BR Ochem book I passage number II

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This passage gives a nucleophilic substitution reaction and then gives multiple data tables of reaction rates in different solvents. One question asks what cannot be concluded from the experiment, and the answer was that "Chirality is retained more significantly in protic solvent than in aprotic solvent."
In the data tables, it gives each nucleophile, the rxn rate for that nuc, and the optical rotation. What exactly does this optical rotation mean in the context of this question? Is it referring to the optical rotation of the final product after each nuc is reacted with the given electrophile? Or is it talking about the optical rotation of that specific nuc?
The answer explanation says that, "In ether solvent, there are greater observed optical rotation values for the product mixture than in alcohol. This means that chirality is retained more in the protic solvent than in the aprotic solvent."
Why would greater optical rotation values = more retention of stereochem in this context?



I put this answer, but my reasoning was that you did not know the optical rotation of the starting compound, so it wasn't possible to make this sort of conclusion....

 

[BerkReviewTeach]

The optical rotation is for the product mixture. What you have to recognize here is that ideal SN1 reactions would result Zero degrees optical rotation and ideal SN2 reactions would result in complete inversion and an optically active compound in solution. The four tables represent changes in temperature and solvent. The reactant is a secondary-alkylbromide in each case, which can go by either an SN1 or SN2 mechanism. If you look at the optical rotation values in Tables 1 and 2, you see numbers around +30 degrees. If you look at the optical rotation values in Tables 3 and 4, you see numbers around +10 degrees. This means that the reaction conditions in Tables 1 and 2 were more conducive to an SN2 reaction than they were in the reaction conditions of Tables 3 and 4.

Add in the fact that this question is a "CANNOT" question, and it's a very challenging question. When going from an aprotic solvent (Tables 1 and 2) to a protic solvent (Tables 3 and 4), the optical rotation went down from around +30 degrees to around +10 degrees, so chirality was racemized more and thus retained less when switching from an aprotic to protic solvent. Choice C is not true according to the data, so it "CANNOT be concluded from the experiment."

You could use background knowledge about SN1 and SN2 reactions, but because the question specifically asks you to base your answer on the experiment, you probably should use the data in the tables.

 

[aldol16]

The answer explanation says that, "In ether solvent, there are greater observed optical rotation values for the product mixture than in alcohol. This means that chirality is retained more in the protic solvent than in the aprotic solvent."

Did you flip these solvents? Ether is aprotic and alcohol is protic and if you have greater optical rotation with ether solvent, that means chirality is retained more for aprotic solvents than for protic solvents.

Why would greater optical rotation values = more retention of stereochem in this context?

You're confusing "retention of chirality" with "retention of stereochemistry." Retention of stereochemistry means that R remains R and S remains S. Retention of chirality just means that the product mixture is chiral - that is, chirality would still be retained if R goes to S and S goes to R (SN2-type process) whereas stereochemistry would not.