Which alkene reacts faster with HCl?

1: methylidenecyclohexane; 2: 1-methylcyclohex-1-ene; 3: 3-methylcyclohex-1-ene; 4: 4-methylcyclohex-1-ene; 5: ethenylcyclohexane

The logic tells me the most stable alkene 2 which is the most substituted would react the slowest.

On the other hand, it would form the most stable tertiary carbocation which would mean it's best for this specific reaction which would make it react faster. Also, one could think about rearrangements because these could take time too. How should I answer this question?

  • $\begingroup$ Future is a bit too much for saying that chemistry predicts - in reasonable extent - what the products of a reaction would be. $\endgroup$
    – Alchimista
    Commented May 3, 2021 at 12:45
  • 3
    $\begingroup$ 1 and 2 will give the same cation $\endgroup$
    – Waylander
    Commented May 3, 2021 at 12:54
  • $\begingroup$ Since the reaction proceeds via carbocation intermediate, stability of the possible carbocation will decide the reactivity here.(Do Keep in mind, Wagner Meerwin Rearrangement) $\endgroup$
    – Rishi
    Commented May 3, 2021 at 14:17
  • $\begingroup$ since the cations are same, Do you think that steric factors might play a role? ig the approach of cation in first case would be less hindered ? But not sure if it applies $\endgroup$
    – Ashish
    Commented May 3, 2021 at 15:07

1 Answer 1


This is an exercise in applying the Hammond postulate (multiple times). To accurately compare reaction rates, we need to have the activation energy. We don't have this, and the actual energy of the transition state is unknowable without experimental data.

Hammond's postulate states that the structure of the transition state is more similar to the intermediate to which it is closer in energy. For these reactions, the formation of the carbocation is endergonic, so the transition states look more like the carbocation than the alkene. We can then make some assumptions about the relative energies of the transition states from the energy change of the mechanism step if some things remain the same.

For example, alkene 3 can be protonated to form two different carbocations: A and B.

enter image description here

Using PM6 calculations in MOPAC, I calculated the heats of formation of alkene 3 and cations A and B.

$$\begin{array}{l|r} & \mathrm{\Delta_r H^\circ \ (kJ/mol)} \\ \hline 3 & -42.1 \\ \mathrm{A} & 710.7 \\ \mathrm{B} & 702.6\\ \end{array}$$

Since we are starting from the same place (alkene 3), we can postulate that cation B forms more quickly than cation A because it is slightly lower in energy.

You need to do similar analysis for all your alkenes and the cations they can form. In some cases you have two alkenes that might form the same cation. In that case, the higher energy alkene would likely translate into the more rapid reaction.

You do not need to go and calculate all the energies. You should have learned some rules by now to assess the relative stability of alkenes and carbocations based on substitution pattern.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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