Question

An optically active compound A undergoes first order conversion to another optically active compounds B and C. From the following data of optical rotation, calculate the time from start at which the solution will become racemic mixture. A→B+C

• A. 20 min
• B. 30 min
• C. 5 min
• D. 100 min

Answer 1

Answer: (A)

20 min

Answer 2

The reaction is a first-order conversion of an optically active compound A to other optically active compounds B and C: A → B + C.

The optical rotation at different times is given:

• Initial optical rotation at t = 0 min, $r_0 = 15^\circ$
• Optical rotation at t = 10 min, $r_t = 5^\circ$
• Optical rotation at t = ∞ min (when the reaction is complete), $r_\infty = -5^\circ$

For a first-order reaction where the progress is monitored by optical rotation, the rate constant (k) is given by the formula:

$k = \frac{2.303}{t} \log \left( \frac{r_0 - r_\infty}{r_t - r_\infty} \right)$

First, let's calculate the rate constant (k) using the data at t = 10 min:

$k = \frac{2.303}{10} \log \left( \frac{15^\circ - (-5^\circ)}{5^\circ - (-5^\circ)} \right)$ $k = \frac{2.303}{10} \log \left( \frac{15 + 5}{5 + 5} \right)$ $k = \frac{2.303}{10} \log \left( \frac{20}{10} \right)$ $k = \frac{2.303}{10} \log (2)$

Next, we need to find the time (t) at which the solution will become a racemic mixture. A racemic mixture has a net optical rotation of $0^\circ$. So, we need to find t when $r_t = 0^\circ$.

Using the same first-order rate equation, rearrange it to solve for t:

$t = \frac{2.303}{k} \log \left( \frac{r_0 - r_\infty}{r_t - r_\infty} \right)$

Substitute the values: $r_0 = 15^\circ$, $r_\infty = -5^\circ$, and $r_t = 0^\circ$. Also, substitute the expression for k we found:

$t = \frac{2.303}{\left( \frac{2.303}{10} \log (2) \right)} \log \left( \frac{15^\circ - (-5^\circ)}{0^\circ - (-5^\circ)} \right)$ $t = \frac{10}{\log (2)} \log \left( \frac{15 + 5}{0 + 5} \right)$ $t = \frac{10}{\log (2)} \log \left( \frac{20}{5} \right)$ $t = \frac{10}{\log (2)} \log (4)$

We know that $\log(4) = \log(2^2) = 2 \log(2)$.

$t = \frac{10}{\log (2)} (2 \log (2))$ $t = 10 \times 2$ $t = 20 \text{ min}$

Thus, the solution will become a racemic mixture after 20 minutes from the start.

Explanation of the solution:

1. Identify the reaction order and relevant formula: The problem states it's a first-order conversion, and data is given in terms of optical rotation. The rate constant (k) for a first-order reaction using optical rotation is $k = \frac{2.303}{t} \log \left( \frac{r_0 - r_\infty}{r_t - r_\infty} \right)$.
2. Calculate the rate constant (k): Use the given data ($r_0 = 15^\circ$, $r_{10} = 5^\circ$, $r_\infty = -5^\circ$) to find k. $k = \frac{2.303}{10} \log \left( \frac{15 - (-5)}{5 - (-5)} \right) = \frac{2.303}{10} \log(2)$.
3. Determine the target optical rotation: A racemic mixture has an optical rotation of $0^\circ$. So, we need to find t when $r_t = 0^\circ$.
4. Calculate the time (t): Rearrange the first-order rate equation to solve for t, and substitute the known values including the calculated k. $t = \frac{2.303}{k} \log \left( \frac{r_0 - r_\infty}{r_t - r_\infty} \right) = \frac{2.303}{\frac{2.303}{10} \log(2)} \log \left( \frac{15 - (-5)}{0 - (-5)} \right) = \frac{10}{\log(2)} \log(4)$. Since $\log(4) = 2 \log(2)$, $t = \frac{10}{\log(2)} (2 \log(2)) = 20 \text{ min}$.