2007-SummerC-Midterm-1-Key

2007-SummerC-Midterm-1-Key - Manashi Chattelfjee,Ph.D. Chem...

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Unformatted text preview: Manashi Chattelfjee,Ph.D. Chem 3 OB — Summer C 2007 Last Name Student ID Number : “ Name Of your TA: Rebecca / Eric / Jesus (Please Circle) Total: Discussion Section - Day : Time: / Chem 30 B Summer C 2007 Midterm # 1 (90 mins) Friday August 17, 2007 Be C N F Ne 1.5 1.2 1.5 G 1.8 2.1 2.5 3.0 C Ge S Br KI 1 0 16 1.8 20 2. 2 8 1.0 1,7 1.8 19 2.1 2.5 l l a ' DO NOT OPEN THIS EXAM UNTIL INSTRUCTED TO DO SO Manashi Chatteljjee,Ph.D. Chem 3 0B — Summer C 2007 1. [12 + 15 points] Give the major product of the following reactions. Please indicate stereochemistry and include stereoisomers Where ever possible. Please indicate “No Reaction” where appropriate. MCPBA CHZCIZ MCPBA:me la-Chl oroperoxybenzoic acid K2CI207 ——> | H2804 0 OH ' [oJ M, o [M L Ohd _.___, Lei-om: NaOEt Cl , EtOH Manashi Chatteijee,Ph.D. Chem 3 OB — Summer C 2007 f. HBr (1 mol) —-——————-——-————> o g NaCN O ethanol:H20 h. .0509 7% 02 OH Hirotr O H _._——_—> H —————-> O OH NaH503 ,sz (Indicate the correct reagents that are required to do the above transformation in the boxes) Manashi Charla/jeaPhD. Chem 3 OB — Summer C 2007 i 2. [12 points]. Only one combination of 1k lhalide d QCOX .de appropriate for the er 5 preparation of each of the following ethers by Williamson et ynthesis. What is the correct combination in each case? halide. Should 21. (En In \ ( I r0 0 rt” NHL-.1] hahr'c (2°) b. CH2=CH-CH2-O-CH(CH3)2 Mamas/117 Chatteljjee,Ph.D. Chem 3 OB — Summer C 2007 3. [6 points]. Given the epoxide of absolute configuration as shown, decide which one of the compounds A through C correctly represents the products of its reaction with sodium methoxide in MeOH. \K ‘ . EFL) )ctd€ GPA no 4 n I} 1,2-epoxy-1-methyl-cyclopropane pwtnung e, G‘F mu LLB/(Trilaleflé A N: n O V . k _ , ' f6 (U’bO‘fl OHCKILS less SttvtkfllU [\MdH“ 5“ > OH H3CO HO CH3 0 OH H A c 4. [6 points] How can IR spectroscopy distinguish between the following compounds? Use information fiom the IR table attached at the end of the paper. 3. __OH ha5 a, bi’GCkCigfl. OH 3 (“TC/Hg K. (A f,” ’3 (a c; I ’35 go and |:> \NIH not - will Shem} © gzg of alKem: 4 c [6 20 m I650 rm VS; Cgc of (XlK-tjfle C 2| 00 7 22,00 2‘) (J C '-:-C"'/l:\ 5 O I O _ V5 3m 5 Manashi Chatterz’eeth. Chem 3 OB — Summer C 2007 5. [12 points]. How will you convert 4-pentyn-1—ol to 4-heptyn-1-ol? Show all the steps and reagents required for this transformation. : N66 0'1 [1' (\dd 2 (or b on f; 0 two kmds (If (lucid P'rcfuns m gtm‘hne} wtka U1 (1“ - L1H 2 (ii-t" LjOM (ficx‘flVWOE r . fl 6 {Ti “SC- NQ C, V P r c. h" of N ( of t , I / OH ( st be“ > h q T‘C 4L Ho 6 A ~ “f 9 Na LPE W;\ rterwvi“ i f 't Br Al Lg la h C’ "\ /l RM Zr ( 'L bond (.Q; nfiuHUfl I s; « o Manashz' Chatleljee,Ph.D. Chem 308 — Summer C 2007 6. [12 points]. Identify and circle the correct compound in each case. Give a reason for the observation (less than15 words): /- [Jim 9 a. Methyl pentyl ether 0 2-methy1—pentano evolves a gas after reacting with NaH. 7 \WI/ y \C"‘/\/\/ V3615 m (4 Cuba ‘H I does not" read tot/{w Nat/ /\,/K/0H + [Va H ,,__._._7 \i/OWNCAT + 9/2» T Ow H 15 +he a (461! C prev/Law H0 (4 f 1:, abS’H’aFr'd by H — g- (1VG‘9 (“n 90.5 (7 U0! UPI/"t ' b. Dimethyl ether or isopropyl alcohol is more soluble in water. ’ OH h a s C H 1% a (t (a n hydmye 1*) bond" more Polar ///<€ dissoivv“) (die) .' SC;/Libi(i with H) Lb," 2L (“30 L413~O—-LL/‘f3 ls Lr750/ub/e SMKK 'Z" ("’Hm‘r H~-b(*0d H—bond donor) a is [(-53 polar khan A/‘ccho/S //—"‘ m—v-—\._ _\\ c. 2—isopropyl-5—methyl-cyclohexanol or (Mthylcyclopropanolddoes not undergo oxidation with chromic acid. W' o. (“I . . If rY\ ("Hajl ’ LHCID YUP” HO! ’ 5 a 3 KA/C‘CJ’W/ 2 - “(05> “a rt L, under 5' U OKAC-LJHOI’) 2 ("Manhole (jH/(‘f [C 6 forth . DH 0 (C2 ) ,JL —~————-7 '2. alcohol" Manashi Chatteljjee,P/1.D. Chem 3 OB — Summer C 2007 7. [15 points]. Propose a mechanism to account for the acid catalyzed dehydration shown below. Show curved arrows to show movement of electrons and include all intermediates that are formed during the course of the reaction. OH st04 _______. + H20 warm DHV < MC for L t J I a , “a g X eltmmal-rcm ((—______ Manashi Chatterjee,Ph.D. 8. Chem 308 — Summer C 2007 [10 points] IR spectrum A and B belongs to compounds with molecular formula C4H802 and C4H100 respectively. Identify the functional group present in the two /4 molecules and draw one possible structure for compound A and B. / lUD lkflv€c (' \i fl *4)“ l‘W/Ct H U ‘1‘ 2 1R W15" L 3 U“ fisc— (utd g c r ” (2 gtfl m: ~C‘H 1" I Fl / A] Functional Group: [Ester AJ/Oh Possible Structure: u—r—fi—fir-rrfifi.. f—w—vrfi .r—w j .—— — T.— r s ,»—w 9—. E l 4 “100 3000 EDD“ HRVENUMERI .Il 1500 1000 5110 . e the V . A 0 A‘ B] Functional Group: Possrble Structure: w (L." “(U “nth V009, oxygen {3 M i '\ h" (txlttlwl / t‘fhcr 53 h (it s Vi C v OH r u 3000 EDDIJ NHVENUHIER' -n| but? ("l “631+”. ullt‘ m ‘ C' Esp s 'y'\(;- km? 5 r plum“ bum Cl 5 LVWNPKJlndS n_l__._._._r_r__._.._._._T_.—,—.—..—r——.—.——.—.—7——.: I —..— a , . —-.————, —1 “100 15110 1000 EEO “Git (3 O . Aldehyde/k (4030c) ‘ a will have idea] LUflgt’lflinlf'lC‘W 2 0C etheV & a have r -v 9 o r (l H (j Cum Inna h on YW (Li C. (a h e r L \Ntll hove M m n 2.16 Infrared Spectroscopy: An Instrumental Method for Detecting Functional Groups 79 Mates] 7 Characteristic Infrared Absorptions of Groups Frequency Group Range (cm—1) intensitya .A. Alkyl C—H (stretching) 2853—2962 (m—s) lsopropyl, —CH(CH3)2 13804385 (5) and 1365—1370 (s) tert-Butyl, —C(CH3)3 1385—1395 (m) and ~ 1365 (s) B. Alkenyl C—H (stretching) 3010—3095 (m) C=C (stretching) 1 1620-1680 (v) Rr-CH:CH2 985—1000 (s) (out—of—plane and 905-920 (s) R2C=CH 2 (C—H bendings> 880—900 (s) cis-RCH=CHR J 675—730 (5) trans—RCH=CHR 960—975 (s) ‘C. Alkynyl EC—H (stretching) ~ 3300 (s) CEC (stretching) 2100—2260 (v) D. Aromatic Ar—- H (stretching) ~ 3030 (v) Aromatic substitution type (C~H out—of‘plane bendings) Monosubstituted 690—710 (very s) and 730—770 (very s) o-Disubstituted 735 —770 (s) m—Disubstituted 680—725 (5) and 750—810 (very s) p—Disubstituted 800—860 (very s) E. Alcohols, Phenols, and Carboxylic Acids O—H (stretching) Alcohols, phenols (dilute solutions) 3590—3650 (sharp, v) Alcohols, phenols (hydrogen bonded) 3200—3550 (broad, s) Carboxylic acids (hydrogen bonded) 2500—3000 (broad, v) F. Aldehydes, Ketones, Esters, Carboxylic Acids, and Amides C=O (stretching) 1630—1780 (s) Aldehydes 1690—1740 (s) Ketones 1680-1750 (s) Esters 1735—1750 (s) Carboxylic acids 1710—1780 (5) Amides 1630—1690 (5) - G. Amines N— H 3300—3500 (m) H. Nitriles C5 N 2220— 2260 (m) “Abbreviations: s = strong, m = medium, w = weak, v = variable, ~ = approximately. Because IR spectra of even relatively simple compounds contain so many peaks, the pos— sibility that two different compounds will have the same IR spectrum is exceedingly small. It is because of this that an IR spectrum has been called the “fingerprint” of a molecule. Thus, with organic compounds, if two pure samples give different IR spectra, one can be certain that they are different compounds. If they give the same IR spectrum, then they are very likely to be the same compound. ...
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This note was uploaded on 09/23/2008 for the course CHEM 30B taught by Professor Rubin during the Summer '05 term at UCLA.

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2007-SummerC-Midterm-1-Key - Manashi Chattelfjee,Ph.D. Chem...

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