TBR Organic, Section VI - Carbohydrates, Passage I - Sugar Conventions

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organesha

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The answer explanations for #8 and 9 are exactly the same as the explanation for #7. Can anyone (@BerkReviewTeach) enlighten me on the actual answers? I have #8 as D and #9 as C, but I obviously can't see if I'm right or wrong!

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7. Choice C is the best answer. Glycosidic linkages involve an anomeric carbon with two OR groups, so the functionality cannot contain the prefix "hemi". This eliminates choices A and B. Because maltose is formed from glucose, an aldose, the functional group is an acetal. This makes choice C the best answer. Aldoses form hemiacetals as monosaccharides and acetals as polysaccharides. Ketoses form hemiketals as monosaccharides and ketals as polysaccharides.


8. Choice D is correct. In ß-D-glucopyranose, all of the substituents on the pyranose ring have equatorial orientation. This is a piece of information you should commit to memory. The structure in this question differs from ß-D-glucopyranose at carbon 3, where the hydroxyl group has axial orientation rather than equatorial orientation. This makes the structure a C-3 epimer of ß-D-glucopyranose, making choice D the best answer. An anomer would vary in chirality at the anomeric carbon. This eliminates choice A. A conformer is the identical molecule rotated or contorted. If only one substituent changes from equatorial to axial while the ring remains in the same orientation, then the structures are not conformers. This eliminates choice B. Enantiomers are mirror images, so they will differ at every chiral center. All of the centers would need to be axial for it to be an enantiomer, but they are not all axial. This eliminates choice C.


9. Choice C is correct. The glycosidic linkage involves carbon 1 of the glycosyl on the left and carbon 4 of the glycoside on the right, so we need to identify the right sugar. Carbon 6 is up in the Haworth projection, so it is a D-sugar. This eliminates choices B and D. The OH groups on the sugar match that of glucose, with the exception of carbon 2, so the sugar is the C-2 epimer of glucose. From memorization, you should know that mannose is the C-2 epimer of glucose. This makes choice C the best answer.
 
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