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Solutions8.3and8.4

Solutions8.3and8.4 - Chapter 8 Section 8.3 Answers to...

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Unformatted text preview: Chapter 8, Section 8.3 - Answers to Review Problems 1) NVA% W/ MOO 7A \ 1—1 CH 2) 5’5”), 3 A H CH3 1-1 CH1 J' m (a) or 1-13C CH3 ch H H H 3) Hoaw sAow-m A4: 8 (IMAM alum: 1-1 CH3 H H (2113:2113 II H3C H X (a) 11 C113 cmfim3 (9 44’5"” CHZCH3 H H H H H _ CHZCH3 H H —> / i H CH3 CH3 CHZCH3 H CH3 H3CH2C H H HsHC H CH x’f 3 H H H H CH -10- 5) 1, 2 -isomer; trans CH3 ll CH3 1, 2 -isomer; cis CH3 XXX/CH3 _ equal stability 1, 3 -isomer; trans CH3 CH3 equal stability -11- CH3 more stable CH3 CH3 CH3 6) 1, 3 -isomer; cis (2H3 IIIIH3 The most stable form always has the maximum number of substituents in the equatorial position. a) Constitutional Isomers C) Constitutional Isomers e) Constitutional Isomers g) Identical Molecules i) Identical Molecules H3C @043 more stable b) Not Isomers cl) Not Isomers f) Identical Molecules h) Identical Molecules j) Constitutional Isomers -12- EU: 33: trans isomer more stable OCH3 NH2 —e_ b) mocm Ball : NH2 trans isomer, more stable Br CHZCH3 AU Br :— cis isomer Both forms have about the same stability 331:2;7: O C c “e /“e\ OCH3 0/ OCH3 trans isomer more stable e ) Cl OH —I—. ~— CI NW cis isomer more stable -13- Cl OH CI NHZ HZN f) — CI — cis isomer HzN Both forms have about the same stability CI CH3 CH3 g) CH3 ======== . OH . trans Isomer ; OH 5 more stable 2 OH h) 00%: l _ 3 w, CH3 cis isomer, more stable H3CO 8) a) CI CI b) CI CI CI CI CI Cl CI CI CI CI 3 Clare axial, 3 Clare equatorial 4 Clare equatorial, 2 Clare axial -14- CH3 9) Net'fhor E 01 Z Map -15- Chapter 8; Section 8.4 - Answers to Review Problems 1. Use the model kit to build molecular models of each of the molecules in each pair illustrated below. For each pair, compare the two models by trying to overlap them. Indicate if they isomers of each other or if they are the same molecule? Jame. &m L and H H H {m was H WM (- fiesta/30mg: flare m3 Omar-S 2% and R — and _ {Hercoi so he refs ems/WIOM/ /5amers -a. H 3%. Hand -16- 2) Chiral molecules contain a chiral carbon; a carbon with 4 different groups. The chiral carbon is marked with an ’5 Chiral molecules are indicated with a check mark. H \/ 3‘ \/ * of Am Ads/rd 3) I 3 I 72 33w’1 3 an; AM FaC—IE:CH H3C {1:3 I H Cl Br 31- - Br 0 OH CH3 5 CH HOHECWOH II H BrHEC {31.1320 131‘ch 2 O B Ffu‘fit . r Hound‘s .’ fine-NI'O’WS 1. (II—13 H Bnéfl'la 1-} H3 H3C H 5 Hsc Cl 5 Cl Mag/cu H [’inm-liom/ ISOWFS V 5 E R 3 H SCI—[3 H IIIIIOI-I HO H a CHZCHa W H -17- 4) a) C1>SH>OH>H b) OH>CHO>CH3>H C) OCH3> N(CH3)2 > CH3 > H d) CPhBr>CIlCh>>CPhOPI>C}k e) —C(CH3)3 > —CH = CH2 > —CH(CH3)2 > —CH3 5) All of these molecules are chiral and they all have the same four substituents on the chiral carbon. The easiest way to decide if the representations shown are all the same molecule is to assign R or S to each molecule. Then Q. 6) is already done. Br H 1) H I II I I 5 13+ (31-13 CH F 3 Br E ET EH3 § .5 § 5 ¥ S H3CI--(_:--IBr Fh-(E‘ICHS Hing-um E F1 ? 6) See above for R or S designations. 7) B /\(5 H 5‘1: c1 2 r a) g b) R 1:} HO CH3 lit) }1 {flag Jr CEiF: -18- O OH 5 d) IIIIBI‘ H0 CH3 HQCI ' c1 I3 IIHCI 12C H303 \H 6‘?" s -19- Cl CH1, ”W 72 8) EH3 Br: 91.0% Br—E-‘H _ HW“ — HFé‘C] — H30 CI E A CH3 9H3 CH i H3O .53 |-|3 _ Hill-(IE‘IBr _ HWH — CIF'Q‘H — Br CI 5 B EH3 H CI 7:. H §II4 “acI--c «llibr H3CWH 5' I Br CH3 “av-gat-l 0 cl A and B are enantiomers. A and C are representations of the same molecule. -20- Br CH3 CI CH3 CH3 Br CH3 9) Molecule A is already completed as an example. do top carbon first and assign the priorities 3 11' t' th butt CH3 EH3 camfiilga :ingleflgngiup 9H3 Br * H _ ai—é—m 31.65..“4 H * CI = | :> I2 H""§""CI H—i|::—r::I CH 5 3 CH3 CH3 A CH E 3 thendobuttunj carbon $213 Er-u—{F‘H flist 3:30:3an the Br—fic—H Hh—g—um tmaflnagé'lfglgpggig’m as Hh-[E—Ilm i 4 R; 1 CH3 :> acHa H3 H Br S Cl H S CH3 B CI :3 H R Han“ Br CH3 R C -21- C113 W5“ M- 11 Cl '75 G H “*5 H i a CI I . E . S {artifigfa’fiiarrflemr} H31; m ‘ 11 |."|L'|-"-1' ll ephedrine HAC [bmdmdila tor} 11) Stereocentres are indicated with a 0 The maximum number of stereoisomers possible in a molecule is 2“, Where ”n” is the J¥=lé b 4 = 61+ HI ficf’CI—I H2N O . HNW-Ng H s 0 HO unrethindmne amoxicfllin (an oral contraceptive) {an antibiotic) number of stereocentres. -22- OH Vitamin E = H OH HO OH <0 OH O NH OH O pancratista tin (an anti-cancer drug) -23- ...
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