Question

Below are shown two reaction pathways in which optically active (R) -1-phenylethanol (F) is converted to ethyl 1-phenylethyl ether (TsCl = p-toluenesulfonyl chloride, DMF = N, N-dimethylformamide). Explain why the optical rotation of the product has the opposite sign from the two reaction paths.

CH3 CH3 CH-ONa CH3CH2OTs DMF CH-OCH2CH3 NaHr [a]= +19,9 CH3 -CH-OH TSCI [a]o = +33,0 pyridin ou un - CH2 CH3 CH-OTS CH3CH2ONa DMF CH3 -CH-OCH2CH3 [a]o = -19,9

Answer 1

Solution:

The conversion of (R)-1-phenylethanol to ethyl-1-phenylether by two pathways (NaH, CH3CH2OTs) and (TsCl and CH3CH2ONa), the products formed are enantiomers to each other. Enantiomers have identical physical properties but they rotate plane-polarized light in equal and opposite directions.

A) Conversion by NaH and CH3CH2OTs follow SN2 mechanism, hence formed product has opposite (S) configuration.

B) Conversion by TsCl/pyridine and CH3CH2ONa follow SN1 mechanism in which retention (R) product is more favours, giving rise form a product having R configuration.

The formed products in A and B are enantiomers, hence they posses equal but opposite rotation.

CH3 cn2-cors CH3 CH2-Ch2OTS SM2 enz enz euzen20NG Tsel OTS 40H Pundine CM3 CM3 ÇADCH2CH+ R.