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J & K CET Medical J & K - CET Medical Solved Paper-2001

  • question_answer
    The specific rotation of (R) (-)\[2-\]bromooctane is\[-\,40\]. What is the percentage composition of mixture of enantiomers of\[2-\]bromooctane whose rotation is\[+20\]?

    A)  The mixture has 25% R and 75% S

    B) (b )The mixture has 20% R and 80% S

    C)  The mixture has 50% R and 50% S

    D)  The mixture has 75% R and 25% S

    Correct Answer: A

    Solution :

                    Specific rotation of (R) (-) 2-bromooctane is \[=-\,40\] Specific rotation of (S) (+) 2-bromooctane is \[=+40\] Let the mixture contains\[x\]%R and\[(100-x)\] % S form. \[\frac{x\times (-40)+(100-x)(+40)}{100}=20\] \[x=25\]and \[S=100-x=75\] Hence, the mixture has 25% R and 75% S.


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