1sttestkey - SSN Name KEY FIRST HOUR TEST 1. a. (10)...

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Unformatted text preview: SSN Name KEY FIRST HOUR TEST 1. a. (10) Reaction of 4-methyl—1,3-pentadiene with hydrogen bromide converts it 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 shown can be formed from the same intermediate. Be sure to show any formal charges that occur on atoms in your structures. CH3 Br CH3 \ A A I \ /C:CH—CH=CH2 + H—Br —" H3C“C|3"CH:CH—'CH3 + /C:CH—(I:H—CH3 CH3 CH3 CH3 Br b- 134-3Fldltl0” PrOdUCt b. 1.2-addition product 0- kinenc PrOdUCt _C. thermodynamic product C53 6) C53 /C=CH-CH-CH3 <74\<s>/(:—CI~I=(:rl—(:H3 CH3 k, :Br' CH3 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 HBr. 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, 4 each) Give the correct IUPAC names for the compounds having each of the following structures. Be sure to use the terms ortho, meta, and para when appropriate. CHQCHZCHg; HzN Br H“ HO Br 9 N02 0 a. l-iodo-Z-nitro—4-propylbenzene b. ortho—aminobenzaldehyde or Q-aminobenzaldehyde c. 3,4-dibromophenol First Hour Test (b), Chemistry 302, Fall 2002, Page 1 of 4 pages 3. (10, 2 each) From the following list of compounds, circle those which would be expected to exhibit aromaticity according to the Huckel rules. You may assume that all will be planar or nearly so. : ;\ h E H 0 sp3 carbon N. uinolirie 10 7‘5 elecmms 12 a electrons I q 6 1t electrons H 10 1c electrons 4. (22) The trimethylammonium group, —+N(Cl-l3)3 , is a strongly deactivating meta director that was listed in class, but not discussed further. This problem will allow you to explain why this group is a deactivating meta director using an anaiysis similar to the ones presented in class concerning nitration of methoxybenzene and benzaldehydc. a. (7) Draw the resonance structures that could stabilize the intermediate leading to nitration para to —-"N(CH3)3. Be sure to show any formal charges that are present on the ring atoms in your structures. @9143 cat“; are are CH3 _N_CH3 CH3 C 3 CH3 N_CH3 + CH3 'N— CH3 (9 I - H + N02 (,9 (B H No2 H NO2 No2 b. (7) Draw the resonance structures that could stabilize the intermediate leading to nitration meta to —-N(CH3)3. Be sure to show any formal charges that are present on the ring atoms in your structures. CH CH CH @le3 CH Ehit—in CH git—bit CH @th—ZJH @913 CH3-N-CH3 3 3 3 3 3 3 +CH3—N——CH3 e ——> <9 e -H + N02 H ‘ ' "' @H‘c—p QH w" 1on2 No2 e No2 No2 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 hybrid. In the other hybrid, all of the contributors you drew are roughly equivalent contributors. If you decide 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 X through it. ONLY ONE STRUCTURE SHOULD BE MARKED out of all of those you drew above. (1. (5) Circle the letter beside the one of the following statements that best explains why a trimethylammonium group on a benzene ring causes it to undergo nitration more slowly than does benzene itself and gives much more of the meta product shown in part b than it does of the para product shown in part a. A. The alkyl (methyl) groups on —-+N(CH3)3 make it an activating ortho/para director. B. Relative to benzene, the [+1 charge on the nitrogen of the ~—*N[CH3)3 group destabilizes the transition state 1T5.) leading to gara substitution more than it destabilizes the TS. leading to meta substitution so meta product forms faster. First Hour Test (b), Chemistry 302, Fall 2002, Page 2 of 4 pages C. The *N02 being added to the benzene ring bearing the —+N(CH3)3 group is a deactivating meta director. D. Relative to benzene, the (+) charge on the nitrogen of the —-+l’\l(CH3)3 group destabilizes the transition state (T.S.) leading to meta substitution more than it destabilizes the TS. leading to para substitution so the para product forms faster. E. There are no unshared pairs of electrons on the nitrogen atom to stabilize the transition state leading to the para product so it forms faster than the meta product. 5. a. (15, 3 each) Below are shown the structures of twenty substituted benzenes that can be made from benzene or its readily available derivatives using the methods studied in this unit. Following the structures are given five multi-step syntheses, each of which will lead to one of the substituted benzenes. In the space provided at the end of each arrow, write the letter that corresponds to the structure of the product which would result if the starting material shown were to be treated with that sequence of reagents in the order shown. Obviously, fifteen structures will not be used. REMEMBER: When a monosubstituted benzene is converted to a mixture of ortho 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. (25) 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. O\\ 0‘ O\ z I Br Br A. B C. D E. Br Br C,N <3 Br F. l" 131" CHZCH2CH3 CHZCHZCH3 CHzCH2€H3 Br Br L. M. N. P. Q. Br Br R. Br OH OH B r Br Br Br Br S. T. V, W. Br Br First Hour Test (b), Chemistry 302, Fall 2002, Page 3 of 4 pages N02 N02 N02 Br ._ QB Q C: G l‘ as Br a Br2 HNO3 (cone) J . ——-—--—D- __.—..__...._> FeBr3 (cat) H2804 (cone) "— 2) Br Cl I b C) CH3CHCH3 ' ——-—-—-)- Amhmmg 1) KMnO4 / H20 ma+ "OH —;—————+ -———-——> A 2)H to pH} FeBr3 (cat) ——-— N02 NHZ No2 Q HNO3 (cone) Br2 1) Fe I HCI C. ——-——-—> ———> —~——-—-—>_ H2804 (cone) F6313 (cat) 2) OH to pH 12 B B 1) I‘ (2) - ' Na+ NOZ' fHCl D <——-—— H20 / ooc |(|) NH; d CH3C 0 Br: , I Pyridine FeBr3 (cat) Br (2) T H20 / A Na+ N02' / H2804 — Hzoxooc O \ 1) :CCH2CH3 mum3 Q C] H2 de/C CHZCHZCHB e. —-q,——-——“-> 2) H20 H r’alcohol © CH2CH2CH3 CH2CH2CH3 B p M 4.“; H2804 / so3 —— H20 FeBr3 (cat) First Hour Test (b), Chemistry 302, Fall 2002, Page 4 of 4 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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1sttestkey - SSN Name KEY FIRST HOUR TEST 1. a. (10)...

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