1.

An organic compound P exists in two enantiomeric forms, which have specific optical rotation values [alpha] = +-100^(@). The optical rotation of a mixture of these two enantiomers is -50^(@). Calculate the percentage of that enantiomer which is in lower concentration in the mixture.

Answer»


Solution :`"Enatiomeirc excess" = ("observed rotation")/("specific rotation of pure ENANTIOMER")= (-50^(@))/(-100^(@)) XX 100 = 50%`
`% " of LAEVOROTATORY isomer " = 50+25%`
`% " of DEXTROROTATORY isomer " = 25%`


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