Question

Like glucose, galactose mutarotates when it dissolves in water. The specific rotation of $\mathit{\alpha }\mathbf{-}\mathit{D}\mathbf{-}\mathit{g}\mathit{a}\mathit{l}\mathit{a}\mathit{c}\mathit{t}\mathit{o}\mathit{p}\mathit{y}\mathit{r}\mathit{a}\mathit{n}\mathit{o}\mathit{s}\mathit{e}$is +150.7⁰ , and that of the β anomer is +52.8⁰ . When either of the pure anomers dissolves in water, the specific rotation gradually changes to +80.2⁰. Determine the percentages of the two anomers present at equilibrium.

Short answer

$\alpha anomer=28\%and\beta anomer=72\%$

Step-by-step solution

Mutarotation

A spontaneous change in the specific rotation of a solution of an optically active compound is known as mutarotation. This implies that the two samples are different but in solution they form an equilibrium mixture.

Anomers

The diastereomers resulting from cyclisation are known as anomers. They differ only in the configuration around first carbon (C1) which is referred to as the anomeric carbon (hemiacetal carbon atom).

Calculation

If the fraction of galactose present as the $\alpha$anomer is role="math" localid="1664865398062" $\left(\left[\alpha \right]=+150.7^0\right)$ is a, the fraction present as the βanomer is $\left(\left[\alpha \right]=+52.8^0\right)$, and the specific rotation of the mixture is +80.2, then we can write as:

role="math" localid="1664865861024" $a\left(+150.7^0\right)+b\left(+52.8^0\right)=+80.2^0$

The fraction present as the $\alpha$ anomer (a) plus the fraction present as β anomer (b) should account for all the galactose.

a+b = 1 or we can write b = 1-a

Now,

$$\begin{gathered} a\left(+150.7^0\right)+b\left(+52.8^0\right)=+80.2^0 \\ \Rightarrow a\left(+150.7^0\right)+\left(1-a\right)\left(52.8^0\right)=80.2^0 \\ \Rightarrow a\times 150.7^0+52.8^0-a\times 52.8^0=80.2^0 \\ \Rightarrow a\left(150.7^0-52.8^0\right)=80.2^0-52.8^0 \\ \Rightarrow a\times 97.9^0=27.4^0 \\ \Rightarrow a=\frac{27.4^0}{97.9^0} \\ \Rightarrow a=0.28 \end{gathered}$$

So,

$\begin{array}{rcl}b& =& 1-a\\ & =& 1-0.28\\ & =& 0.72\end{array}$

Hence, the amounts of two anomers present at equilibrium are $\alpha$ anomer = 28% and β anomer = 72%.