CHEM213exam1S24_key

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1 CHEM 213 Organic Chemistry Spring 2024 Exam 1 Answer Key 1) Why do C ≡ C bonds vibrate at a higher frequency than C=C bonds? a. The triple bond is the stronger “spring” due to its stronger bond. b. Since the C ≡ C atoms are bonded to less atoms, they have a smaller mass than the C=C atoms. c. Vibrations break the  bonds and the triple bond wants to be broken less than the double bond. d. Actually, C=C bonds vibrate at a higher frequency. 2) The peaks in an infrared spectrum correspond to… a. …the IR lightwaves that cause the molecule to slow down. b. …the IR lightwaves transmitted by the sample. c. …the IR lightwaves with frequencies that match vibrations in the molecule. d. IR lightwaves have nothing to do with the peaks. 3) What IR absorption bands are consistent with this structure? a. …3400, 3050, 2900 cm -1 b. …2900, 1650, 3400 cm -1 c. …3400, 1650, 1600 cm -1 d. …3400, 2900, 2250 cm -1 4) B eff , the magnetic field felt by a nucleus, is determined by… a. … only the applied magnetic field. b. … the applied magnetic field plus electron shielding. c. … the applied magnetic field, electron shielding and any nearby magnetic fields, such as the magnetic field induced by benzene ring current. d. … the applied magnetic field, electron shielding, nearby magnetic fields and the vibrations of the nucleus 5) 13 C NMR can tell you… a. …about functional groups present and how shielded some carbons are. b. …the number of non-equivalent carbons in the molecule. c. …whether or not your compound has an alkyl ring. d. …all of the above. e. …only a and b. f. …only a and c. 6) How many carbon atoms does a molecule contain if in its mass spectrum the relative abundance of the M+ peak is 50.1% and of the M+1 peak is 5.5%. a. 10 b. 9 c. 8 d. 7

2 7) List 4 of the 5 non-fingerprint region IR peaks ( range and vibration ) you expect to see for the following molecule. (16 pts.) a. C ≡ N, 2260-2210 cm -1 c. Csp 3 -H, 3000-2800 cm -1 b. C=C, 1620-1680 cm -1 d. Csp 2 -H, 3100-3000 cm -1 e. O-H, 3600-3200 cm -1 8) For each of the following compounds below tell how many signals you would expect the molecule to have in its proton NMR specturm (ignoring splitting) (12 pts.) 9) Predict the splitting patterns for the protons indicated by the arrow. (12 pts.) quartet singlet triplet of triplets triplet 10) For each of the compounds below tell how many signals you would expect the molecule to have in its normal, broadband decoupled 13 C NMR spectra. (6 pts.)

3 11) Rank (1-4) the chemical shifts of the indicated hydrogens from highest  (1) to lowest  (4). (8 pts.) 12) Label all non-fingerprint peaks in the IR below. Based on these peaks, make a separate list of the functional group(s) most likely in the molecule that made this spectrum. (10 pts.) Possible functional groups: alcohol, alkene 13) Provide a molecular formula for the compound with the following mass spectrum. (8 pts.) 77 112 114 C 6 H 5 Cl 3:1 ratio of M+ peak and M+2 peak indicates Cl (35 amu) 112 – 35 = 77 (benzene cation peak) Rule of 13: 77/13 = 5.9 ~ 6 12 x 6 = 72 77 – 72 = 5 = # of H’s ignore this peak O-H Csp 3 -H C=C Csp 2 -H

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4 14) Based on the IR and proton NMR below, determine the structure of the compound with molecular formula C 4 H 9 NO. Calculate the DU. At the bottom of each peak is a letter. Label the hydrogen in your structure with the letters corresponding to their peaks. Label all important peaks in the IR. Show all work for partial credit. (20 pts.) DU = [(4 x 2) + 2 – 9 + 1]/2 = 1 from C=O peak in IR; C=O peak in IR is in the range of amide functional group Peak a, the broad singlet, is the NH 2 group; Peak b is a CH with multiple neighbors; Peak c are two equivalent CH 3 ’s next to a CH 1 1 6, d a b C=O Csp 3 -H Carboxylic acid 4000 3000 wavenumber cm -1 2000 1500 1000 500 2 broad singlet 1 6, d a b NH 2 Csp 3 - H C=O

5 15) Based on the IR and 13 C NMR below, determine the structure of the compound with molecular formula C 4 H 6 O. Calculate the DU. At the bottom of each peak is a letter. Label the hydrogen in your structure with the letters corresponding to their peaks. Label all important peaks in the IR. Show all work for partial credit. (20 pts.) DU = [(4 x 2) + 2 – 6]/2 = 2, comes from triple bond because triple bond peak in IR. Peak at 3300 cm -1 indicates triple bond is terminal C ≡ C bond. No C=O peak or O-H peak means oxygen functional group is an ether. NMR peaks a and b must be groups next to O. Peaks a and c are only peaks that can correspond to triple bond (can’t have triple bond carbon with more than 1 H). Leaves other side of ether as ethyl group from peaks b and d. 4000 3000 wavenumber cm -1 2000 1500 1000 500 C CH 2 CH CH 3 a b c d ignore these peaks Csp 3 - H Csp- H C ≡ C

6 16) Based on the proton NMR below, provide a structure for the following molecular formula: C 6 H 12 O 2 . Calculate the DU.Label each peak in the spectrum with a letter and label the corresponding hydrogen in your structure with the appropriate letter. (15 pts.) DU = [(6 x 2) + 2 – 12]/2 = 1; peak a is CH 3 with no neighbors and C bonded to O (3.7 ppm); peak b CH with many neighbors, peak c CH 2 with many neighbors; peak d CH 3 next to CH; peak e CH 3 next to CH 2 ; accounts for 5 carbons, sixth carbon must be C=O 17) Based on the carbon NMR below, provide a structure for the following molecular formula: C 8 H 7 ClO. Calculate the DU. At the bottom of each peak is a letter. Label the carbon in your structure with the letters corresponding to their peaks. (17 pts.) DU = [(8 x 2) + 2 – 7 – 1]/2 = 5 indicates benzene ring, confirmed by 2 peaks between 7 and 8 (b and c); benzene peaks integrate to 4, so benzene has two substituents. Two doublets each integrating to 2 indicates the substituents will be opposite each other. Peak at 10 (a) is from an aldehyde hydrogen and singlet means it has no neighbors. Peak d indicates there is a CH 2 group with no neighbors and chemical shift indicates it is next to Cl. 3 2, m 1, m 3, d 3, t a b c d e 2, s 2, d 2, d 1, s a b c d

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I am not sure if I'm right on the answers I did. Thanks for helping.

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Which type of bond exists in each compound? KCl a. polar covalent bonds b. ionic bonds c. nonpolar covalent bonds BCl3 a. ionic bonds b. polar covalent bonds c. nonpolar covalent bonds P4 a. nonpolar covalent bonds b. ionic bonds c. polar covalent bonds Br2 a. ionic bonds b. polar covalent bonds c. nonpolar covalent bonds CO a. polar covalent bonds b. ionic bonds c. nonpolar covalent bonds SO2 a. nonpolar covalent bonds b. ionic bonds c. polar covalent bonds

the bond between the nitrogen atoms in N2H4 must be a______ single bond no direct c-c bond in this molecule ionic bond double bond or triple bond

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