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Concept explainers
(a)
Interpretation:
The acetal and hemiacetal in isomaltose should 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:
The 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 is n 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 anomers 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 isomaltose is hydrolyzed should be predicted.
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.
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Chapter 20 Solutions
EBK GENERAL, ORGANIC, & BIOLOGICAL CHEM
- 3. Consider the compounds below and determine if they are aromatic, antiaromatic, or non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly drawn and you should be able to tell that the bonding electrons and lone pair electrons should reside in which hybridized atomic orbital 2. You should consider ring strain- flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti- aromaticity) H H N N: NH2 N Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic TT electrons Me H Me Aromaticity (Circle) Aromatic Aromatic Aromatic Aromatic Aromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic aromatic πT electrons H HH…arrow_forwardA chemistry graduate student is studying the rate of this reaction: 2 HI (g) →H2(g) +12(g) She fills a reaction vessel with HI and measures its concentration as the reaction proceeds: time (minutes) [IH] 0 0.800M 1.0 0.301 M 2.0 0.185 M 3.0 0.134M 4.0 0.105 M Use this data to answer the following questions. Write the rate law for this reaction. rate = 0 Calculate the value of the rate constant k. k = Round your answer to 2 significant digits. Also be sure your answer has the correct unit symbol.arrow_forwardNonearrow_forward
- in which spectral range of EMR, atomic and ionic lines of metal liesarrow_forwardQ2: Label the following molecules as chiral or achiral, and label each stereocenter as R or S. CI CH3 CH3 NH2 C CH3 CH3 Br CH3 X &p Bra 'CH 3 "CH3 X Br CH3 Me - N OMe O DuckDuckarrow_forward1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward
- 1.3 grams of pottasium iodide is placed in 100 mL of o.11 mol/L lead nitrate solution. At room temperature, lead iodide has a Ksp of 4.4x10^-9. How many moles of precipitate will form?arrow_forwardQ3: Circle the molecules that are optically active: ДДДДarrow_forward6. How many peaks would be observed for each of the circled protons in the compounds below? 8 pts CH3 CH3 ΤΙ A. H3C-C-C-CH3 I (₁₁ +1)= 7 H CI B. H3C-C-CI H (3+1)=4 H LIH)=2 C. (CH3CH2-C-OH H D. CH3arrow_forward
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