lecture 10 for bb

lecture 10 for bb - CHEM 231 Stereochemistry III LECTURE 10...

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Unformatted text preview: CHEM 231 Stereochemistry III LECTURE 10 LECTURE 10 1 Determination of absolute configuration of a drawn compound using CIP CH3 1 Cl 3 The counterclockwise trend indicates that this molecule is an (S) molecule This is related to the latin, sinister, which means left handed H 4 F2 counterclockwise trend The molecule is (S)1chloro1fluoroethane LECTURE 10 2 Using models to simplify determinations 1 Br F 4 H3C C H2 2 CH3 H 4 Cl 3 3 2 1 LECTURE 10 3 Assigning absolute configuration to cyclic compounds H CH3 Cl 2 4 H CH 3 3 1 Cl Cl Cl LECTURE 10 4 Assigning absolute configuration to cyclic compounds H3C H H3C H H3C H H H3C H CH3 H CH3 LECTURE 10 5 Chirality with heteroatomic centers Any central atom with 4 different groups is CHIRAL N Cl F Br O P O O NH S H3 C H Si LECTURE 10 6 Chirality with heteroatomic centers LECTURE 10 7 Nitrogen inversion A compound with nitrogen chirality center cannot be isolated as an enantiomericallly pure substance. R1 N R2 R3 Via this sp hybridized pathway 8 LECTURE 10 Compounds with multiple chirality centers For a given number of chirality centers, n, there are a potential of 2 stereoisomers. How do we describe relationships among stereoisomers of compounds with multiple chirality centers? We know that each chiral compound has a single enantiomer With multiple centers in one molecule more relationships must exist for a given stereoisomers LECTURE 10 9 The four Stereoisomers of a compound with two chirality centers CH3 CH3 Br Br CH3 CH3 Br Br 10 LECTURE 10 Diastereomers are stereoisomers that are not enantiomers In large molecules with many chirality centers there are far more diastereomers of the given molecule than enantiomers. CH3 CH3 Br Br Diastereomers CH3 CH3 Br Br LECTURE 10 11 Diastereomers Br CH3 H3 C Cl H H LECTURE 10 12 Stereoisomers that are no longer chiral CH3 CH3 H3C CH3 CH3 CH3 H3C CH3 LECTURE 10 13 Find the meso compounds H3C H H3 C CH3 H CH3 CH3 H3C H Br H Br CH3 CH3 LECTURE 10 14 How are the compound pairs related: Cl H a Br Cl H Br Cl Br H H b Br Cl Br c d CH3 CH3 Br LECTURE 10 15 How are the compounds related CH3 CH3 CH3 H CH3 H3C H CH3 e H3C Br Br H 3C Br CH3 f H3C Br g h LECTURE 10 16 Conformational steroisomers CH3 CH3 CH3 CH3 rotate 120 H3 C CH3 CH3 CH3 CH3 CH3 CH3 CH3 H3 C CH3 LECTURE 10 17 Conformational steroisomers CH3 CH3 CH3 CH3 rotate 60 CH3 H3C H3 C CH3 Because of rapid equlibration between the 2 conformations the compound remains meso just likeoriginal the drawing suggests. LECTURE 10 18 Atropisomers Binol OH OH Our simple drawing makes binol appear meso OH OH HO HO Compounds like this have C symmetry, LECTURE 10 19 Fischer projections b a d c a b c d Take a model of a chiral compound and look at it from above in an orientation that has2 groups pointing up and 2 groups pointing down. LECTURE 10 20 Fischer projections LECTURE 10 21 Fischer projections LECTURE 10 22 Reaction stereochemistry achiral + achiral achiral racemic chiral + achiral chiral racemic achiral LECTURE 10 23 Separation of chiral compounds Enantiomers have identical physical charactersitics in an achiral environment Diastereomers have different properties in an achiral environment. We use these 2 facts to hel LECTURE 10 24 ...
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This note was uploaded on 10/06/2008 for the course CHEM 231 taught by Professor Kissling during the Spring '08 term at Binghamton.

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