Chem 109b final review answer key

Chem 109b final review answer key - Casside CHEM 169B Fina}...

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Unformatted text preview: Casside CHEM 169B Fina} Review-i Please remember that this is NOT all of the material you shauid focus on, this is just practice! K have tried to give a fair representation of the materiai covered but you should revisit the practice midterms I made‘previously as We}! as your midterms and his practice packet. Also look through your iecture notes for ideas for conceptual problems and structures you may have to draw. USE THE BOOK for more practice and guidance with mechanisms. Part E (Materiai from Mifiterm E): 1. Write a structure corresponding to the names below: (a) ciaoid conformer of trans, trans~3,5~octadicne (b) 6—bromo~7~mcthyl~2-octyne W Magma; (c) 3~chioro—butanoi (a) 3-nitro-benzaldehyde Ci 0 - MGM N z\©fi\H 2. Complete the following acid-base equiiibrium equation. (a) Labe} the pKa’s of the wide in this reaction (b) Show with an arrow the direction in which the equilibrium lies. (0) Give the equiiibrium constant, K. (d) Give AG for the reaction at 25°C. Some pka’s that might be needed are: CH3CH2CCH (25), NH4+ (9.2), H20(16), CH3CH2CH2€H3 (60), NH3 (3 5) K = me?” = tom“: 10"“ AC1 = ~l.‘~HogK =~m log 00‘“) = ~e H («:0 :2 1* ISA b! all NH3 W CH3CHZC-fi—CH + :NHZ owa”35 Pm”25 _. .93 cHagHzCr—C. + (6) Draw the 3—D orbital diagram for the aeetylide ion formed above, label the hybridization of the atoms and explain in words (and effects) why 1—butyne is a stronger acid than n~butane (pka ~60). Due 41: W Mbvids‘zwfimnw em?) \“bu’i’ym is as aw act’s! WM “'4” ' H aware‘s’mbkz= “flaw-C 5.... C39 Vs. CHBCHLCHZ" smonaeapvm) @‘9 (i or 1:4) 60:64) j Q2;wa W " S? . . bum TN WWW base o4: i~m+ym WW “3”” OhMBCf-(e'PW) ‘” om 5P 0' ‘ (SD 0/0 $0th} *so :‘+ is W mom Cissy-3+1; W nucuusflhi‘s mam “L mm (Saba. W W Omit/(94134 30m 041 m—bui-ane (5P3... 259/05), 9": 3*. Which of the following reactions would occur fastest (F) or slowest (S). Explain why. ‘WHYI? - ‘ . ‘ 0+ . HEV‘ . . we)? 1 analch (a) 1. W” WNW-P WH ® MW Ham W @5ch 3r I’M/(gut. 20:»ch is mmSl-aJala, .. Her ' H 59 law Mmrfim 5W“ Evans— ”. M M We, @ pomi frgna'pLL peed/FMS MW, v” CEPP (b) i. HO! M C" CF) 3» mu‘cfomd w““""’ + W MW W 2" .. H a 0* Mu‘e we (low 11' W Wu» Waih‘m Shae— / \fiw/ .1... W @ Maw mi'uzka _. 01 0* EPPPW‘m MW 4. Designate with the apprepziate letters which of the following molecules are Aromatic (A), Antiaromatie (AA) or Nonaromatie (NA) 2» m o l \, a) N I A A NA H A A 5. Circle the molecule that is the most stable in this set of compounds and put a square around the one that is least stable. Underlinethe molecule below with the $33351 wavelength UV— visible absorption band and explain how this relatee to the energy difference between HOMO and LUMO. (Use an energy diagram in your explanation). Nam-i- REE. ' tow/\MM QM “Dam yam. longest 2 will Wespwvi em mpound with-HM lonW omit} oct— miuflqj-Ed double bands.m HOMO and LuMD W W5 will knew (1034154 'l'DW (WMJLH W9 A) and awm Wespmol ~hz>~Hma Lmyfi 2 abswlaect. 6: Write the main precinct {3), or missifig reagents, for the fofiowing reactions, showing appropriate stereoehemistry where needed with dotted line and wedges. a)? pmsvu 0 I Q o g owév .mwflfle 9‘ z; r: + E‘: C CHonlo H Part fl. 1. Write the structure eerresponcfing to the names below: a) N,N~dimethy1aeetamide b) ethyl 3-bromopentanoate . I? .. Br 3 CH ’c—‘N'r’w WC‘OW‘LCHz 5% 9.6-3 c) trans—2—3»propy1eyelopentaléecarbexylie acid :1) a phosphoniure ylide ,phEP r“ CR2 . 0W (fir- 2. Which of the following reactions is the fastest (F) and which one is the siowest (S). Drawthe products and explain Why. - I Products Why? : . Hzo 0H , no 5+9yfc hMmflm In a) 1- HKH 1430+ H+OH Mnsm'm 5m “Fwd a“. R H and fireman 09' mainly: +3 ” “’0 H 3' swm mafia/1‘ “V‘” M 0 ‘ H. CH3 0H3 H; 0"” ma/(EE‘CJHE © mm?ng .1. I O V H m. Hfimg “gig-E? H1043 @ pro/$6 3 0H Sloww blc 5mm ,Wm WM saw-m w WW AoH a, SH'I‘M W M“ p . _I , \n 10) LC”; Ola/[3 W CH3 - Owe/W's * HOW?) MMSWC m 3 + @ mm #56 LOSE“ H OH ‘1 CH3jLN-’C+Qg QC??? wstoW'J/HE + N 3 3 T5 and PM“ \U. WWW (5” MW) MAMWM wi—i’hdva (Ram WWS‘ -- .63 £94 £9 3. Write the main product {5) or missing reagents for the following, induciing stereoohemistry where appropriate. Label the product with the name of the compound (or type of compound) and reaction (so you can go back and reference where you need help). H0! Ca1+) A ( 0»; “Wm nan) no MM! 0, W) '0.ch Q “30*0-1‘) 55.0 I 90% sum Ha (Camqu am!) How? awvmfim DiBA’lH 8V Ewfigcmflz Am \. N I THE: “Woe dawn“ C) Pth: «W9. Phgp:m%M3 J3me (R6 w 2! “g “L! . “ (WM) CH WLW3 RD (Mg L__.___. b g n 1 Hair? 0H3 ebwffio (M3 9th28’“, vfifph + HUGH?) (9m9nwol) 2.) How Ph ‘4 H 9 £3 a + W H3a+az+. a , {430+ + Ho/j (MR/0mg gym/gs) LV% 63 W 0H 69 NH4 9» mg, MEN #7» z. ) + H0 0 H W-w—v MOW) > 0 Part- 111. (“10032003 1. Write a structure for the following: a) B—kcto~ester b) a haloform c) LDA Q d) an o,B—unsaturated aldehyde 0 Li bk; K “W 3 >~_ N '< MM F3 a< fl, X=Cljgy~JI 2. Match the following pka’s with each of the following? (pkas: 13, 30,- 17, 9) (a; a com/buns) a) an aldehyde 1)) a B-dike’cone c) a B-diester c) an amide Ki NH1 (3“ ® @ £0 Eeri- /w.? 3. Write an equilibrium for t e acid—base reaction 13 con methyl acetoacetate (pka ~96) and sodium hydroxide (pka~ 16). (21) Label the pka’s and strong acid (SA)fweak acid UNA). (b) Use and arrow to Show the direction ofthe equilibrium and find the equilibrium constant K and AG for the reaction. K: 10194974" : i053 Ag =1 4.4 [93 [ms-3) 2 —I.'-i (523') Jmows + @QH Moo—ts + “£ij ll PMMIOF} e FERN”; SA WA Wu» (0) Why is the pka of methyl acetoacetate higher than 2,4wpentancdione? .wghw \o/c M . m furrow-mfg WW; grevcxis'mfg OOH; VS' go greaH/V SW! WW WSW x OH 951 Lama, 904V» TSWWfioi biu‘ ~0H Jul—mm awful :3. 9613060?th L 10031) W oil lmpcur MKOW M ca 'é 4. Write the main product(s) or missing reagents for the followihg: ‘ I g + 2. N 6r MOH Ecd‘ommt WW’ ctr/ML 53m mammal W9— 0 e)\jLH WK, fiat-1..., W? + wkfib Mia-Mi 0 900143 00+ 00 0043 k W 5 ) d 0mg lMgG” - map, cmism (Evadinng egg/35 Sim addih‘vn 3) CHgoi/ “3°fi°wg “30M 00“; 0W“ WWWM 6(3ng m 90H 0H 39%» Maw—9 X5 H10 ‘ b 1 WW a Wm WM *) I) ,3"0H a. , L W ‘ 66 ~ [Him 2.) H0 ,Hzo 4,, H291, Lobsmm AW“ H H LISP» H S NH3 CH4) W H U K) $0”; .) LDA /THF W W Z) Agv \W ' ? 2;) OH; a?” fi ’9’) Omit/OH CFO?) OJKOH (mfdmm) HLSUH FCC H CHI/ml. (mfdwf'l‘m) 0 a» I W‘) /\/0H hamstng 0*,JL§,—e0°c '\/$LH DMvQNm \fioH 4- Ag” 2. Trie‘l‘i’wlami'm 13H3°+ LfiO/ONH ii 0 . \LMhDu,HD” JKDH + H 0 P .9 WM V0 0 Md. 0H HIGH (:4ka 2" w) W w 0 Hr!!- DOSOL} " - {H I HMS/5r. “JIM; Z) Hvo'z. H 3am /I ‘ 005!" $83C 2) Z‘n,HZO @ WC AW 1 Doha/145m H 214mm! dim”) puffi‘afa, “fiat ar m OwQHEM Reyfew 1g OVER}. GOOD LUOLH WWU ...
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Chem 109b final review answer key - Casside CHEM 169B Fina}...

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