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.

Answer

`alpha-`anomer `=37.2 %, beta-` anomer `=62.8 %`.

Explanation

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`.