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Question 1

The formula for concentration of cane sugar as a function of time is given by

$$k = \left(\frac{2.303}{t}\right) \log\left[\frac{\alpha(0) - \alpha(\infty)}{\alpha(t) - \alpha(\infty)}\right]$$

where $\alpha(t)$ is the optical rotation of the solution at any time t. This was the formula given in my textbook, but I feel it is wrong.

The optical rotation in the formula is the rotation of the solution, which includes rotation due to external factors which is taken care of by the $\alpha(\infty)$ term but it also includes rotation due to glucose and fructose , whereas the formula would only be valid if $\alpha(t)$ denoted only the rotation of cane sugar + external factors.

The reasoning given in the book is that the rotation due to glucose and fructose included in the external factors, but that is just wrong. The fundamental assumption is that rotation due to external factors remains constant and hence gets cancelled out, but if you include glucose and fructose in external factors, then their concentration and hence rotation is changing with time. Also they do not form a racemic mixture, so their rotations do not get cancelled out.

Question 2

Another question given in my textbook was to find the optical rotation if the reaction was 50% complete. Given rotations of cane sugar = 34, glucose = 10 , fructose = -20

The solution was $$\left(\frac{17+5-10}{3/2}\right)$$

Notice the 3/2 which represent the total moles in the denominator. If this is correct then the formula becomes even more incorrect as optical rotation also depends on total number of moles.

What is the correct formula? I am also not able to understand the calculation for total optical rotation as given in second question. Why did they divide by 3/2 and how does the optical rotation depend on the number of moles?

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  • $\begingroup$ Your formula is wrong. It is $\alpha (t) - \alpha (\infty)$ and not $\alpha (0) - \alpha (t)$ in the denominator $\endgroup$
    – Safdar Faisal
    Commented Aug 5, 2020 at 5:37
  • $\begingroup$ Yes edited that $\endgroup$
    – Manit Agarwal
    Commented Aug 5, 2020 at 5:39
  • $\begingroup$ Did you try using $\frac{n}{n-x}$ to prove this formula on your own? The equation for $k$ of a first order reaction $\endgroup$
    – Safdar Faisal
    Commented Aug 5, 2020 at 5:42
  • $\begingroup$ Which text book is this from? $\endgroup$
    – Safdar Faisal
    Commented Aug 5, 2020 at 13:37
  • $\begingroup$ @Safdar this is from P bahadur - Concepts of physical chemistry. I did try to prove this formula but I couldn't figure out how to remove the rotation of glucose and fructose from my expression. $\endgroup$
    – Manit Agarwal
    Commented Aug 6, 2020 at 2:55

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It seems as though your issue with the formula arises due to the fact that there is a feeling of not knowing what is considered to be factors that are internal or if they are external factors. To prove the formula, the method would be as follows.

The formula for the given reaction would be:

$$\ce{C12H22O11 + H2O -> C6H12O6 (glucose) +C6H12O6 (fructose)}$$

Drawing a modified version of a RICE table with optical rotation, and the case at t=$\infty$, we get:

\begin{array}{|l|c|} \hline \text{R} &\ce{C12H22O11 &C6H12O6(glucose) &C6H12O6 (fructose)} &\text{Optical Rotation} \\ \hline \text{I} & n & 0& 0 & \alpha_{s}n+\mathrm E \\ \text{C} & -x & x& x & - \\ \text{E} & n-x & x & x & \alpha_{s}(n-x)+ \alpha_{g}(x) + \alpha_{f}(x)+ \mathrm E \\ \infty & 0 & n & n & (\alpha_{g}+\alpha_{f})(n) +\mathrm E \\ \hline \end{array}

Now, for a general first order reaction, the formula would be: $$k=\dfrac{2.303}{t}\log \frac{n}{n-x}$$

Therefore, we need to find the value of $\dfrac{n}{n-x}$. Using the given data, we get:

$$\frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}=\frac{n}{n-x}$$

Therefore the formula given is true, even considering an external factor. You consider glucose and fructose to be internal.

$$k=\dfrac{2.303}{t}\log \frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}$$

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  • $\begingroup$ I think $\frac{n}{n-x}$ = $\frac{\alpha_s(n)}{\alpha_s(t)}$ $\endgroup$
    – Manit Agarwal
    Commented Aug 6, 2020 at 2:48
  • $\begingroup$ In your expression rotation of glucose and fructose is also appearing whereas it should only be of cane sugar. Also how to calculate total rotation at $E$ (Question 2) $\endgroup$
    – Manit Agarwal
    Commented Aug 6, 2020 at 2:52
  • $\begingroup$ Why should it not appear in the expression? In a solution the net optical rotation is said to be the sum of the all components internal and external. $\alpha$ is observed optical rotation. $\endgroup$
    – Safdar Faisal
    Commented Aug 6, 2020 at 5:06
  • $\begingroup$ Because in first order reaction rate is directly proportional to concentration of solute, which is assumed to be directly proportional to optical rotation of solute $\endgroup$
    – Manit Agarwal
    Commented Aug 6, 2020 at 5:41
  • $\begingroup$ @ManitAgarwal-Elpsycongroo What we are able to observe in a reaction is the final optical rotation of the entire solution. And that is what we use to check the rate. We cannot observe only the rotation due to one component of a mixture in said mixture. $\endgroup$
    – Safdar Faisal
    Commented Aug 6, 2020 at 5:43

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