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5.20: Solutions to Selected Problems

  • Page ID
    191275
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    Exercise 5.2.1:

    Answers to Exercise 5.2.1, a through c, with diastereomers of metal complexes. a is an iridium complex with two amine groups, a chlorine group, and a carbon monoxide group. b is a metal complex with a hydrogen, a bromine, and two phosphine groups. c is a platinum complex with two arsine groups, a carbon monoxide group, and a hydrogen.

    Exercise 5.2.2:

    Answer to Exercise 5.2.2. Caption reads: "Needs to cis. The tmeda group is not long enough to reach the trans position".

    Exercise 5.3.1:

    Clockwise.

    Exercise 5.3.2:

    Counter-clockwise.

    Exercise 5.3.3:

    Counter-clockwise.

    Exercise 5.3.4:

    Clockwise.

    Exercise 5.3.5:

    Enantiomer B has a molecular weight of 126 g/mol, a density of 0.995 g/mL, an optical rotation of [α] = – 26°, a melting point of 65°C and a boiling point of 225°C.

    Exercise 5.3.6:

    Answers to Exercise 5.3.6.

    Exercise 5.3.7:

    A. The plane of the page is a mirror plane. There is also one perpendicular to the page that reflects one H into the other.
    B. The plane of the page contains one P-Cl bond and bisects the other Cl’s.
    C. The plane of the page is a mirror plane.
    D. Mirror plane contains P-Br bond and bisects the Cl’s.
    E. There is no lone pair on the B. Therefore all atoms lie in a mirror plane.
    F. No mirror planes--the molecule is therefore chiral.
    G. There is a plane perpendicular to the page that contains the Br and Cl and bisects the cyclopropane ring.
    H. No mirror planes--the molecule is therefore chiral.
    I. There is a plane perpendicular to the page that contains the Br and Cl and bisects the cyclopropane ring.
    J. The C-C bond can be rotated by 60 degrees so that there is a plane perpendicular to the C-C bond axis.
    K. The C-C bond in one of the chlorine-containing arms can be rotated so that there is a mirror plane that goes through the ethyl group (with no Cl’s) and the P, and one chlorine containing arm is the reflection of the other.
    L. Since a double bond is planar, there is a mirror plane that contains all six atoms.
    M. There is a mirror plane that contains two C’s and bisects the two Cls.
    N. No mirror planes--the molecule is therefore chiral. The rings are not in the same plane due to the CH3 and NH2 groups, which bump into each other. They also prevent rotation around the C-C bond between the rings.

    Exercise 5.3.8

    Picture (a)

    Exercise 5.3.9:

    Picture (b)

    Exercise 5.3.10:

    Picture (d)

    Exercise 5.3.11:

    Picture (c)

    Exercise 5.4.1:

    Priority of groups:
    1 Br (red)
    2 Cl (bright green)
    3 F (pale green)
    4 H (white)

    In the molecule in figure 5.4.2, with the low-priority hydrogen pointed away, bromine is at the top, chlorine is clockwise from the bromine, and fluorine is clockwise from the chlorine. It therefore has an assigned configuration of R.

    In the molecule in figure 5.4.3, with the low-priority hydrogen pointed away, bromine is at the top, chlorine is counterclockwise from the bromine, and fluorine is counterclockwise from the chlorine. Thus, it has an assigned configuration of S.

    Exercise 5.4.2

    Answers to Exercise 5.4.2: two pairs of enantiomers of 1-methoxyethanol and and 1-methoxypropanol.

    Exercise 5.4.3:

    Answers to Exercise 5.4.3.

    Exercise 5.4.4:

    Answers to Exercise 5.4.4.

    Exercise 5.4.5:

    Answers to Exercise 5.4.5, a through f. Groups are numbered from highest to lowest priority, and the molecule is labelled R or S.

    Exercise 5.4.6:

    Answers to Exercise 5.4.6, a through c.

    Exercise 5.5.1:

    Answers to Exercise 5.5.1, a through c.

    Exercise 5.5.2:

    Answer to Exercise 5.5.2: the R and S enantiomers of 2-butanol.

    Exercise 5.5.3:

    Answers to Exercise 5.5.3.

    Exercise 5.6.1:

    A pure sample of A would have \([\alpha] = 75^{\circ}\)

    Optical purity or enatiomeric excess = \(\frac{50}{75} = 66 \% \)

    % major enantiomer = \(66 + \frac{34}{2} = 83 \% \)

    % minor enantiomer = \(100 - 83 = 17 \%\)

    Exercise 5.6.2

    \[ \% \: major \: = 90 \% \nonumber\]

    \( \% \: major \: = 10 \%\)

    Optical purity or enantiomeric excess = \( \frac{X}{-50} = 90-10 = 80 \% \)

    Solve for X. \(X = -40^{\circ}\)

    Exercise 5.6.4:

    Answers to Exercise 5.6.4, with moelcules labelled "yes", "no", or "no, meso".

    Exercise 5.7.1:

    a) \([\alpha] = \frac{\alpha}{(c)(l)}\)

    \(c = (\frac{0.250g}{2mL})(\frac{10mL}{1dL}) = 1.25 \frac{g}{dL}\)

    \(\alpha = \frac{(0.775 ^{\circ} + 0.806 ^{\circ} + 0.682 ^{\circ})}{3} = 0.754 ^ {\circ} \)
    \([\alpha] = \frac{\alpha}{(c)(l)} = \frac{(0.754 ^{\circ})}{(1.25 \frac{g}{dL})(0.5dm)} = + 1.21 ^{\circ}\)


    b)

    – 1.21°

    Exercise 5.7.2:

    \[[\alpha] = \frac{\alpha}{(c)(l)} \nonumber\]

    \([\alpha] = 32 \circ\)

    \(c = (\frac{0.150g}{1mL})(\frac{10mL}{1dL}) = 1.5 \frac{g}{dL}\)

    c = (0.150g/ 1 mL)(10 mL/1 dL) = 1.5g/dL

    \([\alpha] = \frac{\alpha}{(c)(l)} = 32 ^{\circ} = \frac{\alpha}{(1.5 \frac{g}{dL})(0.5dm)}\)

    Solve for α. \( \alpha = + 24^{\circ}\)

    Exercise 5.8.1:

    D-glyceraldehyde is R
    L-glyceraldehyde is S

    Exercise 5.8.2:

    Answers to Exercise 5.8.2, a and b.

    Exercise 5.8.3:

    Answers to Exercise 5.8.3, a and b, with bowtie projection and equivalent skeletal structure.

    Exercise 5.9.1:

    a) left b) right

    Exercise 5.9.2:

    a) right b) left

    Exercise 5.9.1:

    a) right b) right

    Exercise 5.9.2:

    a) left b) left

    Exercise 5.9.5:

    D-threose
    2S, 3R

    Exercise 5.9.6:

    L-threose
    2R, 3S

    D- and L-threose are enantiomers of one another

    Exercise 5.9.7:

    L-erythrose
    2S, 3S

    L-erythrose and L-threose are diastereomers of one another.

    Exercise 5.9.8:

    a) (2)3 = 8 possible stereoisomers

    RRR; SSS; RRS; SSR; RSS; SRR; SRS; RSR

    b) 4 pairs

    c) 12 different possible pairs of diastereomers

    Exercise 5.9.9:

    Answers to Exercise 5.9.9, a and b. a shows enantiomers and diastereomers for 2-methylcyclopentanol. b shows enantiomers and diastereomers for 3-methylcyclohexanol.

    Exercise 5.9.10:

    Answers to Exercise 5.9.10, a and b, showing pairs of diastereomers and enantiomers as Fischer projections.

    Exercise 5.10.1:

    a) D-threitol → 2R, 3S

    b) L- threitol → 2S, 3S

    c) erythritol → 2S, 3R or 2R, 3S (a meso compound)

    Exercise 5.10.2:

    It does not matter which end you start counting from on these compounds since they are constituted the same.

    Exercise 5.10.3:

    Answers to Exercise 5.10.3. a: B and D. b: A and D or A and B. c: A and C or B and C or D and C. d: C.

    Exercise 5.10.4:

    clipboard_e271866d62f6c7394efed6b86f2f4aacc.png

    Answers to Exercise 5.10.4.

    Exercise 5.11.1:

    Answers to Exercise 5.11.1, a and b, showing Haworth projections.

    Exercise 5.11.2:

    Answers to Exercise 5.11.2, showing beta- and alpha-deoxyribofuranose. a is trans. b is cis.

    Exercise 5.12.1:

    Glycine

    Exercise 5.12.2:

    Answer to Exercise 5.12.2, showing two diastereomers of isoleucine.

    Exercise 5.12.3:

    Proline

    Exercise 5.12.4:

    Glutamic acid and aspartic acid

    Exercise 5.12.5:

    Arginine, asparagine, glutamine, lysine; also tryptophan contains an aromatic heterocycle, although it is not basic.

    Exercise 5.12.6:

    Bond-line structure of D-alanine.

    Exercise 5.12.7:

    Cysteine

    Exercise 5.13.1:

    a. Glycine-alanine and alanine-glycine. b. No. They are constitutional isomers of one another.

    Exercise 5.13.2:

    Bond-line structure of an alanine-phenylalanine-lysine polypeptide.

    Exercise 5.14.1:

    Formation of diastereomic salts of two amino acids linked by a peptide bond.

    Exercise 5.15.1:

    A. Cis
    B. Trans
    C. Trans
    D. Neither (There is free rotation around the C-C single bond.)
    E. Neither (C-C triple bonds have substituents at 180 degrees to each other (linear).
    F. cis

    Exercise 5.16.1:

    A. Z
    B. E
    C. E
    D. Z
    E. E
    F. E
    G. Z
    H. E

    Exercise 5.17.1:

    Answers to Exercise 5.17.1, a through e, showing trans and cis isomers of complex ions.

    Exercise 5.17.2:

    Answers to Exercise 5.17.2, a through c, with facial and meridional arrangements of metal complexes.

    Exercise 5.18.1:

    a) Enantiomers

    b) Enantiomers

    c) Identical

    d) Identical

    e) Enantiomers

    f) Identical

    g) Enantiomers

    Exercise 5.18.2:

    a) Δ b) Λ c) Δ d) Λ e) Δ f) Δ g) Λ h) Λ

    Exercise 5.20.1:

    Pure = 125°

    Optical purity = \(\frac{100}{125} = 0.80\)

    \[\% \: Major \: = 80 + \frac{20}{2} = 90\% \nonumber\]

    \[\% \: Minor \: 100-90 = 10 \% \nonumber\]

    Exercise 5.20.2

    Pure = 100°

    Optical purity = \(\frac{95}{100} = 0.95\)

    \(\% \: Major \: = 95 + \frac{5}{2} = 97.5 \%\)

    \[\% \: Minor \: 100-97.5=2.5 \% \nonumber\]

    Exercise 5.20.3

    \[Pure \: = 18^{\circ} \nonumber\]

    \[\% \: Major \: = 60 \% \nonumber\]

    \(\% \: Minor \: = 40 \% \)

    Optical purity = \(\frac{X}{18} = 20 \%\)

    Solve for X:

    \[X = 3.6^{\circ} \nonumber\]

    Exercise 5.20.4:

    \[Pure \: = 25^{\circ} \nonumber\]

    \[ \% Major \: = 80 \% \nonumber\]

    \( \% Minor \: = 20 \% \)
    Optical purity = \(\frac{X}{25} = 80-20= 60 \%\)

    Solve for X:

    \[X = 15^{\circ} \nonumber\]

    Exercise 5.20.5

    \[[\alpha] = \frac{observed \: rotation}{(l)(c)} \nonumber\]

    \(c = \frac{0.050g}{2.0mL} = 0.025 \frac{g}{mL}\)

    Average observed rotation = \(\frac{(0.625 + 0.706 + 0.682)}{3} = 0.671 ^{\circ}\)

    \[[\alpha] = \frac{0.671 ^{\circ}}{(0.025 \frac{g}{mL})(0.5 dm)} = 53.68 ^{\circ} \nonumber\]

    Exercise 5.20.6:

    \[[\alpha] = \frac{observed \: rotation}{(l)(c)} \nonumber\]

    \(c = \frac{0.540g}{2.0mL} = 0.27 \frac{g}{mL}\)

    Average observed rotation = \(\frac{(1.225 + 1.106 + 1.182)}{3} = 1.171 ^{\circ} \)

    \[[\alpha] = \frac{1.171 ^{\circ}}{(0.27 \frac{g}{mL})(1.0 dm)} = 4.34 ^{\circ} \nonumber\]

    Exercise 5.20.7:

    \[[\alpha] = \frac{observed \: rotation}{(l)(c)} \nonumber\]

    \[c = \frac{0.250g}{2.0mL} = 0.125 \frac{g}{mL} \nonumber\]

    \(42 ^{\circ} = \frac{observed \: rotation}{(0.125 \frac{g}{mL})(0.5dm)}\)

    observed rotation = 2.625°

    Exercise 5.20.8

    a) \([\alpha] = \frac{observed \: rotation}{(l)(c)}\)

    \[c = \frac{0.10g}{2.0mL} = 0.05 \frac{g}{mL} \nonumber\]

    \(c = \frac{0.10g}{2.0mL} = 0.05 \frac{g}{mL}\)

    Average observed rotation = \(\frac{(0.995 + 0.904 + 0.936)}{3} = 0.945 \circ \)

    \([\alpha] = \frac{0.945 ^{\circ}}{(0.05 \frac{g}{mL})(1.0 dm)} = 18.9 \circ \)


    b) % optical purity = \(\frac{(100)(18.9)}{25}= 75.6 \%\)

    c) enantiomeric excess = 75.6%

    d) \(\frac{(100-75.6)}{2} = 12.2 \%\)

    12.2 % one enantiomer
    87.8% other enantiomer

    e) Enantiomers differ in how they interact with plane polarized light, but not in other physical analyses.

    Exercise 5.20.9

    \[[\alpha] = \frac{observed \: rotation}{(l)(c)} \nonumber\]

    \[40 \circ = \frac{observed \: rotation}{(l)(c)} \nonumber\]

    \(40 \circ = \frac{observed \: rotation}{(l)(1.1c)}\)

    Observed rotation = 44°

    Exercise 5.20.10

    Answers to Exercise 5.20.10, a through e, with variations of 1,3-cyclohexanediol. a is the same compound. b is enantiomers. c is diastereomers. d is any cyclohexanediol. e shows cis-1,4-cyclohexanediol and cis-1,3-cyclohexanediol.

    Exercise 5.20.11:

    Answers to Exercise 5.20.12, a through d, with cariations on 4-methyl-1,3-cyclohexanediol. a is the same compound. b is enantiomers. c is diastereomers. d shows constitutional isomers.

    Exercise 5.20.12:

    Answers to Exercise 5.20.12, a through c, showing variations on 2-methyl-1,4,5-cyclohexanediol. a shows two of the same compounds. b is enantiomers. c is diastereomers.


    5.20: Solutions to Selected Problems is shared under a CC BY-NC 3.0 license and was authored, remixed, and/or curated by Chris Schaller.