1.

Consider the reaction {:((R)-CH_(3)CHICH_(2)CH_(3)+^(137)T^(-),rarrCH_(3)CHICH_(2)CH_(3),,,),(alpha_(obs)=-15.90^(@),"containing" 2%137T^(-),,,),(,alpha_(obs)=-15.26^(@),,,):} The percentage of racemic from in the product is

Answer»

`10%`
`8%`
`4%`
`12%`

Solution :Note that replacement of `I` with `.^(137)I` does not change he sign or value of the rotation. The optical purity of the recovered iodide is `(15.26//15.90)(100%) = 96%`. HENCE `4%` of the iodide is RACEMIC. This MEANS thet `2%`, (the same percentage that contains isostopic `I)`. has been converted to the `(S)` configuraiton. It is concluded that every REACTION with `.^(137)T^(-)` resulted in an INVERSION at chiral carbon.


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