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Question

What is the percent composition of a mixture of (S)(+)2butanol,[±]25D=+13.520, and (R)()2butanol,[±]25D=13.520, with a specific rotation [±]25D=+.760:

A
75%(R)25%(S)
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B
25%(R)75%(S)
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C
50%(R)50%(S)
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D
67%(R)33%(S)
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Solution

The correct option is D 50%(R)50%(S)

Rotaion for S(+) is +13.52o and for R() is 13.52o given rotation of mixture is +0.76o

Let mixture R be x and of S be 100x

x(13.52o)+(100x)(13.52o)=76o

27.04ox1352o=0.76o

27.04ox=1352.76o

x=50

So mixture has 50 of S(+) & 50% of R()


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