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typo
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Zhe
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This really doesn't answer the question, but the discussion is too long for comments.

The question only makes sense if you start with a mixture that has optical activity.

Let's call the concentration of the starting enantiomer $x$ and the other enantiomer $y$.

Optical activity is unfortunately, not a concentration, so we need a definition to make the two rates comparable.

$$\alpha = \frac{\phi l}{10}(x - y)$$

$$\frac{d\alpha}{dt} = \frac{\phi l}{10}\frac{d(x - y)}{dt}$$

For the rate of loss of optical activity, we can strip off the leading constant to convert to a concentration based rate. But we could just as well replace it with any constant, and I'll address this later. For now, we have:

$$\mathrm{rate}_{\mathrm{optical}}=-\frac{d(x-y)}{dt}$$

This mostly makes sense, since $\ce{[x]}-\ce{[y]}$ represents the concentration of the excess of the major enantiomer. Thus, the rate of change in this value, which should be negative, represents the rate of loss of optical activity. As a sanity check, it has the maximum value at the beginning of the reaction and is equal to zero when the system is racemic.

We also should make clear that the respective rates of $S_{\mathrm{N}}1$ and $S_{\mathrm{N}}2$ are:

$$\mathrm{rate}_{1} = k_{1}\ce{[X]}$$ $$\mathrm{rate}_{2} = k_{2}\ce{[X]}$$

where I've absorbed the assumed constant concentration of salt into the equilibriumrate constant of the $S_{\mathrm{N}}2$ reaction.

The loss of radioactivity is equal to the rate of types of substitution because both types of substitution replace the labelled iodine with an unlabelled iodine.

$$\mathrm{rate}_{\mathrm{label}} = \left(k_{1} + k_{2}\right)[\mathrm{electrophile}]$$

Half the products of the $S_{\mathrm{N}}1$ reaction result in retention while the other half result in inversion. All of the $S_{\mathrm{N}}2$ product results in inversion.

Therefore:

$$\frac{d(\ce{[x]}-\ce{[y]})}{dt}=\frac{k_{1}}{2}[\mathrm{electrophile}] - k_{2}[\mathrm{electrophile}] - \frac{k_{1}}{2}[\mathrm{electrophile}] = - k_{2}[\mathrm{electrophile}]$$

But this means that, at the start of the reaction, the magnitude of the rate of loss of optical activity is smaller than the magnitude of the rate of loss of radioactivity.

The real issue here is that you cannot compare apples and oranges. The rate of loss of radioactivity is a concentration based rate, but the rate of loss of optical activity is an a change in angle over change in time. Under the simplest assumption (using a conversion of unity), this problem cannot be solved.

This really doesn't answer the question, but the discussion is too long for comments.

The question only makes sense if you start with a mixture that has optical activity.

Let's call the concentration of the starting enantiomer $x$ and the other enantiomer $y$.

Optical activity is unfortunately, not a concentration, so we need a definition to make the two rates comparable.

$$\alpha = \frac{\phi l}{10}(x - y)$$

$$\frac{d\alpha}{dt} = \frac{\phi l}{10}\frac{d(x - y)}{dt}$$

For the rate of loss of optical activity, we can strip off the leading constant to convert to a concentration based rate. But we could just as well replace it with any constant, and I'll address this later. For now, we have:

$$\mathrm{rate}_{\mathrm{optical}}=-\frac{d(x-y)}{dt}$$

This mostly makes sense, since $\ce{[x]}-\ce{[y]}$ represents the concentration of the excess of the major enantiomer. Thus, the rate of change in this value, which should be negative, represents the rate of loss of optical activity. As a sanity check, it has the maximum value at the beginning of the reaction and is equal to zero when the system is racemic.

We also should make clear that the respective rates of $S_{\mathrm{N}}1$ and $S_{\mathrm{N}}2$ are:

$$\mathrm{rate}_{1} = k_{1}\ce{[X]}$$ $$\mathrm{rate}_{2} = k_{2}\ce{[X]}$$

where I've absorbed the assumed constant concentration of salt into the equilibrium constant.

The loss of radioactivity is equal to the rate of types of substitution because both types of substitution replace the labelled iodine with an unlabelled iodine.

$$\mathrm{rate}_{\mathrm{label}} = \left(k_{1} + k_{2}\right)[\mathrm{electrophile}]$$

Half the products of the $S_{\mathrm{N}}1$ reaction result in retention while the other half result in inversion. All of the $S_{\mathrm{N}}2$ product results in inversion.

Therefore:

$$\frac{d(\ce{[x]}-\ce{[y]})}{dt}=\frac{k_{1}}{2}[\mathrm{electrophile}] - k_{2}[\mathrm{electrophile}] - \frac{k_{1}}{2}[\mathrm{electrophile}] = - k_{2}[\mathrm{electrophile}]$$

But this means that, at the start of the reaction, the magnitude of the rate of loss of optical activity is smaller than the magnitude of the rate of loss of radioactivity.

The real issue here is that you cannot compare apples and oranges. The rate of loss of radioactivity is a concentration based rate, but the rate of loss of optical activity is an a change in angle over change in time. Under the simplest assumption (using a conversion of unity), this problem cannot be solved.

This really doesn't answer the question, but the discussion is too long for comments.

The question only makes sense if you start with a mixture that has optical activity.

Let's call the concentration of the starting enantiomer $x$ and the other enantiomer $y$.

Optical activity is unfortunately, not a concentration, so we need a definition to make the two rates comparable.

$$\alpha = \frac{\phi l}{10}(x - y)$$

$$\frac{d\alpha}{dt} = \frac{\phi l}{10}\frac{d(x - y)}{dt}$$

For the rate of loss of optical activity, we can strip off the leading constant to convert to a concentration based rate. But we could just as well replace it with any constant, and I'll address this later. For now, we have:

$$\mathrm{rate}_{\mathrm{optical}}=-\frac{d(x-y)}{dt}$$

This mostly makes sense, since $\ce{[x]}-\ce{[y]}$ represents the concentration of the excess of the major enantiomer. Thus, the rate of change in this value, which should be negative, represents the rate of loss of optical activity. As a sanity check, it has the maximum value at the beginning of the reaction and is equal to zero when the system is racemic.

We also should make clear that the respective rates of $S_{\mathrm{N}}1$ and $S_{\mathrm{N}}2$ are:

$$\mathrm{rate}_{1} = k_{1}\ce{[X]}$$ $$\mathrm{rate}_{2} = k_{2}\ce{[X]}$$

where I've absorbed the assumed constant concentration of salt into the rate constant of the $S_{\mathrm{N}}2$ reaction.

The loss of radioactivity is equal to the rate of types of substitution because both types of substitution replace the labelled iodine with an unlabelled iodine.

$$\mathrm{rate}_{\mathrm{label}} = \left(k_{1} + k_{2}\right)[\mathrm{electrophile}]$$

Half the products of the $S_{\mathrm{N}}1$ reaction result in retention while the other half result in inversion. All of the $S_{\mathrm{N}}2$ product results in inversion.

Therefore:

$$\frac{d(\ce{[x]}-\ce{[y]})}{dt}=\frac{k_{1}}{2}[\mathrm{electrophile}] - k_{2}[\mathrm{electrophile}] - \frac{k_{1}}{2}[\mathrm{electrophile}] = - k_{2}[\mathrm{electrophile}]$$

But this means that, at the start of the reaction, the magnitude of the rate of loss of optical activity is smaller than the magnitude of the rate of loss of radioactivity.

The real issue here is that you cannot compare apples and oranges. The rate of loss of radioactivity is a concentration based rate, but the rate of loss of optical activity is an a change in angle over change in time. Under the simplest assumption (using a conversion of unity), this problem cannot be solved.

Source Link
Zhe
Zhe
  • 17.5k
  • 1
  • 39
  • 71

This really doesn't answer the question, but the discussion is too long for comments.

The question only makes sense if you start with a mixture that has optical activity.

Let's call the concentration of the starting enantiomer $x$ and the other enantiomer $y$.

Optical activity is unfortunately, not a concentration, so we need a definition to make the two rates comparable.

$$\alpha = \frac{\phi l}{10}(x - y)$$

$$\frac{d\alpha}{dt} = \frac{\phi l}{10}\frac{d(x - y)}{dt}$$

For the rate of loss of optical activity, we can strip off the leading constant to convert to a concentration based rate. But we could just as well replace it with any constant, and I'll address this later. For now, we have:

$$\mathrm{rate}_{\mathrm{optical}}=-\frac{d(x-y)}{dt}$$

This mostly makes sense, since $\ce{[x]}-\ce{[y]}$ represents the concentration of the excess of the major enantiomer. Thus, the rate of change in this value, which should be negative, represents the rate of loss of optical activity. As a sanity check, it has the maximum value at the beginning of the reaction and is equal to zero when the system is racemic.

We also should make clear that the respective rates of $S_{\mathrm{N}}1$ and $S_{\mathrm{N}}2$ are:

$$\mathrm{rate}_{1} = k_{1}\ce{[X]}$$ $$\mathrm{rate}_{2} = k_{2}\ce{[X]}$$

where I've absorbed the assumed constant concentration of salt into the equilibrium constant.

The loss of radioactivity is equal to the rate of types of substitution because both types of substitution replace the labelled iodine with an unlabelled iodine.

$$\mathrm{rate}_{\mathrm{label}} = \left(k_{1} + k_{2}\right)[\mathrm{electrophile}]$$

Half the products of the $S_{\mathrm{N}}1$ reaction result in retention while the other half result in inversion. All of the $S_{\mathrm{N}}2$ product results in inversion.

Therefore:

$$\frac{d(\ce{[x]}-\ce{[y]})}{dt}=\frac{k_{1}}{2}[\mathrm{electrophile}] - k_{2}[\mathrm{electrophile}] - \frac{k_{1}}{2}[\mathrm{electrophile}] = - k_{2}[\mathrm{electrophile}]$$

But this means that, at the start of the reaction, the magnitude of the rate of loss of optical activity is smaller than the magnitude of the rate of loss of radioactivity.

The real issue here is that you cannot compare apples and oranges. The rate of loss of radioactivity is a concentration based rate, but the rate of loss of optical activity is an a change in angle over change in time. Under the simplest assumption (using a conversion of unity), this problem cannot be solved.