2ndtestkey - SSN Name KEY SECOND HOUR TEST 1. (19, 6 each)...

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Unformatted text preview: SSN Name KEY SECOND HOUR TEST 1. (19, 6 each) In the spaces provided below, give the correct IUPAC names for the compounds having each of the following structures. a. 3,6,7-trihydroxy-3-propyl-4-nonen-2-One 4_mercapto 2,5'CyC100Ctadiene 3,6,7—triL1ydroxy \ carbaldeh de _ CH H HS 4 y % OH 2 one /C;H3 2 I? H0\ I I ,P/ HO—C—CEg—CH— H 5 I a. 7CH-g3H-CH29H—3C—2C b. 3 ‘5 5 c. C\ CH C/H I \CIH3 \ffle’) 6 3 H 9 38 2 (EhcPhCHZ 26-dicyclohexyl 3-propyl = CH3®CH3 7,7—diethy1 b. 2,6-dicyclohexyl-3-heptyne-2,5-diol c. 7,7-diethyl-4-mercaptO-2,5-cyclooctadienegarbaldehyde 2. (12) Using any one alcohol centaining six or fewer carbon atoms as the only source of carbon atoms for the final product, show how to synthesize 1,2-dicyelopentylethanol. You may use any other needed inorganic or organic chemicals and solvents so long as they do not contribute carbon atoms to the final product. About five steps are required. A list of rea cuts on have encountered while stud in or anic chemistr is rovidecl at the end of the test. ANY ONE ALCOHOL several steps 9“ CONTAINING SIX OR _> O—CHZCHO FEWER CARBON ATOMs (only source Of carbon atoms for the final product) I) combine 0 2 H+IH O CH2C12 \H PBI‘3 ether Mg CHz—Br ——> CHZMgBr ether Second Hour Test (b), Chemistry 302, Fall 2002, Page 1 of 6 pages 3. (20) The digestion of sucrose, or table sugar, involves hydrolysis of a glucose acetal in a process analogous to the one shown below except, that the methoxy group is replaced by a fructose molecule and an enzyme is the catalyst rather than acid. In the laboratory, the comparable reaction can be carried out as shown in the following chemical equation. Write the steps in the mechanism that accounts for formation of the indicated products under the conditions shown. REMEMBER: You may not add any other reagents. Be sure to Show: all intermediate structures that occur in the course of the mechanism, any resonance structures that play a role in the process, what, if anything, is added or lost at each step, and allformat charges on the structures that you draw. (EH3 " H+ /CH—QH " + H—“QH CH2 /0_ + CH3-OH / . CHz—CH2“C\ H It is acceptable to protonate this CH oxygen first then open the ring. (EH3 H /CH—QH CH2 ,. _ \ #06) H CH2_CH2'C\ H \ (1mg CH—éH \2 :O—H CH2"CH2—C\ OH - alcohol leaving group + H+ —> Second Hour Test (b), Chemistry 302, Fall 2002, Page 2 of 6 pages 4.(l 8) lsoquinolines are heterocyclic structures related to pyridine that are found in many of the natural products that are neuroactive. Among the most important to humans is papaverine which is used as a smooth muscle relaxant and from which poppy plants make morphine and codeine. Obviously, the laboratory and industrial synthesis of such compounds is of great importance. Formation of a partially reduced isoquinoline analogue of papaverine is represented by the following chemical equation. As shown, the product can easily be oxidized to isoquinoline by dehydrogenation. Write the steps in the mechanism that accounts for formation of the indicated product under the conditions shown. REMEMBER: You may not add any other reagents. Be sure to Show: all intermediate structures that occur in the course of the mechanism, any resonance structures that play a role in the process, what, yanything. is added or lost at each step, and aliformai charges on the structures that you draw. You do not need to give a mechanism for the dehydrogenation shown in brackets as the second step. (lIHZCH3 CHZCH3 CHZCH3 CQOt 5 \ :[q= \ N: .0 u #- CH/CHz—NHz alcohol (dehydrogenation) / 2 : + H__§H l-ethylisoquinoline + H+ ' _ H+ CHZCH3 (IIHZCH3 CHBCH2 o ®C \ zH \ IH \ ‘ <—> O, N—H Q: :NH2 3NH2 (I16) CH2—CH2 CHZ‘CHz + Nu CH3CH2 CH3CH2 OIiII mill! N’s) H H+ - 14% ‘i H CH :” CH CH 0'; C 3 2 0H 3 2 e H + H N; N“H —-——-> H Second Hour Test (b), Chemistry 302, Fall 2002, Page 3 of 6 pages 5. (12) Starting with one alcohol and one alkyne, each having five or fewer carbon atoms, as the only sources of carbon atoms for the final product, show how to prepare 2-methyl-3-octene that is free of any other double-bond isomers. You may use any other needed organic or inorganic chemicals or solvents so long as they do not contribute carbon atoms to the final product. About 6 steps are required. A list of reagents you have encountered while studying organic chemistry is provided at the end of the test. one alcohol and one alkyne, each 53mm] Elms CgéHCH_CHCH CH C C having 5 or fewer carbon atoms Cfi _ 2 2 H2 H3 3 free of double bond isomers P combine CH CH Ph P Q \3 SOCIZ \3 1) [ 3 ] Cug /CH—CH2"OH /CH—CH2—Ci /CH—CH:P—© 3 2 2 2 1 CH3 (9 CH3 [butyllithium] CH3 0133 NOTE: /CECH is NOT a possible alkyne starting material. CH3 1) 3H3 -THF (3 H-CEC-CHQCHZCHg, ————> {C—CH2CH2CH2CH3 2) H202 I, H20 H AN ALTERNATIVE 1) 3H3 -THF Cl NaBH4 / H20 H'CEC'CHZCHQCHQ, —‘“""“"—> C—CH2CH2CH2CH3 2) H202 / NaOH / H20 H/ alcohol P 1) [1311310] soc1 _ Cl-CHQCHZCHZCHZCH3 2 HO~CH2CH2CH2CH2CH3 2) CH3CH2CH2CH2L1 Q [butyllithium] N ©P :CH"CH2CH2CH2CH3 b. (343 com "‘6 /CH-CH=CH—~CH2CH2CH2CH3 o CrO Cl- C133 3C9 PC(33 C133 /0 CH-CHg-OH [ ] CH—C/ / (:chl2 / ~ CH3 CH3 H Second Hour Test (b), Chemistry 302, Fall 2002, Page 4 of 6 pages 6. Norepinephrine is a chemical found in the central nervous system that stimulates constriction of blood vessels and is involved in the “fight or flight” response to an attack. A neurochemist wanting to study the brain distribution of norepinephrine has asked you to synthesize a MC labeled sample of the compound. Potassium cyanide (K+ TEEN) having 1“C at its carbon atom is readily available. Below is summarized a possible synthesis of norepinephrine that begins with the commercially available, but expensive ($3. per gram), 3,4-methylenedioxyacetophenone. The reagent(s) that would be required to complete each step have been replaced by a box. a. (18, 3 each) In each box, write the structure of the reagent(s) required to carry out the indicated transformation. REMEMBER: If you want to add the reagents in separate steps, you must number the steps. If no numbers are shown, it will be presumed you intended to add them at the same time. For any boxes requiring more than one step, the steps will be ones that are typically found together such as addition of a Grignard reagent to an aldehyde or ketone followed by addition of acid and water. A list of reagents you have studied since the beginning of Chem 301 is provided on the next page. b. (I) On the norepinephrine structure shown at the end of the synthesis, circle the carbon atom that will be a l“C. 3,4-methylenedioxy- acetophenone NaBH4 / H20 9 alcohol /0 C~.. CH2 \ O 1) LiftlH4 fether 1) H+/HZ0 /A 2) H szO 2) NaOH to pH = 12 /O CH_CE 3) NaOH to pH = CH2 ———-——~——-—-> CH2 \0 norepinephrine Second Hour Test (b), Chemistry 302, Fall 2002. Page 5 of 6 pages ...
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This note was uploaded on 04/02/2008 for the course CHEM 301 taught by Professor Morrow during the Spring '08 term at New Mexico.

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2ndtestkey - SSN Name KEY SECOND HOUR TEST 1. (19, 6 each)...

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