W06 midterm2 key

W06 midterm2 key - Chemistry 30C Name Examination II 10 March 2006 D Brown Develop answers to each of the following questions in the space

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Unformatted text preview: Chemistry 30C Name Examination II 10 March 2006 D. Brown Develop answers to each of the following questions in the space available only. Do no work that is to be graded on the back sides of the examination pages. Show all work to receive full credit. There are eight pages including this cover page and pages 2 — 8 of questions on this examination. The examination has four questions worth a total of 100 points. There is fifth question that is a replacement question in case you prefer it to one of the first four. This fifih question has less points and your performance is still based on 100 points possible. Do this question only if you are unable to do one on the first four questions. Please circle the name of the teaching associate (and the day and time) whose section you are attending. Amanda Amanda Ryan Travis Travis Mon Mon Tu Wed , Thu 12 2 l 12 1 Points Possible 1 - — — w — v — - l -2- (25 points) 1. Reaction Mechanism—Theory. The reverse Michael reaction mechanism is superior to the E2 and E1 mechanisms when structure permits. ' O O )1 ____.., )% OH ' Develop the acid catalyzed reverse Michael reaction mechanism, You may use H30+ and H20 as your acid and conjugate base throughout. n n “. H‘. A I, I R “+34 6:0 “12‘” P +t'0/ A VP.O~H' D 2‘ e A? 2‘ )3 \i\.. ‘5‘ 6/“ ’ \'o/“\ B/ .c .. .. V “\NIH hie-\- u“ "‘ +il ‘3‘“ p R’o‘u’ b L ‘Kl H n -‘ l 0/ 0 ‘7,“ 4' / :0 an 8“ c ‘m 7 A ‘- 1? H I ll 7 ko/ ~l¥ .. u "i" {6’ :0’ H 1 l Ll l‘l -3- (25 points) 1. Reaction Mechanism-Theory. Continued from page 2. Explain why the reverse Michael mechanism is superior to the E2 and El mechanisms shown below. Criticize these alternative mechanisms to the acid catalyzed system you developed on page 2 explaining how these (E2 and E1) would be poorer. A H . .. .. + I. ‘O :O :5~.+ :O H'V-flr 5‘49‘ 14 p 5 3L? 1 {DH 2 brbe 50"" :fi‘s .6~“. " fix 1‘“ H /" mu. m5ng . 1L1; o oi Ru 0“ is mlw'c “moi/h 0 it it c.le 5- firiimi yoth in“. will New». 5H1 is “V‘WWtML . its ‘87. 51:9 viva“); a “main; wL‘J,., is wipww l" «we we (Mm, Hie). “w Ms M Wat“ M W. l . .. 1"" l, .. l4 :0 4' 0 .0." :0 “t5”, E1 53 SE91 kw“ 555 )5 1* .. 4 3 l 9H; u\?6 K} R . . 8V,\Mf\sk‘\u$wc~;\wiam&&. 59,3 n A 3% Mk2 Mei)me R “09”, W (“5&3 \M-u M 2.“. M NW“ 35‘ i {5 1L 1.5% 9. He. by a» M Willi isodalhmae‘ LIA-AL Gag “~13 Sh, is M “Our A“ on" 3 9.." ‘K‘ (Gk- l.) WNS s.l"w (39‘ M 9‘ woude S 1 WA 13 l“ ‘65 NJ“. H ‘5 . ’ F k f ’ The‘rvingxdyns icall‘y whic‘la‘of e reverse fichaels, E2 and E1 ogl‘d'be the best 5”“ MAM produce elimination product? Explain. ‘ Sim Sir-J; Mule a. {Mafia} in «Law «a; skkvj a... K; a“; 31m um‘wufiafin “u, (an—Hui FLMM Kmfipu-aimm sh... TL» AR (3 Hi 51"“ n. mu. “TL. M ‘KKV‘MI’HM'WI‘k-‘k “H.511”. \‘s intuLi-ai' -4- (25 points) II. Diene Chemistry. lwlmoh a. When 2-methy1—l,3-butadiene, I, reacts with HCl the two pi bonds have different basicities. Protonation at the 1,2-pi bond is thermodynamically better than at the 3,4-pi .bond. As a result two addition products are produced Via this conjugate acid. cH3 (EH3 CH2=c—CH=CH2 CH2=C—CH=CH2 1 2 3 4 N Develop appropriate structures to show why C1 is is thermodynamically more basic than / C4' “Wt Ml “3‘03 haul; __s N «a (9* u M ‘W 3 Q! N <‘ sf" H Ci \‘5 W Sam on ii’ix awach 3° l' . cAC‘ MW gtthJ 5m“. M 5+ 1’: Null»... m Arm—J l" C- “4&0 (,4 W :3 N 6—) L/g/l i W Wu. H4 1° Ml” 54' C; - \ cm P .l. {ILL c4 alth Maud-‘41:- cAct «‘x Mum shkkwil b3 WM R:- B; ' Develop the mechanisms to yield the two addition products. a A (34 . /l\¢ A 4V pm 1' Q //J\// (L? 4;}: =9.“ .5 “D “E3 H“ ‘ I gfi", ‘g ‘ 1 ’ &q i" N PM :9". .. . __ ‘9’.“ a; é C» ‘ Predict the principal product if kinetics controls the chemistry. ' v .3 / RENEW/mid ltw'b‘wu-thafiaddi'KuM-MW Mdtofluhia) "Jim m “I MWVC c - in A k.) ‘4". mamas want it» «MR.» 11st% with 9 Mt 30c, “with N‘le ["c. j 3&5 Miw‘mtm ML») c. M ‘Wxfichh‘m ’n {M M ksifihufié (“Li Wm»: _‘ Kilan NU. FEM“ hath “41,30 C M “NB 3" C.“ M cg P -5- (25 points) II. Diene Chemistry. Continued from page 4. Predict the principal product if thermodynamics controls the chemistry. ll haw.“ WM lL wskb in M Mr Md: will u no“ ml ‘ WJIMMMWKMJ he.me vhk'nmzx (Kt/mm yaw mm‘h M9» at“). \ makhmMMdh-xtWMSH5wths Pm: f? U _ DIM: M GA» W? Milaghnbcs “A is “MEAME wk 4“} W Miwvu . pug Which of the #2- and 1,4— acdi’diti 11 product comes from the more stable intermediate ‘ carbocation. M Mimi; wen m 5‘... ,mm mg mums k) ‘ I ‘ 2' + / "3: 67V ‘ ML Mu, Mu QM (A. \J . x 2W “SHMH‘M . Ci )0” 2': W A) u .5 :é/l'.’ / /l§/"r (25 points) III. Aromaticity. Imidazole is aromatic. N 1/ l ll H Develop a three dimensional sketch for imidazole locating the maximum number of nuclei in the xy plane as possible. Show in plane sigma bonds with lines and in plane atomic orbitals or hybrid atomic orbitals with lines. Pi bonds should be shown with overlapping parallel p orbitals. Non-bonded electron density should be shown in appropriate orbitals. xy plane With specific reference to your sketch identify the aromatic set of electrons that is involved in cyclic ring resonance. 4H1 = (2 um. n = [ .-. ' 1t. «Cumin—t. C-c 92-h Tl, human—N nghfiIh-Jl'luh'llv pymwnmmum. ' Develop the complete set of good resonance structures for imidazole showing all non- bonded electron density and carefully drawn curved arrows. will: a 4“): by (wins-9 .941 99 "a I [4' H :4 -7- (25 points) IV. Electrophilic Aromatic Substitution. Furan is aromatic. B [3. O Predict whether position on or [3 positiOn is the principal site for electrophilic aromatic substitution if kinetics (starting material electron density) controls the chemistry. Explain. ___ f "' CE? 4—) 499') e, éij‘J’e—v fig; e-) I 11' I n ' m 'i wit it I: Bu+—Md~wb( It M; I m wn‘www-m km awe-ct. i “A “I (Mn N wvmiwwf Mawlkwuiwkm 111,13 a watt... 3in .0 (at yranh'm othsww s». my. 1k M M 91”“ iv. 3,: Wu: Wt W 85- Ws Lin/Mo dbl-whim M it): imp}; ta m .E‘. A: «m M» h M kaw w Mm twat-M it my}; m- e’M MM WNW "t h * mm M o» em...“ mama» M MM. mm Predi t whether position 0!. or [3 position is the principal site for electrophilic aromatic p.555“. substitution if thermodynamics (intermediate stability) controls the chemistry. Explain. A ‘TLHM‘WB *i ‘ “5 “X5 W“. l (2* ,,. /.. 4. 4” l/ /o so Vt. idhbml) (a "' __. " t-i NV ma 6 ) '5‘ e9 4.4.. )(A. .4-9 6 Ci < E 09 E file, [CYMLIMD - .. a G '1 Omaha twill» Marat) a. in paw. NEW-lair a W 591%, a. at W Xi ball-J... ihitnwfltak is. mm: V3 IWDWL ; M W mimibulivs, ‘H-L t- "M inth-aJh/h 1‘s sfikkwt to (ma wt, 1%. mt WNW“, 3L1 Mkml W www Mm to buttxnwm mash mm Masha fikhfiwmbvimiflw {LL %—¢kik’~‘lhi'O~N-&A‘ok- M hm J-JM W a...qu M ii" Malawi“ Mu. J. a.” “ -3- This is a replacement question for any one of questions I — IV but the point value is reduced and your exam is still out of 100 points possible. If you elect to do this question you must select a question for it to replace and you must X out the question on the appropriate page or pages on the exam. (15 points) V. Reaction Mechanism. Develop a complete mechanism (to include all intermediates, all non-bonded electron density, all formal charges and carefillly drawn curved arrows) to account for the following transformation. NO 0 HONO -—-————> k HA ‘5 A HA is a strong acid A vb .- _ 4 {\_ I —+ #47-»sz % 4-6-3:}; g Mr WW: 5%, W Aab' J. I ,. .. “£0 “:0 The NO group is an ortho, para director. Explain why this is reasonable. “alum w w. u mama Mia W a}; am?) \qu my ,3 '. 1 ' Ha Wbmaum. «Jr M c n [HEW-s. CI’ g . "‘01,. ' “cheafil' «a @evED’,‘ Would you expect the NO group to activate as much as the NH; group? Explain briefly. m, 1.. Manual...” Mrs mountain.ng mac/.mqkqu , . {’15 “Marie‘s- e—p (“3; El $1 (9 8 LL‘ L\/ .4. ...
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This test prep was uploaded on 04/06/2008 for the course CHEM 30C taught by Professor Kwon during the Winter '08 term at UCLA.

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W06 midterm2 key - Chemistry 30C Name Examination II 10 March 2006 D Brown Develop answers to each of the following questions in the space

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