Exam_3_Green_Key - CEM251/Fall 2010/Third Hourly Exam(page...

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Unformatted text preview: CEM251/Fall 2010/Third Hourly Exam ....... (page 1) (12 points total) 1. Draw the structure of the major organic product of the following reactions: a) 2 pts. ‘ CH—CEC— / CH3 b) 2 pts. CH3CEC—CHQCH3 C) 2 pts. 0% CH— CEC- CH3 H2‘ Pd’ / Pb(OAc)2, Quinoline CH3 d) 2 pts. CH3 Pt —|> 02 (deuterium) H show stereochemistry 2. to co co CHQCH3 CH: CH2 CHZCHS 6) Which of the above olefins (A, B or C) has (E) (2 pts') the highest heat of hydrogenation? b) Which of the above olefins (A, B or C) has CEM251/Fall 2010/Third Hourly Exam ....... (page 2) (15 points total) 1. Draw the structure of the major Organic product of the following reactions: a) 2 pts. CHZCHZIOH Crog, H2504 . (excess) 0 b) ll CHzC—OH 1) LiAlH4 —D- . (excess) CH3 2) H30+ C) CH ('3 H _ 2 pts. 2 pts, 2 1) NaBH4 , H35“ 16“ 0'03. 0* ICOLH —> 2) H30+ 6 H2304 @ (excess) d) . 2 ts. (I)H CO HCI P r'd‘ne p r ' a y' I CH3CHZCH2—CH2 3 - (p00) 2pts. ‘ e) \\ /O\ /H \\ /O\ O + \/C=C< —> + H CH3 CI CI 4. Draw the structure of X in the box based on the following ozonolosis results: show slereochemistry 9 E? CH3C—CH2-C—CH3 + O d) _ CEM251/Fall 201.0/Third Hourly Exam ....... (page 3) (6 points total) H0 5 H ., CH3 H0 '. CH3 H0 2 CH3 Inuuo H K _/ """O H 7 <"m, H CH3 H CH3 . H CH: OH R 5 A B C Which one of the above (A, B or C) is the product of the following reaction? 2 pts. ois-2—butene fl. NMO 6- ( ) O O ' II I . /H | — /H R—C—N\ 4—h- R—C—N\ amide H H The above resonance is important in the description of an amide. Such resonance is clearly not possible with the carbonyl of a ketone. O ) g RI \R Based on this, would you expect the IR carbonyl stretching vibration of a ketone to appear at higher or lower frequency compared to an amide? 2 pts. lJn‘ 3 h em 7., The presence of an M+2 peak of roughly 1/3 the intensity t 2 ms- of the M peak in the mass spectrum is indicative of the Cl" low ”Q” presence of what element? CEM25‘l/Fall2010flhird Hourly Exam.......(page 4) (18 points total) 8. (2 pts. each). Indicate the expected multiplicity (singlet, doublet, etc...) of the 1HNMR signal for the indicated protons (assume equality of all coupling constants): 0 ll . a) CH3CH2CH2c;—oc;H3 d) CH3CHZOCHZCH3 l l b OH <—' . 2pts. 9) o 2 _ ) l 4cm 9W1} CH: \CHF 5 Fugleji' pts CH30H2 I _ I (slow exchange) CH2 /CH2 ‘0 Br C) H H ‘— 5'. M ‘93? 29‘5- f) >CH—O—CH “3‘ " cm, 3- M. H H Br S Q P *_ ef+ 2 pts. 9. (2 pts. each). How many signals are present in the 13CNMR spectrum of the following? . CH3 0 ii a) CHzc—H 0 CH CH CH3 3 QCCH3 2 pts. 2 pts. 2 pts- CEM251/Fall 2010fThird Hourly Exam.......(page 5) (16 pts. total) 10. Deduce the structure of the following compounds: a) IR: 1740 cm" 1HNMR; 6 1.23 (3H, triplet) 6 2.00 (3H, singlet) 6 4.02 (2H, quartet) 12‘: (3th. 3 Elk/J § 9 W (.113 O C c1)» 16.1.13 C4H502 (4 p15.) IR: 1710 cm'1 ‘HNMR: 6 1.2 (6H, doublet) 6 2.1 (3H, singlet) 6 2.8 (1 H, septet) 05H“) O (4 pts.) IR: 1750 cm-1 1HNMR; 62.1 (3H, singlet) R 6 2 8 (2H triplet) 3 6 4 2 (2H triplet) 6 7. 3 (5H brad singlet) Q54; CH. 01 @244 Sal-161i: (fi—vEWC‘VFLC—M " . OCH—3t 3 ‘ C10H1202 (4 pts.) d’_ (”H‘- Lc-H' ___ §_‘b (*Mgé’ 3 EL lR: 1709 crn‘1 ‘HNMR: 6 2.2 (3H, Singlet) 6 3.5 (2H, triplet) 6 4.5 (2H, triplet) CEM251/Fall 2010/Third Hourly Exam ....... (page 6) (33 pts. total) 11. How many constitutional isomers would be formed by monochlorination of the following hydrocarbons? CH CH CH _ 3 a) \3 -CH CH -CH/ 3 CH3 /CH 2 2 \ b) C) CH3CH2CH2CH2CH3 CH3 CH3 3 pts. 3 pts. .3 pts. 12. Draw the structure of the organic product of the following reactions: a) 3 pts. major monobromination product 3 p15 4-m 3 p15 -3 p15 carbon radical intermediate major mono— bromination products c) 3 pts. HBr ' _ (No peroxides) CHZCHa d) 3 pts. fl ——I- CH CH peroxides 2 3 8) CEN _ CH _ C peroxrde 2_ —> \CH3 A ‘ an olefn polymer 502 Table 13.2 __ ;. lmportantiR APE‘HEYO“; Bondtype O—H N-H C-H - Csps-H - Cspz-H I - Csp-H Hi “I 0000 H 0020 © I Approximate V (cm‘1) Intensity 360043200 strong, broad 3500—3200 medium ~3000 3000—2850 strong 3150-3000 medium 3300 medium 2250 medium 2250 medium 1800—1650 (often ~1700) strong 1550 medium medium 1600. 1500 Chapter 14 Nuclear Magnetic Resonance Spectroscopy _r__-.-' -. -""" "'u Table 14'] Type of proton Chemical shift (ppm) Characteristio Chemic'al Shifts of Common Types of Protons 0 .9—2 ~09 ~13 ~1.7 1.5-2.5 2.5—4 . '---~____-- ‘—"1. :..-..-~. Type of proton Chem \c=c/H épzf. \ HO-—H or R~N-H icai shift (ppm) 4.5-6 5.5—8 9—10 10—12 ...
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