CHM238 1. Fall 2002 Practice Exam 2 Complete 5 out of the 6 following reactions, showing the main organic products. Cross one out or graded in order. CH3 O CH3 O AlCl3 O CH3 H3C + H3C CH3 Cl (a) Friedel-Crafts Acylation, CH3 is o,p director H3C CH Br H3C CH3 CH3 + Br2 NBS (b) benzylic substitution CH3 CH3 CH3 Br OH o 350 C + high P + NaOH H2O OH (c) reaction thru benzyne intermediate NO2 Cl + NO2 O OH NaOH - N + O - OH Cl from Meisenheimer (d) Nucleophilic Aromatic Substitution CH3 CH3 H3C + Br2, FeBr3 Br Br (e) o,p directed electrophilic aromatic substitution. N C N + Br2, FeBr3 (f) meta directed, EAS Br C 2. Fill in the proper reagents A- F +A O Br Br CH3 +C +B O2N A Br2, FeBr3 B O2N HNO3/H2SO4 C Na+ CH3O- O O +D +E +F OH O CH3 AlCl 3 DCl 3. E H2, Pt F KMnO4 In the following reaction of chlorobenzene with nitric/sulfuric acid, complete the resonance structures below with the correct bonds. Then answer the questions underneath. Cl Cl Cl Cl + Cl HNO3 H2SO 4 H H NO2 NO2 H NO2 H NO2 . (a) Why is the sulfuric acid necessary? Nitric acid is already acidic, isn’t it? (Note : pKa (Nitric Acid) = -1.5, pKa (Sulfuric Acid) = -10). O H + O - + O H + N N OH O - O + O + N H + H 2O O (b) What is the reactive electrophile in the above reaction? NO2+, nitronium ion. (c) If we used only pure (fuming) sulfuric acid, what would be the product(s)? mostly sulfonation of Cl benzene, both o and p, because SO3H+ becomes the superelectrophile and there is not as much protons for the dehydration of nitric acid. (d) Chlorine is o,p directing group but chlorobenzene is slower to react with nitric/sulfuric acid than benzene is. Why? Chlorine is o,p directing because of resonance. Note the bonus structure above. 4 resonance structures vs 3 for meta subst. Chlorine is EWG through sigma bonds because of its high electronegativity. Chloro-substituted benzene has less electron density to donate towards the NO2+ group. 4. Starting with toluene, design syntheses, providing the correct reagents for the following transformations. Both processes can be accomplished with 2 steps, but there is more than one correct answer for each. Assume that ortho and para isomers can be separated. OH O CH3 Br Toluene is reacted with KMnO4 to make benzoic acid. Benzoic acid is then brominated, Br2, FeBr3 to give metabromobenzoic acid. Carboxylic acids are EWG. OH O CH3 Br Toluene is first brominated with Br2, FeBr3. Then KMnO4 turns the methyl group into a carboxylic acid. This is the only way to make para substituition, because Br is o,p director. 5. Assign the NMR spectrum below to the molecule with the formula C8H8O2 that reacts with nitric/sulfuric acid to make the meta product. Draw arrows to peaks. This is the reactant not the product. IR peak near 1700 cm-1, 1620 cm-1 and C-H sp2 Molecule is single substituted benzene (5H) with a C=O (1700 cm-1 IR peak) and EWG-meta director. O H CH3 O H H 3H 3H 2H 8 7 6 5 4 6. Draw the following alcohols: HO H C H3C CH CH3 CH2 (S)-2-butanol H 3C OH CH CH2 OH CH 2 H 2C H 3C allyl alcohol isobutyl alcohol 7. Rank the order of acidity from l most acidic Æ 3 least acidic. 1 2 3 phenol methyl alcohol isopropyl alcohol 2 3 1 phenol p-methylphenol p-chlorophenol 8. Write the reagent that you would use to reduce the following carbonyl compounds over the arrow. Predict the product. O CH3 O LAH CH HC C HC CH2 O NaBH LAH OH 4 OH CH CH 9. Write reagents over the arrows for these transformations: PCC O LAH KMnO4 CH2OH CH3 C OH O OH CH CH3MgBr CH CH3 Acid Workup 10. What would be the product of Br H (a) 1-propanol + PBr3 Æ 3C (b) cyclohexanol + POCl3 Æ H H3C (c) (S)-2-butanol + TosCl/pyridine followed by NaBr in DMF? H 3C CH2 C Br C CH2 CH3 O O H3C CH2 C O H (e) 1-propanol + PCC Æ H (d) 1-propanol + Jones Reagent (CrO3-H2SO4) Æ 11. Predict the products of the reaction of Phenyl MgBr with the carbonyl compounds below (assume acid workup): OH O H3C H H3C O OH O NR Benzene + alkoxide HO 12. The alcohol in 11.c would cause trouble. Why? How could we get around this problem? 12. The problem is the alcohol will react with Grignard to make benzene and alkoxide. R pKa OH + R MgBr R O - + R 16 H 45 Keq = 1030 The solution is to protect the alcohol, perform the Grignard and then deprotect. OH OH O HO TMSCl pyridine CH3 TMSO O PhMgBr CH3 TMSO CH3 HO CH3