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I have been assigned with the task of figuring out whether a reaction is $\ce{S_{N}1}$ or $\ce{S_{N}2}$. The information that I have is, that the yield is dependent on solvent polarity. More specifically, the higher the solvent polarity, the higher the yield. The problem is that both $\ce{S_{N}1}$ or $\ce{S_{N}2}$ products are the same. Both result in charge separation (therefore both are favored by polar solvents).

I proposed to repeat the experiment. I proposed to make the site of substitution chiral. That way it is possible to check for $\ce{S_{N}2}$ inversion.
Another proposal was to use the fact that the rates of $\ce{S_{N}2}$ and $\ce{S_{N}1}$ were different. However, I am not sure how to use this fact and perform an experiment.

What should I do in this situation? The people that I am working for do not want to repeat the experiment at all. Using the rates approach seems to be the best option since there won't be a need to obtain new materials. But I have no idea on how to do this. What are your suggestions? How can I figure out whether the reaction proceeded through $\ce{S_{N}1}$ or $\ce{S_{N}2}$ by using rates approach?

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  • $\begingroup$ In praxis there is no such thing as a purely $\ce{S_{N}1}$ or $\ce{S_{N}2}$ mechanism, therefore it is hard to figure it out in the first place. Why do you need to know this? Is it of any implication for the continuance of your study? $\endgroup$
    – Martin - マーチン ♦
    Commented Aug 18, 2014 at 12:45

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As you have the information that the yield depends on solvent polarity. It is the key point to be considered in analysis of type of reaction. In case of a $\ce{S_{N}1}$ reaction, the polar solvent favors the reaction. $\ce{S_{N}1}$ is a two stepped reaction, in which the 1st step is carbocation formation, such that there appears the dipole dipole interaction between the carbocation and polar solvent which favors the reaction.

Whereas in the case of $\ce{S_{N}2}$ reaction, the dipole dipole interaction develops between polar solvent and the nucleophile rather than with intermediate product. It is because the charge is dispersed in two centres in intermediate while charge is centered in single center in nucleophile. This deactivates the $\ce{S_{N}2}$ reaction and the reaction cannot proceed forward smoothly.

Since you have another specific information that the higher the solvent polarity, the higher the yield. Since higher solvent polarity favors the $\ce{S_{N}1}$ reaction, the higher will be the yield through a $\ce{S_{N}1}$ mechanism. So the reaction with tendency with higher solvent polarity and higher yield is a $\ce{S_{N}1}$ reaction.

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    $\begingroup$ Good argument. In addition, if the leaving group is an anion (which is likely) then this will be also stabilised by a more polar solvent, as it is favouring the dissociation of carbocation and leaving group. $\endgroup$
    – Martin - マーチン ♦
    Commented Aug 18, 2014 at 12:57
  • $\begingroup$ Yeah. The polar solvent makes the carbocation more stable by creating dipole dipole interaction which will favor the formation of carbocation and as the formation of carbocation is the rate determining step , use of polar solvent also increases the rate of SN1 reaction. $\endgroup$
    – CCR
    Commented Aug 22, 2014 at 13:38
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As the other answers state, kinetic analysis is one way to go, especially if you want to avoid the hassle of inserting a chiral label. If you're interested at looking at the relative rates, your best bet for a practical technique is probably NMR, depending on your reactants and solvent, so long as the reaction isn't too fast. It's easy enough to get 30 s resolution for arrayed proton NMR experiments.

ESI-MS may also be an option if your compounds ionize well and you're using a compatible solvent. A reaction in a 500 µL syringe should be good for an hour or two of measurements.

Depending on your exact reaction, optical techniques could potentially be used, as well. If you're lucky, it may be as simple as absorbance measurements.

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You should be very careful about changing the reagents or solvent, because then you are looking at a different reaction than the one proposed which may go by a different mechanism.

Your idea of looking at rates is a good one. Since an SN2 reaction depends on the concentration of nucleophile, while SN1 does not, set up two experiments exactly the same (same concentration of electrophile, same solvent, same temperature, etc) but double the amount of nucleophile in one of the experiments. If the reaction rates differ (by a factor of two), then the mechanism is SN2. If the rates are the same, then the mechanism is SN1.

Another method that chemists use to learn about reaction mechanisms is by isotopic labeling. Depending on the exact electrophilic substrate, you could consider labeling a proton as deuterium (or other atom as necessary), rendering the substrate chiral. Then you could take advantage of the difference in stereochemical outcomes of SN1 and SN2 mechanisms. However, whether this can be utilized depends on the exact substrates of the reaction being considered.

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Another proposal was to use the fact that the rates of SN2 and SN1 were different.

SN2, as its name suggests, is a bimolecular mechanism and its rate depends on not just substrate concentration but also nucleophile concentration. SN1, however, is solely dependent on substrate concentration, as carbocation formation has a high activation energy barrier, and overcoming this barrier takes time.

I proposed to repeat the experiment. I proposed to make the site of substitution chiral.

How would the results of this proposed experiment apply to the results of your previous experiment with the achiral substrate?

The information that I have is that the yield is dependent on solvent polarity. More specifically, the higher the solvent polarity, the higher the yield. The problem is that both SN1 or SN2 products are the SAME. Both result in charge separation (therefore both are favored by polar solvents)

Well, try thinking about the intermediates, and try thinking about the nucleophile. Nucleophiles can be aided or hindered by the solvent.

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  • $\begingroup$ How would the results of this proposed experiment apply to the results of your previous experiment with the achiral substrate? - In SN2 the stereochemistry is inverted, in SN1 a racemic mixture is to be expected. Based on %ee you can propose the mechanism to be more like SN1 or SN2. $\endgroup$
    – Martin - マーチン ♦
    Commented Aug 18, 2014 at 12:49

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