Experiment 13 Discussion - Experiment 13 Discussion:...

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Experiment 13 Discussion: Possible Stereoisomeric Products Benzion: NaBH 4 adds a hydride to the sp 2 hybridized carbonyl carbon in benzoin. This means benzion and NaBH 4 react in a 1:4 ratio. NaBH 3 is then regenerated to NaBH 4 by the solvent, 2- propanol. There are 3 possible products for this reaction. The carbonyl carbon is sp 2 hybridized and the benzoin is a racemic mixture of both its R and S configuration. The hydride can add to the carbonyl from either direction and so can make it R or S. This means the product can be a racemic mixture of (R,R) (S,S) or symmetrical meso product with opposite configurations. Reduction of Benzil: NaBH 4 adds a hydride sequentially to each of the two sp 2 hybridized carbonyl carbons in benzil. This means benzil and NaBH 4 react in a 1:2 ratio. NaBH 3 is then replenished into NaBH 4 by the solvent, 2-propanol. There are also 3 possible products of benzil. Both carbonyl carbons are sp 2 hybridized. The hydride can add from either side and so is capable of creating either the R or S configuration. This means that the product, like benzoin, can either be a racemic mixture of (R,R) and (S,S), or a symmetrical meso product of offsetting configurations. Products formed: Hydrobenzoin is the preferred product. This can be seen through the TLC data. The R f values decrease throughout the reaction. This
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This note was uploaded on 02/21/2011 for the course CHEM 3b taught by Professor Pederson during the Fall '08 term at University of California, Berkeley.

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Experiment 13 Discussion - Experiment 13 Discussion:...

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