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Unformatted text preview: Last 4 digits of SSN Name FIRST HOUR TEST 1 . a. (9) Reaction of 2-methyl-l,3-cyclohexadiene with hydrogen chloride converts it, primarily, to a mixture of two products as shown in the following chemical equation. Write the steps in a mechanism which shows how the two products given can be formed from the same intermediate. Be sure to show all non—zero formal charges in the structures you draw, what is added or lost in each step, and any resonance structures that play a role. CH3 H—Cl c1 CH3 CH3 H /r: U H H H H + H—Cl __.. H + C] H H H H + . 1’2. Pr9duct 1,4 product + H +Cl lunenc _ . + thermodynamic product product b. (2, 1 each) Identify which product would have been formed from a 1,2 addition and which product would have been formed from a 1,4 addition of the HCl. c. (4, 2 each) Identify which product would be expected to be the “kinetic product” and which would be expected to be the “thermodynamic product.” (They may or may not be the same product.) 2. (12, 3 each) Below are shown three Diels—Alder reactions. In each, either the product or one or both starting materials has been replaccd by a box. In each box, write the structure of the chemical that correctly completes the equation. Not activated for Diels Arrows show reverse of bond formation by the Diels- Alder reaction First Hour Test (b), Chemistry 302. Fall 2004. Page 1 0” pages 3. (9, 3 each) Give the correct IUPAC names for the compounds having each of the following structures. Be sure to use the terms ortho meta and am when a ro riate. OH /CH3 H0 NHZ Cl I C\H a. b. c. CH3 0 CHZCH2CH3 a. meta-aminophenol b. 4—chloro-5-hydroxy-2-propylbenzoic acid c. 4-iodo-2-isopropyl-l -nitroben26ne 4. (20) The ester group, —C(O)OCHJ , is a deactivating meta director that was listed but not discussed further in class. This problem will allow you to explain why this group is a meta director using an analysis similar to the ones presented in class concerning nitration of methoxybenzene and benzaldehyde. a. (6) Draw the resonance structures that could stabilize the intermediate leading to nitration meta to a ~C(O)OCH3. Be sure to show any formal charges that are present on the ring atoms in your structures. 59¢ IOCH3 5-0\ ,ocn3 5—0\ ,OCH3 5-0\ ,OCH3 0: ’0CH3 \C \C \C C 3+ 6+ 5+ G) H H _H+ ‘1? N02 No2 No2 PIX/:13 H H H N02 b. (6) Draw the resonance structures that could stabilize the intermediate leading to nitration part: to a —C(O)OCH3. Be sure to show any formal charges that are present on the ring atoms in your structures. c. (3) Upon examining your responses to a and b, you should be able to make a case that one of the resonance structures you drew in either part a or b above is a more or less important contributor to its hybrid than the other structures in that CONTINUED ON NEXT PAGE First Hour Test (b), Chemistry 302, Fall 2004, Page 2 of5 pages CONTINUED FROM PREVIOUS PAGE hybrid. In the other hybrid, all of the contributors you drew are roughly equivalent contributors. If you decide that the resonance structure you have identified is a more important contributor to its hybrid, draw a circle around it. if it is a less important contributor to its hybrid, draw an K through it. ONLY ONE STRUCTURE SHOULD BE MARKED out of all of those you drew above. d. (5) Circle the letter beside the gig of the following statements that best explains why, when an ester group is on a benzene ring, nitration of the ring gives much more of the meta product shown in part a than it does of the para product shown in part b. A. The (5+) charge on the carbonyl carbon of the —-C(O)OCH3 group causes the transition state (T.S.) leading to para substitution to be less stable than the TS. leading to meta substitution so the meta product forms faster. B. The methoxy (-OCl-Is) group on —-C(O)OC H3 make it a moderately activating artho/para director. C. The +NOZ being added to the benzene ring bearing the -~C(O)OCH3 group is a deactivating meta director. D. There are unshared pairs of electrons on the oxygen atoms of the C(O)OCH3 group which stabilize the transition state leading to the para product so the para product forms faster than the meta product. E. The (5+) charge on the carbonyl carbon of the —-C(O)OCH3 group causes the transition state (T.S.) leading to meta substitution to be less stable than the TS. leading to para substitution so the para product forms faster. 5. (10, 2 each) From the following list of compounds, circle those which would be expected to exhibit aromaticity according to the Hiickel rules. You may assume that all will be planar or nearly so. 16 It electrons 8 7E electrons 14 7: electrons 14 n electrons 14 r: electrons 6. a. (12, 3 each) On the last page of the test are shown the structures of twenty-three substituted benZenes that can be made from benzene or its readily available derivatives using the methods studied in this unit. On page 4 are given four multi-step syntheses, each of which will lead to one of the substituted benzenes. In the space provided at the end of each seguence, write the letter that corresponds to the structure of the product which would result if the starting material shown were to be carried through the given sequence of steps in the order shown. (Obviously, l9 of the structures will not be used.) YOU MAY REMOVE PAGE 5 FROM THE TEST IF YOU WISH. REMEMBER: When a monosubstituted benzene is converted to a mixture of artth and para disubstituted benzenes, only the para product is available for going on to the next step. Other than that, you should assume that the major product formed in each step is to be the starting material for the next step. b. (22, points shown in each box) Within each synthetic scheme the product formed at each step (except the last one) has been replaced by a box. In each box, write the structure of the chemical that would be present at that point in the synthesis. To be correct, the structure you write must be both the product of the preceding step and the starting material for the subsequent step. First Hour Test (b), Chemistry 302, Fall 2004, Page 3 of5 pages CH3CH3 9 3‘ 1) KMI‘IO4 /H20 I/A C-HOH C12 2) H+—- pH 1 0/ Fer J— Cl b CH3(':HCH3 ' AIC13 H2804 / SO3 (fuming sulfuric acid /0 / 1) CH3CH2C\ C. AIC13 C1 H2 / Pd/C 2) H20 alcohol / H HNO3 (conc.) (2) H2804 (conc.) CH CH CH - 2 2 3 Na+ N02 CH2CH2CH3 N CuCN H2504 l)Fe /HCl/H20 _‘— H o /0 oc A. H50? 2 2)Na0H pH 12 (2) @NEN CH3 N \H 0 C12 (excess) d \ ' :S \ F 6Cl3 HO ‘0 H2304 / A Na+ N02' H20 / A —‘ HCI/ H20 0 0C First Hour Test (b), Chemistry 302. Fall 2004, Page 4 of5 pages THIS PAGE MAY BE SEPARATED FROM THE REST OF THE TEST IF YOU WISH I recommend saving it for when your test is returned. Cl C] 0 c1 c1 (I? H C] N\O@ N\Oe A. B. c. D. 9 E. a c1 C1 C1 C1 Cl c1 0 0 0 0 r, H II & c1 c1 C1 N N N “I m III c c c Cl OH OH CH2CH2CH3 RU (J M N- P- c1 CHQCHZCH3 CHZCHZCH3 NI ‘fH3 E? e.) E? C NO CH N.‘ e a CS 2 crfi ® 0 N C O‘CHQCH3 OH Q- R, 00 S. T- CH3CH2CH2 0,,S-0H N02 N02 0 e o 0 0 Eur/0 O H E fir e IuI 9 V C‘OH ‘OH @‘0 Y (9‘0 w x. /CH3 2 ND2 (‘31-! CH N02 CH3 / \ CH CH3 CH3 CH/3 \CH3 First Hour Test (b). Chemistry 302, Fali 2004, Page 5 of5 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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