10/4/2010. Chapter 5 Stereochemistry at Tetrahedral Centers. Handedness. 5.1 Enantiomers and the Tetrahedral Carbon

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1 John E. McMurry Chapter 5 Stereochemistry at Tetrahedral Centers Richard Morrison University of Georgia, Athens Handedness Right and left hands are not identical Right and left hands are mirror images of each other they are nonsuperimposable mirror images Almost all the molecules in the human body are handed Handedness primarily arises from the tetrahedral stereochemistry of sp 3 -hybridized carbon atoms 5.1 Enantiomers and the Tetrahedral Carbon Molecular handedness Molecules CH 3 X and CH 2 XY are identical to their mirror images Molecular images can superimpose on their mirror images Molecule CHXYZ is not identical to its mirror image Molecular image can not superimpose on its mirror image 1

2 Enantiomers and the Tetrahedral Carbon Enantiomers From the Greek enantio, meaning opposite Stereoisomers in which molecules are not identical to their mirror images Result whenever a tetrahedral carbon is bonded to four different substituents CHXYZ (one need not be H) Lactic acid (2-hydroxypropanoic acid) has four different groups (-H, -OH, -CH 3, -CO 2 H) bonded to the central carbon atoms and exists as a pair of enantiomers Enantiomers and the Tetrahedral Carbon Enantiomers of lactic acid (+)-lactic acid Occurs in muscle tissue Found in sour milk (-)-lactic acid Found in sour milk Enantiomers and the Tetrahedral Carbon A molecule of (+)-lactic acid can not superimpose on a molecule of (-)-lactic acid Regardless of how the molecules are oriented, they are not identical When the H and OH substituents match up, the CO 2 H and the CH 3 substituents do not When CO 2 H and the CH 3 match up, -H and OH do not 2

3 5.2 Chiral From the Greek cheir meaning hand Molecules that are not identical to their mirror images, and thus exist in two enantiomeric forms A molecule is not chiral if it has a plane of symmetry Plane of symmetry A plane that cuts through the middle of an object (or molecule) so that one half of the object is a mirror image of the other half a) A laboratory flask has a plane of symmetry One half of the flask is a mirror image of the other half b) A hand does not have a plane of symmetry One half of the hand is not a mirror image of the other half Achiral A molecule that has a plane of symmetry in any of its possible conformations must be identical to its mirror image Propanoic acid, CH 3 CH 2 CO 2 H Has a plane of symmetry and so must be achiral 3

4 Lactic Acid Has no plane of symmetry in any conformation and is chiral center Most common cause of chirality in an organic molecule is the presence of a carbon atom bonded to four different groups The central carbon atom in 5-bromodecane is a property of the entire molecule Methylcyclohexane Achiral because there is no carbon atom in the molecule that is bonded to four different groups Has a plane of symmetry passing through the methyl group and through C1 and C4 of the ring 4

5 2-Methylcyclohexanone Chiral because C2 is bonded to four different groups: a CH 3 group, an H atom, a COCH 2 ring bond (C1) and a CH 2 CH 2 ring bond (C3) Has no plane of symmetry Note: Carbons in CH 2, CH 3, C=O, C=C, and C C groups cannot be chirality centers * denotes a chirality center Worked Example 5.1 Drawing the Three Dimensional Structure of a Chiral Molecule Draw the structure of a chiral alcohol. 5

6 5.3 Optical Activity Stereochemistry Study originated in the early 19 th century during the investigations by the French physicist Jean-Baptiste Biot into the nature of plane-polarized light A beam of ordinary light consists of electromagnetic waves that oscillate in an infinite number of planes at right angles to the direction of light travel Optically active organic substances Biot observed that when a beam of plane-polarized light passes through a solution of certain organic molecules, the plane of polarization is rotated Optical Activity Polarimeter Measures the amount (angle) of rotation A solution of optically active organic molecules is placed in a sample tube Plane-polarized light is passed through the tube Rotation of the polarization plane occurs Light goes through a second polarizer called the analyzer The new plane of polarization and degree of rotation can be found by rotating the analyzer until the light passes through it Angle of rotation is denoted and is expressed in degrees Optical Activity Rotation The amount of rotation observed in a polarimetry experiment depends on the number of optically active molecules Number of optically active molecules depends on sample concentration and sample pathlength the pathlength is the length of the sample tube Assigning direction of rotation Levorotatory molecules Optically active molecules that rotate polarized light to the left (counterclockwise) Given the symbol (-) as in (-)-morphine Dextrorotatory molecules Optically active a molecules that rotate polarized light to the right (clockwise) Given the symbol (+) as in (+)-sucrose 6

7 Optical Activity The specific rotation, [ ] D Optical rotation expression under standard conditions The observed rotation when light of nanometer (nm; 1 nm = 10-9 m) wavelength is used with a sample pathlength l of 1 decimeter (dm; 1 dm = 10cm) and a sample concentration C of 1 g/ml Light of nm, sodium D line, is the yellow light emitted from common sodium lamps Observedrotation (degrees) [ ] D 3 Pathlength, l (dm) Concentrat ion, c (g/cm ) l c Optical Activity When optical rotation data are expressed in the standard way the specific rotation, [ ] D, is a physical constant characteristic of a given optically active compound (+)-lactic acid has [ ] D = (-)-lactic acid has [ ] D = Two enantiomers rotate the plane-polarized light to exactly the same extent but in opposite directions Worked Example 5.2 Calculating an Optical Rotation A 1.20 g sample of cocaine, [ ] D = -16, was dissolved in 7.50 ml of chloroform and placed in a sample tube having a pathlength of 5.00 cm. What was the observed rotation? 7

8 Worked Example 5.2 Calculating an Optical Rotation Strategy Since[ Then ] D l l c c [ ] D where [ ] D = -16 l = 5.00 cm = dm and C = 1.20 g/7.50 ml = g/ml Worked Example 5.2 Calculating an Optical Rotation Solution l c [ ] D = (0.500) x (0.160) x (-16) = -1.3º 5.4 Pasteur s Discovery of Enantiomers Louis Pasteur discovered enantiomers in 1848 when he began his study of crystalline tartaric acid salts derived from wine He observed that two distinct kinds of crystals precipitated from a concentrated solution of ammonium tartrate The two kinds of crystals were mirror images Pasteur separated the crystals into piles of lefthanded crystals and righthanded crystals 8

9 Pasteur s Discovery of Enantiomers Solution of ammonium tartrate The original mixture, a 50 : 50 mixture of right and left, was optically inactive Solutions of crystals from each of the sorted piles were optically active Their specific rotations were equal in amount but opposite in sign Enantiomers, also called optical isomers Have identical physical properties, such as melting and boiling point Differ in the direction in which their solutions rotate plane-polarized light 9