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Exam 2 Answer - CHEMISTRY 322aL/325aL [email protected]— EXAM NO 2...

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Unformatted text preview: CHEMISTRY 322aL/325aL Please @— EXAM NO. 2 print L a S Name OCTOBER 9, 2002 First Name SSN TA's Name Grader (1) (15) __ (2) (20) __ __ (3) (10) __ __ (4) (20) __ __ (5) (15) __ —.— (6) (15) __ __ (7) (05) __ __ (100) Lab Day 8: Time first letter of last name I will observe all the rules of Academic Integrity while taking this exam. Chemistry 322a / 325a -2~ Name Exam No. 2 (1) (15) (A) (12) Supply the information requested for each of the following reactions. (See information in the Appendix, as needed). Write yes or no in the box above the arrow to indicate if reactions (i) - (iv) are spontaneous in the direction shown. In (v) identify t e Lewis acid and Lewis base for the reaction shown. I j m 14' , “1 WM 4}, Mr? S nontaneous? _ + CH3CH3 ‘1 (1) CH3CH2- Ll+ + CH3OH [L 6 conjugate acid conjugate base [0 S ontaneous? W + (ii) CH3CECH + NaH H“ H Z 9 M conjugate acid conjugate base (iii) C6H5CH2COO‘Na+ + CH30H CH CH COOH + CH g M g!» 2. 3 9L conjugate acid conjugate base H (M S - ontaneous? +g/ it — CH3 0 CH3£CH3 + CH3OI-I CH3 + CH30H Identify the Lewis acid and Lewis base in reaction (v). + .. CH3CH20H + an12 CH3CH2'Q'ZUC12 L: S) \ § Eggs _M‘¢ ; d H (V) (B) (3) From information in the Appendix, calculate the pr value of ammonia, NH3. Show your simple calculation and place your answer in the box below. L PKG. + PHD":— "$0 “Fer (Om- Panes [We :4 C(“nCl NH3 an corny Paxr$J prole-I3 Chemistry 322a/ 325a -3- Name Exam No. 2 (7-) (20) (A) (12) Provide systematic or IUPAC names for the structures, or provide structures inside the boxes for the names, as requested below. 3 Mozv H3 $H2CH3 CH CH HCH HCH ' / CH3CHCH2CH=CH2 3 2&1 2 3 1A 3’6! ’ 5‘. Mac, 1 /ml Lf- moi—k l~— \“ l‘vk c K j—les 91:9 ~ 1-_l~_’\<'\-\o( l lrusgckmk feis—Z—methyI-B-heptene 5—methyl-4—hexen—1-ol 1 (B) (8) Provide structures for the following constitutional isomers of C4H100 that are identified by their common or alkyl group names. 0 H \ CH 3‘CHZQH C H3 sec-butyl alcohol (l S’l/Nb ("/4 0K . butyl alcohol tert-butyl alcohol Chemistry 322a/ 325a -4- Name Exam No. 2 (3) (10) Answer the questions below about the following reaction of acetoae‘an 4 CH 6 L ethyllithium. MW \\ omzL‘L- 6(a)!” 3 l d’ th 1 th (Ii- w .+ _1_e__y_e__eL, CH3_(F_CH3 6, . CH3-C-CH3 +i CH3C'H :L1 (A) (4) In the above reaction, O ’L‘ ' the nucleophile is CH 3C3 [All and the electrophile IS CH3 8 CH 3 £46k In the above reaction, ‘ the Lewis base is fight gu IL; 2' and the Lewis acid is CHQE C(ii . / (B) (2) Show the movements of all electron pairs in the above reaction using the curved arrow formalism. Add the curved arrows to the equation in the above box. (C) (4) This reaction is run in diethyl ether, a nonaqueous solvent. Clearly explain (using information in the Appendix) why this reaction cannot be run in water. Write a chemical equation in the box below that describes what happens in water. The vie—3 9mm“) “2ng CH sew zr'L LM- \t ‘ho‘l‘ Survwvx (M Wuhfi The 5K1 0%) ‘l‘kc (ombus c..+( acid MQHgCt-FS \g ’I/ 780, So mafia“ urkk wm‘txr, Pch Ho) \3 / \Jcr‘ C—osT Tkisqg can «youngshi oé T\,\K‘.‘ Attac-t 0? Luann! \‘V‘thmt reaction in water Chemistry 322a / 325a -5- Name ' Exam No. 2 (4) (20) Circle the correct answer for the following statements or questions. (A) The intensity of an infrared band is often given in absorbance units (A) where A = Log (10 / I). When an ir band shows an absorbance of A = 1 at a specified wavelength, the sample has absorbed approximately what percentage of the incident light at that wavelength? « . 1% 10% 50% (B) Potential Lewis acids among the structures below are . . + __ (circle answer) CHS’N’CHB C1130 CHB‘CF'CHB =Br: only 111 land 1v 1 II 111 1v ’1“ I (C) According to the pKa values, the order of acid strength (strongest to weakest) of the compounds below is H1 H20 CHBNH3+ CHSCOOH (circle answer) I>II>III>IV II>III>IV>I PK -10 15.7 10.6 4.7 .- I 11 111 IV IV>III>II>I I>IV>III>II (D) The order of base strength (strongest to weakest) of the chemical species below is (Hint: Refer to the information in the Appendix). (circle answer) NH3 CH3CH26 CH3OH l—\IH2 11>1v>m>1 IV>I>II>III 1 11 III IV II>III>IV>I (E) The changes in standard free energy at 250C for the acid dissociation of three carboxylic acids are given below. According to the AG0 values, the order of acid strength (strongest to weakest) is OOH OOH OOH (circle answer) -III>II>I Cl 4 CH3 OH O II>I>III I>III>II AG° (kcal/mol) 5.4 5.9 6.2 1 II 111 Chemistry 322a / 325a -6- Name Exam No. 2 (4) Contd. (F) Among the structural representations for C6H14 below, the two structural formulas that are actually the same constitutional isomer are CH3CH25342042042043 CH3CH2CH2QHCH3 CH3CquHCH2CH3 (circle answer) " “I IV and v III and V IV and 9'13 qHs C 3 VI CH3QHCHCI-I3 CH3CH2§HCH3 CH3CHZQCH3 CH3 CHZCH3 CH3 1v V VI (G) What different types of H are in the structure below? (circle answer below) 1°, 2° and 3°H land 2°H °and 3° H. 2°and 3° H. (H) The constitutional isomer of C6H12 that has no ring strain is (circle structure below) use (I) The number of different groups of equivalent H in the alkane below is $31-13 (circle answer below) CHaczcriZCHZCH3 two five six CH3 (J) In the alkane below, the C-C bond with the largest barrier to rotation is - (circle er below) 1\—\ 1 3 CH3“CH2—(EH—-CHf—CH3 CH3 3 Chemistry 322a / 325a -7- Name Exam No. 2 (5) (15) The heats of combustion of three constitutional isomers of methylcyclopentene have been measured in the liquid phase: 1-methylcyclopentene, AHC= -896.85 kcal/mol 3-m'ethy1cyclopentene, ABC: -900.22 kcal/mol. 4-methylcyclopentene, ABC: -901.68 kcal/mol (A) (2) In the box below, write a balanced general equation for the combustion reaction of these isomers. , balanced equation for the combustion reaction c... t—k o i- 2 .so L “a 1C6H10 + ‘7 CL (B) (6) Provide the structures for the isomers inside the proper boxes, as requested below. gm ...... - C;;3§;:{§°“- _________ «5Q 1—methylcyclopentene : 4-methylcyclopentene / CH3 3-methylcyclopentene (B) (7) Show how the heats of combustion provide a measure of the relative stabilities of the three isomers by drawing an energy state diagram below. The relative energy scale on the Y—axis is broken for convenience. Draw and label the starting and product states for the combustion reactions of the three methylcyclopentenes. Use an arrow to connect each starting state with the product state and indicate the enthalpy change. Clearly indicate the isomer associated with each starting state, and show the energy differences that measure the gelatiye s‘tabiliéi Uof the isomers. M - - 900 _ EL»:- L «filtyci 92%; xi-Q‘HQ *‘ 3" Oz relative \\b¢i\\u! (I c o P¢m+g MQ * (KS 0 energym : kcal/mol 890 — ¢ MM KCaI/Mc,‘ MW .2 33 The most stable isomer is \‘hzj—fiigcxc! h thji'k K Chemistry 322a / 325a -8— Name Exam No. 2 (6) (15) Rotation around the QC bond in a disubstituted ethane of general structure XCHz-CHZY gives rise to different conformations with different energies. (A) (4) For example, during a complete 3600 rotation around the C2-C3 bond in butane, the potential energy changes as shown below. The diagram shows that conformations A and B differ by 0.9 kcal/mol. Draw Newman Projection formulas for conformations A and B in the boxes. BfiLA * , “<— l l 4.§‘kcallmo \ \/ 3.x kcal/mol 7‘ (I? Relative Potential Energy —--> 0 p 60 120 180 240 300 360 degrees Rotation (B) (2) Briefly, why is conformation A more stable than conformation B? SM CMOMEDmJQiVQ'Pu‘S‘VQ lm‘l‘ToéfiOM'b 51+qu ~Hne ’hztfilnlborwxfi Qfis fimups \m B nous-a. +\~k («u-‘7 exp +\~T> COM-hachcfi-x‘o—n. Pr \3 $4T‘Ck‘wv‘g‘12. (C) (4) In 1,2-ethanediol (HOCHZCHzOI-I), rotation around the C-C bond yields anti 3 and gauche conformations where the gamma than the anti conformation by 2.7 kcal/mol. Draw the Newman projection ormulas of the anti and, a gauche conformation for 1,2-ethanediol in the correct boxes below. Newman projection H formulas of l H HOCHZCHzOH 1,2—ethanediol H A ‘\ anti gauche Problem (6) continues on the next page. Chemistry 322a / 325a -9- Name . , Exam No. 2 (6) Contd. (D) (5) Studies on 1,2-dimethoxyethane (CHgOCHZCHZOCHg) indicate the anti conformation is more stable than the gauche. What factor could explain the energetic preference for the gauche conformation in 1,2-ethanediol? Draw a Newman projection formula in the box below clearly illustrating your explanation. \‘n-\v&r~°\<<m\c~r~ hidrbsi’vx \covkehms occer (m ”FR @ Emudnk COW-forrmxflbm- H‘ b°WA3 3-539}th Sfiughus b1 2 ~s tux/ml. The res-FM si‘BHLcc—Hom Cmcfl— 1)»st b[K \m +kk max-H (Om-Lormbfi'~\\ (Oh?¢’7\\¢g*s 'Cv“ 5*‘Y‘K 5+roC\v-V\. (7) (5) I am a hydrocarbon with the formula C5H8. In my infrared spectrum there are characteristic absorptions near 2900 cm'l, 3300 cm'l, and 2200 cm'l. I react with hydrogen gas over a platinum catalyst to produce another hydrocarbon with the formula C5H12. This hydrocarbon has all three types of hydrogens: primary, secondary and tertiary. Draw my structure in the box below and write my IUPAC name on t e ine below the box. cc 'hCMKMc-A <L\K\(.,\Q: So +ka m vs C3“? ~Q=3_c——H. and 12°C <rv:\ memo-5K Paw-HQ S‘kruukxpg Lom+6ktmg 1‘: 1° 3° H The cowupke‘u s+mg4~cwk Musr \DA QHg—EFCH CH 3/4 Cayuga-«ecu % ct; award/f; MW( 1/],er flu; m @{g’i‘ ...
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