14C_s11_lec01_finalA_key

14C_s11_lec01_finalA_key - Chem 14C Lecture 1 Spring 2011...

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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 well. 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 molecular force. (b) Ion-dipole. (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 H N 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 = 2.1. (b) Electronegativity and inductive effect. The inductive effect is caused (in part) by electronegativity differences. (c) Fluorine decreases/reduces/delocalizes/stabilizes the electron density at difluorococaine's nitrogen atom. (d) Cocaine > difluorococaine due to the electron-withdrawing inductive effects of the two fluorine atoms. 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 effect. 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 O 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 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 (b) Adenine 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 carbohydrate. 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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