CH-341, HW #2 Key, Summer 2010

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CH-341, HW #2 Key, Summer 2010

1. Circle and label neatly the functional groups listed in chapter 2 in the following structures. Do not include Alkanes.

a penicillin

Cicutoxin: a poisonous compound isolated from the water hemlock, which has shown anti-leukemic activity.

2. What are the hybridization states of the indicated atoms in the following structure?

3. Which of the following structures have the possibility of cis/trans isomers? For those that do, draw and label the cis and trans isomers.

a. /
No cis or trans isomers possible, when there are two of the same atom on a double-bonded carbon. / b. /
cis trans
c. /
no cis or trans isomers possible, since there are two CH2CH3’s on the right C. / d. /
CH3 is cis to O CH3 is trans to the O
e. /
no cis or trans isomers possible: no C=C.

4. Which compound in each of the following pairs would have the highest boiling point? Briefly explain your reasoning. Draw any hydrogen-bonding interactions.

a. / Butane / Octane
Octane has the higher boiling point. Because Octane has more atoms, it has more London dispersion attractive forces than butane. No hydrogen-bonding is possible, since C and H have very similar electronegativities.
b. / 1-butanamine / Pentane
1-butanamine has the higher boiling point. The molecules are the same size and shape, so they have the same London dispersion attractive forces between molecules. Since 1-butanamine is an amine, there can be additional hydrogen-bonding attractions between amine groups in different molecules. These attractive forces require more energy to overcome: more energy needed implies a higher boiling point.

5. Which compound in each of the following pairs would have the highest water-solubility? Briefly explain your reasoning. Draw any hydrogen-bonding interactions.

a. / 1-propanol / 1-heptanol
1-propanol is more water-soluble. Both alcohols can hydrogen-bond with water, since the O-H group is polar. The hydrocarbon portion of both alcohols is not attracted to water, since C and H are similar in electronegativity. 1-propanol has a smaller hydrocarbon portion than 1-heptanol, so a greater proportion of the 1-propanol molecule is attracted to water than 1-heptanol.


b. / 1-butanamine / Pentane
1-butanamine is more water-soluble, since the amine group can hydrogen-bond to water. Pentane has only C’s and H’s, so it has no polar bonds to be attracted to water. /

6. Predict whether each compound below has a zero or non-zero dipole moment. Explain your reasoning.

a. / NH4+ / / Ammonium has a zero dipole moment. The H’s are arranged tetrahedrally around the N, so all of the bond moments cancel out.
b. / BF3 / / BF3 has a zero dipole moment. The F’s are in a trigonal planar arrangement around the B, so all of the bond moments cancel out.
c. / (CH3)2C=O / / This has a non-zero dipole moment, in the direction of the O, since the O is the most electronegative atom in the structure.
d. / / This has a zero dipole moment. The C-F bond moments are in opposite directions, so they cancel out, as do the C-H bond moments.