322bs07_e2_key - CHEMISTRY 322bL 325bL PleaseM‘ EXAM NO 2...

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Unformatted text preview: CHEMISTRY 322bL / 325bL PleaseM‘ EXAM NO. 2 Print La Name FEBRUARY 21, 2007 First Name First four digits of USC ID _ _ _ _ -XXXX-XX TA's Name Grader (1) (15) ____ (2) (15) _. Lab Day 8: Time (3) (10) (4) (10) __ _. (5) (10) __ ____ (6) (10) __ (7) (15) (8) (15) __ ____ (100) I will observe all the rules of Academic Integrity while taking this exam. Chemistry 322bL, 325bL -2- Name Exam No. 2 3 Zack (1) (15) Complete the reactions below by providing the missing products inside the boxes. Be careful to give all the important organic products, as requested. It is not necessary to give the byproducts or minor products of the reactions. "Workup" means treatment with water and bringing the reaction mixture to neutral pH. H3 (1) HNO3, H2504, heat (A) .. (11) workup OOH (I) Brz, Fe .._.. (B) (11) workup i? (C) ©CH2©NOZ (i)CH3CCl/A1C13 (ii) workup H3 (i) hot aqueous KMnO4, KOH (D) (ii) workup Br HF (E) + H2C2CHCI'12CH3 00C large excess Chemistry 322bL, 325bL -3— Name Exam No. 2 3 £4: L (2) (15) Complete the reactions below by provim products inside the boxes. Be careful to give all the important organic products, as requested. It is not necessary to give the byproducts or minor products of the reactions. "Workup" means treatment with water and bringing the reaction mixture to neutral pH. (1) Zn<Hg) 1130+ CCHzCHzCOOH ' (ii) workup CHzCH3 1),Na liq NH ,, trace EtOH (ii) workup OOEt ).1 0 eq (i-- -Bu,)2AlH -78°C (ii) H20 HZCH3 i() 1. D eq NBS/ CC14 peroxides, heat )NaCN in acetonitrile (i) CH3MgBr in ether 3 “Wk—7“?) 11 wor up a7 an (E) C10H120 Chemistry 322bL, 325bL —4— Name Exam No. 2 (3) (10) Answer the questions below. (A) (5) The cyclopentadienyl anion (I) is unusually stable, but the corresponding cation (11) is not. (i) Use the polygon inside a circle method to determine the number and the relative energy alignment of the at molecular orbitals in I and 11. (ii) Show below the electron occupancies of these molecular orbitals for I and II. (iii) Briefly explain their difference in stability in the comment box below. approximation of at MO schemes 0 Q . '1' II B relative It electron inscribe polygon inside circle above cmwufinimwr “NW“ 13 \s k c —s M “(F MD 5‘13“?“ sad—k 0-\\ Ti." the-hon”; \‘K “\Ds firh't‘ an. Shiv-EM} . (ydopqad—«AMmTl C1L¥ WK (9:) \S 03“ OE'K— skit] “Tr V10 4; 51th + +137w+ QWV‘L'l'TK‘ k {—kg, HUcbsK r—u L '1 ’tr'mkuk-NM}. (B) (5) The heats of hydrogenation for all cis-1,3,5,7-cyclooctatetraene ([8]annulene) and cis — cyclooctene are given below. \§ CHN*§ k AHH2 = -24.3 kcal/mol AHH2 = -97.9 kcal/mol all as cyclooctane cis Does this data reveal any special stability for [8]annulene relative to a tetraene electronic structure? Explain your answer using a simple calculation. Is this result consistent with the HfiCkelgif? Biplai: LVOKU‘A ‘0‘ “Yd-”5" «ursd-Dk o {1 C‘Idocf‘i‘i‘n‘l'm— Ztk \; Low?”w¥‘K0 +0 '(LL LLLT N0\U¢- k‘é F L{ [5103\th Aouhsh. bun/d3 ""Udefi +Ln. ADW\°\¢ bod/6L L'v\ (jg.— ..- A . WWW w :~ mam/m g... L! WWW Ll ska-‘63 Keck/W\\ ~ (0%? are-591k ‘l'b "47-? KQ4/N\ ‘61,? ct.\\ ¢7S-¢V¢(oocfi4‘l‘k‘trq&u Tl-ws obmvv~ofio~~ v3 °°~\S\S'\"’~m"‘ LOCTk’t’llu Wth\ FUKK ' Chemistry 322bL, 325bL —5- Name Exam No. 2 (4) (10) Answer the questions below. (A) (5) Provide structural formulas inside the boxes for the names below the boxes. skit—17‘ CH O3No Z 1 ar‘. 8‘» 3,5-dinitrobenzoic acid —bromoan111ne 4 chloro 2—n1trotoluene 3—chloro-2-hexanone (B) (5 ) SSketch a free energy diagram for the general reaction of electrophilic aromatic substitution of benzene by an electrophile (E+ ). Draw and circle the hybrid structure and show the location of the key intermediate 1n the diagram. Also id rate-determining step. relat1ve free energy Chemistry 322bL, 325bL -6- Name____________ Exam No. 2 A 2 W (5) (10) Circle the correct answer for each question or statement below. (A) The order of reactivity (fastest to slowest), ignoring position of substitution, of the substituted benzenes below in nitration is CH3 (circle answer below) II>IV>III>I IV>I>II>III III (B) The more stable arenium ion in each pair below is C1 C1 C1 (circle answer below) 2K? A I, B I A I, B II GL)NO A II, B II N02 mW—J CLW_H_J A B (C) Which chemical species below are potential electrophiles or Lewis acids? + (circle answer below) BF3 NOE CH3C =0: 503 II and III I, II and III I II III IV 11, III and IV (D) In the nitration of A and B below, the major product in eac case will be (circle answer below) CH3 CH3 c1113 AI, BI AI, BII H3 AII, BI orOZN CIEOI3 (E) The addition of H1 toI p- -subs1 tituted styrenIes (phenylethenes) will proceed fastest with (circle answer below) [email protected]=cHH H H ...... - H: H @ 02H- Chemistry 322bL, 325bL -7— Name Exam No. 2 (6) (10) Answer the questions below. (A) (5) Carefully draw all the resonance structures of the arenium ion intermediate for the para-bromination of anisole. The hybrid structure is not required. for para substitution product anisole resonance structures of the arenium ion (B) (5) One mole (1.0 mole) of a 50:50 mixture of propene (I) and 3,3,3—trideuteriopropene (II) reacts with 0.1 mole of N—bromosuccinimide (NBS) in CC14 with a small amount of added peroxides. The 0.1 mole of allylic bromination product is analyzed. a .1 1 CH2=CHwCH3 + CH2=CH-LD3 0 mo eNBS allylic brornination product I 11 / CC}, peroxides 1.0 mole of 50/50 mixture Analysis of the recovered 0.1 mole of product reveals the major product is 3-bromopropene (III) (no deuterium). A minor product is dideuterio-3-bromopropene (IV). Answer the questions below within the context of the propagating steps of this allylic bromination reaction. (i). Why aren't III and IV produced in equal amounts? Th“— \5 3~ kw \5 V“ JV)“. H€Orb§ °~\>s—)«~.—c’Vo—v\ She?’ Qkkck \S Fa-‘ifl—AO—‘RFM\1\4\S) “\Ds‘lfi‘mcjfi‘ocw 0% $4“ lower “’35.H \g fiflf. Q cu~cH~§Bf~H~~ c' ———' CHzLQWLCh 1‘ “”3? “Rh“? 3". (5‘. M _' __ [CthcH~tDiv-bm3r1 «a cut—oa—r—oz + (343‘— 3\N..r~ 0.1 mole recovered (ii) Draw the structure or structures of the dideuterio—3-bromopropene minor product in the box below. What does this result reveal about a key intermediate in this reaction? mg KC], \M‘er it 94% l S +\u\ duke can L i LLA‘. o S \\ g \ r ... A i Q c...\ . structure or structures of C3H3DzBr ST (.qu :C—D “AA*\’V"‘0~\ O'Q‘ 8“ Q'K'ow <¢M OCCUF \Bof/ 2‘. ac‘r 169m» Hrmtxgk («she-t“ Bra-Br/ .L Chemistry 322bL, 325bL -8- Name Exam No. 2 (7) (15) Provide the missing structures A-E 1n the following synthetic scheme. ©+ Ki) \ AlCl _worku2 3%Zx, A C10H1003 (i) Zn(Hg), HCl, acetic acid (ii) workup (i) A1C13 B *‘P / cq‘cucu 0:4 (1) (CH3)2CuLi/ether,-780C Sf w 2 ________, K/ (ii) workup C (i) LiAlH(O-Bu-t)3/ ether, —78°C (ii) workup CD @«suekfi H 1 E Chemistry 322bL, 325bL -9- Name Exam No. 2 (8) (15) Provide reasonable syntheses of A and B, as requested below. OOH (A) (6) from toluene and any other needed reagents CH3 Lo ou CC): (-. 3 lno’r kmq‘p‘ ,KOP: H no; M1505; © _.—-—-—> M NO (fl) H30 Q5 Chemistry 322bL, 325bL -10— Name Exam No. 2 (8) (Contd.) “H2CH2CH2CH3 from benzene and 1-butanol with any needed reagents ...
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