(a)
Interpretation:
The given molecule is D or L should be identified.
Concept introduction:
Monosaccharides: (single sugar unit) simple sugars units are the most basic units of carbohydrates.
Structure and nomenclature: Monosaccharide has this chemical formula:
Monosaccharides can be classified by the number x of carbon atoms. If they contain three carbon atom and is called triose. If they contain four carbon and is called tetrose, if they contain five carbon and is called pentose, if they contain six carbon and is called hexose and so on.
Aldose: A monosaccharide in which carbon serves as a backbone and contains an
Ketose: A ketose is a simple sugar unit (monosaccharide) containing one
D and L enantiomers: L isomers have the hydroxyl group attached to the left side of the asymmetric carbon furthest from the carbonyl.
R isomers have the hydroxyl group attached to the right side of the asymmetric carbon furthest from the carbonyl.
To find: Type of enantiomers.
(a)
Answer to Problem 42PP
Answer
The given compound is D-aldotetrose (a).
Explanation of Solution
The OH group connected to C3 is pointing to the right side, so it is called as D-sugar. The
(b)
Interpretation:
The given molecule is D or L should be identified.
Concept introduction:
Monosaccharides: (single sugar unit) simple sugars units are the most basic units of carbohydrates.
Structure and nomenclature: Monosaccharide has this chemical formula:
Monosaccharides can be classified by the number x of carbon atoms. If they contain three carbon atom and is called triose. If they contain four carbon and is called tetrose, if they contain five carbon and is called pentose, if they contain six carbon and is called hexose and so on.
Aldose: A monosaccharide in which carbon serves as a backbone and contains an aldehyde group at the beginning.
Ketose: A ketose is a simple sugar unit (monosaccharide) containing one ketone group per molecule.
D and L enantiomers: L isomers have the hydroxyl group attached to the left side of the asymmetric carbon furthest from the carbonyl.
R isomers have the hydroxyl group attached to the right side of the asymmetric carbon furthest from the carbonyl.
To find: Type of enantiomers.
(b)
Answer to Problem 42PP
Answer
The given compound is L-aldopentose (b).
Explanation of Solution
The given compound is L-aldopentose.
The OH group connected to C4 is pointing to the left side, so it is called as L-sugar. The functional group at Cl is an aldehyde group, so the compound is an aldose. Finally, the compound has five carbon atoms, so it is a pentose. The given compound is an L-aldopentose
(c)
Interpretation:
The given molecule is D or L should be identified.
Concept introduction:
Monosaccharides: (single sugar unit) simple sugars units are the most basic units of carbohydrates.
Structure and nomenclature: Monosaccharide has this chemical formula:
Monosaccharides can be classified by the number x of carbon atoms. If they contain three carbon atom and is called triose. If they contain four carbon and is called tetrose, if they contain five carbon and is called pentose, if they contain six carbon and is called hexose and so on.
Aldose: A monosaccharide in which carbon serves as a backbone and contains an aldehyde group at the beginning.
Ketose: A ketose is a simple sugar unit (monosaccharide) containing one ketone group per molecule.
D and L enantiomers: L isomers have the hydroxyl group attached to the left side of the asymmetric carbon furthest from the carbonyl.
R isomers have the hydroxyl group attached to the right side of the asymmetric carbon furthest from the carbonyl.
To find: Type of enantiomers.
(c)
Answer to Problem 42PP
Answer
The given compound is D-aldopentose (c).
Explanation of Solution
The given compound is D-aldopentose.
The OH group connected to C4 is pointing to the right side, so this is a D-sugar. The functional group at Cl is an aldehyde group, so the compound is an aldose. Finally, the compound has five carbon atoms, so it is a pentose. The given compound is a D-aldopentose.
(d)
Interpretation:
The given molecule is D or L should be identified.
Concept introduction:
Monosaccharides: (single sugar unit) simple sugars units are the most basic units of carbohydrates.
Structure and nomenclature: Monosaccharide has this chemical formula:
Monosaccharides can be classified by the number x of carbon atoms. If they contain three carbon atom and is called triose. If they contain four carbon and is called tetrose, if they contain five carbon and is called pentose, if they contain six carbon and is called hexose and so on.
Aldose: A monosaccharide in which carbon serves as a backbone and contains an aldehyde group at the beginning.
Ketose: A ketose is a simple sugar unit (monosaccharide) containing one ketone group per molecule.
D and L enantiomers: L isomers have the hydroxyl group attached to the left side of the asymmetric carbon furthest from the carbonyl.
R isomers have the hydroxyl group attached to the right side of the asymmetric carbon furthest from the carbonyl.
To find: Type of enantiomers.
(d)
Answer to Problem 42PP
Answer
The given compound is D-aldohexose (d).
Explanation of Solution
The given compound is D-aldohexose.
The OH group connected to C5 is pointing to the right side, so this is a D-sugar. The functional group at Cl is an aldehyde group, so the compound is an aldose. Finally, the compound has six carbon atoms, so it is called as a hexose. The given compound is a D-aldohexose
(e)
Interpretation:
The given molecule is D or L should be identified.
Concept introduction:
Monosaccharides: (single sugar unit) simple sugars units are the most basic units of carbohydrates.
Structure and nomenclature: Monosaccharide has this chemical formula:
Monosaccharides can be classified by the number x of carbon atoms. If they contain three carbon atom and is called triose. If they contain four carbon and is called tetrose, if they contain five carbon and is called pentose, if they contain six carbon and is called hexose and so on.
Aldose: A monosaccharide in which carbon serves as a backbone and contains an aldehyde group at the beginning.
Ketose: A ketose is a simple sugar unit (monosaccharide) containing one ketone group per molecule.
D and L enantiomers: L isomers have the hydroxyl group attached to the left side of the asymmetric carbon furthest from the carbonyl.
R isomers have the hydroxyl group attached to the right side of the asymmetric carbon furthest from the carbonyl.
To find: Type of enantiomers.
(e)
Answer to Problem 42PP
Answer
The given compound is D-ketopentose (e).
Explanation of Solution
The given compound is D-ketopentose.
The OH group connected to C4 is pointing to the right side, so this is a D-sugar. The functional group at C2 is a ketone group, so the compound is a ketose. Finally, the compound has five carbon atoms, so it is a pentose. The given compound is a D-ketopentose.
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Chapter 24 Solutions
ORGANIC CHEMISTRY-STD.WILEY PLUS CARD
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- For each reaction below, decide if the first stable organic product that forms in solution will create a new C - C bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. NH2 tu ? ? OH Will the first product that forms in this reaction create a new CC bond? Yes No Will the first product that forms in this reaction create a new CC bond? Yes No C $ ©arrow_forwardAs the lead product manager at OrganometALEKS Industries, you are trying to decide if the following reaction will make a molecule with a new C-C bond as its major product: 1. MgCl ? 2. H₂O* If this reaction will work, draw the major organic product or products you would expect in the drawing area below. If there's more than one major product, you can draw them in any arrangement you like. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry. If the major products of this reaction won't have a new CC bond, just check the box under the drawing area and leave it blank. Click and drag to start drawing a structure. This reaction will not make a product with a new CC bond. G marrow_forwardIncluding activity coefficients, find [Hg22+] in saturated Hg2Br2 in 0.00100 M NH4 Ksp Hg2Br2 = 5.6×10-23.arrow_forward
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