What is the configuration of the product yielded in the following reaction? Will

Question

What is the configuration of the product yielded in the following reaction? Will the relative configuration of the product be cis or trans? A) (2R, 3R) B) (2S, 3S) C) (2R, 3S) D) (2S, 3R) E) A 1:1 mixture of (2R, 3R) and (2S, 3S) F) (2R, 3S) and (2S, 3R) are identical, called meso Knowing that a compound has an enantiomeric ratio (er) of 72:28. Calculate enatiomeric excess (ee). It is known that the mechanism of nucleophilic substitutions (S_N 1 or S_N 2) determines the stereochemical outcome. Based on the diagram below and the enantiomeric ratios (er) shown for the starting material and product, propose a reason why the product of this reaction deviates from typical examples. Calculate enantiomeric excess (ee) for the for the starting material and product.

Explanation / Answer

8. the addition of H2 on alkene is in syn-fasjion that is both the hydrogens get added from the same side.

the product thus formed would be a trans-product with repect to the -CH3 on either side or the Phenyl on either end.

the product configuration would thus be,

A) 2R,3R

9. An enantiomeric ratio of 72:28 for a compound

enantiomeric excess (ee) = (72 - 28) = 44%

10. The reaction shown here is majorly undergoes by an Sn2 mechanism as we get the major product with inversion of configuration. NaOMe is a strogn nucelophile and Br- is a good leaving group. These factors support the reaction to go by an Sn2 mechanism. A small amount of reaction however undergoes an Sn1 pathway to form the other enantiomer of the product because the carbocation generated by loss of Br- is benzylic and is stabilized by resonance with the benzene ring. thus allowing for an Sn1 reaction to proceed as well.

Enantiomeric excess (ee) = 80 - 20 = 60%

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