Concept explainers
(a)
Interpretation: The number of carbons in
Concept introduction: Open chain structures of monosaccharides which contain five or more carbon atoms exist in equilibrium with two cyclic structures. These cyclic structures are the dominating forms at equilibrium. The carbonyl group in the open-chain structure of monosaccharide reacts intramolecularly with a hydroxyl group which results in the formation of cyclic forms. These cyclic compounds are called cyclic hemiacetals.
(b)
Interpretation: The number of carbons in
Concept introduction: Open chain structures of monosaccharides which contain five or more carbon atoms exist in equilibrium with two cyclic structures. These cyclic structures are the dominating forms at equilibrium. The carbonyl group in the open-chain structure of monosaccharide reacts intramolecularly with a hydroxyl group which results in the formation of cyclic forms. These cyclic compounds are called cyclic hemiacetals.
(c)
Interpretation: The number of carbons in
Concept introduction: Open chain structures of monosaccharides which contain five or more carbon atoms exist in equilibrium with two cyclic structures. These cyclic structures are the dominating forms at equilibrium. The carbonyl group in the open-chain structure of monosaccharide reacts intramolecularly with a hydroxyl group which results in the formation of cyclic forms. These cyclic compounds are called cyclic hemiacetals.
(d)
Interpretation: The number of carbons in
Concept introduction: Open chain structures of monosaccharides which contain five or more carbon atoms exist in equilibrium with two cyclic structures. These cyclic structures are the dominating forms at equilibrium. The carbonyl group in the open-chain structure of monosaccharide reacts intramolecularly with a hydroxyl group which results in the formation of cyclic forms. These cyclic compounds are called cyclic hemiacetals.
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Chapter 18 Solutions
GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP
- The disaccharide α,β-trehalose differs from the α, α structure as shown by having an (α1→ β1) linkage. Draw its structure as a Haworth projection.arrow_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. (The sugars shown here are not all of the possible five-carbon )arrow_forwardWrite the name for the trisaccharide depicted below (for example, of the form a - D - xylofuranose). Be sure to include if the sugars are in the pyranose or furanose conformations in the name and include the linkages between sugars (for example, (2 - - > 4)). CH₂OH ОН OH н ОН -CH₂ H\H ОН н н ОН CH2OH н ОН но н н ОНarrow_forward
- Does gentiobiose (Problem 20.66) have an acetal grouping? A hemiacetal grouping? Do you expect gentiobiose to be a reducing or nonreducing sugar? How would you classify the linkage (x or β and carbon numbers) between the two monosaccharides?arrow_forwardDraw the two possible Haworth structures (both alpha and beta anomers) for the following monosaccharides and give their corresponding systematic names. [Show the stepwise process]arrow_forwardConsidering the formation of glycosidic linkages between monosaccharides, how many possible trisaccharides can be produced by combining glucose, mannose and galactose? How does this differ from the number of possibilities for a tripeptide composed of three different amino acids?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 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_forwardDraw the two possible Haworth structures (both alpha and beta anomers) for the following monosaccharides and give their corresponding systematic names. [Show the stepwise process] (a) D- Fructosearrow_forwardFollowing are Fischer projections for a groupof five-carbon sugars, all of which are aldopentoses. Identify thepairs that are enantiomers and the pairs that are epimers. (Thesugars shown here are not all of the possible five-carbon sugars.)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 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_forwardFollowing are Fischer projections for a group of five-carbon sugars, all of which are aldopentoses. Identify the pairs that are enantiomers. CHO сно H-C- OH H-C-OH H-C- OH но-с — н н-с—он но- ČHOH ČH,OH сно CHO Н-с—он но—с— н H-C- OH H-C-OH но—с—н Н-с—он ČHOH ČH,OH сно сно н-с—он но—с —н но—с— н но -с — н H-C- OH но- C-H ČH,OH ČH,OHarrow_forwardThe monosaccharide shown below is the CH₂OH OH OH beta- L- D- alpha- Он OH anomer.arrow_forward
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