Concept explainers
Interpretation:
The disaccharides formed by the
Concept introduction:
Glycosidic linkage forms disaccharide by reaction of the
Reducing sugar: Sugars that contain
Anomeric position is carried by an
Anomeric position is carried by an alkoxy group for acetal sugars.
Hemiacetal sugars are in equilibrium with their open chain form.
Acetal sugars are not in equilibrium with their open chain form.
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Fundamentals of General, Organic, and Biological Chemistry (8th Edition)
- Following are Fischer projections for a group of five-carbon sugars, all of which are aldopentoses. Identify the pairs that are enantiomers and the pairs that are epimers. (The sugars shown here are not all of the possible five-carbon sugars.) 1 СНО 2 СНО СНО Н-С—ОН Н-С—ОН Н—С—ОН Н-С—ОН НО —С—Н Н-С—ОН Н- С—ОН НО—С— Н НО—С—Н CH̟OH CH̟OH ČH̟OH в iоchemistry |61 STATE ATION Republic of the Philippines Romblon State University Romblan, Philippines 4 СНО 5 СНО 6 СНО НО—С—Н Н-С—ОН НО -С—Н Н—С—ОН НО -С—Н НО -С—Н H-C–OH Н-С—ОН НО—С—Н CH,OH ČH,OH CH,OHarrow_forwardFollowing are Fischer projections for a group of five-carbon sugars, all of which are aldopentoses. Identify the pairs that are enantiomers and the pairs that are epimers СНО СНО СНО H-C-OH H-C-OH H-C-OH H-C-OH Но-С—н H-C-OH H-C-OH Но-ҫ—н Но—с—н CHOH CHĻOH ČH,OH СНО СНО СНО Но —С— н H-C-OH HO-C-H Н—С—он Но -С—н HO-C-H H-c-OH H-C-OH HO-C-H CH̟OH ČHĻOH ČHĻOH Pairs of Enantiomers Pairs of Epimersarrow_forwardWhy is sucrose a non-reducing sugar with Benedict’s solution while lactose is a reducing sugar under the same condition?arrow_forward
- Following are Fischer projections for a group of five-carbon sugars, all of which are aldopentoses. Identify the pairs that are enantiomers and the pairs that are epimers. (The sugars shown here are not all of the possible five-carbon )arrow_forwardDraw the structural formula for -D-glucosyl-(1n6)-D-mannosamine and circle the part of this structure that makes the compound a reducing sugar.arrow_forwardWhat is the pyranose/furanose ring structure of this sugar?arrow_forward
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- Draw the structure of malibiose a sugar found in plants and answer the following question: a) What two monosaccharides are formed on hydrolysis of melibiose? b) Is melibiose a reducing sugar? Explain. c) Describe the glycosidic linkage between the two mono- saccharide units.arrow_forwardThe amount of branching (number of (α1→6) glycosidic bonds) in amylopectin can be determined by the following procedure. A sample of amylopectin is exhaustively methylated—treated with a methylating agent (methyl iodide) that replaces the hydrogen of every sugar hydroxyl witha methyl group, converting —OH to —OCH3 . All the glycosidic bonds in the treated sample are then hydrolyzed in aqueous acid, and the amount of 2,3-di-O-methylglucose so formed is determined.(a) Explain the basis of this procedure for determining the number of (α1→6) branch points in amylopectin. What happens to the unbranched glucose residues in amylopectin during the methylation and hydrolysis procedure?(b) A 258 mg sample of amylopectin treated as described above yielded 12.4 mg of 2,3-di-O-methylglucose. Determine what percentage of the glucose residues in the amylopectin contained an (α1→6) branch. (Assume that the average molecular weight of a glucose residue in amylopectin is 162 g/mol.)arrow_forwardUse the two Fischer projections shown below to draw the Haworth projection of these two sugars linked by an a(1-2) linkage, with the ß isoform of the ketohexose. H H ㅎ HO H H OH OH ОН Н НО CH₂OH 애,애애 OH CH₂OH OH H OH OH OH I OH CH₂OH OH OH OH Eng са CH2OH CH2OH OH OH ОН ОН CH₂OH Н OH Н Н OH Н T I CH₂OH CH₂OH OH OH OH CH₂OH CH₂OH Н CH₂OH CH₂OH ОН OH OH НО OH H Н OH CH₂OHarrow_forward
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