F07 midterm2 key

F07 midterm2 key - Manashi Chatterjee, PhD. Fall 2007 Chem...

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Unformatted text preview: Manashi Chatterjee, PhD. Fall 2007 Chem 30 B Last Name Student ID Number : m Name Of your TA: Melanie / Eric / Rob / Chris (Please Circle) Total: Discussion Section - Day: Time: / Chem 30 B Fall 2007 Midterm # 2 (50 mins) Friday November 21, 2007 - DO NOT OPEN THIS EXAM UNTIL INSTRUCTED TO DO SO M+2 777/5 page was AOL gram Mamas/1i Chatterjee, PhD. Fall 2007 Chem 30 B 1. Some spectroscopic data for unknown compound (A) with molecular weight 133.2 are shown below. Use these data to answer the questions on the next page. (17 points) IR Spectrum: / “’ " 00101 mag s. : odd number (ml-\vv/ ‘ O F \. N I z , cc no N 00/1 0 H 5 3 c 3 1" N F .m ' - “a. "mmwfim mu _'_ M - M +1 . . . r) h/ Mass Spectrum: The base peak (100 % mtensrty) is at m/z = 54. The peaks at 133, 134, _._,135 and 136 have the following relative intensity (49:2:4922) respectively. 103—l mwflmu meflt has B?" '1‘ H 2 ~ Pregame OF B“ < M. * ' ‘ I; 40*; g + \=\) M- M+1 207 q OF M+l P6“K '3 o ‘ ’ M ‘30 75 100 125 150 175 me ———> 4/1.; « 3-6 l I lHN1VIRam1'3CNMR 3 Carbons 9‘ 3 C 3 x c = 3é % L 3y : 767 1 w _ -1 vet mi were-e r r v 1;. 1:1; at. a I 33 ’ I I 5 :' I 8 a up, ‘‘‘‘‘‘‘ v onc N : I17!- M 1H_NMR Spectrum 13C —NMR Spectrum (proton decoupled) 4 5 = 3.0 (triplet, integration 2 1) ppm 6 = 22’ 25’ 117 ppm 6 = 3.5 (triplet, integration = 1) ppm K9 who l:| Br—CH2»CHZ'C£N “H79 "CHZ" 1H wok ofunSat Mopformz CsHs C3H4 BrN Manashi Chatteijee, PhD. Fall 2007 Chem 30 B a. What is the molecular formula of compound A? (3 points molecular formula and 5 points for reasoning using information from different spectra) IR: “0 'OH l—IIVH /‘NH2 C no [scabs above 4 BIOOUV‘ ) Pray/nee 019 CEC/CEN C 225,0 W4) 0ch Mass : Nitrogen (odd nurn/ocr) Mass sped: 1 133:135 [1:0 Bromine present a (mil) : (M12) r) ‘5; ———3 shows 3CWL’O'"S »CEN@ 0—0/me «H —-9 «wircHz—(lhipl BXC : 36 133—1141 Br 2 7‘? 3 if H ‘7 3 L 01 S ( l " 1) / -' - Frag M: n l' c. Give the structure of the fragment that gives rise to peak at m/z = 95 (2 points) magi: but we (if Manashi Chatterjee, PhD. Fall 2007 Chem 30 B (1. Give the structure of the fragment that gives rise to peak at m/z = 136 (2 points) 2. Indicate if the protons in bold are Homotopic, Enantiotopic or Diastereotopic and if they must appear at the same chemical shift in proton NMR or not. Circle the correct answer. W +0 D, you 35% am, Cmml center (9130mm) -"”"‘“\_ enanHoMer \ .v Diastereotopic Could be Different 6 Homotopic / Enantiotopic / Dia stereotopic Same 5 / ' uld be Different 5 Enantiotopic Dia stereotopic Could be Different 5 Manashi Chatterjee, PhD. Fall 2007 Chem 30 B 3. A compound with molecular formula CngoO (MW: 134.18) shows the following peaks in mass spectrum: m/z 134, 119, 91 (base peak), 43. Deduce the structure of the compound and account for the mass fragments. (12 points) Manashi Chatterjee, PhD. Fall 2007 Chem 30 B 3. Describe the origin of splitting pattern that would occur for the (b) protons of CH3CH2CH2Z if Jab is much higher than he. Use Jab= 3ch (12 points) a C i i CH3CH2CH2Z i Manashi Chatterjee, PhD. Fall 2007 Chem 30B 0H 7. ” 4a. Compound X with Molecular Weight 164.2 shows the following peaks in the mass spectrum: m/z 165 (2.4 %), 164 (21.5 %), 119 (base peak), 91 (34 %). 1 m '1'! M 2' H NMR and IR spectrum are given below. Propose a reasonable structure for compound X. Show your work and explain the peaks in all 3 spectrums. (20 points) CH v 1HNMR: 655 - 1"! )olo’D . k— /. 51: MN ’06“ 94:5 -: Ll" rte-015C : “VI-t : I0 ’I W \ 0 | I I I I ' I I 7 I I | I I ll 10 8 E 7 5 5 4 3 Z 1 0 use—mass ppm IR Spectrum. W (00‘ @ I L, H 1 drawbe 2 dame/ts d1 sabsfiwfijbfiifi) g . S WWI“ 5 (Provide answer on next page) g-flvs faflc was noIr grade/war 7 Manashi Chatterjee, PhD. Fall 2007 Chem 30 B InformationfromMass 2 Even HUMbE" of WV Mr (Hr/WHOM -/~ 69; M+l :2 114% 710' J} C m]z; 4/ PYWcJZré] InfonnationfromIR a 'YLU 70", /'NH /'NHL . “5:0 172.0 pram/v5 I Information from NMR Molecular f onnula Manashi Chatteljee, PhD. Fall 2007 Chem 30 B 4b. Another isomer of oompound X (from qs 4a) shows the IR and NMR shown below. Deduce the structure of the isomer. (5 points) IR Spectrum: U: a l TlflNSHITYHNtEI '4: ' F" " j ' ' ' ' ' ' ' “an 6— 5157 0? lHNMR: \ unsah/alrion \ 2 0,4,4 {gt‘njlc’f __L a i | | | l 1" | l 10 9 8 7 6 5 4 3 Z 1 O HPh-OD- 023 ppm Manashi Chatterjee, PhD. Fall 2007 Chem 30B 5. Using an alcohol of no more than four carbon atoms as your only organic starting material outline synthesis of compound A. You will need more than one step. You can use any suitable inorganic reagents to carry the transformations. (10 points) Kata“; ( W to: made From L0 algal/101$) compound A £1 l alcohol T\O H )tpsb/D at Hgof 0H ~~9Ln§ “LACS Lamina A NHJM‘ 10 Manashi Chatterjee, PhD. Fall 2007 Chem 30B 6. Propose the structure of the products including stereochemistry where appropriate. (15 points) infrared Spectroscopy: An instrumental Method for Detecting Functional Groups Characteristic Infrared Absorptions of Groups . Frequency Group Range (cm-1) Intensitya A. Alkyl C— H (stretching) 2853 — 29 62 lm - 5) lsopropyl, ~— CH(CH 3)2 1380“ 1 385 (S) and 1365—1370 (5) tert—Butyl, ——C(CH 3)3 1 385-1395 (m) and ~ 1365 (s) B. Alkenyl C— H (stretching) 3010 —3095 (m) C=C (stretching) 1620-1680 (v) I R--CH=CH2 985—1000 (s) (out-of-plane and 905 —920 (s) R2C=CH2 C-‘H bendings) 880—900 (S) cis-RCH==CHR 675—730 (s) transRCH=CHR 960—975 (5) C. Alkynyl EG—H (stretching) ~3300 (5) CE C (stretching) 21 00 — 2260 (v) D. Aromatic Ar— H (stretching) ~ 3030 (V) Aromatic substitution type ‘3 (C—H out—of—plane bendings) Monosubstituted 690—7 1 0 (very s) and 730—770 (very s) o—Disubstituted 735—770 (5) rn—Disubstituted 680-725 (5) and 750—810 (very s) ‘ p—Disubstituted 800—860 (very s) E. Alcohols, Phenols. and Carboxylic Acids a 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=C (stretching) , 1630—1780 (5) Aldehydes 1690—1740 (s) Ketones 1 680 ~1750 (s) Esters 1735—1750 (5) Carboxylic acids 1710—1780 (5) Amides 1630—— 1 690 (s) - G. Amines Nr-‘H 3300—3500 (m) H. Nitriles CEN 2220—2260 (m) aAlobrevisticms: s = strong, m = medium, w = weak, v = variable, ~ = approximately. Because IR spectIa 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 difierent compounds. If they give the same IR spectrum, then they are vary likely to be the same compound. RCHER acyclic RCHZR cyclic (X = Cl, Br, I) Chemish'y__303 13c NMR Chemical Shifts nge of Carbon R—CH3 R—CHz-R «211211 (n > 3) R3CH CH3-OR (OH) RCHz-OR (OH) R3C—OR (0H) R-CHz-I R—CHz—Br R-CHz-Cl R—CHz-F R2CH-NR2, RCHz-NR2,C H3-NR2 R2CH—SR, RCHz-SK CH3-SR (SH) R—CsC-R RQC(OR)2 R—CEN C=C Aromatic C=C RC=O(OR) R&O(OH) RC=OCNR2), Rc=o(NHR> RCHO R2C=O Chemical Shift 5 10-40 15-55 20-27 20-60 50-80 55-90 65-100 0-40 25-60 35-80 55-90 25-65 20-35 65-90 90-110 115-125 100-150 110—160 160-180 175—185 165-180 180-205 190-210 ...
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F07 midterm2 key - Manashi Chatterjee, PhD. Fall 2007 Chem...

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