This is from one of the practice papers for the JEE , asks for possible hydrolysis products of (chloromethyl)cyclopropane .

products of hydrolysis  of (chloromethyl)cyclopropane

Product A happens to be due to stability of cyclopropyl methyl carbocation.

Product B happens to be due to relaxation of ring strain of cyclopropane ring (though A is major even this is possible)

Pathway for C doesn't seem to be anywhere in my books , I tried googling it but found nothing.

So, is but-3-en-1-ol a possible product of hydrolysis of (chloromethyl)cyclopropane?

  • $\begingroup$ Bad news. Cyclopropylmethyl cation doesn't exist in a single form. It's nonclassical. It's equivalent to the cyclobutyl cation and equivalent to homoallyl cation (which gives product C). $\endgroup$ – Zhe Jun 16 '20 at 21:28
  • $\begingroup$ I didn't know this , hence I didn't know the answer , thanks for your help $\endgroup$ – Fallen Grenade Jun 17 '20 at 2:44

I'd say, yes it is possible. The pioneering work on interconversion reactions of cyclobutyl, cyclopropylcarbinyl, and allylcarbinyl derivatives has been done by John D. Roberts and coworkers (Ref.1). In these works, they have checked reactions under both $\mathrm{S_N1}$ and $\mathrm{S_N2}$ conditions. They have observed the reactions expected to involve carbonium ion intermediates have shown interconversion, which have been correlated to factors such as carbonium ion stability, ease of interconversion, and stability of possible products. One such interconversion have been shown in the reaction of cyclopropylcarbinyl-, cyclobutyl-, and allylcarbinylamines with nitrous acid, which gave relevant carbonium ions. Similar approach was recently taken by Arca, et al. 2014 using solid catalysts (Ref.2). They have proposed a bicyclobutonium cation transition state to explain the product formation:

Cyclopropylcarbinyl Ion Rearrangement

The coorelation to liquid phase reaction is well explanied in their abstract:

Rearrangement of cyclopropylcarbinyl bromide over proton and ammonium-exchanged mordenite ($\ce{H}$-MOR and $\ce{NH4}$-MOR) was studied at different temperatures. The product distribution analysis revealed a similar profile to what is found in solution, with prevalence of cyclobutyl over allylcarbinyl bromide. The activation parameters showed that compared with 80% aqueous ethanol solution, the zeolites present lower enthalpy of activation and higher entropy of activation, but yielding a significantly lower $\Delta G^‡_{298}$. The results may be explained in terms of the ability of zeolites to serve as solid solvents, providing a polar nanoenvironment for ionic reactions to occur.

Therefore, we can assume the reaction in hand have also unedergone with similar mechanism, giving same three products, $\bf{1}$, $\bf{2}$, and $\bf{3}$ (probably in different ratios). The formation of $\bf{3}$ can also explained by the possible resonance contribution as depicted in above $\color{red}{\text{Red Box}}$. Note that two open resonance structures are essentially the same, but shown because it would make a double contribution due to the symmetry.

It's noteworthy that the open-chain derivatives have been achieved from the reaction of cyclopropylcarbinyl halides with amines (Ref.3).


  1. John D. Roberts, Robert H. Mazur, “Small-Ring Compounds. IV. Interconversion Reactions of Cyclobutyl, Cyclopropylcarbinyl and Allylcarbinyl Derivatives,” J. Am. Chem. Soc. 1951, 73(6), 2509–2520 (https://doi.org/10.1021/ja01150a029).
  2. Hugo A. Arca, Gabriel C. C. Gomesab, Claudio J. A. Mota, “Solid solvents: activation parameters for the rearrangement of cyclopropylcarbinyl bromide on mordenite zeolite,” New Journal of Chemistry 2014, 38(7), 2760-2762 (https://doi.org/10.1039/C4NJ00287C).
  3. Michael B. Smith, Robert T. Hrubiec, Charles A. Zezza, “Reaction of amines with cyclopropylcarbinyl halides: $\mathrm{S_N2'}$ or solvolysis?,” J. Org. Chem. 1985, 50(24), 4815–4821 (https://doi.org/10.1021/jo00224a033).
  • 1
    $\begingroup$ Thanks for the help sir $\endgroup$ – Fallen Grenade Jun 17 '20 at 2:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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