TBR Orgo: Carbohydrates

[purplelightning]

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Treatment of an aldohexapyranose with 1 equivalent of HIO4 breaks the sugar between:

A. Carbon 1 and 2
B. Carbon 5 and 6
C. Carbon 2 and 3
D. Carbon 3 and 4

The answer is D


I'm not sure as to why this is 😕

Thanks!

 

[TieuBachHo]

I don't know what explanation you have but I have a suspicion that the Oxygen atom in the ring play a role in lesser the negative charge of OH in C2. The OH(s) of C3 and C4 are further away so they have lesser impact than that of OH in C2.
Since I looked up the 3D of the glucose ring, the OH(s) distances between C2 and C3, C3 and C4 are very similar, and no steric hindances that can affect more to one than the other.
Since 1 equivalent of HIO4 can only oxidise 1 equivalent C-bond. Therefore, it must be an oxidation of C3-C4.

 

[purplelightning]

Your explanation does make sense. The key says that apparently C3 and C4 are the least sterically hindered so HIO4 breaks that bond. It doesn't go on to explain -why- that is. Thanks for your help! 🙂

 

[TieuBachHo]

I don't see why steric hindrance plays a role in this; but that can be if they are talking about the alpha or beta sugar-ring. If it's an alpha-ring, then cis- of C1andC2 can hinder its ability to get oxidized.

 

[BerkReviewTeach]

Treatment of an aldohexapyranose with 1 equivalent of HIO4 breaks the sugar between:

A. Carbon 1 and 2
B. Carbon 5 and 6
C. Carbon 2 and 3
D. Carbon 3 and 4

The answer is D


I'm not sure as to why this is 😕

Thanks!

The steric hindrance stems from the ability for the large HIO4 molecule to bind adjacent OH groups, and that is best done away from the anomeric site.

But perhaps more important is that I can't seem to find that question anywhere in my book. I'm looking at the latest version of the BR book, so maybe it's in an older copy. I think they probably removed it after AAMC announced changes to what they test in organic chemistry.