Section Bank C/P #85

[deleted647690]

I got the correct answer, but I want to make sure my reasoning was okay. I decided it couldn't be B because a disaccharide can't be reducing if both anomeric carbons are tied up in the glycosidic linkage. For the same reason, you can cross off C because it would be reducing if only one anomeric carbon was tied up in the linkage.

Both A and D are accurate in their descriptions of reducing and non reducing sugars, but I only chose A because I knew that maltose was made up of two glucose sugars (from the passage), and I knew glucose was reducing, so I kind of just guessed.

EDIT: Actually, I just realized it told you in the passage it was a 1--->4 bond, so that tells you the answer since you know that in glucose, carbon 1 is the anomeric carbon

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[theonlytycrane]

If the sugar can be oxidized, it's reducing. The trick is to see if there's a free -OH at the anomeric carbon. Choices (B) and (C) are both logically wrong so they can be eliminated. Then, as you said, the passage tells you that the linkage is 1 -> 4, so you can eliminate choice (D) and go with choice (A).

Your logic is good and is the kind of thinking that is needed for your test. Keep at it

 

[aldol16]

Your reasoning for B and C is correct. To distinguish A and B, you simply need to be able to identify the anomeric carbon. This is the one that is the aldehyde in the linear form. In other words, if you can envision a detachment (or a retron, if you will) where the attacking OH comes off to reform the aldehyde, you can 'see' the aldehyde carbon and therefore the anomeric carbon. A sugar is 'reducing' because the aldehyde can serve as an electron source and thus get oxidized, which provides a reducing equivalent.

 

[quickaskz]

I'm confused because I thought glycosides were acetals which are nonreducing sugars. Why is C ruled out and why is A the answer?

 

[PlsLetMeIn21]

The glycosidic linkage is not the reducing part, it's the hemiacetal end of the glucoside that can be oxidized.

Carbon 1 of the first alpha-glucose is 'protected' by the glycosidic bond, so it cannot react. But carbon 1 of the second alpha-glucose is an 'unprotected' anomeric carbon, so that hemiacetal can be oxidized. This makes it a reducing sugar.