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CH343, Exam 1a Answers, Spring 2010
______
Name
This exam consists of several parts of unequal difficulty. Look over the whole exam right now, and start on something you know well. If you don’t know something, skip it, and come back to it later. Use your time wisely.
Keep Your Eyes On Your Own Paper!
Grading Summary
Problem1. ______(15) Problem 4. ______(25)
Problem 2. ______(11) Problem 5. ______(12)
Problem 3. ______(25) Problem 6. ______(12)
Total ______
A Partial Periodic Table of the Elements
1A / IIA / IIIB / IVB / VB / VIB / VIIB / VIII / IB / IIB / IIIA / IVA / VA / VIA / VIIA / VIIIA1
H
1.01 / 2
He
4.00
3
Li
6.94 / 4
Be
9.01 / 5
B
10.8 / 6
C
12.0 / 7
N
14.0 / 8
O
16.0 / 9
F
19.0 / 10
Ne
20.2
11
Na
23.0 / 12
Mg
24.3 / 13
Al
27.0 / 14
Si
28.1 / 15
P
31.0 / 16
S
32.1 / 17
Cl
35.5 / 18
Ar
40.0
19
K
39.1 / 20
Ca
40.1 / 21 / 22 / 23 / 24 / 25 / 26 / 27 / 28 / 29 / 30 / 31
Ga
69.7 / 32
Ge
72.6 / 33
As
74.9 / 34
Se
79.0 / 35
Br
79.9 / 36
Kr
83.8
37 / 38 / 39 / 40 / 41 / 42 / 43 / 44 / 45 / 46 / 47 / 48 / 49 / 50 / 51 / 52
Te
128 / 53
I
127 / 54
1. Multiple Choice: Choose the BEST ANSWER for each of the following questions.
1. ____B____ A strong peak at 1700 cm-1 in an IR spectrum is most likely due to:
A. C≡C stretch
B. C=O stretch
C. C=C stretch
D. O-H stretch
2. ____D____ A peak at 140 ppm in a 13C-NMR spectrum is most likely due to:
A. a C≡C carbon
B. a C=O carbon
C. an sp3 carbon attached to an electronegative atom
D. a C=C carbon
3. ____C____ A set of peaks in a proton NMR spectrum consists of three lines in a ratio of 1:2:1. This set of peaks is most likely due to:
A. a methyl group.
B. a type of hydrogen next to a methyl group.
C. a type of hydrogen next to a CH2.
D. four hydrogens; one of one type, two of another type, and one of a third type.
4. ____B____ A compound contains 1 carbon, 3 hydrogens, and a nitrogen. The easiest way that mass spectrometry would indicate that this compound contains a nitrogen is:
A. There are two peaks of equal height at m/z = 31 and 33.
B. The molecular ion is odd.
C. A peak at m/z = 14 is present.
D. There is only one peak in the mass spectrum.
5. ____D____ A strong peak at 3300 cm-1 in an IR spectrum is most likely due to:
A. C≡C stretch
B. C=O stretch
C. C=C stretch
D. O-H stretch
2. Answer the questions about the structure and spectrum below.
A. Draw arrows from the H’s in the structure to the peaks they produce in the spectrum.
B. How many peaks will the carbon spectrum of the compound show? Label each kind. Explain your reasoning.
There are eight types of carbons, due to symmetry through the benzene ring.
Problem 3. Propose a reasonable structure for the compound that produced the following sets of spectra. Write comments on the spectra to tell me what information you obtain from each one. Write your structure on the proton NMR spectrum, and indicate which hydrogens are producing each set of peaks. MW = 88.
OH sp3 C-H stretches
No apparent Cl or Br. Even MW implies no N’s, or an even number of N’s.
Problem 3, continued. MW = 88
Signal at 3.8: 6 peaks, for 5 adjacent H’s. Position puts it next to the O. The two CH2’s are the peaks at 1.4. The left end methyl is the two peaks at 1.2, since there is only 1 adjacent H. The right methyl is the 3 peaks at 0.9, since there are 2 adjacent H’s. The OH is at 3 – it could be anywhere.
Five types of C. peak at 68 is a C-O carbon: the rest are sp3 carbons. No C=C, benzene, C=O’s.
Problem 4. Propose a reasonable structure for the compound that produced the following sets of spectra. Write comments on the spectra to tell me what information you obtain from each one. Write your structure on the proton NMR spectrum, and indicate which hydrogens are producing each set of peaks. MW = 212.
sp2 C-H sp3 C-H C=O
Two peaks two mass units apart with equal size shows a Br. No apparent Cl or N. MW = 212, for 79Br-contanining molecule.
Problem 4, continued. MW = 212
Two doublets in the aromatic region says para-disubstituted. Both doublets are left of 7.24 ppm, which indicates electron-withdrawing groups are attached. The 2H quartet and the 3H triplet indicate an ethyl group.
C=O 4 benzene C’s (symmetry) 2 sp3 C’s
Problem 5. Consider the proton NMR spectrum below. For which of the following structures does it best correspond? Briefly explain your reasoning.
Only 2 types of H’s 5 types of H’s and no 12H doublet
4 types of H’s, and no 12H doublet
Diidopropylamine fits best. By symmetry, all of the methyl groups are equivalent, and would be a doublet, due to one adjacent H. The CH should be 7 peaks. The NH could be anywhere.
Problem 6. Consider the proton NMR spectrum below. For which of the following structures does it best correspond? Briefly explain your reasoning.
Only 2 types of H’s No peak at 4 ppm for CH2-O
No peak at 4 ppm for CH2-O, and wrong splitting patterns.