)CH(OH)CH(OH)CHO has four stereoisomers. 3/25/2024 Dr. Madlen B. Labib 59  NUMBER OF STEREOISIOMERS CH2(OH)*CH(OH)*CH(OH)CHO I II III IV 4 stereoisomers (I, II , II , IV) I (R,R) and II (S,S) → Enantiomers III (S,R) and IV (R,S) → Enantiomers I (R,R) and III (S,R) → Diastereomers I (R,R) and IV (R,S) → Diastereomers II and III → Diastereomers 3/25/2024 II and IV → Diastereomers Dr. Madlen B. Labib 60  NUMBER OF STEREOISIOMERS N.B. In meso compounds, number of stereoisomers =.three isomers rather than 4 isomers I II III (Meso tartaric acid) optically inactive no mirror image 3/25/2024 Dr. Madlen B. Labib  QUIZ Determine R&S configuration of previous Tartaric acid isomers and find the relationship between them? 2S, 3S 2R,3R MESO 3/25/2024 Dr. Madlen B. Labib 62  Thank You 3/25/2024 Dr. Madlen B. Labib 63