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.
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.
Below are shown two reaction pathways in which optically active (R) -1-phenylethanol (F) is converted to...