Addition of acid to 3-methyl-2,3-dihydropyran

[![structure of compound to which acid is added](https://i.stack.imgur.com/ynt52.png)](https://i.stack.imgur.com/ynt52.png)


When a general acid, H-X reacts with the alkene in an addition reaction, which regiochemistry is expected?


In addition reactions X attaches to the most substituted carbon whilst H bonds to the carbon with fewest alkyl groups. In this case I would believe the **third** carbon to be the most substituted carbon due to it being bonded to three other carbons. And the **second** carbon to be where the hydrogen will bond. However, apparently it should be the other way around. And this would be the compound formed:


[![product after addition of acid](https://i.stack.imgur.com/MWIzL.png)](https://i.stack.imgur.com/MWIzL.png)


Why is the second carbon the most substituted in this case? Does it have to do with oxygen being more electronegative and thus it is seen as a higher prioritized substituent? 




Apparently, you're using the Markovnikoff's rule here. Recall that it is only a *basic* rule intended for simple unsaturated hydrocarbons, and **fails** at more advanced examples involving heteroatoms or rings.


In this case, the first step is the addition of $\ce{H+}$ ion from the acid to the pi bond, which leads to the formation of a carbocation. The carbocation's positive charge is most stable next to the oxygen atom. Can you tell why?


In the second step, the $\ce{X-}$ ion will simply attach at the position where the positive charge was the most stable, leading to a product as you've shown in the second image.


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