Determining the stereochemistry of the product(s) when an alkyl shift in a carbocation generates a chiral carbon

Question

Consider this carbocation:

We know that this carbocation would immediately rearrange - by a methyl shift - to form the more stable carbocation.

Now, notice that the migration will produce a chiral center at carbon 1. Assuming that the leaving group was a chloride, will the configuration at 1 - after the methyl shift - be retained, inverted, or racemized?

Interestingly, assuming the nucleophile is $\ce{EtOH}$, in the final product - 2-d-3-ethoxy-3-methylpentane, carbon 2 is also a chiral center. I assume that it will be racemized, as in usual S_N1 reactions, but I am not sure. Am I correct?

At a very detailed level, I was also taught that in S_N1 reactions, 100% racemization does not occur. The inverted product is slightly more than the retained product. This is because - while the leaving group is still leaving - backside attack is preferred. So, does this phenomenon also have an effect on the molecule above?

To conclude, I wish to know the stereochemistry at both chiral carbon 1 and 2. Are they retained, inverted, or racemized? If racemized, then is the racemization 100%, or is the inverted product formed more than the retained one?

Answer 1

Let's take your question back a step to a single chiral diazonium salt (R)-1, prepared by treatment of the corresponding amine with nitrous acid. This reaction (1 --> 2) is typified by the Demjanov ring expansion. Although there are examples of direct S_N2 displacement of N₂ by water, they are not on such hindered primary carbons. A quick Chemical Abstracts search did not reveal any concerted alkyl migrations in this reaction type which would cause inversion at C₁.

When the primary carbocation 2 is formed, it is renderd achiral with enantiotopic faces at C₁. There is also free rotation about the C₁ - C₂ bond. The migratory aptitude of alkyl groups is tertiary > secondary > primary > methyl. Thus, racemic 3 would lead to racemic ether (alcohol) 4. [E₁ products are ignored.] While migration of ethyl is preferred over methyl, methyl does have a statistical advantage in that there are two of them. In this case, racemic carbocation 5 is formed, which captures solvent to form racemic 6 as a near equal mixture of diastereomers. There is always a chance that the solvolysis of the cationic site in 5 may not be totally racemized. A priori, it is difficult to predict that outcome.

Answer 2

Since this problem is about the stereochemistry, I'd try to address it according to sophomore undergraduate Organic Chemistry text books used in the USA.

Briefly, any nucleophilic substitution reaction at primary carbon predominantly undergoes with S_N2 mechanism while those reactions at tertiary carbon undergo S_N1 mechanism exclusively. Meanwhile, reactions at secondary carbon would undergo either mechanism, based on reaction conditions, solvents, etc.

The "carbocation" given is a primary carbon even though it is a chiral carbon due to the presence of deuterium atom in place of hydrogen. Note that the stereo-effect caused by deuterium can be considered similar to that of hydrogen. Nonetheless, the bond energy difference (the $\ce {C–D}$ bond dissociation energy listed as $\pu{341.4 kJ/mol}$ while that of $\ce {C–H}$ as $\pu {338.4 kJ/mol}$) may cause the isotope-effects in kinetic of reaction mechanisms (that is a different avenue than what in discussion). Since $\ce{-CH2-Cl}$ and $\ce {-CHD-Cl}$ are similar and $\ce {C}$ atom in both cases is primary, the mechanism of the reaction should be S_N2 exclusively. Therefore, the stereochemistry will be $\ce {100\%}$ inverted.

For reading for substitution reactions involving alkyl halides: article. Please also note that, one of example given on that website, the compound $\ce {Ph-CH(CH3)CH2-Cl}$ does not undergo methyl group migration to give benzylic carbocation during its nucleophilic substitution reaction.