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Topic: Aldol product treated with NaIO4 to yield..  (Read 3416 times)

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Offline Alkene

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Aldol product treated with NaIO4 to yield..
« on: November 17, 2014, 01:42:00 PM »
Hi all of you, kind internet strangers!
I read this forum regularly, but I had never posted before:

The thing is, I've just done an organic chemistry exam, and I think everything was allright, but I am not so sure about the final step of one of the exercises... And I've been struggling all day thinking about it (I will probably ask the professor later on, but that would have to wait at least a few days, so I decided to try luck here).

The exercise was about an aldol condensation, and the product of that reaction was the syn aldol (comming from a cis enolate), after hydrolisis and deprotection of the -OH group of the image bellow

But the last step was to treat the product with NaIO4. As far as I remembered, NaIO4 is an oxidant which can cause C-C bonds where those carbons are bonded to oxygen to easily cleave, yielding two carbonylic compounds.

A hint was also provided: The final product is a beta-hydorxiacid.

So I supposed that the most reasonable thing to occur was to yield the product that is also drawn in the image right bellow:





What do you think, fellas?
Is it all right or did I miss something important?

I didn't specify a mechanism, but I suppose that indicating the product will do it.

I think I aced the rest of the exam, so it would be a shame to fail in something like this.


Kind regards, and all the best :)

Offline OrgXemProf

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Re: Aldol product treated with NaIO4 to yield..
« Reply #1 on: November 17, 2014, 02:45:24 PM »
Alkene wrote: "As far as I remembered, NaIO4 is an oxidant which can cause C-C bonds where those carbons are bonded to oxygen to easily cleave, yielding two carbonylic compounds.

A hint was also provided: The final product is a beta-hydorxiacid."

* * *

Periodic acid (HIO4) is used to oxidize vicinal diols, e. g., RCH(OH)-CH(OH)R, to alpha-hydroxycarbonyl compounds, e. g., RCH(OH)-C(=O)R, which are further oxidized to RCH=O + RCO2H.

Oxidation of vicinal diols proceeds via formation and subsequent fragmentation of a cyclic intermediate. This process is closely analogous to the mechanism of OsO4 promoted bis-dihydroxylation of alkenes.

Periodic acid is used to differentiate between pyranose and furanose structures and also to distinguish between alpha- and beta-anomers of pyranosides and furanosides.

Your instructor didn't just give you a hint - - he gave away the farm! So, what other product was formed via the indicated reaction in addition to 3R-hydroxy-2S-methyl-4-phenylbutyric acid?

Then: Can you explain why reaction of HIO4 with the methyl alpha-D-pyranoside of ANY D-aldohexose affords the same compound (in addition to formic acid)?



Offline Alkene

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Re: Aldol product treated with NaIO4 to yield..
« Reply #2 on: November 17, 2014, 04:50:50 PM »
Periodic acid (HIO4) is used to oxidize vicinal diols, e. g., RCH(OH)-CH(OH)R, to alpha-hydroxycarbonyl compounds, e. g., RCH(OH)-C(=O)R, which are further oxidized to RCH=O + RCO2H.

I understand that the treatment with HIO4 is equivalent that the treatment with NaIO4, am I right? (The only reaction condition that was provided was that, "NaIO4").
So, in my molecule, is like if the hypothetical vicinal diol is already oxydized to an apha-hydroxycarbonil compound, which is my compound on my drawing, and then peryodate will yield the acid that I have drawn.


Your instructor didn't just give you a hint - - he gave away the farm! So, what other product was formed via the indicated reaction in addition to 3R-hydroxy-2S-methyl-4-phenylbutyric acid?

I guess you are referring to the other fragment that is obtained, i.e. acetone?
I also put that on my exam, but forgot to draw it here.


I would really appreciate the confirmation "Yes, your product is correct" so I can move on with my life again  ;D



Then: Can you explain why reaction of HIO4 with the methyl alpha-D-pyranoside of ANY D-aldohexose affords the same compound (in addition to formic acid)?

Yeah, I understand, you eliminate one terminal carbon of the chain, in which the difference resides, giving always the same product.

Offline OrgXemProf

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Re: Aldol product treated with NaIO4 to yield..
« Reply #3 on: November 17, 2014, 05:52:12 PM »
1. The I(+7) in periodate serves as oxidizing agent. So, should NaIO4 work in much the same way as HIO4 in this regard?

2. Yes, acetone is the other fragment.

3. "Yes, your product is correct." However, please see #4, below, before you move on with your life again:

4. NO, periodate oxidation of any methyl alpha-D-glycopyranoside does NOT "eliminate one terminal carbon of the chain, in which the difference resides"!

Let's consider methyl alpha-D-glucopyranoside vs. methyl alpha-D-allopyranoside. These are methyl pyranosides of diastereoisomeric D-aldohexoses, but the difference between their respective structures does NOT lie in the nature or identity of a "side-chain". (You might ask: What structural features render these two aldohexoses diastereoisomeric? How does reaction of each methyl pyranoside with HIO4 affect those structural features?)

You'll know that you've solved the problem correctly after (i) you have adopted a mechanistic approach to the reactions of HIO4 with each of the methyl pyranosides of D-glucose and D-allose and then (ii) you have examined the products thereby obtained. Try it, you'll like it!

On another topic: I suppose that someday I'll figure out how to use SMILES to enter structural formulae properly....but not today. (This gets us back to moving on with one's life again...老狗学不会新把戏 !)

PS: Handy mnemonic to help you to recall the absolute configuration of each of the eight D-aldohexoses: "All altruists gladly make gum in gallon tanks".


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