Chemistry 14C Spring 2016 Final Exam Part B Page 1

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1 hemistry 14 Spring 2016 Final Exam Part B Page 1 In lecture we discussed the possibility that the first cells may have been formed in boiling mud puddles, which have been shown (in the lab) to produce fatty acids, carbohydrates, and primitive RA. When conditions are right (or so it is hypothesized) fatty acids assembled to form a micelle that just so happen to encapsulate RA. Voila! Life! Questions 1 20 concern the molecules involved in this origin-of-life hypothesis. In a recent paper it was shown that melamine (M) and barbituric acid (BA) (both of which can be formed via the boiling mud scenario) self-assemble into hydrogen-bonded structures resembling DA nucleobase pairs Melamine (M) Barbituric acid (BA) 1. (4) omplete this statement by writing a number in the blanks: Barbituric acid has sp 2 atoms and lone pairs. 2. (2) omplete this statement by writing an element symbol (U, Ag, etc.) in the blank: The least electronegative element in barbituric acid is. 3. (2) Draw in the box one additional resonance contributor for melamine. Make this contributor as significant as possible. Less significant contributors will earn less credit. Show all pi bonds, lone pairs, formal charges, and other details (4) Draw in the box the resonance hybrid for melamine. This may require consideration of some resonance contributors not drawn in question (previous). Include all pi bonds, lone pairs, formal charges, and other details in this hybrid structure. 5. (3) n the structure shown here circle all atoms of barbituric acid that are conjugated when the molecule is in its most stable conformation. Page 1 score =

2 hemistry 14 Spring 2016 Final Exam Part B Page 2 6. (2) omplete the following statement by adding no more than ten words. If multiple answers are possible write the best/most significant answer. onjugation provides stabilization to a molecule by providing (2) omplete the following statement by adding no more than ten words. If multiple answers are possible write the best/most significant answer. Aromaticity is an important feature of molecular structure because aromaticity causes (4) Let's decide if melamine is aromatic. For each question below write "Y" if the answer is yes, or "" if the answer is no. (a) Does melamine have a closed loop of p orbitals? (b) Do the atoms hosting the possible closed p orbital loop lie in the same plane? (c) Does melamine obey ückel's rule? (d) Is melamine aromatic? Questions 9 14 concern ribose, the carbohydrate portion of our barbituric acid/melamine DAlike structure. 2 Ribose 9. (1) omplete this sentence by writing 'D' or 'L' in the blank: The structure shown above is -ribose. 10. (1) omplete this statement by writing 'pyranose' or 'furanose' in the blank: The cyclic form of ribose is more likely to be a. 11. (2) omplete this sentence by writing a number in the blank: Including the stereoisomer shown above, there are stereoisomers of the ribose structure. 12. (2) In the box draw an enantiomer of the ribose structure shown above. Page 2 score =

3 hemistry 14 Spring 2016 Final Exam Part B Page (2) ompare molecule A to the ribose structure shown above, then in the space below write all of the answers which describe the relationship between ribose and molecule A. Answer choices: onstitutional isomers, conformational isomers, configurational isomers, enantiomers, diastereomers, meso. 2 Molecule A 14. (2) omplete this statement by writing 'more', less', or 'same' in the blank: ompared to ribose, glucose has carbon atoms. 15. (4) DA (as we know it) requires nucleobase pairing via hydrogen bonding. omplete this statement by writing no more than six words in each space. A hydrogen bond donor requires a hydrogen atom......whereas all hydrogen bond acceptor atoms have (2) The boiling mud puddle hypothesis uses molecules such as water, urea, and carbon dioxide as precursors to melamine, barbituric acid, and ribose. It is important that these molecules do not evaporate quickly from the boiling mud. Rank the relative ease of evaporation of these precursor molecules by writing 'most', 'middle', and 'least' in the ease of evaporation boxes. There are no ties. 2 Precursor molecule: Water (Urea) arbon dioxide 2 Ease of evaporation: 17. (3) In the box write the name or draw the structure of a molecule whose boiling point is clearly less than the boiling point of any of the precursor molecules shown in question (2) Acidity of a hydrogen bond donor is critical. If the donor is too acidic it will function as an acid and protonate the acceptor instead of just forming a hydrogen bond. omplete this sentence by writing 'greater than', 'equal to', or 'less than' in the blank: The pk a of melamine's bond is the pk a of aniline's bond. Aniline Page 3 score =

4 hemistry 14 Spring 2016 Final Exam Part B Page (2) Barbituric acid gets its names from its acidic properties. onsider this equilibrium: Barbituric acid + 2 Barbiturate anion pk a 4.0 pk a 15.7 pk a 20.0 pk a -1.8 omplete this statement by writing 'greater than', 'equal to', or 'less than' in the blank: For the equilibrium shown K eq is (6) By adding, subtracting, or changing into another element exactly one atom, redraw in the box the structure of barbituric acid to make its bond obviously more acidic. In the space below write the name of one structural feature such as resonance that is the most significant reason your new molecule is more acidic. Miscellaneous questions: These questions are not part of our barbituric acid/melamine origin of life hypothesis. 21. (2) The definition of 'lipid' contains two key ideas. List these ideas using one word for each idea: and. 22. (2) In the box write the general lipid category (fatty acid, etc.) to which the molecule belongs. If the molecule is not a lipid write 'none'. 23. (2) In the box draw the basic molecular structure that is characteristic of all steroids. 24. (5) By adding any number of any atoms except carbon, complete this drawing of a standard amino acid having a hydrophobic, nonacidic side chain. Page 4 score =

5 hemistry 14 Spring 2016 Final Exam Part B Page (3) omplete the statement by writing one letter in the first two blanks, and then a number in the third blank: ne of the DA base pairs occurs between nucleobases and. This base pair involves hydrogen bonds. 26. (1) omplete this statement by writing element symbol(s) (U, Ag, etc.) in the blank. In DA, phosphorus is bonded to. Work area. othing below this line will be graded. Page 5 score =

6 hemistry 14 Spring 2016 Final Exam Part B Page (33 points) Deduce the structure that corresponds to the spectral data on pages 6 8. Write your final answer in the box. A correct answer is worth full credit. If the answer is incorrect, your analysis of the spectra can be worth significant partial credit, so show your work clearly in the space below each set of data only. Answers outside of these places will be ignored. Final Structure Box (4) Mass spectrum: m/z = 141 (M; 100%), m/z = 142 (9.36%) and m/z = 143 (0.25%). The molecule does not contain fluorine or iodine. Write in the box the one formula that is consistent with the mass spectrum, and is not rejected due to other reasons. Page 6 score =

7 (13) IR: hemistry 14 Spring 2016 Final Exam Part B Page cm cm cm -1 IR workspace: 3000 cm cm cm cm cm cm -1 Page 7 score =

8 hemistry 14 Spring 2016 Final Exam Part B Page 8 Anything written outside the boxes on this page will be ignored. Write only 1 -MR implications and 13 -MR conclusions in these boxes. (10) 1 -MR: hemical shift Splitting Integral # Implications 3.47 ppm singlet ppm triplet ppm pentet ppm singlet ppm triplet 2 (2) 13 -MR: ppm (singlet), 76.0 ppm (singlet), 34.3 ppm (triplet), 32.8 ppm (triplet), 27.4 ppm (quartet), 25.4 ppm (quartet), and 17.2 ppm (triplet). 13 -MR conclusions: Page 8 score =

9 atural Abundance of Important Isotopes 1 (99.985%), 2 (0.015%) 12 (98.893%), 13 (1.107%) 14 (99.634%), 15 (0.366%) 16 (99.759%), 17 (0.037%), 18 (0.204%) 19 F (100%) 32 S (95.0%), 33 S (0.76%), 34 S (4.22%) 35 l (75.77%), 37 l (24.23%) 79 Br (50.69%), 81 Br (49.31%) 127 I (100%) Typical Infrared Stretching Frequencies Functional Group Bond Stretching, cm -1 Intensity alcohol strong; broad (-bonded) amine, amide medium; often broad alkyne strong aryl, vinyl sp variable carboxylic acid strong; often very broad alkyl sp variable; usually strong aldehyde sp 2 ~2900 and ~2700 medium; 2 bands nitrile variable alkyne variable ester = strong aldehyde = strong ketone = strong carboxylic acid = strong amide = strong alkene = variable benzene ring = 1600 and variable; 1600 often 2 bands Typical Proton MR hemical Shifts (ppm) R R 2.5 benzene R 2 R acyclic 1.3 R aldehyde R R 2 R cyclic 1.5 R 2 X (X = l, Br, I) 3.5 R R R 2 R benzene R R R R 2 = benzene 4 7 benzene R=R R benzene 2 R 2.6

10 Typical arbon MR hemical Shifts (ppm) ( 3 ) 4 Si = TMS = 0.00 ppm R R 2 l benzene ring sp R 2 R R = ester R R 3 R = amide R 2 I 0 40 R R = carboxylic acid R 2 Br R 2 =R = aldehyde, ketone