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2-iodo butane (having radioactive iodine) reacts with KI (having non radioactive iodine). Rate of loss of optical activity was 1.96 times the rate of loss of radioactivity. What percentage of reaction proceeds via SN1 mechanism? Assume that 100% inversion takes place in SN2 reaction while in SN1 there is 50% retention and 50% inversion.

I proceeded with KI substituting the iodide ion with the radioactive iodide ion but couldn't understand how would 2-iodo butane lose its optical activity(it still has a chiral C). So I couldn't go any further.

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  • $\begingroup$ It has to do with an (equimolar) Racemic mixture being formed! Note that a racemic mixture still has chiral atoms but cannot contribute to any net rotation of the plane polarized light making it optically inactive as a combined product $\endgroup$ May 22 '20 at 17:03
  • $\begingroup$ chemistry.stackexchange.com/questions/132885/… $\endgroup$
    – user94580
    Jun 6 '20 at 6:16
  • $\begingroup$ This question does not make sense. At the beginning of the reaction, the rate of loss of radioactivity is the rate of both types of substitution. But the rate of loss of optical activity is tempered by the fact that only one of the two mechanisms produces more of the other enantiomer to progress towards racemization. By that logic, how could the rate of loss of optical activity be greater than the loss of radioactivity? $\endgroup$
    – Zhe
    Jun 7 '20 at 1:36
  • $\begingroup$ @Zhe As I have written in my answer, rate of loss of optical activity is proportional to extent of racemisation. If I take a substrate with optical activity having only one enantiomer, then initially, there is no racemisation. Suppose 10% of substrate reacts and undergoes inversion, then the solution will be 20% racemic since a racemic mixture consists of equimolar amounts of two enantiomers. So rate of loss of optical activity is twice of rate of reaction for SN2 mechanism. $\endgroup$
    – Indian135
    Jun 12 '20 at 1:32
  • $\begingroup$ "Suppose 10% of substrate reacts and undergoes inversion, then the solution will be 20% racemic" That statement is unfortunately not correct. As soon as any of the initial starting material reacts, you generate the other enantiomer which reacts at the same rate as the initial enantiomer. $\endgroup$
    – Zhe
    Jun 12 '20 at 3:11
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First, for simplicity, I am assuming that the initial substrate (2-iodobutane with radioactive iodine) consists of only one enantiomer.

How would 2-iodobutane lose its optical activity (it still has a chiral $\ce{C}$)?

In $\mathrm{S_N1}$ mechanism, there is 50% inversion and 50% retention, so a racemic mixture starts to form which decreases the optical activity of reaction mixture due to external compensation.

In easy terms: In a racemic mixture, although the two enantiomers rotate plane-polarized light in opposite directions, the rotations cancel because they are present in equal amounts. And in a partially racemic mixture or still undergoing reaction mixture with racemisation, the optical activity is decreased (rotation of plane polarised light is less).

So, 2-iodobutane doesn’t lose its optical activity, it’s optical activity is only countered by its inverted enantiomer.

What percentage of reaction proceeds via $\mathrm{S_N1}$ mechanism?

The rate of loss of radioactivity is proportional to amount of substrate that has reacted (2-iodobutane with radioactive iodine), that is rate of reaction.

And, the rate of loss of optical activity is proportional to rate formation of racemic mixture.

Now suppose,

$x =$ rate of reaction of substrate at an instant in moles per unit time.

Then, $1.96x$ is the rate of racemisation.

So, during a small time period $t$, $xt$ moles of substrate has reacted and $1.96xt$ moles of racemic mixture is obtained.

The racemic mixture consists of 50% inverted product and 50% with original configuration. Note that the reaction mixture is not purely racemic. It has substrate and retention product (both of which have original configuration) and inversion product. So, by racemic mixture I mean all the inverted product and equal amount of molecules with original configuration.

Since no inverted molecule was present before the reaction (assumption at start of answer), $0.98xt$ moles of substrate proceeded via inversion and the rest $0.02xt$ moles via retention ($x$ moles had reacted)

So, only $\mathrm{S_N1}$ can cause retention and that also only 50% of reaction via $\mathrm{S_N1}$ gives retention product, $0.04xt$ moles of substrate must have proceeded via $\mathrm{S_N1}$ mechanism.

Thus, percentage of $\mathrm{S_N1}$ mechanism $= \dfrac{0.04x}{x} \times 100\% =4\%$.

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  • $\begingroup$ See my comment to the question on why I don't believe you should be able to solve this problem... $\endgroup$
    – Zhe
    Jun 10 '20 at 14:41
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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 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.

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