Consider this carbocation:

enter image description here

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 SN1 reactions, but I am not sure. Am I correct?

At a very detailed level, I was also taught that in SN1 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?

  • $\begingroup$ Isn't it a primary carbonium ion to begin with (in addition, in here, $\ce {C-D}$ is smaller than $\ce {C-H}$)? In that account, I'd say $100\%$ inversion because reaction is $\ce {S_N2}$ rather than $\ce {S_N1}$. $\endgroup$ Mar 28, 2018 at 15:15
  • $\begingroup$ As a general rule, if you start with an optical inactive solution, you'll end up with an optically inactive solution. Since the carbocation is not optically active, then you'll get a racemate. $\endgroup$
    – ralk912
    Mar 29, 2018 at 4:23
  • $\begingroup$ Stereochemical information is lost once you form a carbocation, since in general it presents a planar geometry and $sp^2$ hybridized geometry; barring some special cases involving bulky substituents etc. there is no preferred face of attack. $\endgroup$
    – getafix
    May 28, 2018 at 12:32

2 Answers 2


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 SN2 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.

enter image description here

  • $\begingroup$ (not my downvote) But I am finding it difficult to infer what your final conclusion is regarding the stereochemistry of the products. Are they racemic or diaseteromers? You've mentioned both in your answer. $\endgroup$ Mar 29, 2018 at 9:46
  • $\begingroup$ I feel there are two options. Either 4 and/or 6. Structure 6 as a mixture of diastereomers. I don't have a precise answer, haven't done the experiment. $\endgroup$
    – user55119
    Mar 29, 2018 at 12:39

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 SN2 mechanism while those reactions at tertiary carbon undergo SN1 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 SN2 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.

  • $\begingroup$ Thanks for your answer. First of all, I am not sure why you're assuming that the reaction proceeds by $\ce{SN2}$ instead of $\ce{SN1}$. I have said a carbocation was formed, that means it is $\ce{SN1}$, so, please don't assume $\ce{SN2}$. Thanks $\endgroup$ Mar 29, 2018 at 9:42
  • $\begingroup$ Thank you for your editing and clarification about the carbocation. I'll think about tomorrow. :-) $\endgroup$ Mar 29, 2018 at 9:48
  • $\begingroup$ Did you happen to think about it again? :) $\endgroup$ Apr 28, 2018 at 10:30

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