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Transcript
Principles of Drug Action I, Spring 2004 Sample Exam Questions 1. What is the likely bonding arrangement (neutral state) between chlorine (atomic number 17) an atom with an atomic number of 12? An atom with an atomic number of 12 will have 2 electrons in its valence shell (3 shell). Such an atom will most likely "give up" those two electrons in ionic bonding to achieve and octet (in it's 2 shell) A. B. C. D. E. 1 covalent bond 2 covalent bonds 6 covalent bonds 1 ionic bond 2 ionic bonds # 2. What is the formal charge on an atom with an atomic number of 5 when it forms four covalent bonds with hydrogen? This atom has 3 electrons in its valance shell. Thus in the neutral form it will form 3 bonds. If it forms an additional bond it will be sharing an additional electron and have a -1 charge. A better way to solve such problems is using the formal charge equation: FC = Valence - [1/2BEs + NBEs] which for this atom is: 3- [1/2(4) + 0] = -1 A. B. C. D. E. 0 +1 +2 –1 # –2 3. The amine compound below is not optically active because: it is sp3 hybridized, tetrahedral and contains a chiral atom (N) A. B. C. D. there is no sp3 hybridized atom in this compound there is no chiral atom in this compound there is rapid inversion about the nitrogen atom # the compound is not a tetrahydral N H CH3 1 4. What is the stereochemical relationship between the two compounds shown below: Same configurations about the double bond (cis) and chiral carbon (S). Only difference is rotational position about a C-C single bond HO CH3 CH3 C CH3 H A. B. C. D. E. H3C CH2CH3 and H HO C CH3 C CH2CH3 C H3C CH3 CH3 Positional isomers Conformational isomers # Geometric isomers Optical isomers Tautomers 5. Which order (A-E) correctly ranks the relative water solubility (from highest to lowest) of compounds I-V below? A. Generally diols > alcohols > ethers > hydrocarbons for water solubility of comparable structures. Also, secondary alcohols (branched) are more water soluble than their corresponding primary alcohols! A. B. C. D. E. V > II > III > IV > I V > III > II > IV > I V > IV > II > III > I II > III > V > IV > I IV > V > III > II > I I III II OH IV OCH3 OH V OH OH 2 6. Which compound shown below (A-E) would be the major product formed from the electrophilic addition reaction shown below? B. The reaction involves addition across an alkene bond and will proceed by addition of H+ to give the most stable intermediate carbocation (tertiary) to which Br- will add. CH2 HBr Product A, B, C, D or E? CH3 CH2 Br CH3 CH3 Br Br CH3 CH2 CH3 B A C CH2 CH2Br Br CH3 D E 7. How many different (regioisomeric and stereoisomeric) electrophilic addition products could form in the following reaction? Two regioisomers are possible by addition of OH- and Cl+ across the double bond. During addition a chiral center is created at each addition site (for each regioisomer). Thus two chiral centers are generated for each regioisomer, meaning there can be four optical isomers for each regioisomer, or a total of 8 products. Note that the chiral center present in the starting material is not involved in the reaction and does not change (it is fixed!). A. B. C. D. E. 2 4 6 8# 12 H H Cl CH2CH3 HOCl Number of Products? H 3 8. The nucleophilic substitution reaction shown below would proceed most rapidly with which electrophile (variable X group)? I is a better leaving group than Cl (larger, more polarizable halogen) and halogens in general are better leaving groups than amines and alcohols. Alkenes (and other hydrocarbons) do not function as leaving groups typically (too unstable) A. B. C. D. E. Cl I # NHCH3 OH CH=CH2 X H2N + Nucleophilic Substitution Product 9. Which order (A-E) correctly ranks the relative acidity (from most to least) of compounds I-V below? C. Phenols are more acidic than ethers, so II is least acidic. Phenols with electron withdrawing groups in conjugation with the site of ionization (phenol O) enhance acidity to a greater extent by stabilizing the negative charge of the conjugate base. OH OCH3 I II OH O CF3 OH H3C III A. B. C. D. E. OH CH3 IV OCH3 V III > IV > I > V > II V > III > I > IV > II III > I > IV > V > II I > III > IV > V > II III > IV > V > I > II 10. How many electrophilic mono-substitution products are possible in the following reaction (include even those less likely to form)? Did this in the review session. There are four possible positions for substitution and each one yields a different regioisomeric product! A. B. C. D. E. 1 2 3 4 # 5 OH CH3Cl AlCl3 Total Products Possible? Cl 4 11. Which compound below (A-E) would be the primary elimination product formed in the following reaction? B. The reaction by definition involves loss of halogen (Br-) by removal of a proton on the carbon atom adjacent to that carbon bound to Br. In this case there are two such carbons, but the "interior" one will lose a proton faster because it gives the more stable alkene (more substituted), and generates and alkene where the C=C is in conjugation with the aromatic ring!!!! CH3 HO - Primary Elimination Product? Br CH3 CH2 CH3 OH C B A CH3 HO CH3 Br OH D E 12. What is the stereochemical relationship between the two compounds shown below? Note the asymmetric carbons at the site of restricted rotation (C=C) and that the priority groups are oriented in different directions! A. B. C. D. E. Positional isomers Conformational isomers Geometric isomers # Optical isomers Tautomers H H Br and H H CH3 H H CH3 H CH3 H CH3 H H Br 5 13. Based on the mechanism shown below, which X substituent listed below (A-E) would this enhance this reaction to the greatest extent? This reaction proceeds by formation of a carbocation, so any functional group that stabilizes + charge, especially by resonance donation (+R) will enhance this reaction to the greatest extent. Methoxy is the only +r group shown. Methyl is +I (so it helps, but weakly), H is "neutral" and F and NO2 are withdrawers CH3 CH3CH2SH Br X A. B. C. D. E. CH3 CH3 X X Br- SCH2CH3 X=H X = OCH3 # X = NO2 X=F X = CH3 14. How many optical isomers are possible for the compound shown below? C. This compound has two chiral centers, but is symmetrical. Thus while RR, SS, RS and SR isomers are possible, the SR and RS isomers are the same compound!!!! (meso) A. B. C. D. E. 0 2 3 6 8 CH3 CH3 6 15. Which order (A-E) correctly ranks the relative acidity (from most to least) of compounds I-V below? Relative acidity trends: Acids > phenols > alcohols > alkynes > alkenes/alkanes. Furthermore, phenols with electron withdrawing groups (V) are more acidic than those with donating groups (IV) since they stabilize negative charge better! Br C CH Br CH2OH II I CH3O OH III Br OH IV A. B. C. D. E. Br COOH V II > V > IV > III > I # V > IV > II > III > I II > IV > V > I > III II > III > V > IV > I III > II > V > IV > I 16. Which compound below (A-E) will undergo electrophilic aromatic substitution most readily? C. That aromatic ring with the best electron donating group will be most reactive toward electrophilic substitution. In this case only one compound has an electron donor (C), and it is a weak donor by +I. If an electron donor by resonance (such as OH) was included in this probloem, it would be the correct choice! O NO2 C N CH3 C CH3 N(CH3)3 A B C D E 7 17. Which description below (A-E) accurately describes the regiochemical or stereochemical relationship between compounds I and II below? Note the asymmetric carbons at the site of restricted rotation (C=C) and that the priority groups are oriented in different directions! A. B. C. D. E. Tautomers Regioisomers Enantiomers Geometric isomers # Conformers H3C H H CH3 H H H CH3 H H3C I II 18. Which order (A-E) correctly ranks the relative ease of oxidation with CrO3 (from most to least) of compounds I-V below? Should be III > I > II (no correct choice present…sorry) A. B. C. D. E. I > II > III II > I > III III > II > I II > III > I I > III > II OH I H3C CH2OH OH III II 19. Which order (A-E) correctly ranks the relative acidity (from most to least) of compounds I-V below? A. Sulfonic acids are more acidic than COOH which > thiphenols > phenols > alcohols. See class discussion notes A. B. C. D. E. V > IV > III > I > II IV > V > I > III > II V > IV > I > III > II V > IV > II > III > I V > IV > III > II > I OH CH2OH SH COOH SO 3H NO 2 NO 2 II NO 2 III NO 2 IV NO 2 V I 8