Why does a proton prefer to attack a double bond instead of protonating an alcohol in a reaction of unsaturated alcohol with sulphuric acid?

Question

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 ($\ce{H2SO4}$):

The mechanism provided by the book is as follows:

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:

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.}

Answer 1

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")