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After forming the carbocation adjacent to the cyclopropyl ring I had two options,

  1. $\ce{Cl-}$ then and there without any changes
  2. Expand ring, form a new carbocation and add $\ce{Cl-}$ there.

I chose the first option because of exceptional stability of cyclopropyl methyl carbocation. However, answer given is option C which follows mechanism (2).

Can someone explain how to judge and decide how to proceed after the carbocation formation?

  • $\begingroup$ Cyclopentyl ring does not let the cyclopropyl ring to provide stability. Think why? The reason is due the cyclopentyl ring the p-orbital of carbon (containing positive charge) becomes perpendicular to p-oribital on carbon of cyclpropyl ring. So the double bond character cannot be achieved there is a image in my answer see it. $\endgroup$ – Saurabh Singh Apr 20 at 6:32
  • 2
    $\begingroup$ This is a challenging question because you can form a nonclassical carbocation (homoallyl) by protonating farther from the cyclopropyl ring... $\endgroup$ – Zhe Apr 20 at 17:31

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In case 1. You can see it is a 2° carbocation (or secondary carbonation) with angle strain of ~25° on cyclopropane

In case 2. You can see it is 3° carbocation (or tertiary carbocation) with angle strain of ~10° on cyclopropane

It is obvious that case 2 will be major intermediate yielding $\ce{1-chlorobicyclic [3,2,0] heptane}$ as main product that is C.

Edit1: After reading your question more thoroughly I found that I missed one thing about 'exceptional' stability of cyclopropyl methyl carbocation.

The stability that cyclopropyl provides to carbocation is due to interaction of p-orbitals which is not possible in your case because p-orbitals of carbon having '+' charge becomes out of plane with respect to p-orbitals of cyclopropyl.

Edit2: See image below for reference

enter image description here

Then in case of cyclopropyl methyl carbocation the p-orbitals of both carbon containing '+' charge and the carbon on ring are parallel

In your case the p-orbitals of carbon containing '+' charge is perpendicular to plane of cyclopentyl ring but the p-orbitals of carbon on cyclopropyl ring is parallel to plane cyclopentyl ring, so stability is not provided by cyclopropyl ring.

To sum up you can say that the cyclopentyl ring does not let cyclopropyl ring to provide stability to carbocation which makes ring expansion a preferred option.

Edit3 (Bonus): I asked to my professor about this, he told me that there are two conformational isomers of cyclopropyl methyl carbocation. One is bisected conformation and other is perpendicular conformation. The bisected one allows the carbocation to be exceptionally stable but the perpendicular one isn't that much stable (like in your case) due to less interaction of p-orbital and the partial double bond character cannot be achieved.


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