CHEMISTRY 213 MIDTERM EXAM 1 - VERSION A KEY

FALL 2014

Answer multiple choice questions on the green computer sheet provided with a PENCIL. Be sure to encode both your NAME and Registration Number (V#) You will calculate the number of double bond equivalents (DBE) and draw the structure of A-1 on the Separate STRUCTURE ANSWER SHEET. Be sure to put your NAME on that sheet as well, in the location indicated. There should be 9 pages in this examination AND you should have a green Scantron form PLUS a STRUCTURE ANSWER SHEET. Multiple choice questions 2-21 are worth 1 mark each while multiple choice questions 22-29 are worth 2 marks each; the structure sheet is worth 9 marks for a TOTAL of 45 MARKS. There are 50 minutes available for this examination so ration your time accordingly. YOU MAY ONLY USE A Non-Programmable CALCULATOR SOME GENERAL DATA YOU MAY NEED DURING THIS EXAM:

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1. This is exam VERSION A, mark A on the computer sheet as the answer to question 1 2. If a chemical bond has a bond strength of 200 kJ mol-1, what is the longest

wavelength of light that could break this bond? A: 599 m B: 59.9 m C: 0.12 m D: 23.9 km E: 300 nm F: 599 nm G: 1198 nm H: 12.5 km I: 30.0 m J: 599 cm 3. If your microwave oven leaks radiation, what is the most likely response of the

molecules in your body if you get too close to it? A: electronic transition B: nuclear spin flip C: bond stretch D: bending vibration E: bond rotation F: bond dissociation G: ionization H: photon emission Compound X has energy levels (J mol-1) given by En = ½ (n3) for n = 0, 1, 2, 3, 4 4. What lines will appear in the spectrum if all energy levels are populated and the

selection rule is n = ± 2? Use the following energies of the observed lines in J mol-1 to answer this question:

A: 0 B: 4 C: 32 D: 4, 28 E: 4, 32 F: 13, 28 G: 0, 4, 32 H: 4, 13, 28 I: 0, ½, 3½, 9½, 18½ J: 0, ½, 4, 13½, 32

5. If there are 1000 molecules in E0, how many molecules will there be in E4 at 20 K? A: 0 B: 175 C: 681 D: 825 E: 922 F: 976 G: 992 H: 999 I: 1000 J: 1212 For the diatomic molecule 75As-2H (i.e. As-D) having a fundamental frequency of

960 cm-1, answer the following questions. 6. What is the zero point energy of this molecule in kJ mol-1? A: 10 B: 23.0 C: 11.5 D: 5.75 E: 0.230 F: 0.0575 G: 0.115 H: 1.9 x 10-10 I: 3.8 x 10-10 J: 7.7 x 10-10 7. What is the fundamental frequency of 75As-1H (i.e. As-H) in cm-1? A: 246 B: 486 C: 948 D: 1349 E: 1920 F: 480 G: 683 H: 973 I: 1895 J: 3742

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8. Knowing that the infrared active diatomic HX has a zero point energy of 12.2 kJ mol-1, what is an appropriate energy for the photons being absorbed in the first hot line in kJ mol-1?

A: 12.2 B: 24.4 C: 23.4 D: 36.6 E: 35.6 F: 12.0 G: 25.4 H: 37.6 I: 6.1 J: 5.1 The molecule BrF3 is T-shaped (see picture at right). Use this fact to predict the answers to the following questions: 9. How many vibrational modes would you expect for BrF3? A: 1 B: 2 C: 3 D: 4 E: 5 F: 6 G: 7 H: 9 I: 12 J: 15 10. Which of the following statements correctly indicates the vibrational modes that are

IR ACTIVE? A: sym Br-F stretch B: asym Br-F stretch C: sym Br-F bend D: asym Br-F bend E: sym Br-F stretch and sym Br-F bend F: ALL G: asym Br-F stretch and asym Br-F bend H: NONE 11. Assuming the resolution is very good in the gas phase, which of the following

statements about the branches observed for the rotational fine structure of the sym Br-F stretch in BrF3 is true?

A: PQR observed B: PR observed C: P observed D: R observed E: Q observed F: none observed The bending vibration of the linear molecule HCN in the gas phase shows rotational fine structure with the first three lines of the R branch at 723, 726 and 729 cm-1. 12. What is the fundamental frequency, 0, in cm-1 for this vibrational mode? A: 717 B: 718.5 C: 720 D: 721 E: 721.5 F: 722 G: 722.5 H: 723 I: 730.5 J: 732 13. What change in J state is represented by the third line of the R branch? A: J = 0→1 B: J = 1→0 C: J = 1→2 D: J = 2→1 E: J = 2→3 F: J = 3→2 G: J = 0→3 H: J = 3→0 I: J = 2→2 14. Is an absorption actually observed at the fundamental frequency, o, in for this

bending vibration? A: YES B: NO

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Answer the questions about NMR in this box using the following choices: A: singlet B: doublet C: triplet

D: quartet E: pentet F: decet G: doublet of doublets H: doublet of triplets I: doublet of quartets J: doublet of pentets Give the pattern you would observe for: 15. 31P in FP(CH2Cl)2 J 16. 1H in FP(CH2Cl)2 G 17. 31P in O=P(CF3)3 F 18. 19F in O=P(CF3)3 B 19. What is the intensity pattern for the lines of the 13C resonance in 13CD2Cl2? A: singlet B: 1:1 C: 1:2:1 D: 1:1:1 E: 1:3:3:1 F: 1:2:2:1 G: 1:4:6:4:1 H: 1:2:3:2:1 I: 1:1:1:1:1 J: 1:5:10:10:5:1 20. What is the resonance frequency of 2H (D) on an instrument that detects 19F at 470 MHz? [Note: γ1H = 267.51, γ2H = 41.06, γ19F = 251.80] A: 41.1 B: 49.5 C: 72.1 D: 76.6 E: 251.8 F: 332.3 G: 442.4 H: 500.0 I: 2882 J: 3062 21. Which of the proton resonances in molecule X (at right) disappear if a drop of D2O is added to the NMR sample? A: 1 B: 2,7 C: 3 D: 4,5,6 E: 8 F: 9 G: 1,3,8,9 H: 1,3,8 I: 3,8 J: 3,8,9

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Use the following structural choices to answer Questions 22 and 23: __________________________________________________________________________ 22. Which of the structures above is most consistent with the IR spectrum shown below? F __________________________________________________________________________ 23. Which of the structures above is most consistent with the IR spectrum shown below? B

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Use the following structural choices to answer Questions 24 and 25: __________________________________________________________________________ 24. Which of the structures above is most consistent with the IR spectrum shown below? G __________________________________________________________________________ 25. Which of the structures above is most consistent with the IR spectrum shown below? B

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Use the following structural choices to answer Questions 26 and 27: __________________________________________________________________________ 26. Which of the structures above is most consistent with the NMR spectrum shown below? F __________________________________________________________________________ 27. Which of the structures above is most consistent with the NMR spectrum shown below? D

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Use the following structural choices to answer Questions 28 and 29 (2 marks each): __________________________________________________________________________ 28. Which of the structures above is most consistent with the NMR spectrum shown below? A __________________________________________________________________________ 29. Which of the structures above is most consistent with the NMR spectrum shown below? F

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Molecule A-1 has the formula C7H12O3

Calculate the number of double bond equivalents (DBE) for this molecule and enter that number in the appropriate space on the separate Structure Sheet (PUT YOUR NAME ON THE STRUCTURE SHEET). [1 mark]

Determine the structure of Molecule A-1 using the formula, IR spectrum and 1H NMR spectrum given below and DRAW THE STRUCTURE on the structure sheet showing ALL HYDROGEN ATOMS explicitly. [8 marks]

IR Spectrum (KBr disk):

1H NMR Spectrum (300 MHz, CDCl3):

[END]

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Structure A-1 Solution DBE = 2 IR tells us we have an alcohol (3430, confirmed by one exchangeable H in NMR), a conjugated ketone (1688, no aldehyde 2750 or CH in the NMR so ketone) and a trans alkene (968, confirmed by 17Hz pair of doublets in NMR). Both DBE accounted for and one missing O so we also have an ether. We can rule out an ester due to the much too low carbonyl stretch. NMR shows 3H singlet at δ3.7 which must be a CH3 group on oxygen -OCH3. The remaining resonances in the NMR are coupled to one another and are a triplet at δ3.6 (must be a CH2O), a 2H triplet and a 2H pentet consistent with an -OCH2CH2CH2-X chain. Thus we have these pieces:

But we know a few additional things: for one, the group X cannot be as deshielding as an O since the triplet is only at δ2.8 ppm. It also cannot be connected to the alkene because if it was the alkene H would not be a simple pair of doublets. This must mean it is connected to the ketone carbonyl and note that the chemical shifts are about right for this. We also have too many O so the oxygen in this chain must be either the -OH end group or the -OCH3 end group giving us two possible solutions. Both of these are worth full points.* * If you have done some organic chemistry you might realize that the enol structure on the left is unstable and would probably tautomerize to an aldehyde.