Summer 2007 Test 2

Summer 2007 Test 2 - 2 UGO Chemistry 228. Summer II , Dr....

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 UGO Chemistry 228. Summer II , Dr. Wells, Test 51 Name:_ _ KC l _ __ Score ——— Opscan * 9024 Jr essay = * 90 f 24 + l0 =LQO complementary [UL one! i + I 5 7L (nm) color color N‘ k UK MW < 400 UV 09,...» 5c“- C“ I 425 Violet yellow I / 470 blue orange 0&2 ’lmm‘)‘ MM CHJ’ ' 520 green red 570 yellow violet 620 orange blue 680 red green >700 1R : Useful Equations: E = by : hc/2. h 6.626 x 10 '34 m .o: 3.00 X 1081n/s__=____?w HONOR CODE: You are an adult at a Christian university. You are on your honor to neither provide information to nor acquire information is? \ from other students during the test. There Ef‘should be no talking or looking around during the test. There should be no use of notes of any form: electronic, paper, or other. Summary: don’t cheat or look like you are cheating. The penalty for cheating is a course grade of “Fx”. “F” means fail; “x” means honor code Violation. Essay (10 points) Consider the electrophilic aromatic substitution of methylbenzene. Draw the areniurn cation intermediate and resonance structures that result from meta attack and para attack of the Cl)r electrophile that explain the experimental observation that the methyl substituent is not a meta director, but is a para director. Be sure to label your drawings and use words in your explanation. [A C 3 CH3 Examine The reaction sequences below for ques‘rions 1 - 6. 0 || /C\ CH3 Cl 1303 Clz/FcCl3 /———> 1 ———>— 2 CH3CHCICH3 / BCl3 3 CH3CH2CH2CI / BCI3 4 l\: kaN 5f YLL(/6~W‘7L he“ HNO3 “112304 H2804 5 *> 6 1 . Compound 1 is most likely Cl OH H CH3 Cl—B —Cl CH3—C—Cl CH3—C:O C CH3—C-—OH 2. Compound 2 is most likely OH CH3 H H (RAH ' l l l c 3CH3—C—Cl CH3—C—OH CH3—L:O CH3— —0 ll dc. (1 Q C— 3 . Compound 3 is most likely Cl | c1——B —c1 a all} CH3— C—Cl CH3—c——H to CH3 4. Compound 4 is most likely CH3 CH3— C— CH3 l CHg— c——<:H3 l CH3 9 5. Compound 5 is most likely W 6 . Compound 6 is most likely N03 3‘” ml!‘ 3m 3 s a “five 1 m “a Cl Cl—B —Cl CH2CH2CH3 QM CH3 CHQCHQCH3 CH3 NOS; :5 N08 (:1—13 CH2CH2CH3 CHg—C*CH3 CH3 1 CH2CH2CH3 CH3— C—H / ® l \ B —Cl Cfla—CI‘r-H l C] CH3 30314 S 0N2 SO3H Consider reactions below for guestions 7 - 9. reaction X ‘ reaction—‘1’ l CH3 H CH3 CH3 (71“; H i t- ——I- 7 l + l CH3 H Cl l3 / (‘11:, H CF3 6—. {tab affix)" 7. The compound 7 would most likely be _ _ .1 C113 { H} '\\\I{ (:1 [1 (:1 in, "7m: ll CH; D 8. 9. Which of the following statements is the most accurate? . . . . . ~. ow; . . ® Reaction Y 15 faster than reaction X because electron no &a cth-cndily Wllll electron \Il‘t 3'3 3 S deficient dienophiles/ I‘ \III. latte-Mg; «,i . 1 QI “W011 dcfigfiéfwi'apidly with B a f”— ? . ‘; I ‘ fi'tfi'lhffim I . q . . a . - ' an reaction X b c crs-dieneophiles react taster than tram— We‘lflfl / E. Reaction Y is faster than reaction X because trithggnnethyl groups slow the Diels-Aldcr reaction. 10. You can determine which isomer ofdiethylbcnzene you have by _ __ . A. counting the number oftriplets near l ppm in the 1H NMR spectrum 1‘ R. number of broad humps near 300 nm in the u\-'-"'\'is spectrum 7\ @the number of peaks near 700 cm"1 in the IR spectrum 1/ D. two ofthese E. all of these relate); one. or“ 11 . Compound X does not undergo Diets-Alder reactions nith ethene beeause__ A. electron deficient dienes are not reactive 7’~ B. ethene is a good lewis acid *- C. ethene is a good leaving group 7‘ E. the s-z‘rmzs conformation of‘X is not obtainable 7k @. compound X is not a conjugated diene X 12 . The methyl orou is an orthot ara director because - r: P P __ _ ’\. it is an electron density withdrawing group ® it is an electron density donating group / C. of steries . t l L I). of induction —.. hr» buff [5 o ‘7; W" a“ li. of resonance 13 . As the wavelength of electromagnetic radiation increases @ its energy decreases v/ C 13. its frequeneyincreaSes 7Q r; .— ln c, '3 s—r (3. its propagation rate decreases K “F ‘7):- 2/ 1). two ofthese “x I hi. all of these \Yflflmnu‘x}; -.-\;\\\\'\tllll%“ “‘ w t a «tube a r . . l“"\¥\\\u\° 21 fire 14. When an organic molecule abso his}. I 1% _- a _ who Iggéafia fl "\‘t'i'lrf‘w‘ ‘16ng energy tW r ital of higher energy . 1 purple light . absorb purple light C. emit yellow light 3 absorb yellow light G Consider the addition of deuterium bromide (DBr) t0 1,3,S,7-nonatetrene for questions 16 and 17. / / / / __... 16 . Whieh structure below best represents the resonance hybrid of the earhoearion intermediate 1’ 5+ 8+ 6* / 17. Which compound below would NOT he a product ofthe DBr addition ‘? @D / / / i[L\/ Br D Br 18 . iodide is an orthofpara director because _ A. it is an electron density withdrawing group 7< B. it is an electron density donating group 1K C. of steries D. oi‘induetion of resonance 19. inc low yield ofthc ortho chlorinated product can be explained by 0, m, p products A. the inductive effect oi‘carbonyl group B. considering the “para-searching" ability ofCl' C. the acidity of the I] on nitrogen . resonance stabilization by nitrogen’s lone pair he steric bulk of the Ni 1(COJCH; group 20. Which ofthe ions below is not a legitimate areniunt intermediate“? CH3 ii® —+ / / M/ CH3 \/ CH3 CH3 CH3 CH3 H E E H E H H H E A B C D E V715) H /1/ '2, D 21 . The pKa ofthe indicated proton is than the pKa oftypieal H A protons on sp'3 hybridized carbons because m/‘DWN” ink. _ .. \ K g p a . Q less an aromatic ion is formed by its removal B. greater an aromatic ion is formed by its removal C. less a non aromatic ion is formed by its removal D. greater cyclopentadiene is aromatic 17.. less eyelopentadiene is aromatic CE'ClUpemadicnc 22 . Lewis acids catalyve eleetrophilic aromatic substitution by A. stabili7ing the areniuni intermediate B. sterieally directing the cleetrophile C. warping the ring enough to make it reactive 63) forming a reactive cationic species l3. pulling electron density out ofthe ring 23 . Which graph below is the most accurate? KHZ)” arolnatieity g arornaiieity M DJ 4.; UI CN 11 3 I) aroniatieity i n / aromaticity _. M t...) 4:. m (3“ ._. be: u) _;_.‘ U'i O 1'] Ii 24. The hydroxyl group (—OH) is an orthor‘para director because A. it is an electron density withdrawing group ire,th ' / B. it is an electron density donating group /\ '3 C. of steries D. ofinduction L efswfi ’2, ® of resonance //___,_._ a D ...
View Full Document

Page1 / 9

Summer 2007 Test 2 - 2 UGO Chemistry 228. Summer II , Dr....

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online