# Loss of aromaticity vs low entropy of activation

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.