[blueblueblue]

Hello!  I found a similar post to this from about 3 years ago, however the attachment was dead and it was never resolved.  So I hope someone can help me crack this one! Here is the problem:

"The structure formed from the Markovnikov addition of HI to the C=C bond in the cyclohexene ring in Tamiflu is shown.  The mechanism in the formation of this product involves the entire C=C-C=O moiety in the original Tamiflu molecule.  
Please provide a detailed mechanism that explains its formation."


I am thinking that H+ and I- will attach to the double bond in the cyclohexene, according to the Markovnikov rule.
After that, carbonyl group (Oxygen with a double bond) will take Hydrogen away from the cyclohexene.
This will create a carbocation.  And furthermore, Cl will rearrange the location with the carbocation.  

I would really appreciate anyone who could help me solve this problem, or point out some flaws in my idea, especially regarding Markovnikov addition.

Thanks!

[discodermolide]

I would think that HI protonates everything.
When the ester carbonyl oxygen is protonated iodide attacks in a conjugate fashion giving the product you describe

[blueblueblue]

Thank you very much for your explanation.

So, do you think that the oxygen with a double bond (in the ester carbonyl group) will be protonated as the first step?

I came up with this method and asked my teacher.  But, she told me to come up with a way of attaching HI to the double bond in the cyclohexene as the first step.  That is proving to be the difficulty for me.

Do you also think that since iodine is relatively big and "lazy"/slow, that might be part of the reason why iodine does not attach to the place where it is supposed to according to Markovnikov (Where the carbonyl group is attached to the cyclohexen)?

[discodermolide]

Thank you very much for your explanation.

So, do you think that the oxygen with a double bond (in the ester carbonyl group) will be protonated as the first step?

I came up with this method and asked my teacher.  But, she told me to come up with a way of attaching HI to the double bond in the cyclohexene as the first step.  That is proving to be the difficulty for me.

Do you also think that since iodine is relatively big and "lazy"/slow, that might be part of the reason why iodine does not attach to the place where it is supposed to according to Markovnikov (Where the carbonyl group is attached to the cyclohexen)?

Imagine that HI protonates the double bond. You get a three membered ring as an intermediate (imagine an epoxide where the oxygen atom has been replaced by an H+), the I- the attacks at the beta position to open this "ring" thus adding HI across the double bond. Remember the partial charges of such an alpha-beta unsaturated system, where 1-4 attack of nucleophiles is the usual course of reaction.

[macman104]

I agree with discodermolide, the expected way of this mechanism would be to protonate ester carbonyl, and add I^- in conjugate fashion.

However, if she wants you to add HI across the double bond first, I think you could also explain the regiochemistry by looking at the electronics of forming that carbocation.  You indicate that markovnikov would point to the I^- attacking at the tertiary center, that would normally be true.  But consider the functional group adjacent, the ester (specifically the carbon adjacent to the tertiary center), what do you know about the electronics and dipole of that group?

[jake.n]

This is a classic 1,4 conjugate addition.  I'm unsure what your instructor wants regarding the addition across the double bond, you can see that the alpha-C is not hydrogenated in enol product shown.

I imagine they would like you to show the protonation of the carbonyl, followed by addition of the iodide to the second order resonance structure of the protonated ester.

[blueblueblue]

Thank you very much for letting me know about 1,4 addition.  It makes a lot of sense, and I just found it in my textbook.  It surprised me because we are only in chapter 4, but 1,4 addition is not mentioned until chapter 7...so it was the first time for me to hear it.  However, I feel like I understand that process well.

Now, I am still trying really hard to figure this out.  So, why does the H+ proton attach to C=C like epoxide instead of O?

Secondly, I see ester inducing electrons from the side with a very strong negative charge, but I am lost on how it connects...

If the Carbon with I attached to it is #4 Carbon, then #1 Carbon will be the protonated oxygen? 
I am sorry for my amateur-level questions, but I really appreciate all of your expertise and effort!

[discodermolide]

Thank you very much for letting me know about 1,4 addition.  It makes a lot of sense, and I just found it in my textbook.  It surprised me because we are only in chapter 4, but 1,4 addition is not mentioned until chapter 7...so it was the first time for me to hear it.  However, I feel like I understand that process well.

Now, I am still trying really hard to figure this out.  So, why does the H+ proton attach to C=C like epoxide instead of O?

Secondly, I see ester inducing electrons from the side with a very strong negative charge, but I am lost on how it connects...

If the Carbon with I attached to it is #4 Carbon, then #1 Carbon will be the protonated oxygen? 
I am sorry for my amateur-level questions, but I really appreciate all of your expertise and effort!

The attachment of the H+ to the double bond is only a proposed transient intermediate, I only mentioned an epoxide to try to get you to visualise what I meant.

[jake.n]

I see ester inducing electrons from the side with a very strong negative charge, but I am lost on how it connects...

If the Carbon with I attached to it is #4 Carbon, then #1 Carbon will be the protonated oxygen? 
I am sorry for my amateur-level questions, but I really appreciate all of your expertise and effort!

In this case the ester functionality is not necessary for the mechanism, but it does allow you to draw another resonance structure, as it donates into the carbonyl.  This should let you know that the carbonyl is more basic than a typical ketone.

And your numbering is correct, 1 = O, 4 = beta carbon