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The specific rotation of two glucose anomers are `alpha = + 110^(@)` and `beta = 19^@` and for the constant equilibrium mixtures is `+52.7^@`. Calculate the percentage compositions of the anomers in the equilibrium mixture. |
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Answer» Correct Answer - `alpha-`anomer `=37.2 %, beta-` anomer `=62.8 %`. Suppose `%` composition of `alpha= x %` Suppose, `%` composition of `beta = 100 -x` `:.` Optical rotation of equimixture `= ((x xx 110)+(110-x)19)/(100)=52.7` `:. x=37.2 %` `:. %` for `alpha= 37.2` `:. %` for `beta =62.8`. |
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