mit-o1m5-ans

mit-o1m5-ans - PAGE 1 Short questions 1-8; 3 points each)...

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Unformatted text preview: PAGE 1 Short questions 1-8; 3 points each) 1. Estimate the pKal of the proton indicated. A. 50 3.30 D.5 same/‘4 [0&0 H H<= O {W (9 am mzh‘c noanC'H 2. Which of the following compounds is likely to adopt a planar conformation? H angFE/DiALVC ammzbcc’ N Mirabeng Clan/3" \OQ, V01“ $0 8 0 «CA. Pl any Planu’ “On "P\MZV OH V\ A “a none é] *ohgf} “211 MM; C afkr loss &HL0 4. What would be the major organic product of the reaction scheme shown? W 1. Excess MeMgBr —————> Ca 2. pH? OH O 0—? O—(O 0—?“ O—< OMe H Me Me A B c D PAGE 2 5. Which of the following statements about benzene is INCORRECT? A. It undergoes electrophilic substitution reactions rather than electrophilic addition reactions. B. It is less reactive than hexatriene (CH2=CH-CH=CH-CH=CH2) C. The carbon-carbon bond lengths are all the same. a he carbon-carbon bond lengths are longer than ethane. gkocw 6. (3 points) Which of the following is the best description of a "catalyst"? Catalysts alter the kinetics of a reaction by: A. Making the products more stable B. Making the reaction more exothermic @Lowering the energy of activation for the reaction . Providing a source of fee radicals to initiate a reaction 7. What is the approximate magnitude of aromatic stabilization achieved by a simple compound like benzene? A. 5 kcallmol B. 15 kcallmol C. 25 kcallmol 8. Which of the following would be most reactive with Br2 in the presence of FeBr3? E: E: 7?: PAGE 3 (9-11; 4 points each) 9. Rank the following compounds in order of increasing oxidation level (1 lowest, 4 highest) CH3CH20H CH3CH0 CH3002H CH3'CH3 __;___3_4Ll“__'_ 10. 0f compounds I and II, M is the stronger acid and of compounds Ill and IV [I | is the stronger base. | O II G Ill CW IVQ A. | and III B. I and IV C. II and Ill D. II and IV 11. Rank the following in order of reactivity in electrophilic aromatic substitution CF3 CH3 CF2H CH2F L}— MOST LEAST REACTIVE REACT'VE Page 4 12. (30 points) Design syntheses of compounds I and II (15 points each). 0 I ' )1: ‘5“:j §) You may only use the carbon containing starting materials in the square bracket Pay careful attention to directing effects in the key reactions that you plan to carry out :11 on the benzene ring. A clear retrosynthetic analysis will help in planning the synthetic strategy and will be given partial credit if you don't get all the details of the synthesis completed. 0 + C‘ )Lig 2>HD’\E@ K wch \ Q I swam mug \ @A n axes 5‘ #0 73 ; Cl _ l L TARGET (i) {a 6) Ti Sflmu’esw / \ MHCWA '* "El/u L ’ A a 3/ @r2 flier $4 EEK/M3 @va Page 0 Page 5 13. (26 points) The outcome of electrophilic aromatic substitution on functionalized benzene rings can be predicted by considering the effect of substituents on the intermediate formed in the rate determining step of the reaction. HNO3IHZSO4 OMe l For the reaction shown above carry out the following: a) (3 points) Identify the active ELECTROPHILE in the reaction. ® OZ‘L‘NZO b) (9 points) Show the stepwise mechanism that accounts for the formation of a product in which the position indicated by the arrow is substituted with an -N02 group. You must identify all the important resonance structures in the positively charged intermediate 0% H _Q \‘l N Q5 Co H N®Q /N@ // § @n 9’ Ca 0 q, fr %W H N® // \ 0 De Page 6 c) (6 points) Construct an energy diagram for this electrophilic aromatic substitution reaction. On the diagram you should indicate the starting material, the high energy intermediate (6 complex), the product and the energy of activation for the rate determining step. Energy Reaction coordinate d) (8 points) Explain why a single major product is obtained by considering the features of the 6 complex when attack occurs at the other sites on the ring. eom Hate Th ® \3 4/ @ Atlacin 03V ® SMLLKXW lru‘ normed OAWO 1’9 OW‘l‘ l 3 W ~§ ; oMe \i n ® \A u o r 0 (PO H @ -£ gt Page 7 13. (8 points) Explain why the reaction shown affords only a single addition product upon treatment with HBr. In your answer you need to explain, by illustrating the intermediate formed in the reaction why none of the other double bonds in the staring material react and why the bromine adds to the carbon indicated in the scheme. Br <—_—. x H60 PK 0 0 u QT E e fruseroel) -r HBr QroM‘g‘dCi .IW’" V /, . n (9 19k) mk‘wwqr) \ .39 C‘ ‘ \ _. ¢———» {S «.———~ 5 , , ——- monmrg rgp \"fm ‘ V\ NOT 2° _no NSQAE’V‘CG $‘I‘Lg\\lfi~s‘ NONE. OF ADDETHDJC) pQODMCTS To GH~92_EV‘~.)*: {Zqu DOMSLZ wouu’) PE- {3894sz 3"; gEC/‘V‘Di AQOMAT\‘CiT\1 WORLD 13%;. tom.” M3 THE P Roam: T ...
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mit-o1m5-ans - PAGE 1 Short questions 1-8; 3 points each)...

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