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6.15: Solutions to Selected Problems

  • Page ID
    191306
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    Exercise 6.3.1:

    Table showing gauche (top row) and anti (bottom row) conformations of, from left to right, 1,2-dichloroethane, 1,2-butanediol, and 2-chloro-1-ethanamine.

    Exercise 6.4.1:

    Several chair conformers and Newman projections. Eclipsed bonds circled in red.

    Exercise 6.4.2:

    Several chair conformers and Newman projections.

    Exercise 6.4.3a:

    Graph of torsional energy and degree rotation of 2-methylpropane. 0 kcal/mol at 0, 120, 240, and 360 degrees. 3.2 kcal/mol at 60, 180, and 300 degrees.

    Exercise 6.4.3b:

    Graph of torsional energy and degree rotation of 2,3-dimethylbutane. 3 kcal/mol at 0, 240, and 360 degrees. 2 kcal/mol at 120 degrees. 4.7 kcal/mol at 60 and 180 degrees. 6 kcal/mol at 300 degrees.

    Exercise 6.4.3c:

    Graph of torsional energy and degree rotation of pentane. 0 kcal/mol at 0 and 360 degrees. 1 kcal/mol at 120 and 240 degrees. 3.2 kcal/mol at 60 and 300 degrees. 4.5 kcal/mol at 180 degrees.

    Exercise 6.8.1:

    From top to bottom: Template of chair cyclohexane. A: Ethane. B: Propane. C: Butane, gauche conformation. D: Butane, anti conformation.

    Exercise 6.9.1

    Table. Left column: skeletal structures of monosubstituted cyclohexanes. Middle column: chair projection with group equatorial. Right column: group in axial conformation. Top row: methylcyclohexane. Middle row: ethylcyclohexane. Bottom row: isopropylcyclohexane.

    Exercise 6.9.2:

    A: cyclohexanol in equatorial and axial conformation. Estimate = (0.33)(2 kcal/mol) = 0.66 kcal/mol. B: cyclohexanamine, axial and equatorial. Estimate = (0.66)(2 kcal/mol) = 1.33 kcal/mol. C: 1-fluorocyclohexane, axial and equatorial. Estimate = 0 kcal/mol.

    Exercise 6.9.3:

    Cyclohexane chairs with groups labelled equatorial or axial and up or down.

    Exercise 6.10.1:

    Chair and skeletal structures of five molecules. top to bottom: trans-1,2-dimethylcyclohexane, (1S,2S,4R)-1-bromo-2-methyl-4-phenylcyclohexane, (1R,2R,4S)-1-bromo-2-methyl-4-phenylcyclohexane, and (1R,2S,5R)-2-bromo-5-isopropylcyclohexan-1-ol.

    Exercise 6.10.2:

    Top to bottom: identical, enantiomers, diastereomers, constitutional isomers, conformational isomers, enantiomers, enantiomers.

    Exercise 6.10.3:

    Axial and equatorial chair conformations for (1R,2R)-2-methylcyclohexan-1-ol and cis-1,3-cyclohexane.

    Exercise 6.10.4:

    trans-1,2-dimethylcyclohexane. Methyls equatorial: 1 kcal/mol. Methyls axial: 4 kcal/mol.

    Exercise 6.10.5:

    Axial and equatorial conformations of cis-1,3-dimethylcyclohexane, trans-1,3-dimethylcyclohexane, cis-1,4-dimethylcyclohexane, and trans-1,3-dimethylcyclohexane.

    Exercise 6.11.1:

    Skeletal structures of cis-decalin and trans-decalin.

    Exercise 6.11.2:

    cis-decalin (2 kcal/mol) and trans-decalin (0 kcal/mol).

    Exercise 6.11.3:

    1. The substituents are always trans along the junctions between each pair of rings. The steroids resemble a series of trans-decalin structures.
    2. The overall structure would be more wide and wavy like a trans-decalin, rather than curled or boxy like a cis-decalin.

    Exercise 6.11.4:

    Bicyclo[2.2.0]decane

    Exercise 6.11.5:

    a) Bicyclo[2.1.1]hexane b) Bicyclo[3.2.1]octane c) Bicyclo[2.1.0]pentane (more commonly called "housane")

    d) Bicyclo[2.2.2]octane e) cis-Bicyclo[3.3.0]octane

    f) cis-Bicyclo[1.1.0]butane g) Bicyclo[1.1.1]pentane h) Bicyclo[4.3.3]dodecane

    Exercise 6.11.6:

    Adamantine, a three-ring system.

    Although we could sketch out many rings using adamantane, just three rings are needed to include all the carbon atoms in the structure. Thus, adamantane is considered a tricyclic system. The systematic nomenclature of tricyclic systems gets a little more complicated, so we won't worry about that.

    Exercise 6.11.7:

    If cyclodecane adopted a regular diamond lattice conformation, there would be a whopping 8 atom interaction in the middle of the ring. That interaction isn't even included in our basis set. It would cost at least 6-7 kcal/mol. As a result, the cyclodecane adopts a twisted structure to avoid this interaction.

    Cyclodecane in chair conformation. Internal hydrogens on C1 and C6 collide with each other.

    Exercise 6.11.8:

    Four-ring system with cis methyl and acetyl groups.

    Exercise 6.11.9:

    Four-ring system with cis methyl and hydroxyl groups.

    Exercise 6.12.1:

    Pyranose with all equatorial hydroxyl groups (0 kcal/mol) and all axial hydroxyl groups (4 kcal/mol).

    Exercise 6.12.2:

    Beta-D-glucose; expected major isomer, 0 kcal/mol. Alpha-D-glucose: 0.66 kcal/mol.

    Exercise 6.12.3:

    a) The β-D-glucose isomer should be the more stable isomer. The β-D-glucose isomer places the C1 hydroxyl group in the equatorial position.

    b) The β-D-glucose isomer should be the more abundant isomer.

    c) This is due to something called the anomeric effect. In solvents of modest polarity, such as dicholoromethane, the α-D-glucose isomer is not as polar as the β-D-glucose isomer. In the α-D-glucose isomer the dipoles of the ring oxygen and the C1 hydroxyl group opposing each other (therefore the overall effect is the molecule is less polar). In addition, the α-D-glucose isomer is stabilized by hyper conjugation of the ring oxygen and C1. For more information see http://en.Wikipedia.org/wiki/Anomeric_effect

    d) A more polar environment would promote having more of the β-D-glucose isomer around. In the β-D-glucose isomer, the dipoles of the ring oxygen and the C1 hydroxyl group align each other (therefore the overall effect is the molecule is more polar).

    Exercise 6.13.1:

    Hydrogen bonding between adenosine and thymine.

    Exercise 6.13.2:

    Hydrogen bonding between cytosine and guanine.

    Exercise 6.13.3:

    Hydrogen bonds between guanine and thymine (top) and cytosine and thymine (bottom).

    Exercise 6.13.4:

    Adenosine binding to a nucelotide with two fluorine groups.

    Exercise 6.13.5:

    Hydrogen bonding between cytosine and an artificial nucleotide.

    Exercise 6.14.1:

    a)

    Complex organic molecule with horizontal bonds circled.

    b) about 9 Å

    c)

    Complex organic molecules with functional groups labelled and highlighted.

    d) Stereochemistry. Because biological receptors are generally proteins containing unique, and chiral, binding sites, they hope to find a stereoisomer that fits the GluN2B receptor, but not the hERG receptor.

    e) These compounds are diastereomers of each other.

    f)

    Complex molecule with priorities and R/S assignment for each chiral center.

    g)

    Left: new ring amide bond is circled in an organic molecule. Right: complex organic molecule.

    h)

    Newman projections of butane and cyclobutane. Cyclobutane cannot freely rotate as butane can.

    The ring introduces conformational rigidity by preventing the bond from undergoing complete rotation; it is tied back by the connection on the other side of the ring.

    i) They were trying to decrease the number of possible conformations, or shapes, that the compound could adopt. That way, it would be less likely to bind in both the GluN2B receptor and the hERG receptor.

    j) The substituents around the flexible six-membered ring could adopt a number of conformations. We will choose ways that minimize additional steric interactions so that we can focus on the most basic interactions.

    Note that amines do not usually have fixed stereochemistry; the fourth substituent on an amine nitrogen is a lone pair, which does not occupy a fixed position in space. Unlike a carbon, the nitrogen can readily switch from one stereochemistry to another.

    Steric hindrance from equatorial and axial conformations of a complex molecule with a phenol functional group. Equatorial is 2 kcal/mol. Axial is 7 kcal/mol.

    The one on the left, with more equatorial substituents and thus fewer steric interactions, is the more stable one.

    Exercise 6.14.2:

    Original organic molecule with enantiomer (hydroxyl and hydrogen chirality reversed), diastereomer (both hydroxy and hydrogen on far side of cyclohexane ring), and other diastereomer (both hydroxy and hydrogen on near side of cyclohexane ring).

    b) The original and its enantiomer; the first two above, from the left.

    c)

    Comparison of axial and equatorial positions of anisole on chair conformation of cyclohexane. Anisole equatorial: 4 kacl/mol. Anisole equatorial: 8 kcal/mol.

    d) The one on the left, with both groups equatorial and less steric strain, is more stable.

    e) If the one on the left is active, a lower dose can be used because it will spend more time in the active form and be able to bond more receptors. If the less stable one, on the right, is the active form, a higher dose will be needed because only a small fraction of the molecules will be ready to bind the receptor at any given time.


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