Concept explainers
(a)
Interpretation:
Fischer projection of the given molecule is to be drawn.
Concept introduction:
A molecule containing a chain of carbon atoms is frequently represented in its zigzag conformation. If it contains multiple asymmetric centers, it is more convenient to draw its Fisher projection. Fischer projections are generally drawn with the longest carbon chain vertical. The two groups attached to each carbon except the first and last are shown on horizontal bonds. The carbon atoms in the middle of the chain are represented by the intersections of vertical and horizontal lines. The vertical lines represent bonds that are oriented away from the viewer while the horizontal bonds are oriented toward the viewer. The most oxidized group must be present at the top of the vertical line.
In the zig-zag conformation, if two groups on adjacent carbon atoms are on the same side of the plane and are shown by wedge bonds, then those groups will be on the opposite sides of the vertical chain in the Fischer projection.
In the zig-zag conformation, if two groups on adjacent carbon atoms are on the opposite sides of the plane and are shown by a wedge and a dash bond, then, in the Fischer projection, those groups will be on the same side of the vertical chain.
In the zig-zag conformation, if two groups on alternate carbon atoms are on opposite sides of the plane and are shown by a wedge and a dash bond, then, in the Fischer projection, those groups will be on the opposite sides of the vertical chain.
Answer to Problem 5.52P
The Fischer projection of the given molecule is
Explanation of Solution
The given molecule is
The given molecule consists of a five carbon chain, each of which is numbered. C2, C3, and C4 are the asymmetric carbon atoms. We begin to draw the Fischer projection with the framework shown below:
The asymmetric centers are denoted by asterisks. Two of the bonds on each asymmetric center have been left with question marks because two substituents must still be added to each so that the stereochemical configuration at those carbon atoms in the Fischer projection matches with what was given in the dash-wedge notation.
Notice that in the zig-zag conformation, the OH groups on C2 and C3 carbon atoms lie on the same side of the plane and are shown by a wedge bond. Thus, these two groups must lie on opposite sides in the Fischer projection, as shown below:
The OH groups on C3 and C4 carbon atoms lie on the opposite side of the plane and are shown by a wedge and a dash bond. Thus, these two groups must lie on the same side in the Fischer projection, as shown below:
The OH groups on C2 and C4 carbon atoms lie on the opposite side of the plane and are shown by a wedge and a dash bond. Thus, these two groups must lie on the opposite side in the Fischer projection.
Thus, the OH groups on C3 and C4 carbon atoms must lie on the same side of the vertical chain, and OH group on C2 must lie on the opposite side. Then fill in the remaining groups on C2, C3, and C4 carbon atoms to complete the structure as below:
Thus, the structure above is the correct conversion from a given zig-zag conformation to a Fischer projection.
In the zig-zag conformation, if two groups on adjacent carbon atoms are on the opposite sides of the plane and are shown by a wedge and dash bond, then those groups will be on the same side of the vertical chain in the Fischer projection. In the zig-zag conformation, if two groups on adjacent carbon atoms are on opposite sides of the plane and are shown by a wedge and dash bond, then those groups will be on the same side of the vertical chain in the Fischer projection.
Interpretation:
(b)
Fischer projection of the given molecule is to be drawn.
Concept introduction:
A molecule containing a chain of carbon atoms is frequently represented in its zigzag conformation. If it contains multiple asymmetric centers, it is more convenient to draw its Fisher projection. Fischer projections are generally drawn with the longest carbon chain vertical. The two groups attached to each carbon except the first and last are shown on horizontal bonds. The carbon atoms in the middle of the chain are represented by the intersections of vertical and horizontal lines. The vertical lines represent bonds that are oriented away from the viewer while the horizontal bonds are oriented toward the viewer. The most oxidized group must be present at the top of the vertical line.
In the zig-zag conformation, if two groups on adjacent carbon atoms are on the same side of the plane and are shown by wedge bonds, then those groups will be on the opposite sides of the vertical chain in the Fischer projection.
In the zig-zag conformation, if two groups on adjacent carbon atoms are on the opposite sides of the plane and are shown by a wedge and a dash bond, then, in the Fischer projection, those groups will be on the same side of the vertical chain.
In the zig-zag conformation, if two groups on alternate carbon atoms are on opposite sides of the plane and are shown by a wedge and a dash bond, then, in the Fischer projection, those groups will be on the opposite sides of the vertical chain.
Answer to Problem 5.52P
The Fischer projection of the given molecule is
Explanation of Solution
The given molecule is
The given molecule consists of a three carbon chain, each of which is numbered. C1 and C2 carbon atoms are the asymmetric carbon atoms. We begin to draw the Fischer projection with the framework shown below:
The asymmetric centers are denoted by asterisks. Two of the bonds on each asymmetric center have been left with question marks because two substituents must still be added to each so that the stereochemical configuration at those carbon atoms in the Fischer projection matches with what was given in the dash-wedge notation.
Notice that in the zig-zag conformation, the OH groups on C1 and C2 carbon atoms lie on the same side of the plane and are shown by a wedge bond. Thus, these two groups must lie on opposite side in the Fischer projection, as shown below:
Then fill in the remaining groups on C1 and C2 carbon atoms to complete the structure as below:
Thus, the structure above is the correct conversion from a given zig-zag conformation to a Fischer projection.
In the zig-zag conformation, if two groups on adjacent carbon atoms are on the opposite sides of the plane and are shown by a wedge and dash bond, then those groups will be on the same side of the vertical chain in the Fischer projection. In the zig-zag conformation, if two groups on adjacent carbon atoms are on opposite sides of the plane and are shown by a wedge and dash bond, then those groups will be on the same side of the vertical chain in the Fischer projection.
Interpretation:
(c)
Fischer projection of the given molecule is to be drawn.
Concept introduction:
A molecule containing a chain of carbon atoms is frequently represented in its zigzag conformation. If it contains multiple asymmetric centers, it is more convenient to draw its Fisher projection. Fischer projections are generally drawn with the longest carbon chain vertical. The two groups attached to each carbon except the first and last are shown on horizontal bonds. The carbon atoms in the middle of the chain are represented by the intersections of vertical and horizontal lines. The vertical lines represent bonds that are oriented away from the viewer while the horizontal bonds are oriented toward the viewer. The most oxidized group must be present at the top of the vertical line.
In the zig-zag conformation, if two groups on adjacent carbon atoms are on the same side of the plane and are shown by wedge bonds, then those groups will be on the opposite sides of the vertical chain in the Fischer projection.
In the zig-zag conformation, if two groups on adjacent carbon atoms are on the opposite sides of the plane and are shown by a wedge and a dash bond, then, in the Fischer projection, those groups will be on the same side of the vertical chain.
In the zig-zag conformation, if two groups on alternate carbon atoms are on opposite sides of the plane and are shown by a wedge and a dash bond, then, in the Fischer projection, those groups will be on the opposite sides of the vertical chain.
Answer to Problem 5.52P
The Fischer projection of the given molecule is
Explanation of Solution
The given molecule is
The given molecule consists of a six carbon chain, each of which is numbered. C2, C3, C4, and C5 carbon atoms are the asymmetric carbon atoms. We begin to draw the Fischer projection with the framework shown below:
The asymmetric centers are denoted by asterisks. Two of the bonds on each asymmetric center have been left with question marks because two substituents must still be added to each so that the stereochemical configuration at those carbon atoms in the Fischer projection matches with what was given in the dash-wedge notation.
Notice that in the zig-zag conformation, the OH groups on C2 and C3 carbon atoms lie on the opposite side of the plane and are shown by a wedge and a dash bond. Thus, these two groups must lie on same side in the Fischer projection, as shown below:
Notice that in the zig-zag conformation, the OH groups on C3 and C4 carbon atoms lie on the same side of the plane and are shown by dash bonds. Thus, these two groups must lie on the opposite sides of the vertical chain in the Fischer projection, as shown below:
In the zig-zag conformation, the OH group and chlorine atom on C4 and C5 carbon atoms lie on the same side of the plane and are shown by dash bonds. Thus, these two groups must lie on the opposite sides of the vertical chain in the Fischer projection, as shown below:
Then place the remaining groups on C1 and C2 carbon atoms to complete the structure as below:
In the zig-zag conformation, if two groups on adjacent carbon atoms are on the opposite sides of the plane and are shown by a wedge and dash bond, then those groups will be on the same side of the vertical chain in the Fischer projection. In the zig-zag conformation, if two groups on adjacent carbon atoms are on opposite sides of the plane and are shown by a wedge and dash bond, then those groups will be on the same side of the vertical chain in the Fischer projection.
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Chapter 5 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
- Draw a chair conformation of this molecule with (a) all CH3 groups in axial positions and (b) all CH3 groups in equatorial positions.arrow_forwardDraw the two chair conformations for each of the following compounds and indicate which conformer is more stable.arrow_forwardFor each of the following pairs of compounds, determine which compound is more stable (may be more helpful to draw out the chair conformations)arrow_forward
- draw the two chair conformers for the following structures. If one is lower energy, circle the one of each pair .arrow_forwardDetermine the # of chiral centers. Determine the most stable chair conformation then indicate which are equatorial and which are axial.arrow_forwardDraw all the conformers of the molecule on the right as Newman projections sighted along the C3– C4 bond. Label the interactions present in each conformer. Rank the conformers from lowest to highest energy (some may be equal).arrow_forward
- Draw the structure of the following molecules that result from the flipping of the existing chair conformation. See attached structures:arrow_forwarddraw the most stable chair conformation. Where two or more conformations are equally stable, draw both.arrow_forwardDraw both chair conformers for the following molecule and circle the most stable conformation.arrow_forward
- There are four isomers of the compound shown below. Draw the chair conformation of each of the four isomers and its conformation isomers. Draw the most stable of the conformation on the left. Be sure your positioning of groups is clear.arrow_forwardDraw the Fischer projection from the chair conformation shown below.arrow_forwardDraw the first compund in Fischer structure and 3D perspective structure( as if it was a molecular model) Draw the second compund in Chair conformation structure and Inverted chair conformation structurearrow_forward
- Chemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning