Nucleophilicity

[nancy9494]

Question: which OH is more likely to attack the aldehyde? I think it would be the one the tertiary C because the Positive charge that is created of the O will be stabilized through hyper conjugation. 

[hypervalent_iodine]

I would believe that, though I think you ought to look at the conditions and what sort of functional groups are involved a bit, as your reasoning implies an incomplete answer. What sort of products do you expect to get out of this (i.e. what functional group(s))?

Could you also attach a snapshot of the entire question?

[nancy9494]

There are 3 OH and two of them are secondary so I think that's supposed to be a clue? I mean the end product will be the same, but they love taking points off. 

[hypervalent_iodine]

Ah, great. I didn't realise that they had explicitly told you that you were making the acetal, which is what I was hinting towards. 

A couple of things. In part i), your mechanism is mostly good except for the last step. I don't like the way you've drawn the arrow from A^- to the proton attached to the positively charged oxygen, as you've drawn it through another carbon making it look as though the arrow is actually a bond. My second issue with this bit is that it is somewhat incomplete. You haven't drawn an arrow that would show the breaking of the H-O⁺ bond (i.e. movement of electrons in the bond onto the oxygen). 

For part ii)., I am happy that you have used the correct OH group in this section, however I don't know why you've redrawn the hemiacetal. Surely the question wants the arrow pushing on the boxed structure, otherwise why would the have reaction arrows coming away from it? My second issue is the next structure: the arrow pushing doesn't make sense. The ethereal oxygen in the hemiacetal is nucleophilic, so it is unlikely that it would be willing to accept any electrons in the way you have drawn. In the proceeding structure you have also drawn this without the water molecule, but nowhere in the previous step have you indicated that this has happened. I think you have gotten a bit confused in this section. 

Some questions to help you along in part ii: Why did you form the water molecule in the first step after formation of the hemiacetal? Are water groups good leaving groups or bad leaving groups? 

Oh, also, the fact that there are two secondary OH groups is irrelevant here. More important is the stereochemistry of the two OH groups that have reacted. Why do you think the reaction would have given the product shown and not the product corresponding to the other secondary OH? 

[nancy9494]

part ii) If the OH group is protonated with an acid, it can be kicked off by the nucleophilic oxygen when it forms a double bond with the C. 

 

Would it react with the OH that are on dashes faster because they are closer together and most of this process is intramolecular? 

 

Thank you so much for your help. I really appreciate it. 

[hypervalent_iodine]

Yep, that all looks great.

[nancy9494]

out here saving lives! thanks again 

[hypervalent_iodine]

One thing I would say looking at it again is that you would probably have the water leaving as a separate step, giving a carbocation which equilibriums between that and the double bonded structure you draw in the next structure. I don’t believe that it is a concerted process, so drawing it that way would be more correct. When I did this at uni, it didn’t matter overly if I did it your way or the one I’ve mentioned, except if I was drawing it out in an exam for final year Organic. Better to be safe though.