Unformatted text preview: Chemistry 30B Winter 2010 Exam 2 Solutions Page 1 Statistics: High score, average, and low score will be posted on the course web site after exam grading is complete. Some questions have more than one answer, even though only one answer may be listed here. To see the projected course grade cutoffs, consult the grading scale on the Chemistry 30B course web page. 1. (a) An x-ray diffraction pattern is caused by constructive and destructive interference of scattered x-ray photons. (b) Studying the structure of DNA by x-ray crystallography is more difficult than studying the structure of NaCl by x-ray crystallography because it is harder to grow high quality crystals of DNA. (c) X-ray crystallography can be used to study hydrogen bond distances in water below 0oC but not a room temperature because below 0oC water is a crystalline solid whereas it is a liquid at room temperature. 2. 3. 4. 5. 6. The role of the x-ray diffractometer component indicated with the arrow is to hold the crystal at a specific orientation within the x-ray beam. The structure of NaCl can be studied by x-ray crystallography because NaCl is a crystalline solid. Cr Electron diffraction power increases with increasing atomic number. Carbon-carbon bond lengths and carbon-carbon-carbon bond angles (a) PBr3 cannot convert (CH3)3COH into (CH3)3CBr because the carbon bearing the OH group is tertiary. PBr3 converts ROH into ROPBr2, which undergoes nucleophilic substitution with Br-. This tertiary carbon is too sterically hindered for SN2 substitution. (b) PBr3, SOCl2, aqueous HBr, and TsCl are similar because in each case they convert an alcohol's OH group into a better leaving group.
NaH 7. (a) H3C OH H3C O Reaction with Na or NaOH is not as efficient as reaction with NaH. (b) H3C OH PBr3 H3C Br The stereochemistry is inverted due to Brattack on the backside of the ROPBr2 bond. A Williamson ether synthesis route is less efficient due to significant competition by E2. Epoxidation with CH3CO3H will not selectively epoxidize internal alkene. (c) H3C OH 1. Hg(CF3CO2)2 H2C=CHCH3 2. NaBH4 H3C O (d) OH Sharpless epoxidation O OH (e)
OCH3 1. BH3 2. NaOH, H2O2 OH OCH3 Hydroboration-oxidation achieves anti-Markovnikov addition of water to an alkene or alkyne. Chemistry 30B Winter 2010
I Exam 2 Solutions
+ CH3I Page 2 8. OCH3 aqueous HI 9.
Ph H OCH3 OH2 Ph H O CH3 H I CH3I + H OH Ph OH2 I Ph Ph I Ph OH2 Ph Some variations on this mechanism may be acceptable. The carbocation rearrangement causes a significant increase in stability (2o 3o with resonance), and is therefore highly likely. 10. Major product: A
H O OH OSO3H HO OH HOCH2CH3 CH3CH2O CH3CH2O OH HOCH2CH3 OH OH OH H The mechanism can also start with H2SO4 protonating CH3CH2OH to give CH3CH2OH2+. CH3CH2OH2+ then protonates the epoxide. CH3CH2OH and the epoxide are of roughly equal basicity, so both molecules are protonated to the same extent. 11. The product I circled in question 10 is the major product because when the nucleophile is weak and the epoxide is protonated, nucleophilic attack occurs at the epoxide carbon most able to handle the largest + charge. The allylic carbon can handle the largest + because is has resonance stabilization absent in the other epoxide carbon. 12. When the nucleophile is strong and the epoxide not protonated, the reaction is much like an SN2 reaction, and steric effects dominate: The epoxide carbons have very nerly equal steric hindrance, so A and B are prouced in roughly equal amount. 13. An oxonium ion has three covalent bonds to an oxygen atom with a positive formal charge. There are only three oxonium ion possibilities among the correct answers on exam page 3, all in question 10.
HO OH CH3CH2OH2+
CH3CH2O OH H Chemistry 30B Winter 2010
HO HO H HO O HO HO H HO O PCC O H Na2Cr2O7 aq. H2SO4 HO O Exam 2 Solutions
O HO OH Page 3 14. (a) O O HO H (b) O 15. The difference between parts (a) and (b) that causes different products to form in question 14 is the presence of water in (a) and the absence of water in (b). Many students had significant difficulties with the synthesis problems simply because they did not know their reactions, both from Chem 30B as well as Chem 30A. In the same way that you need a wide variety of tools to handle a wide variety of repair tasks (how many things can you fix with just a screwdriver?), you need to know a wide variety of reactions to handle a wide variety of synthesis problems. For questions 16-18, other routes may be possible. 16. Retrosynthesis:
Catalytic hydrogenation Ph Alkyne anion alkylation Ph Br PhCH2C CCH2CH3 Forward direction:
C Ph Br CCH2CH3 H2 Pt Ph PhCH2C CCH2CH3 17. Retrosynthesis:
OH Oxymercurationdemercuration Elimination Free radical Br halogenation Forward direction:
Br2 h Br NaOH 1. Hg(OAc)2, H2O 2. NaBH4 OH 18. Retrosynthesis:
O Oxidation O Addition to epoxide O O Epoxidation OH Chemistry 30B Winter 2010
Forward direction: Exam 2 Solutions
O Page 4
O CH3CO3H O H2SO4 HOCH2CH=CH2 OH PCC O ...
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This note was uploaded on 03/02/2010 for the course CHEM CHEM30B taught by Professor Hardinger during the Winter '10 term at UCLA.
- Winter '10