Sample Exam Solution CEM 263, Section B2 1 (6 pts) Aromaticity Circle those of the following structures which you would expect to show aromaticity. - N - N 2. (12 pts) Structure and Nomenclature Draw structures for which names are given, and name the given structures by any accepted system a) m-bromonitrobenzene b) 3-bromo-4-nitrobenzoic acid COO N 2 c) 2,4,6-tribromoaniline d) 1-naphthalenesulfonic acid SO 3 3 C e) C C 2 2-(1-bromopropyl-)benzene f) O 2 N 2,6-dinitrotoluene or 1,3-dinitro-2-methylbenzene
3. (18 points) Reactions Draw the structure of the major organic product(s) of the following reactions. Show only the monobrominated products for a) and b) and the mononitrated products for c) and d) 3 points each COOC 2 COOC 2 a) 2 /Fe 3 I I b) 2 /Fe 3 c) NO 3 / 2 SO 4 small amount d) NO 3 / 2 SO 4 O 2 N small amount C 2 C 2 O C 2 CO e) pyridine, CrO 3 3 C O f) 3 PO 4 120 o C C 3 C C3 C 3 C
B. List the reagents which will accomplish the following transformations. In all cases, more than one step is required. 4 points each a) 1) B 3 3 C C C C 2 3 C C C C 2 O 2 C 2) 2 O 2, O -, 0 o C 3 C 2 C 2 C 2 C 2 C 2 b) 1) (C 2 ) 4 COCl, AlCl 3 N 2 N 2 or Sn/Cl COO c) 1) KMnO 4, 2 SO 4 2) NO 3, 2 SO 4 d) 1) Fe 3, 2 2) Mg 3) CO 2 4) 3 O 4 (8 pts) Reactivity COO a) Rank the following compounds according to their reactivitiy towards electrophilic aromatic substitution by numbering them from 1 to 4 where 1 is the most reactive and 4 the least reactive. (4 pts) NO C 2 3 4 2 3 1 b) (4 pts) Chlorobenzene is less reactive towards electrophilic aromatic substitution than benzene, but incoming substituents are directed into ortho and para positions. iefly explain these two facts The chlorine is electronwithdrawing so pulls electrons away from the aromatic ring and Cl makes is less reactive towards electrophilic substitution. owever, when the intermediate in the substitution reaction in the ortho or para positions is formed, the carbocation is stabilized by sharing of a lone pair on chlorine to give an extra resonance structure, e.g. for ortho substitution: Cl Cl Cl Cl Cl Similar stabilization of the resonance structure is not available for meta substitution This extra stabilization is not available for meta substitution
5. (12 pts) Reaction Mechanisms a) (6 pts) The Friedel Crafts reaction of benzene with 1-chloro-2-methylbutane yields 3-pentylbenzene. Indicate why this product is formed by showing the mechanism of the reaction. C 2 CC 2 AlCl 3 3 C C 2 C C 2 Cl 3 C C 2 C 3 - C C 2 Cl AlCl 3 3 C C 2 C C 2 AlCl 4 3 C C 2 C carbocation rearrangement; both rearrangements may occur 3 C C 2 C C 2 - AlCl 4 3 C C 2 C C 2 - AlCl 4 C 2 CC 2 - C 2 CC 2 b) (6 pts) The hydroboration of alkenes followed by hydrogen peroxide oxidation of the trialkyl borane yields an alcohol. Show how the mechanism of this reaction with 1-methylcyclohexene accounts for the fact that anti-markovnikov and syn addtion of and O occurs across the double bond. B 3 3 C B 2 O 2, O - 0 o C 3 C complex formed with borane adding across the double bond O
6. (10 pts) NMR Spectroscopy. b a) (4 pts) Sketch the NMR spectrum you would expect for 1-ethyl-4-nitrobenzene. c d C 2 b d a a a b c 10 ~7 δ ~2.5 ~1 0 b) (3 pts) iefly explain how you predicted the δ values of the signals in your spectrum. The a and b protons are aromatic so the signal will be at about 7δ. The a protons are further deshielded deshielded by the adjacent nitro group so are slightly downfield from the b protons. The c protons are on a benzylic carbon so their signal is at about 2.5δ. The d protons are aliphatic, so their signal is at about 1δ c) (3 pts) Use diagrams to explain any splitting of the signals in your spectrum. The degree of splitting of an NMR signal is n 1, where n is the number of protons on adjacent carbons. The signal due to the a protons is split into a doublet of relative peak height 1:1 by the b protons. Each b proton can be aligned either with the magnetic field or against it resulting in two slightly different fields experienced by the a protons. Similarly, the signal due to the b protons is split into a doublet by the a protons. or possible alignments of a and b protons with the field or against the field The signal due to the c protons is split by the three d protons, thus is a quartet of relative peak height 1:3:3:1. possible alignments of the c protons possible alignments of the d protons The signal due to the d protons is split into a triplet of relative peak height 1:2:1
7. (12 pts) Synthesis Methoxychlor has replaced DDT as a pesticide. Give reactions for the preparation of methoxychlor from anisole and any other reagents required. Indicate the mechanism of the reaction. 3 CO 3 CO C O Methoxychlor Cl 3 CC(O) 2 2 SO 4 chloral hydrate O anisole O C O C 3 CO C O other reagents required are chloral hydrate and sulfuric acid 2 SO 4 O C 3 CO C O stabilized by resonance 3 CO C O O 3 CO C O methoxychlor 8. (10 pts) You have three bottles containing the three isomeric dibromobenzenes; they have the melting points 87 o C, 6 o C and -7 o. By a great deal of work, you prepare six dibromonitrobenzenes (C 6 3 2 ) and find that of the six, one is derived from the dibromobenzene of m.p. 87 o C, two are derived from the isomer of m.p. 6 o C, and three are derived from the isomer of m.p. -7 o C. Identify each isomer as o, m or p. O 2 N the meta compound can form three mononitro derivatives, has mp -7 o C the ortho compound can form only two mononitro derivatives, has mp 6 o C the para compound can form only one mononitroderivative, has mp 87 o C