Organic Chemistry Lecture (4) PDF

Lecture (4) 1  2) For Compounds containing Two Chiral Carbons: T R1 R1 L1 * R1 L1 T...

Lecture (4) 1  2) For Compounds containing Two Chiral Carbons: T R1 R1 L1 * R1 L1 T R2 B L2 * R2 R2 B T B L1 L2 L2 Fisher Sawhorse Newman projection projection projection T: Top, B: bottom, R1: first right, R2: second right, L1: first left, L2: second left. 2 3 In Sawhorse projection, rotation around the asymmetric carbon with any angle and at any direction does not affect the configuration of the compound. D D F * F E F E E D C 120o 120o * B B B A A A C C (A) (B) (C) A, B and C are identical and so, their Fisher and Newmann projections are also the same. 4 5  Absolute configuration (R & S) 1) using Fischer projection: Rules: 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. 2) If atom 4 (smallest atom) is vertical and the order of arrangement from (1 2 3) is anticlockwise ( ), so configuration is S (as in A). 3) If atom 4 (smallest atom) is vertical and the order of arrangement from (1 2 3) is clockwise ( ) , so configuration is R (as in B). 2 2 3 * 1 1 * 3 Clockwise (R) AntiClockwise (S) 4 4 6 (A) (B) 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  (A) (B) (C) (D) Show relation between: A&B A&C A&D B&D B&C C&D 8 ❑ 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 2S, 3R  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 N.B. Diasteromers: stereoisomers that have same molecular formula, structure but different spatial arrangement & different physical & chemical properties & not mirror images. Enatiomers Diastereomers 1- Mirror images & non 1- not mirror images. superpossible. 2- Identical physical and chemical 2- different physical and chemical properties. properties. A doubly or triply bonded atom A is considered to be equivalent to two or three A atoms. 4 H H O H O 3 2 H3 C C H C C O OH H O OH OH 1 (S) Lactic acid.  ❑ Another example: 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 13  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 14  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.] 15  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. 16  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 17  Remember: * R & S isomers of the same compound are Enantiomers. * Equimolar mixture of R & S enantiomers is called Recamic Mixture and it is optically inactive due to external compensation. NUMBER OF STEREOISIOMERS 1) Number of stereoisomers = 2n, where n is number of chiral carbons. 2) Compounds containing One chiral carbon give two isomers (R&S). 3) Compounds containing Two chiral carbons give four isomers. N.B. In meso compounds, number of stereoisomers = three isomers rather than 4 isomers. 18  Review questions I- Determine the chiral center in the following compounds: CH3 H3C CH CH C2H5 CH CH2 CH3 H3C CH CH2 CH3 CH3 CH3 CH2CH3 H3C CH2CH3 CH3 (H3C) 2HC H CH3 CH CH2CH3 19