CHEM 346 PROBLEM SET 1 ANSWER KEY 1 Ch 15 45 C6H4Br2 4 degrees of unsaturation suggestive of aromatic ring 6 x 2 2 4 2 2 4 C H Br 3 disubstituted benzene possibilities Br Br Br ortho substitution meta substitution para substitution What happens in nitration of these three compounds Br Br O2N Br C Br C Br Br NO2 B Br Br NO2 B Br O2N Br Br D Br D Br O2N O2N O2N Br Br A D are the same B C are the same 2 different isomers Br Br HNO3 H2SO4 Br Br Br HNO3 H2SO4 Br HNO3 H2SO4 Br Br Br NO2 A Br NO2 Br A NO2 Br A Br B D are the same 3 different isomers Br Br Br All four compounds are the same only one isomer NO2 O2N Br B Br C Br D is the original compound Br Ch 15 47 a H H b H Ch 15 51 a O H H2SO4 H B H B OH H O H B H O H O H O H O b O H H O H O H O OH H O B H Ch 15 53 Na is a single electron donor e HO e H H H H H H H H OH Ch 15 56 Consider the intermediate prior to the final product Has three resonance forms H H H H H H H H The three resonance forms tell us what the average bond order is of each C C bond 1 1 67 Bond order 1 All three resonance forms have C C single bond Bond order 1 33 Two resonance forms have C C single bond One resonance form has C C double bond 1 33 Bond order 1 67 Two resonance forms have C C double bond One resonance form has C C single bond H H 1 1 67 Three possible sites for protonation to yield two different products two products are equivalent 1 2 H H 1 H H 1 2 1 This product has the lesser amount of reorganization of atoms from the resonance stabilized intermediate 1 H H 1 vs H H 1 2 2 1 2 Br HNO3 H2SO4 Br O N O OH H Br Br NO2 O N O OH2 for example O N O Br Br Br Br Br H N O O B NO2 Br 3 O O 4 A B O O H HBr O O Br Product doesn t disrupt aromaticity Br 3 carbocation stable O O O O O O O O highly stabilized intermediate 3 carbocation 3 carbocation O O oxygen stabilization The high level of stabilization of the carbocationic intermediate lowers its energy makes reaction facile Cl AlCl3 O Cl AlCl3 C D O AlCl3 Cl Cl AlCl3 5 A B C D E F O 1 0 O OH 2 0 0 O 1 1 0 0 O 2 0 Reduction 1 0 O 0 0 O 0 2 H O O 2 Neither O Oxidation 0 2 0 0 0 0 Reduction 1 0 HO H 1 Reduction 2 1 2 O OH 6 9 deoxygelsemide J O F G O H I O HO E C B A H D A B C D E F 1 0 0 0 1 1 0 O 2 O Neither G H I J 0 0 2 3 O berkeleyamide A N O L M P Q F E G I J K NH H O R O OH A B D C A B C D E F 1 0 1 0 1 0 G H I J K L 0 2 0 1 0 0 7 Ch 16 44 The more electron rich the aromatic group is the more basic the amine will be A 1 0 0 0 0 3 M N O P Q R epiquinamide K J NH D H E N C B G H F I A B C D E F 1 0 0 0 1 1 G H I J K 0 0 1 3 0 NH2 NH2 H3CO H3C Br NH2 H3C NH2 O
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