DAT Bootcamp Orgo Practice Test #2, Question...

[RamsFan1991]

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I am having some difficulty understanding the explanation.

How is Reaction II considered Markovnikov? -OH is connected to a Carbon that is connected to TWO carbons in the product, while if -OH was connected to the Carbon connected to Ph and the other TWO carbons, that would be the most substituted product? I am thoroughly confused here, do I not have the right understanding of Markovnikov product? Thank you.

 

[KineticFlow603]

You can't just randomly put the -OH to Ph because this is E1. You can only choose between the two carbons that have the double bond. The first reaction is different because of hydride shift. Between the two carbons for the II, the secondary carbon will be more substituted than the primary carbon.
Edit: didn't see the oxymercuration-demurcuration reagents.

 

[jwan14]

In reaction II the mercurinium ion is formed as the intermediate. This prohibits rearrangement to form the most substituted alcohol. However, it is still considered markovnikov because it forms the most substituted product that is POSSIBLE to be formed. Hope that helps

 

[easyas123]

All Markovnikov reactions deal with pi bonds of some sort (usually alkenes). Now since you know this, the -OH should end up on the most substituted end of the alkene since it is Markovnikov. Also, as jwan had said, a mercurinium ion is formed and therefore no carbocation is formed for rearrangement. Hence, the -OH group will not end up on the carbon bonded to the phenyl group. The first reaction undergoes rearrangement due to the carbocation that forms in the intermediate step. Hope that made sense.