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Concept explainers
a)
Interpretation:
The other monosaccharides that are reduced to the alditol obtained from the reduction of given monosaccharides are to be stated.
Concept Introduction:
Monosaccharides:
Monosaccharides are made from the
A numbers of isomers are obtained from biomolecules due to the chiral nature of carbon atoms.
Chiral molecules are those molecules that consist of different groups or atoms around the central atom. Isomers are those compounds that have same molecular formula but different arrangement of atoms, groups or substituents in a compound.
b)
Interpretation:
The monosaccharides that are reduced to two alditols and out of two, the alditol obtained from the reduction of the given monosaccharides are to be stated.
Concept Introduction:
Monosaccharides:
Monosaccharides are made from the biomolecules that comprise carbon, oxygen and hydrogen atoms. These molecules cannot be broken into the simpler compounds.
The disaccharides and polysaccharides are further formed from monosaccharide units. A numbers of isomers are obtained from biomolecules due to the chiral nature of carbon atoms.
Chiral molecules are those molecules that consist of different groups or atoms around the central atom. Isomers are those compounds that have same molecular formula but different arrangement of atoms, groups or substituents in a compound.
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Chapter 20 Solutions
EBK ORGANIC CHEMISTRY
- Nonearrow_forward3. 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_forward
- 1. 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_forward1.3 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_forward
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