One of the following diastereomeric chlorides undergoes racemization on heating whilst the other is returned unchanged. Explain.

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From the reaction conditions I can see that there must be an elimination of $\ce{HCl}$, presumably via E2 with the base being the amine group on another molecule. Then addition back to the double bond.

However, I have a couple of questions-

Firstly, wouldn't $\ce{HCl}$ add back onto the alkene the other way around so as to form the more stable tertiary heteroatom stabilized carbocation? Which would then be attacked by the chloride ion thereby inducing racemization, and thus it should be impossible to be "returned unchanged".

Secondly, why would one isomer undergo racemization and the other remain the same?


1 Answer 1


Upon heating, there is an intramolecular $\mathrm{S_N2}$ reaction where the nitrogen lone pair attacks the chlorine-bearing carbon, which gives an aziridinium ion. The stereoelectronic requirement for a "backside attack" results in the formation of a specific diastereomer, illustrated here for the first of the two compounds:

Compound 1

Since there is absolutely no difference between both carbons in terms of sterics or electronics, the free $\ce{Cl-}$ ion can now attack either carbon with equal probability to give a mixture of two products:

Opening of ring

The second compound reacts in exactly the same way:

$\hspace{20 mm}$Second compound

However, no matter which carbon the $\ce{Cl-}$ ion attacks, the same product is obtained - you can verify this yourself. The product is the same as the starting material.


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