30A_W05_Midterm_1_Key

30A_W05_Midterm_1_Ke - Total Score Student ID Number Circle the name of your TA MIKE ROB Discussion Section Day Time Chem 30A Winter 2005

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Unformatted text preview: Total Score Student ID Number: Circle the name of your TA: MIKE ROB Discussion Section - Day: Time: Chem 30A Winter 2005 MIDTERM #1 (50 Min) Weds February 2nd INTERPRETATION OF THE QUESTIONS IS PART OF THE EXAM — DO NOTASK FOR THE QUESTIONS TO BE EXPLAINED TO YOU ONLY ANSWERS WRITTEN IN THE BOXES PROVIDED WILL BE GRADED ***DO NOT OPEN THIS EXAM UNTIL INSTRUCTED TO DO 50*“ um n a 06 n ab Total 5 / (f‘ \ —l O O "A common mistake that people make when trying to design something completely foo/proof is to underestimate the ingenuity of complete foo/s" - Douglas Adams Q1. (a) Using LINE FORMULAE draw the four different CONSTITUTIONAL isomers with the molecular formula C4H9Cl (2 pt each) (b) Name each isomer using SYSTEMATIC naming rules, in the knowledge that chloro groups are higher in priority than methyl groups (i.e., they get the smaller positional number if a choice has to be made) (2 pt each) A B Name Name 1-CMLoao-Q-MeTu‘ILPaoPM€ 2-¢uwao-1.METu~tLFaoPm\€ (c) Which of the constitutional isomers of C4H9Cl (A—D) has the highest boiling point (79 °C)? (2 pt) E] (d) Which of the constitutional isomers of C4H9Cl (A—D) has the lowest boiling point (51 °C)? (2 pt) E} (e) One of the constitutional isomers of C4H9C1 can exist in two stereoisomeric forms (enantiomers). Draw these two non-superimposable mirror images in the appropriate boxes below. (2 pt each) (R)-Enantiomer (S)-Enantiomer C L Q2. (a) Above the 0° line on the graph below, fill in the groups on the Newman projection of the MOST UNSTABLE eclipsed conformation (E1) of 2,3-dimethy1butane, as viewed from C2 to C3, i.e., C2 is in the front. By rotating C2 (and the groups attached to it) 60° CLOCKWISE, the first staggered conformation (SI) is reached, fill in the groups on this Newman projection above the 60° line. Continue this clockwise rotation, and complete the Newman projections of the other eclipsed (E2, E3) and staggered ($2, $3) 2,3-Dirne’rhylbufane conformations. (2 pt each) Me, H Ma, “Q— Me Me. 04¢ Ma ‘* tile—Me W «Ach M” Me.“ H MgHe’ E1 E2 E3 E1 : Me , Me, I Me, I H Mei Me a He He— H Me, H Me; A Mai Me M» i S1 5 $2 5 S3 > E E i i i m : : : : : 3 : : : : : = : : : : : "J : : : : : O 60 1 20 180 240 300 360 Dihedral Angle (°) (b) Draw short horizontal bars on the GRAPH ABOVE (like those drawn for E1 at 0° 8: 360°), indicating the relative energies of 81, E2, 52, E3, and S3, and complete the graph by drawing in a curve that shows how the energy changes relative to the dihedral angle. (6 pt) 2 Q3. (a) Methylcyclohexane can exist in two different chair conformations, one of which is 1.8 kcal/ mol more stable than the other, i.e., the A value for the methyl group is 1.8. In each of the two boxes below, draw in a bond to one methyl (CH3) group in the appropriate position. (2 pt) CH3 —1.8 kcal/mol Least stab/e chair- Most stable chair (b) Chlorocyclohexane also exists in two different chair conformations, one of which is 0.6 kcal/mol more stable than the other, i.e., the A value for the chloro group is 0.6. In each of the two boxes below, draw in a bond to one chloro (Cl) group in the appropriate position. (2 pt) CI AG = 1 —0.6 kcal/mol Least stable chair ~ Most stab/e chair (c) For trans-1-chloro-4-methy1cyclohexane (shown below), draw in bonds to CH3 and Cl groups as appropriate, to indicate the least and most stable chair conformers. Assuming that A values are additive (and hence subtractive if necessary...), predict what the AG value will be. (6 pt) Cl AG = - :2 .4 CH3 _ Least stab/e chair kcanOl M ost stable chair Question 3 is continued on the next page... 3 (d) For cis-1-chloro-4-methylcyclohexane (shown below), draw in bonds to CH3 and Cl groups as appropriate, to indicate the least and most stable chair conformers. Assuming that A values are additive (and hence subtractive if necessary...), predict what the AG value will be. (6 pt) Cl AG = 1 __‘ m a v as — L2 CH3 - k l/ l - Least stab/e chair ca "'0 Most stab/e chair (e) For the isomer of 1-chloro-3,5—dimethy1cyclohexane shown below, draw in bonds to CH3 and Cl groups as appropriate, to indicate the least and most stable chair conformers. Assuming that A values are additive (and hence subn'active if necessary...), predict what the AG value will be. (8 pt) CII--- ~3.¢ CL .. Least stab/e chair- kcal/mo Most stab/e chair OLD H Cyc/ohexylcyclohexane (f) In the appropriate boxes below, draw the three most stable conform- ations of cyclohexylcyclohexane (shown to the right) in which the indicated H atoms maintain an ANTI relationship, i.e., are 180° apart. Once you have drawn these three different conformers, use the box below them to briefly explain your answers. (12 pt) H ex H at H out Most Stable Conformer' 2nd Most Stab/e Conformer 3rd Most Stable Conformer' Explanation: ALL. RIMCrS Age CMAMLS, BUCT cm BC: C,on acne—b mpg Return)! I GQuATomA L. —~ erAToQAfi L, \5 MOST s-rAece/ BXUQL-AXIAL. ls crops—r STABLE, AND erATomAu—Axmu Fflccs \MBETVDee-m) Q4. (a)—(f) In each box, write down either R, S, or X, indicating that the carbon atom to which the arrow is pointing has the R configuration, the 8 configuration, or is (X) not a stereocenter. (3 pt each) (g) The two enantiomers of limonene have EXACTLY the same boiling point, melting point, vapor pressure, and appallineg bad solubility in water. Interestingly, one of them smells of oranges (the R isomer) and the other (the S isomer) smells like lemons. Explain why they smell different. (4 pt) ‘IourL nose V5 CJ—HQAL‘. “ OR WT Lbfifi‘? THQ: \SMeu,‘ aeceptrozs N16. So, Ac—tuouG-H no ani AUMQAL. evaLomMNf (Me, WNP, H10 “6, THE (9.3 LL33 \SoMeaS Hme exam—wt rme 5mg ?(Zo?€:0—’T\e$i HQ A wxzm. emxhethNbJ-r THet RESPOND 091') DWCWCf (R)-Iimonene (S)-Iimonene Q5 (BONUS). Explain Why trans-decalin systems cannot undergo chair flips, whereas cis-decalin systems can (examples are shown in the box below). (5 pt) Trans-deca/in |:|'> CH3 6?, cis-decalin {2:} m‘ ' v HHS NOTHmfi— TO DO wwq Tue MQTyNLCrROOP LOOK AT THE pep-r 2N0 0? Terms puma, u: ‘IOKL FL|P \T, flow G—E‘T ‘THG. ELLQWHQC» / CH1 6\ These Nae MOQ 'Too Fag O43 APART To (54. sca‘ooeb on! B"! (aw—‘1 Tfl {~4er CH,~ 63/ Gaout’s, 50 TH& Rqu— 0:0 The KtG—HT LOCJLSTHG zucc— m We LQG‘T MD wee. veabA, THE‘l WRIST FLA?, Q34 HOWW: \F V00 LOOK AT C15 DeQAuu), Tue LEFT (lnNGr comes 0??? :w THe AjuALL eQuA‘romAL. Posmom'é, 50 \T Cir-4 FLP, H'ND erA—rorame Beowes Axmw. PrND wag: veQSA... Q6 (BONUS). You’ve already seen that a chloro-substituent on a cyclohexane ring prefers to be in the equatorial position, i.e., the conformer with the C1 in the EQUATORIAL position is favored by 0.6 kcal/mol. Suggest Why in the case of 2-chlorotetrahydropyran (shown below), the conformer in which the C1 atom is in the AXIAL position is favored by over 2 kcal/ mol. (10 pt) 0 Cl AG = U :> -2.2 kcal/mol 2—ch/or'o— Cl Te frahydropyran FAVORED MOT A Steam. EFGQQT’ As THAT m0 FmorL SQUATOR\AL. over; Axum... -— \T \5 AN €:L&QTQOtQ\C_ eE‘CQCfl u 4’— LONE ‘ Pair). ow 01Her é\‘\D<>I<M4¥T\<3I€J fizoM fikLQD ORE\TAL. To ewxP‘N we g) S‘Fnsmxzfi’how é— mph; 0”“ orbi’r0~\ 06 C‘C‘ EOND Cl @ Tms \5 meccao 'THe Moment; QQCQQT, and \M 13413 CmgmA-nom \5 THE. Lme PPuQ 0N3 OX‘iG-em Lxmeo up UQVTH 1w; m-nemm-Acr Q—¢\ o~*’ 0R2>\TA\_ «0 ovazLAPJ. spectron swam.on Le-Abs '1'0 STAexu'zA-noro, ‘Tms CNm—r HAPPEN) IQ THC: oTHeR CHAIR emgkmeca (memo Game-UN) SMmAR To THE eecec—r ‘I‘HAT STaeuAaes “me. STAG-Gm CoMFoeMkT\ct\3 OC e—ma—Ne Luau»; BACK. AT HOUR NOTE$> LED Tms EFFECT 032D To Be eprpmget) D‘s-peaekrufi B‘l Loowwcr AT DIPOLE, 7 MoMe-QTS 0C saw—t WFoW, Bu‘r ’rHe (Lem, REASON) ‘5 THAT Agave no enm- (D ...
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This note was uploaded on 01/23/2010 for the course CHEM Chem 30A taught by Professor Chattergee during the Winter '08 term at UCLA.

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30A_W05_Midterm_1_Ke - Total Score Student ID Number Circle the name of your TA MIKE ROB Discussion Section Day Time Chem 30A Winter 2005

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