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Write your name and date on the cover page Do not open exam until instructed to do so Name: Date: Exam III hem. 210 Do not open exam until told to do so. Get out your pencil, eraser, and scientific nongraphing calculator. Put everything else under the desk or on the floor. Turn off or silence then stow all electronic devices. You may leave the exam room after turning in your exam. You may not return to the exam room after leaving (until the exam is over). Provide the best answers as requested. You must show work for credit. Label your work should you use the back side of the paper. If you have any questions during the exam, write them on the exam. When told to do so, check that exam has all of its pages. 1 of 12

1) ow does M + form in mass spectrometry? 2) Draw the product(s) expected from this reaction. + N3 2S4 3) Which species gives a 1:1 ratio for the M+ to M+2 peak in the mass spectrum? 4) Where would the M+2 peak occur for bromoethane? 5) Describe the - absorption band in infrared spectroscopy. 6) A 1 NMR spectrum at 300Mz records a signal at 307 z downfield from tetramethylsilane. What would be the position in hertz (z) on a 90Mz instrument? 7) ow many degrees of unsaturation are in this compound: 614 8) The sharp peak at 1700cm -1 in this infrared spectrum corresponds to what type or class of compound? 9) Ignoring spin-spin splitting, how many types of 1 NMR signals should appear from 2? The answer is the same as the number of chemically equivalent hydrogens. 10) Draw a structure having the formula 511 that is consistent with this 300Mz 1 NMR spectrum. 2 of 12

9 2 ()4Si -10-9 -8-7 -6-5 -4-3 -2-1 0 δ, ppm 11) Which compound gives an integrated 1 NMR signal with a 1:1:2:2:3 ratio? Ignore spin-spin splitting. A. 3-chloro-2-methyl-1-propanol B. 1-bromo-2-methylbutane. 3-bromo-1-chloro-2,3-dimethylbutane D. 3-bromo-1-chloro-2,2-dimethylbutane E. 5-bromo-1-pentanol 12) Using Planck s constant as 6.626x10-34 Js and the speed of light as 3.00x10 8 m/s, convert a bond vibration of 2900cm 1 into kj/mol. 13) ow many peaks would be in the proton-decoupled 13 NMR spectrum of? 14) Given that the UV range is from 200-400nm, Planck s constant is 6.63x10-34 Js, the speed of light is 3.00x10 8 m/s, and Avogadro s number is 6.02x10 23, what is the approximate range of photon energies in J for UV light? 15) A spectrometer indicates that a compound absorbs UV-visible radiation at 325nm. Given that Planck s constant is 6.63x10-34 Js, the speed of light is 3.00x10 8 m/s, and Avogadro s number is 6.02x10 23, what is the approximate frequency in z (s -1 ) required for this electronic transition? 16) Given that Planck s constant is 6.63x10-34 Js, the speed of light is 3.00x10 8 m/s, and Avogadro s number is 6.02x10 23, calculate the energy in kj/mol for the transition that occurs at λmax? 3 of 12

Y2 Y3 0.15 450 20 0.1 525 25 2 0 0 800 200 0 0 6 2.5 Absorbance 200 300 400 500 600 700 800 Wavelength (nm) 17) What type of spectrum is this? P NaB4 4 r3lial, 22 LiAl4, 18) Predict the primary product of this reaction sequence: 1) Mg, ( 2)72, NaAl Acid 23r 23 2S4, 2Base Fe3 Fe3 Al 3 F 40 NaB4, 2 2S4, 100 NaB 3+ Ξ F 2 + NaB4 + N E N2 Py+, r3-, 22 3, 22 Pyridinium r chlorochromate P 2 N2 + N N2 Pyridinium chlor Py+, r3-, 22 Li LiAl +, 2LiAl 3 4, ( 2)23 3 4 2 2) 3) Na2r27, 2S4, 2 19) What is a typical minimum number of conjugated double bonds for an organic compound to absorb in the visible region? + 20) What is λmax in UV-visible spectroscopy? 10 3+ +, 2 + LiAlD D+, D2 in what type 21) Samples for UV and visible spectroscopy are usually dissolved of4solvent?liald4, (2)2 ++ 22) Provide an acceptable name for this compound: 23) A 1 NMR spectrum at 300Mz records a signal at 617 z downfield from tetramethylsilane. What is this shift in δ (ppm) units? 24) Explain whether 2330N2 should have an odd mass number for the molecular ion. 4 of 12

25) Which species has a molecular ion at 30.026? A. 24 B. 2. 4N D. Si E. N2 26) Assigning values in the order that the hydrogens appear in the molecule, which one gives 1 NMR chemical shifts in ppm for the hydrogens in? A. 5.55, 1.58, 5.55, 1.58 B. 5.55, 1.58, 1.58, 5.55. 1.58, 5.55, 5.55, 1.58 D. 1.58, 5.55, 1.58, 5.55 E. 1.58, 5.55, 1.58 27) Draw (R)-2-bromobutane 28) Predict the product and dominant mechanism of this reaction. ()3 222 + ()3K 29) Provide the nucleophilic substitution product(s) for the reaction below + 2 Bond D (kj/mol) Bond D (kj/mol) Bond D (kj/mol) 192 22 272 292 351 F 460 423 I 234 393 243 F F 159 431 435 498 205 I I 151 30) Using the chart of D values, calculate Δ (in kj) for the reaction below. 22 + 2 31) Show the stereochemistry of starting material needed to get the product shown by SN2. 2 + N3 + Ethanol + N3 2 5 of 12

32) Which methylcyclohexane conformer is more stable? Explain why. 33) Predict the major product of this reaction. 22 Na2 34) Given the following energy diagram for a hypothetical reaction, which statements would be true of the reaction? A. Product B will be formed faster, but product A would predominate at equilibrium if both reactions are reversible. B. Product A will be formed faster, and product A would predominate at equilibrium if both reactions are reversible.. Product B will be formed faster, and product B would predominate at equilibrium if both reactions are reversible. D. Product A will be formed faster, but product B would predominate at equilibrium if both reactions are reversible. E. Product B will predominate whether or not the reactions are reversible. 35) Which one is the bond line drawing for this condensed structure? 2()2 A) 6 of 12

B) ) D) E) 36) Given that free-radical bromination proceeds with a selectivity ratio of 1700:80:1:0.002 (tertiary:secondary:primary:methyl), predict the major product of the following reaction. + 2 hν A. B.. D. E. 37) Provide the product(s) for the free-radical chlorination of the following compound at 5. 2 22 {(S)-1,3-dichlorobutane 38) alculate the formal charges on each atom in N. 2 22 39) Predict the product(s) of this reaction. + 2 40) Explain the aspects of the mechanism that leads to the results shown below. ()3 + Na + The rate + of reaction depends + on two reactants. 2 2 2 ()3 + + + 2 + ()3 I 22 22 ()3 + ()3 BMIM hexafluorophosphate Solvent ()3 Fast Slow N N 7 of 12 N I I N N + 222 PF6 22 I

()3 + Na + The rate of reaction depends on one reactant. Slow ()3 41) Rank these alcohols in the order of increasing acidity, starting with the least acidic first: A B D,,, 42) The following structure would be classified as containing what functional group? A. ester B. aldehyde. alcohol D. carboxylic acid E. ketone F. amide 43) Predict the product(s) of this reaction sequence. (2)2 2Mg + 1) 2) +, 2 44) alculate the enthalpy of reaction for the free radical fluorination of methane. Table of bond-dissociation energies. Entries are kj/mol F I 435 569 431 364 297 439 460 356 293 238 (2)n 423 464 352 293 234 370 ()2 412 464 352 297 234 ()3 404 460 356 297 230 F 569 159 431 243 364 192 I 297 151 8 of 12

45) Explain the mechanism(s) producing these products: ()2 + NaI Acetone ()2I + Na 100% 46) Terpenes are made from which unit? A. B.. D. E. 9 of 12

Answers 1) Loss of an electron from a molecule (ionizing) N2 2) 3) bromine 4) Use 81 to calculate the molecular mass as 110. 5) a broad peak at 3300cm -1 6) 92z 7) none or zero 8) The sharp peak at 1700cm-1 occurs with a carbonyl functionality. It could be a ketone, aldehyde, ester, amide, anhydride, or carboxylic acid. (In this case the spectrum is for butanone, a ketone.) 9) 3 10) 1-chloro-2,2-dimethylpropane 11) A. It is best to draw each of the molecules out to determine the numbers of equivalent protons. 12) 34.7kJ/mol 13) 3, the number of chemically different carbon atom environments based on symmetry. 14) You need to know that UV light has a wavelength of 200nm to 400nm to get 5.0x10-19 J to 9.9x10-19 J 15) 9.2x10 14 z 16) Using λmax of 450nm from the spectrum gives an energy of 270kJ 17) It is a UV-visible spectrum (you can tell by the wavelength axis), with a λmax of 631nm. 18) 19) 8 20) The wavelength with maximum absorbance 21) Solvents with no absorption peaks above 200nm. It is better if solvents do not absorb radiation in the region of interest. 22) 1-bromo-3-ethenylbenzene (IUPA) or m-bromostyrene (common). 23) 2.06ppm 24) The mass number should be even because the molecule has an even number of nitrogen atoms 25) Answer B, calculate the exact mass of the molecule using the periodic table provided 26). The - in comes first, then work your way across the carbon chain to the right. Remember that a - attached to a carbon containing to a double bond is further downfield than one that is not. The exact magnitude (i.e. the fact that it is 5.55) of the downfield shift is less important. 3 27) 28) 22 + ()3 + K The mechanism is E2 29) 2 + 10 of 12

30) [272+431]-[423+351] -71kJ 31) R starting material gives S product by SN2 inversion of configuration. 32) The second one is more stable, because the first has 1,3 diaxial interactions with the methyl group. The second form does not. 33) 22 34) D 35) D 36) E 2 37) at 5 the product, 1,3,5-trichlorobutane is not optically active. 38) has a -1 charge4-2-0.5*(6), and N has a 0 charge5-2-0.5(6) in : N: 39) + 2 (minor amounts of 1,2 product) 40) The cis isomer, cis-1-bromo-4-(1,1-dimethylethyl)cyclohexane, has the correct anti orientation of and to occur quickly by E2. The trans isomer does not. The conformer with the tert-butyl, and groups in the axial position is energetically not favorable, so that reaction is slow by that mechanism. The observed product probably occurs by E1 starting with dissociation of. 41) ( < < < ) ( B < < D < A) 42) B 22 22 43) 44) alculate the energy for the two progagation steps Initiation: F2 2F Propagation 1: F + 4 + F Propagation 2: + F2 F + F Termination steps: 2F F2 ; 2 26 + F F 11 of 12

Fluorination: Δ -431kJ/mol Propagation step 1: (439) F (569) -130kJ/mol Propagation step 2: F F (159) F (460) -301kJ/mol 45) SN2 reactions are favored by polar, aprotic solvent, a good nucleophile that is a weaker base than, a good leaving group, and a secondary substrate. Na precipitates from acetone, helping to drive the reaction to form products. 46) E 12 of 12