
Concept explainers
(a)
Interpretation:
The acetal and hemiacetal in sophorose needs to be labeled.
Concept Introduction:
In a hemiacetal, an alcohol and ether attached to the same carbon.
A hemiacetal with an alcohol forms an acetal.
(b)
Interpretation:
Each monosaccharide ring should be numbered.
Concept Introduction:
Monosaccharides or simple sugars are the simplest carbohydrates. Generally, have three to six carbon atoms in a chain, with a carbonyl group at either the terminal carbon, numbered C1, or the carbon adjacent to it, numbered C2.
(c)
Interpretation:
Glycosidic linkage needs to be classified as a or β and its location should be designated using numbers.
Concept Introduction:
Disaccharides are carbohydrates composed of two monosaccharides.
Disaccharides are acetals, compounds that contain two alkoxy groups (OR groups) bonded to the same carbon.
A disaccharide results when a hemiacetal of one monosaccharide reacts with a hydroxyl group of a second monosaccharide to form an acetal. The new C-O bond that joins the two rings together is called a glycosidic linkage.
The two monosaccharide rings may be five-membered or six-membered. All disaccharides contain at least one acetal that joins the rings together. Each ring is numbered beginning at the anomeric carbon, the carbon in each ring bonded to two oxygen atoms
An a glycoside has the glycosidic linkage oriented down, below the plane of the ring that contains the acetal joining the monosaccharides.
A β glycoside has the glycosidic linkage oriented up, above the plane of the ring that contains the acetal joining the monosaccharides.
(d)
Interpretation:
Whether the hemiacetal drawn as an a or β anomer should be predicted.
Concept Introduction:
Anomers are cyclic monosaccharides or glycosides that are epimers, differing from each other in the configuration of C-1 if they are aldoses or in the configuration at C-2 if they are ketoses. The epimeric carbon in anomer is known as anomeric carbon or anomeric center.
Depending on the orientation of carbon number 1 when the carbon number 5 hydroxyl bonds to it, two different forms can result.
These two forms are identical except for the configuration around C1. These two forms are called anomers.
C1 is called the anomeric carbon. If the hydroxyl group on C1 and the -CH2OH group on C5 are on opposite sides of the six-membered ring, C1 is known to be the α anomer.
If they are on the same side, C1 is known to be the β anomer.
(e)
Interpretation:
The monosaccharide formed when sophorose is hydrolyzed needs to be determined.
Concept Introduction:
Disaccharides are carbohydrates composed of two monosaccharides.
Disaccharides are acetals, compounds that contain two alkoxy groups (OR groups) bonded to the same carbon.
A disaccharide results when a hemiacetal of one monosaccharide reacts with a hydroxyl group of a second monosaccharide to form an acetal. The new C-O bond that joins the two rings together is called a glycosidic linkage.
The two monosaccharide rings may be five-membered or six-membered. All disaccharides contain at least one acetal that joins the rings together. Each ring is numbered beginning at the anomeric carbon, the carbon in each ring bonded to two oxygen atoms.
The hydrolysis of a disaccharide cleaves the C-O glycosidic linkage and forms two monosaccharides.

Want to see the full answer?
Check out a sample textbook solution
Chapter 20 Solutions
General, Organic, & Biological Chemistry
- The acid-base chemistry of both EDTA and EBT are important to ensuring that the reactions proceed as desired, thus the pH is controlled using a buffer. What percent of the EBT indicator will be in the desired HIn2- state at pH = 10.5. pKa1 = 6.2 and pKa2 = 11.6 of EBTarrow_forwardCUE COLUMN NOTES (A. Determine Stereoisomers it has ⑤ Identify any meso B compounds cl Br cl -c-c-c-c-¿- 1 CI C- | 2,4-Dichloro-3-bromopentanearrow_forwardThe acid-base chemistry of both EDTA and EBT are important to ensuring that the reactions proceed as desired, thus the pH is controlled using a buffer. What percent of the EBT indicator will be in the desired HIn2- state at pH = 10.5. pKa1 = 6.2 and pKa2 = 11.6 of EBTarrow_forward
- What does the phrase 'fit for purpose' mean in relation to analytical chemistry? Please provide examples too.arrow_forwardFor each of the substituted benzene molecules below, determine the inductive and resonance effects the substituent will have on the benzene ring, as well as the overall electron-density of the ring compared to unsubstituted benzene. Molecule Inductive Effects Resonance Effects Overall Electron-Density × NO2 ○ donating O donating O withdrawing O withdrawing O electron-rich electron-deficient no inductive effects O no resonance effects O similar to benzene E [ CI O donating withdrawing O no inductive effects Explanation Check ○ donating withdrawing no resonance effects electron-rich electron-deficient O similar to benzene © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accesarrow_forwardUnderstanding how substituents activate Rank each of the following substituted benzene molecules in order of which will react fastest (1) to slowest (4) by electrophilic aromatic substitution. Explanation HN NH2 Check X (Choose one) (Choose one) (Choose one) (Choose one) © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Aarrow_forward
- Identifying electron-donating and electron-withdrawing effects on benzene For each of the substituted benzene molecules below, determine the inductive and resonance effects the substituent will have on the benzene ring, as well as the overall electron-density of the ring compared to unsubstituted benzene. Inductive Effects Resonance Effects Overall Electron-Density Molecule CF3 O donating O donating O withdrawing O withdrawing O no inductive effects O no resonance effects electron-rich electron-deficient O similar to benzene CH3 O donating O withdrawing O no inductive effects O donating O withdrawing Ono resonance effects O electron-rich O electron-deficient O similar to benzene Explanation Check Х © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Centerarrow_forward* Hint: Think back to Chem 1 solubility rules. Follow Up Questions for Part B 12. What impact do the following disturbances to a system at equilibrium have on k, the rate constant for the forward reaction? Explain. (4 pts) a) Changing the concentration of a reactant or product. (2 pts) b) Changing the temperature of an exothermic reaction. (2 pts) ofarrow_forwardDraw TWO general chemical equation to prepare Symmetrical and non-Symmetrical ethers Draw 1 chemical reaction of an etherarrow_forward
- Introductory Chemistry: An Active Learning Approa...ChemistryISBN:9781305079250Author:Mark S. Cracolice, Ed PetersPublisher:Cengage LearningOrganic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage LearningIntroduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning
- Chemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage LearningChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning





