CHEM 2211 1st Edition Lecture 9 Outline of Current Lecture I II III IV V Rotation about the Double Bond Conformation vs Configuration Chiral Molecules Isomer with one Asymmetric Center Drawing Enantiomers Current Lecture Chapter 4 Isomers Stereoisomers Cis Trans Isomers I Constitustional Isomers Isomer with Asymetric Center Rotation around Double bonds a Rotation can NOT occur around a double bond unless the bond is broken first b Leads to cis trans isomers aka geometric isomers i The cis isomer has substituent on the same side ii Trans isomers have substituents on the opposite sides c Each molecule is different and has a different reactivity d Example Cis Pentene vs Trans Pentene cis Pentene trans Pentene II III IV V Conformation vs Configuration a Conformations are different spatial arrangements for the same molecule b Configurations are different compounds from each other and can be separated from each other in a mixture Chiral molecules a Chiral molecules are nonsuperimposable mirror images of themselves much like the left and right hands b Achiral molecules are superimposable mirror images c Chiral molecules are the result of a carbon bonded to four different substituents aka a asymmetric center Isomers with one Asymmetric Center a Enantiomers are molecules that cannot be superimposed on each other b Stereocenters are both asymmetric centers and the sp2 carbon of an alkene or the sp3 carbon of a cyclic compound Drawing Enatiomers a Perspective formula shows the bonds in the plane of the paper i A wedge means it s coming out of the plane ii A hatched wedge means its receding away from the plane Both configurations of 2 bromobutane b Fischer projections show that the bonds intersect at the central carbon i Vertical lines are bonds that extend backwards ii Horizontal lines are bonds that extend forward Configurations of 3 chlorohexane VI The R S Naming System for Enantiomers a Rank the substituents in order of highest to lowest priority Priority is determined by the atomic number of the atom directly bonded to that carbon The higher the atomic number the higher the priority b If the lowest priority group is bonded by a hatched wedge then draw an arrow from the highest priority substituent to lowest priority substituent If it turns clockwise right then it gets the R configuration If it turns counterclockwise left then it gets the S configuration R 2 bromobutane S 2 bromobutane c If the lowest priority group is not bonded by a hatched wedge then interchange group 4 with the group with the hatched wedge and do just like in step two The CH3 was interchanged with the H
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