Even if reaction has low entropy of formation does that mean it is favorable to loss aromaticity and form highly strained cyclopropane ring? Why this reaction happens? Maybe someone can explain that in more details? Is this equilibrium reaction? I assume cyclopropane ring can be as easily opened with nucleophile as it closed? It does not make any sense for me here. Thank you for comments.

This is quoted from "The Art of Writing Reasonable Organic Reaction Mechanisms", Second Edition by Robert B. Grossman, DOI: 10.1007/b97257.

The reaction proceeds by an $\mathrm{S_N2}$ mechanism. The reaction has a very low entropy of activation, so it proceeds despite the loss of aromaticity. The product is a model of the antitumor agent duocarmycin. DNA reacts with duocarmycin by attacking the $\ce{CH2}$ group of the cyclopropane ring in an $\mathrm{S_N2}$ reaction.

enter image description here


1 Answer 1


There are several factors we can point to before resorting to entropy of activation.

1) At the end of the "chain" is displacement of an iodide ion. Iodine, with its diffuse valence orbitals, forms rather weak bonds with carbon making this displacement thermodynamically and kinetically favorable. Presumably the nucleophile that re-opens the cyclopropane ring forms a stronger bond with the carbon.

2) We may no longer have an aromatic ring, but a carbonyl group flanked by two conjugating carbon-carbon bonds still has plenty of resonance stabilization. We certainly do not lose all that much stabilization. Moreover...

3) Upon further review, there is still an aromatic contribution after all. The cyclopropane ring is well known to stabilize carbocations by hyperconjugation, and with the carbonyl oxygen tending to withdraw electrons we may argue that the quinone-like ring has some carbocation character. Adding a pair of pi electrons to this ring through hyperconjugation will then give aromatic contributing structures in which the cyclopropane ring is positively charged and the oxygen regains its negative charge.

All told, we don't have to reach for "entropy of activation". The reaction is better off than it looks just in terms of energy.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.