A certain aldose is added to an aqueous sodium hydroxide solution. After equilib

Question

A certain aldose is added to an aqueous sodium hydroxide solution. After equilibrium is reached it is found that there are 3 open chain sugars in the solution. It is believed that the original aldose tautomerizes to the enediol which then tautomerizes back to a carbonyl containing open chain sugar. One of the three is L-xylulose. A) Select the two other isomers. D-gulose D-allose D-mannose D-galactose D-idose D-lyxose D-xylose D-glucose D-altrose D-talose D-arabinose D-ribose L-gulose L-allose L-mannose L-galactose L-idose L-lyxose L-xylose L-glucose L-altrose L-talose L-arabinose L-ribose The answer is L-lyxose and L-xylose B) Select all of the isomer designations that apply to the isomeric relationship between the two sugars you selected in part A. x epimers x diastereomers _ tautomers _ conformational isomers _ enantiomers _ constitutional isomers x stereoisomers The answer is epimers, disastereomers, and stereoisomers C) Select all of the isomer designations that apply to the isomeric relationship between one of the sugars you selected in part A and L-xylulose. _ stereoisomers _ constitutional isomers _ conformational isomers _ diastereomers _ epimers _ enantiomers _ tautomers I tried tautomers only, but that was wrong? D) Draw the enediol that is believed to be the intermediate Any tips?

Explanation / Answer

C. constitutional isomers

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