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enter image description here

first thought was protonating alkene to give carbocation which does EAS on the phenyl group followed by attack of oxygen on the 3 deg carbon to break the newly formed 4 membered ring. problem with this was i wasn't sure which C-C bond would break (i tried breaking it both ways but i couldn't see how mentioned product was forming either way) any hints / solution would be appreciated

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    $\begingroup$ I think the double bond isomerizes to conjugate with phenyl ring by acid catalysis first. Then, acid catalyzed addition of hydroxyl group happens to give the product. $\endgroup$ Commented May 17 at 7:06
  • $\begingroup$ could you provide a source for this isomerization reaction? i thought of the same thing yesterday but couldn't find any evidence regarding that so discarded that theory $\endgroup$ Commented May 17 at 7:22
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    $\begingroup$ @MathewMahindaratne What is the balanced equation? Is elemental hydrogen the second product? $\endgroup$
    – Karsten
    Commented May 17 at 11:23
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    $\begingroup$ The suggestion of @Mathew is fine except that the product should have no double bond. $\endgroup$
    – user55119
    Commented May 17 at 14:10
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    $\begingroup$ There's an error in this pic. Indeed, there should be no double bond. $\endgroup$
    – Mithoron
    Commented May 17 at 16:18

1 Answer 1

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Per your request Bromocresol: Because the reaction is acid-catalyzed, both alcohol 1 and the product, tetrahydropyran 5, must have the same molecular formula $\ce{C14H18O}$ and the same degree of unsaturation, D.U.= 6. The formation of unsaturated tetrahydropyran 6 ($\ce{C14H16O}$; D.U. = 7) would, at some point in the transformation, require a 2-electron oxidation. Given the conditions of the reaction, unsaturated tetrahydropyran 6 cannot be the product.

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  • $\begingroup$ hm, thank you! however, i have a doubt in step 2 as initial carbocation formation should favor the most stable carbocation which would be on the carbon away from phenyl group rather than towards it (checking hyperconjugation and inductive effect), right? even if we do consider that the carbocation is formed towards the phenyl ring, why wouldn't EAS happen? the four member ring thus formed could easily be broken by the -OH group and shouldn't that be favored? no doubts if 2-3 is valid though $\endgroup$ Commented May 18 at 8:40
  • $\begingroup$ @user55119: Good job. I have done this in my mind when I suggested the mechanism, but forgot the double bond given in the given product, which was erroneous. :-) $\endgroup$ Commented May 18 at 13:35
  • $\begingroup$ I'm not sure what you mean by step 2. In the protonation of cmpd. 1, either end of the double bond may be protonated but the other mode of protonation gets you nowhere. Protonation of the double bond in 3 as shown leads to the benzylic, resonance stabilized carbocation 4 that undergoes cyclization to ether 5. Also, I am uncertain about EAS (Electrophilic Aromatic Substitution?) and the 4-membered ring to which you allude. $\endgroup$
    – user55119
    Commented May 18 at 17:49
  • $\begingroup$ @user55119 yes sorry, I'm used to using EAS for electrophilic aromatic substitution by the step 2 thing, i basically meant that shouldn't the major product be the given compound itself as you mentioned that "the other mode of protonation gets you nowhere" or is it that this formed compound is the major product? also i do not know how to add pictures in comments so can't exactly explain what i meant by the 4-membered ring formation but i'll try explaining again in words when the carbocation is formed in step 2, why wouldn't the phenyl group being fairly activated form a wheland $\endgroup$ Commented May 18 at 18:37
  • $\begingroup$ intermediate with the carbocation (which causes a formation of 4-membered ring) after this, the Oxygen atom could attack at the tertiary carbon which causes the 4-membered ring to break. sorry, if the doubt seems silly, i'm just trying to understand the mechanism thoroughly so i don't have to ask similar doubts later $\endgroup$ Commented May 18 at 18:41

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