enter image description here

In this question, I need to find the order of decreasing rate of solvolysis of the above molecules with aqueous ethanol.

I think this can be done by finding the species that can form the most stable carbocation intermediate (and consequently, solvolysis will easily take place).

Now, the third molecule is obviously going to produce the least stable carbocation (since it can only form a secondary carbocation and no hydride shift is possible).

That leaves with us the first two molecules.

In the first molecule, the carbocation can be stablised by resonance. I believe that hyperconjugation from the methyl group will also help.

In the second structure, the carbocation formed will be tertiary and it too is relatively stable.

I had this notion that resonance in the first structure would make the carbocation formed by it the most stable. So, my answer was 1>2>3.

But it seems that I am wrong. The correct answer is 2>3>1 (ie. C).

Could someone tell me why this is the correct answer (and also clarify the mistakes in my assumptions)? Thank you for the help.

  • $\begingroup$ Vinyl halides really don't undergo solvolysis so c is the only possible answer $\endgroup$
    – Waylander
    Oct 5, 2021 at 6:54

1 Answer 1


The reaction can be safely assumed to be an SN1 reaction since, the solvent (aqueous C2H5OH, which is the nucleophile as well) is a polar species.

(2) shall yield a tertiary carbocation and (3) shall yield a secondary carbocation by the cleavage of the carbon-halogen (C-Br) bond.

However, in case of (1) cleavage of the said bond to yield a carbocation would not be that easy. Acquiring of double bond character is responsible for the strengthening of the bond and hence, difficult cleavage.

Further, the more electronegative a species is, the worse it is at handling positive charge. Hence, even if the carbocation does manage to form, it would be unstable. That's because the positive charge would reside on an sp2 hybrid carbon atom and as the 's' character of hybridisation in an atom increases, so does its electronegativity.

The stability of carbocations so formed would be: (2)>(3)>(1). And this order shall also mirror the decreasing rate of solvolysis of the given molecules.


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