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A solved problem in Organic Chemistry by Solomons, Fryhle and Snyder asked to write the mechanism for the following transformation in the presence of sulphuric acid ($\ce{H2SO4}$):

Target transformation

The mechanism provided by the book is as follows:

Book's mechanism

It can be seen that a proton from sulphuric acid attacks the double bond thereby forming a tertiary carbocation as an intermediate. Later the lone pair on the oxygen atom of the hydroxyl group attacks the carbocation to form a cyclic intermediate which after deprotonation gives the required product. I understood the mechanism given in the book.

However, when I attempted the problem with the given reactant and the reagent, instead of attacking the carbon-carbon double bond, I protonated the alcohol. I attained at a much different product by following the mechanism given below:

My mechanism

The reason I chose to protonate the alcohol was I felt that electron density near oxygen is much higher than that in the double bonds. Moreover, I also end up with a conjugated diene, which is resonance stabilized.

Further, the following statement given as an explanation for a different problem supports my method of protonating over attacking the alkene:

Protonation of an alkene forms a carbocation with an open octet, whereas protonation of methanol does not form an open octet. Therefore, we predict methanol to be a stronger base than alkene. The mechanism starts with the reaction of a stronger base with sulphuric acid.

Now, my question is, why does a proton prefer to attack a double bond rather that protonating an alcohol? Under what conditions, does a proton protonate the alcohol instead of attacking the double bond? Are there any well-defined rules to predict which attack will be preferred over the other in such cases?


Image source: I made them using ChemDoodle.

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  • $\begingroup$ That would probably depend on situation chemistry.stackexchange.com/questions/41836/… $\endgroup$ – Mithoron Apr 3 at 16:13
  • $\begingroup$ The alcohol is protonated but reversibly. It is a unproductive event. Protonating the alkene is a productive process. $\endgroup$ – user55119 Apr 4 at 1:53
  • $\begingroup$ Don't fret about your ChemDoodle drawings: they are actually very clear! $\endgroup$ – matt_black Apr 15 at 10:07
  • $\begingroup$ @matt_black: Thanks for the comment :) I've removed that from the footnote. $\endgroup$ – Guru Vishnu Apr 15 at 10:11
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Both the reaction mechanisms are equally correct. Reaction takes place at numerous sites in most of the organic reactions, many kinds of products are formed, but the thing which actually give the major product is one of the intermediate, which is relatively stable than its peers.

We know, that 3° carbocation is a more stable intermediate than an allyl carbocation (irrespective of the fact that the latter is going to give a conjugated product), so the former will be present in adequate amount in order to give the major product.

Also, the formation of conjugated alkene takes place through multiple intermediates, and therefore it becomes important to care for the stability of each intermediate. I especially doubt the 3rd step, where along with a 1° carbocation, alkene is present, which acts as an EWG, and hence lowering the stability of the intermediate.

So, taking account of these factors, I would likely go with the given solution.


Edit: As per Yusuf Hasan's points on the stability of THF in comments, the given product, being a THF derivative is likely going to solvate the other intermediates and it is itself stable. That's why, THF is an unreactive compound and hence it is used as a solvent in many organic reactions. (source, under the heading "Reactivity of cyclic ethers - Oxacyclopropanes")

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    $\begingroup$ I guess you didn't reflect on the product viability of both the pathways. One of the paths is leading to the formation of a THF derivative, and being a cyclic ether,it would tend to be much more stable than the dialkene. Also, the ether formed as a product would also be much better at solvating the intermediates and also itself than the aqueous medium,so the reaction would again be driven forward by Le Chatlier's principle $\endgroup$ – Yusuf Hasan Apr 6 at 6:17
  • $\begingroup$ Thanks for your answer. I don't understand how you decided that a tertiary carbocation is more stable than an allyl carbocation. In the chapter "Radical Reactions", the same book says that allyl radicals are more stable than tertiary radicals. And since, the stability of carbocation parallels the stability of free radicals, I face a contradiction with the second paragraph of your answer. It would be helpful if you could clarify this. $\endgroup$ – Guru Vishnu Apr 6 at 6:20
  • $\begingroup$ @GuruVishnu: in general, hyperconjugation (hypcj) is relatively weaker effect than resonance. But, in case of that 3° C+, there are two things: 8 hypcj structures + inductive effect of surrounding carbons, which overrules a single resonance (as you know, hypcj is also a form of resonance). $\endgroup$ – Rahul Verma Apr 6 at 6:34
  • $\begingroup$ @YusufHasan: thanks for your comments. I'll edit my answer to include it. $\endgroup$ – Rahul Verma Apr 6 at 6:37
  • $\begingroup$ @RahulVerma: I agree with your comment explaining the stability of tertiary carbocation over an allyl carbocation. So does it mean the stability of radicals don't always parallel the stability of carbocations and vice versa? I thought both being electron deficient must be governed by the same stability rules. $\endgroup$ – Guru Vishnu Apr 15 at 10:27

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