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Unformatted text preview: Chem 14C Lecture 1 Spring 2011 Final Exam Part A Solutions Page 1
Statistics: High score, average and low score will be posted on the course web site after exam grading is
complete. The exam is ready to be picked up when these numbers are posted.
A note about exam keys: The answers presented here may be significantly longer than expected from a
student taking the exam. An exam key serves not only to reveal what was expected, but to instruct you as
To see the final course grade cutoffs, consult the grading scale on the Chem 14C course web page. 1. (a) Evaporation of propane. Propane is CH3CH2CH3. It lacks O–H and N–H bonds, so consequently,
hydrogen bonding is also absent.
2. (b) London forces < hydrogen bonding < anion-cation.
3. (a) NaCl. The only substance in this set with anion-cation interactions.
(b) β-D-Glucopyranose. Has the greatest number of hydrogen bond donors and hydrogen bond
acceptors, as well as the largest surface area.
In this set only London forces operate. All have equal polarizability.
This choice (anthracene) has the largest surface area. (c) 4. HOCH2COOH has the greatest number of hydrogen bond donor and hydrogen bond acceptor sites.
5. (a) Uneven sharing or distribution of electron density (resulting in charges), and (b) Interaction of
opposite charges (all noncovalent forces result from electrostatic attractions).
6. H2O and F-. Oxygen is more electronegative than nitrogen, so the hydrogen of water has a larger δ+
charge than the hydrogen of ammonia. Fluoride ion has a negative formal charge whereas the oxygen
of water has only a δ- charge. Larger charges = stronger attraction.
7. In order for hydrogen bonding to occur, the hydrogen bond donor molecule must have a hydrogen atom
attached to a highly electronegative atom (fluorine, oxygen, or nitrogen), so that this hydrogen can have
a sufficiently large δ+ charge.
In order for hydrogen bonding to occur, the hydrogen bond acceptor atom must have a lone pair, and
must be a small atom (oxygen or nitrogen) or must have a negative formal charge.
8. (a) Dipole-dipole, hydrogen bonding, ion-dipole, and London forces. Resonance is not a noncovalent
(c) Dipole-dipole, hydrogen bonding, and London forces.
(d) Octane. Octane has no polar bonds or hydrogen bonding capacity. These features are present in all
of the other choices.
H H 9. R2NCH3 + H O R
R CH3 + H2O Chem 14C Lecture 1 Spring 2011 Final Exam Part A Solutions Page 2
10. Cocaine-HCl. As stated in the problem, the pKa of cocaine-HCl (8.6) is higher than the pKa of sodium
bicarbonate (6.5). Higher pKa indicates a weaker acid.
11. Keq < 1. A proton transfer equilibrium favors the weaker acid and weaker base.
12. Reason #1: Bicarbonate ion (HOCO2-) has resonance stabilization of its negative formal charge. This
feature is absent in hydroxide ion. A more stable base makes for a stronger conjugate acid.
Reason #2: Bicarbonate ion has an electron-withdrawing inductive effect by the HO group, resulting in
stabilization of its negative formal charge. This feature is absent in hydroxide ion. A more stable base
makes for a stronger conjugate acid.
13. (a) ...is more electronegative. EN fluorine = 4.0, EN oxygen = 3.5, EN carbon = 2.5, and EN hydrogen
(b) Electronegativity and inductive effect. The inductive effect is caused (in part) by electronegativity
(c) Fluorine decreases/reduces/delocalizes/stabilizes the electron density at difluorococaine's nitrogen
(d) Cocaine > difluorococaine due to the electron-withdrawing inductive effects of the two fluorine
H H H O O C 14. CH2OH O C or OH
HO OH O HO O CH2OH This conjugate base has the
greatest number of stabilizing
resonance contributors, and is
therefore the most stable conjugate
base. H O O H O C CH2OH OH
HO OH You might have predicted this conjugate base because it is aromatic.
That's a very good prediction, but for reasons beyond the scope of
our introductory course, this is not the most stable conjugate base.
This prediction also earned full credit. 15. (a) H2Se Selenium has a large atomic radius.
(b) CH3OH Oxygen has a high electronegativity.
(c) HOF Fluorine has high electronegativity, resulting in the strongest electron-withdrawing inductive
F (d) OH
F Fluorine has high electronegativity, resulting in the strongest electronwithdrawing inductive effect. O
O (e) OH
CH3 This molecule's conjugate acid has the most resonance contributors
that also delocalize the negative formal charge. Chem 14C Lecture 1 Spring 2011 Final Exam Part A Solutions Page 3
CH3 16. O H CH3(CH2)6 (CH2)6COOH
H H HO Fatty acid Steroid
OH O CH3(CH2)24 (CH2)3CH3 O(CH2)28CH3 HO O OH Wax Prostaglandin O
O CH3(CH2)16 O CH3(CH2)16 O O CH3(CH2)16 O O
O O P OCH2CH2NH3 (CH2)16CH3 (CH2)16CH3
O O Phospholipid Triglyceride 17. Property #1: Hydrophobic, or, soluble in nonpolar solvents and insoluble in polar solvents.
Property #2: Biological origin.
18. There are eight acceptable answers: Alanine, valine, leucine, isoleucine, proline, tryptophan,
phenylalanine, and methionine. Glycine is hydrophobic and neutral (neither acidic or basic), but it does
not clearly demonstrate the stereochemistry.
19. Amide (or peptide bond) and disulfide.
20. Coil, α-helix, and β-sheet are all examples of protein secondary structure.
21. (a) Alcohol, amide, amine, aromatic ring, carboxylic acid, disulfide, and thiol.
(b) Amide (peptide bonds), amine (at the N-terminus), and carboxylic acid (at the C-terminus).
(c) Increased barrier to rotation, partial pi bonds, planarity, and resonance.
(d) In the absence of this conjugation, a protein is more flexible/floppy/disordered.
22. (a) Purine
23. In the DNA double helix, the base that most closely resembles molecule A pairs with thymine using
two hydrogen bonds. Chem 14C Lecture 1 Spring 2011 Final Exam Part A Solutions Page 4
24. In DNA, a phosphodiester group bridges carbon 3' of one carbohydrate and carbon 5' of the next
25. Assuming an equal distribution of G/C (with three hydrogen bonds per base pair) and A/T (with two
hydrogen bonds per base pair) = an average of 2.5 hydrogen bonds per base pair x 1000 base pairs =
2500 hydrogen bonds.
26. Assuming an equal distribution of pyrimidines (one ring) and purines (two rings), plus one ribofuranose
ring per base. = an average of 2.5 rings per base x 2 base per base pair = 5 rings per base pair x 1000
base pairs = 5000 rings. ...
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