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Compare the reactivity order towards SN1 reaction for:

  1. (2-chloropropan-2-yl)cyclopropane
  2. $\ce{CH3OCH2Cl}$
  3. $\ce{Ph3CCl}$

Owing to the exceptional stability of cyclopropyl ethyl carbocation (see), obviously it should be on the top.

We did this question in class and my teacher told me that option 2 is more reactive than option 3, because the carbocation formed in the case of 2 is more stable. However, I couldn't quite understand the reason for it.

Anyway, I searched more for this and found the following table in March's 7th edition, page 425:

enter image description here

According to this table $\ce{Ar3CX > CH3OCH2X}$ in terms of reactivity towards SN1.

Today, I went to my teacher with this data and asked him about it. He told me that things in March can be wrong sometimes, I should rather check some other sources and prove it to him.

So I'd like to know what is the actual order, is 2 more reactive than 3, or vice-versa? (please mention your source) And why?

I have also checked Peter Skyes' and Morrison Boyd's, Clayden's book but it's not given there.


This issue has been discussed previously on the site but no concrete conclusion has been reached.

Jan's answer suggests that $\ce{CH3OCH2Cl}$ should be more reactive than trityl chloride.

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  • $\begingroup$ Unfortunately, a table like this is not super helpful. What are the reaction conditions? Solvent? What is your nucleophile? All of these factors can influence the rate of reactivity. Also, it is incorrect to say that a stabilized cation increases the rate of an SN1 reaction. That is a shortcut obtained by applying the Hammond Postulate. Here, you are assuming that dissociation is rate limiting and endothermic, and therefore, the transition state is late and cation-like. Therefore, factors that stabilize cations will lower the energy of the transition state and increase rate. $\endgroup$ – Zhe Apr 27 '18 at 2:42
  • $\begingroup$ The reaction the rate with 2 is higher than you thing is because the intermediate transition state is stabilized by donation of electron density from the adjacent oxygen. By the Hammond Postulate, the transition state is late and electron density will lower the barrier to reactivity. $\endgroup$ – Zhe Apr 27 '18 at 2:43
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    $\begingroup$ In the table under Sn2, shouldn't the first three groups be in the opposite order because of steric hindrance? $\endgroup$ – FreakyLearner May 10 '18 at 18:03

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