Monosaccharide's have C=O?

[jsmith1]

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TBR Orgo chapter 2 passage 2 question says: The IR spectrum for a straight chain monosaccharide has all of the following absorbance values EXCEPT:
A 3300
B 2980
C 2300
D 1715

I thought it was D because monosaccharides dont have c=o groups right??

 

[sakabato93]

TBR Orgo chapter 2 passage 2 question says: The IR spectrum for a straight chain monosaccharide has all of the following absorbance values EXCEPT:
A 3300
B 2980
C 2300
D 1715

I thought it was D because monosaccharides dont have c=o groups right??

When a monosaccharide (lets say glucose) is in its linear conformation, the would-be anomeric carbon has a carbonyl group attached to it. When the OH on the other side of the molecule nucleophilically attacks the partial positive on the carbonyl carbon, a hemiacetal forms, reducing the carbonyl group. You were probably thinking of the this cyclic, hemiacetal form of glucose.

 

[Vurday]

One more way of going about this problem is to use POE:

A 3300 - in IR, this is probably an Alcohol absorbance which is likely; eliminate A
B 2980 - This is probably representing C-H bonds off a sp3 hybridized carbon. This makes sense so eliminate it

C 2300
D 1715

Now between C and D, we know D represents a carbonyl but we're not sure if the structure contains one or not. So let's take a look at C. An IR absorbance of 2300 represents either a Nitrile (CN) or an alkyne (CC triple bond).

It seems much more likely to have a carbonyl than an alkyne/nitrile, so I would pick C as my answer

 

[Deadlifts]

All monosaccharides have open-chain aldose/ketose forms. It's the reason they all give positive Benedict's tests. That isn't necessarily relevant to spectrometry but it should reinforce the presence of carbonyl groups in monosaccharides.