In my book this reaction is given without any mechanism:

enter image description here

I thought that simple dehydration reaction is going to take place with rearrangement of the carbocation but I have no idea as to how toluene turned out to be the product.


Seeing the reaction given here, I assume that is an on-paper reaction that takes place and that all three alcoholic groups are dehydrated.

enter image description here

The mechanism (on paper) seems to be as follows

Dehydration $1$

For the first dehydration, the carbocation formed is stabilised via ring expansion and then forms a double bond as follows:

Dehydration 1

Dehydration $2$

Similar to the first dehydration, a ring expansion occurs to stabilize the carbocation by forming a 6-member ring:

Second Dehydration

Dehydration $3$

In the third dehydration, a hydride shift takes place concomitantly forming a tertiary carbocation stabilized by conjugation with the double bonds. The double bond is then formed giving the compound aromaticity.

Third dehydration


mechanismThe mechanism should provide for the successive processes of carbocation formation, rearrangement and proton loss.

Observations(Points to consider):

  • The tendency to attain stablity by aromaticity.
  • The high ring strain in the cyclobutane ring.

The first step step should involve protonation of each hydroxy group attached. Protonation produces hydronium ion from hydroxide, which is a good leaving group. It leaves forming carbocation.

The carbocations, although formed in a single step, can be dispayed using successive steps for ease.

Each step of hydronium ion loss will produce a primary carbocation, which is unstable(due to lack of any prominent stabilising effect like resonace or hyperconjugation). Hence, rearrangement is the only method emplyed for efficient stablization of catbocation.

The neighbouring migration may be of -CH or -H. Generally, the migratory amplitude of -H is higher than -CH. But hydride migration will not solve the problem of enormous ring strain. Therefore, to ease ring strain, it is -CH which migrates.

Clearly, now:

  • The carbocation is secondary and more stable than the initial primary.
  • The extreme strain in the ring is also relaxed. This will be followed by a deprotonation step, forming double bond.

This migration of -CH group and successive step of deprotonation occurs twice, in two -OH losses.

In the third -OH loss the ring is expanded enough to ease ring strain. Thus, hydride, having lower migratory amplitude, is preferentially, migrrated to transfer positive charge. But this does not cause any ring size change. Next deprotonation step causes, double bond formation, to generate toluene. This is also enhanced due to the tendency to acquire aromaticity, as multiple double bonds are quite unstable.

  • $\begingroup$ @Safdar When in acidic medium, the protonation of atoms with lone pairs almost always occurs. The hydroxide and hydronium are conjugates and in equilibrium. While 3 positive charges are unstable, they do not exist simultaneously and are in equilibrium. Therefore, when the reaction begins at one hydroxide, the Kp of the protonation steps also increases at another hydroxide. It is just to show this in the first intermidiate. $\endgroup$ – Aditya Roychowdhury Jul 30 '20 at 9:19
  • $\begingroup$ Nice depiction of the mechanism, but forgive me for nitpicking. I can't really see how your answer adds anything substantially different from the previous answer by Safdar other than a running commentary of the arrow pushing diagram(which are quite self-explanatory) $\endgroup$ – Yusuf Hasan Jul 30 '20 at 10:06
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    $\begingroup$ @YusufHasan When I started writing, there was no answer provided. But by the time I finished, there was already an answer. Hence, they overlap. $\endgroup$ – Aditya Roychowdhury Jul 30 '20 at 10:30

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