Chemistry 30A Summer 2004 Final Part B Solutions Page 1 Statistics High score 97 Average 74 9 Low score 29 Standard Deviation Irrelevant as it does not control grade distribution in this class A note about exam keys The answers presented here are usually 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 30A course web page 1 Phenol or benzene ring and alcohol amine carboxylic acid 2 C9H11NO4 3 L dopa has nine lone pairs 4 109 5o but 120o 5 Strain A decrease in molecular stability due to some factor of bonding or geometry 6 Ar aromatic ring The greatest strain is caused by the gauche Ar and CO2H groups H CO2H Ar H2N H H 7 L dopa does not have significant ring strain H 8 HS CO2H HO H HN HO O 9 Reaction product 10 Z 11 a Same b Cannot determine c Different Chemistry 30A Summer 2004 Final Part B Solutions Page 2 12 The molecule could be made achiral by inverting one stereocenter making it meso H H3C H3C O PPh2 PPh2 O H 13 The deoxy analog is more acidic so its conjugate base is less basic and therefore has less electron density at the oxygen atom This is because the second oxygen atom increases electron density due to resonance In fact oxygen and nitrogen atoms bonded to sp and sp2 atoms are electron donators due to resonance even though they are electronegative atoms The second OH group reduces electron density due to the inductive effect so this cannot be the explanation 14 Sulfur is a larger atom than oxygen so in general an SH group is more acidic than the analogous OH group CO2CH3 HS NH2 HS 15 We only know two anti Markovnikov reactions Hydroboration and HBr H2O2 so there are only two acceptable answers 1 BH3 2 HO H2O2 OH or HBr H2O2 Br 16 Catalytic hydrogen gives a syn product both of the new C H bonds are formed on the same face of the ring This means the starting pi bond cannot be between the two methyl groups Possible starting alkenes CH3 CH3 or CH3 CH3 Hydrogenation occurs at the more sterically accessible face of the alkene so hydrogenation of an exocylic outside the ring alkene would give the cis dimethyl product 17 Mechanism A step by step account of all the electron and bond changes in a chemical reaction Chemistry 30A Summer 2004 Final Part B Solutions Page 3 NH3 18 CO2CH3 CO2CH3 R R NH3 Br CO2CH3 19 R Br 20 DMF polar aprotic 21 a Slower OH is not an SN2 leaving group b Faster H is smaller than the ester so steric hindrance is reduced c Slower CF3 3N is a significantly poorer nucleophile due to the inductive effect 22 We consider SN2 before SN1 because SN2 is energetically less expensive lower energy of activation An SN2 reaction requires a moderate nucleophile NH3 is a pretty good one moderate leaving group Br is moderate and the carbon undergoing substitution cannot be tertiary it is secondary Since the SN2 requirements are met we predict SN2 to be feasible H 23 HOCH3 CO2CH3 CO2CH3 CO2CH3 R R R Br H HOCH3 OCH3 H3CO CO2CH3 CO2CH3 R R The stereochemistry of this product is R and S because the carbocation intermediate can accept methanol from either face 24 The major product is the more stable alkene the E isomer in this case CO2CH3 R H 25 E2 OCH3 CO2CH3 CO2CH3 R R Br Chemistry 30A Summer 2004 H E1 Final Part B Solutions HOCH3 CO2CH3 CO2CH3 R Page 4 R CO2CH3 R Br 26 The most probable mechanism is E2 CH3O is a strong base Br is a moderate leaving group and the molecule has the necessary hydrogens Cl 27 HCl CO2CH3 CO2CH3 Cl 28 H Cl Cl CO2CH3 CO2CH3 CO2CH3 29 The major product is derived from the more stable carbocation CO2CH3 Secondary with resonance More stable CO2CH3 Secondary without resonance Less stable Carbonyl groups are actually electron withdrawing due to the inductive effect of the carbonyl oxygen atom