Oxidizing Sugars Mildly & Aggressively

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Description

https://joechem.io/videos/85 for video on jOeCHEM and attached worksheet + solution (below video on jOeCHEM aka the link) In this video, we'll talk about the two ways you can oxidize sugars. In the more mild manner, you can selectively oxidize an aldehyde to a carboxylic acid in an aldose, but if you need to oxidize the "top" and "bottom" of a sugar you can use nitric acid.


Comments:


  • Will • 5 years ago
    So in this, you said that "D" only has to do with the 4th carbon from the aldehyde carbon, but you later go on to say the "L" configuration has every stereocenter different? So my question is, if you change one of the other stereocenters in D-Glucose does it become something other than D-glucose? Is the Epimer that you drew still D-Glucose?

    REPLY
    • Joe Del Nano • 5 years ago
      So here's what it boils down to: D versus L depends on the LAST carbon when looking at a correctly drawn Fischer Projection of a sugar. If the OH is on the "right" side, it has a D prefix; if it is on the left side, it is an L prefix. To take a sugar that is D (for example, D-glucose) and transform it into its L configuration, you need its exact mirror image (aka, you need to draw its enantiomer). Said another way, to go from D to L (or L to D), you need to flip every stereocenter (which will, in turn, change the last OH from being on the right side to the left side

      REPLY
    • Joe Del Nano • 5 years ago
      What I need to revise about this video, is that to have an epimer of a sugar (take D-glucose for example), you just need another structure that differs AT LEAST ONE OR MORE stereocenters. So if you took D-glucose and flipped any of the OH's to be on a different side (it could carbon #1, #2, #3, etc) that resulting new Fischer Projection would be an epimer. But it could also be more than one stereocenter that gets flipped, and you will still end up with an epimer: does that make sense?

      REPLY