Choice A is incorrect since the mass spectrum indicates only the masses of atoms and groups and not bonding relationships.. Because this is proton NMR, no information about carbon atoms
Trang 1ORGANIC CHEMISTRY TOPICAL:
Molecular Spectroscopy
Test 1
Time: 23 Minutes*
Number of Questions: 18
* The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit
Trang 2DIRECTIONS: Most of the questions in the following
test are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions
PERIODIC TABLE OF THE ELEMENTS
1
H
1.0
2
He
4.0 3
Li
6.9
4
Be
9.0
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
39.9 19
K
39.1
20
Ca
40.1
21
Sc
45.0
22
Ti
47.9
23
V
50.9
24
Cr
52.0
25
Mn
54.9
26
Fe
55.8
27
Co
58.9
28
Ni
58.7
29
Cu
63.5
30
Zn
65.4
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
Rb
85.5
38
Sr
87.6
39
Y
88.9
40
Zr
91.2
41
Nb
92.9
42
Mo
95.9
43
Tc
(98)
44
Ru
101.1
45
Rh
102.9
46
Pd
106.4
47
Ag
107.9
48
Cd
112.4
49
In
114.8
50
Sn
118.7
51
Sb
121.8
52
Te
127.6
53
I
126.9
54
Xe
131.3 55
Cs
132.9
56
Ba
137.3
57
La *
138.9
72
Hf
178.5
73
Ta
180.9
74
W
183.9
75
Re
186.2
76
Os
190.2
77
Ir
192.2
78
Pt
195.1
79
Au
197.0
80
Hg
200.6
81
Tl
204.4
82
Pb
207.2
83
Bi
209.0
84
Po
(209)
85
At
(210)
86
Rn
(222) 87
Fr
(223)
88
Ra
226.0
89
Ac †
227.0
104
Unq
(261)
105
Unp
(262)
106
Unh
(263)
107
Uns
(262)
108
Uno
(265)
109
Une
(267)
*
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
(145)
62
Sm
150.4
63
Eu
152.0
64
Gd
157.3
65
Tb
158.9
66
Dy
162.5
67
Ho
164.9
68
Er
167.3
69
Tm
168.9
70
Yb
173.0
71
Lu
175.0
†
90
Th
232.0
91
Pa
(231)
92
U
238.0
93
Np
(237)
94
Pu
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
Md
(258)
102
No
(259)
103
Lr
(260)
GO ON TO THE NEXT PAGE.
Trang 3Passage I (Questions 1–6)
A chemist carried out the reaction of tert-butyl
chloride with sodium methoxide, resulting in the
formation of two products:
Reaction 1 After separation by distillation, Products A and B
were analyzed using infrared spectroscopy (IR), proton
nuclear magnetic resonance spectroscopy (NMR), mass
spectroscopy (MS), and ultraviolet spectroscopy (UV)
The results are as follows:
PPM (δ)
PPM (δ)
9H ∫
3H ∫
6H ∫
2H ∫
NMR
Spectra
Product A Product B
UV
Spectra
ε
Figure 1
Table 1 Product A Product B IR
Major peaks (cm–1)
3050 2900 1650
2900 1200
MS Molecular weight (amu) Base peak (amu)
56
41
88
73
Product A was found to contain only carbon and hydrogen, while Product B contained carbon, hydrogen, and oxygen
1 Based on the mass spectra data given in Table 1, it
can be determined that:
A both products contain a double bond.
B there is an ethyl group present in each product.
C both products contain at least one methyl group.
D no useful information can be obtained.
2 All of the NMR signals in Figure 1 are singlets
because:
A the hydrogen atoms are bonded to carbon atoms
that have no other hydrogen atoms on adjacent carbon atoms
B the proton resonances occur at the same magnetic
field
C the carbon atoms within each molecule are
identical
D the hydrogen atoms in each molecule are
identical
GO ON TO THE NEXT PAGE.
Trang 43 The NMR signal at 5.5 ppm for Product A most
likely indicates the presence of:
A vinyl protons.
B aromatic protons.
C an aldehyde proton.
D the proton of an alcohol group.
4 Which of the following is responsible for the IR
stretching peak at 1200 cm–l for Product B?
A C–H
B C=O
C C–O
D C–C
5 Which of the following correctly represents the
structure of Products A and B, respectively?
A
CH 3 CH 2 CH 2 and CH3 C CH 2 CH 3
O
B CH3CH2CH2CH2CH3 and O
C CH2 CH CH CH2 and CH3CH2OCH2CH3
D
CH3 C CH2
CH3
and CH3 C CH3
CH3
OCH3
6 Which two processes are likely to occur in the
reaction of tert-butyl chloride with sodium methoxide?
A Elimination and substitution
B Rearrangement and hydrolysis
C Photolysis and elimination
D [4+2] Cycloaddition and substitution
GO ON TO THE NEXT PAGE.
Trang 5Passage II (Questions 7–12)
Three students were each given an unknown
compound and asked to identify it by using only
spectroscopic techniques They were told that their
compound was one of those listed below
O
Cyclohexyl ethyl ketone
O
Propiophenone
O H 3–Phenylpropionaldehyde
OH 1–Phenyl–2–propen–1–ol
The results of the students’ experiments are shown
below
Student 1 Student 2 Student 3
IR peaks
(cm–1)
3050 2900 1710
–
1700-1740
NMR
(ppm)
1.0; 3H triplet 2.0; 2H quartet 7.0; 5H
9.5; 1H
MS peaks
(amu)
134 (M+) 105 77 57
7 From the data given in Table 1, what compound did
Student 1 have?
A Cyclohexyl ethyl ketone
B Propiophenone
C 3-Phenylpropionaldehyde
D 1-Phenyl-2-propen-1-ol
8 The peak at 57 amu in the mass spectrum of Student
l’s compound can be attributed to loss of which of the following fragments?
A
C O
B
C
C 2 H 5
D CH 3
9 Which of the following correctly describes the signal
at 2.0 ppm in the NMR spectrum of Student l’s compound?
A It is a quartet because there are 3 protons on the
adjacent carbon atom
B It is a quartet because there are 4 protons on the
adjacent carbon atom
C It is due to the aromatic protons in the
compound
D It has an area twice as large as the area of the
signal at 7.0 ppm:
GO ON TO THE NEXT PAGE.
Trang 61 0 The signal at 2900 cm–l in Student l’s IR spectrum is
the result of:
A protons absorbing radio waves as they change
spin state
B electrons changing to a higher energy
antibonding orbital
C a C–H bond absorbing energy as it goes to a
higher energy vibrational state
D several C–H bonds breaking as it absorbs infrared
light
1 1 The NMR spectrum of Student 2’s compound
revealed the presence of which of the following
functionalities?
A A ketone
B A phenyl group
C A vinyl group
D An aldehyde
1 2 Which of the following could be Student 3’s
compound?
I Cyclohexyl ethyl ketone
II Propiophenone
III 3-Phenylpropionaldehyde
IV 1-Phenyl-2-propen-1-ol
A I only
B III and IV only
C I, II, and III only
D I, II, and IV only
GO ON TO THE NEXT PAGE.
Trang 7Questions 13 through 18 are
NOT based on a descriptive
passage
1 3 Which of the following is NOT true?
A Ultraviolet spectroscopy involves electronic
transitions
B Mass spectroscopy involves the fragmentation of
molecules
C NMR spectroscopy utilizes radio waves.
D IR spectroscopy utilizes light of wavelength
200-400 nm
1 4 Which of the following compounds would give only
one signal in its 1H NMR spectrum?
A Ethanol
B Phenol
C 1,2-Dichloroethane
D tert-Butyl alcohol
1 5 What type of spectroscopy would be the LEAST
useful in distinguishing diethyl ether from
chloroethane?
A Infrared spectroscopy
B Nuclear magnetic spectroscopy
C Ultraviolet spectroscopy
D Mass spectroscopy
1 6 Which of the following compounds is NOT infrared
active?
A CH3Cl
B CH3CH2OH
C CO
D Br2
1 7 A student recorded the NMR spectra of each of the
following compounds Disregarding chemical shifts, which one would possess a spectrum significantly different from the others?
A CH3CH2CH2OCH2CH2CH3
B
C CH3CH2CH2Cl
D
CH3CH 2 CCH2CH3 Cl
Cl
CH3CH2CH 2 C Cl
Cl
Cl
1 8 If a compound has an NMR spectrum with 3 signals
in an area ratio of 1:3:4, which of the following compounds could it be?
I Cl III.
CH 3
CH 3 CH 2 C Cl
II.
CH
3 C C CH
3
Cl Cl
H H
IV CH
2 Cl H CH
3
C ClCH
2
CH 3 C CH 3
CH
3
Cl
A III only
B IV only
C I and II only
D II and IV only
END OF TEST
Trang 8ANSWER KEY:
Trang 9MOLECULAR SPECTROSCOPY TEST 1 EXPLANATIONS
Passage I (Questions 1–6)
from the molecule that is to be analyzed, creating a positively charged molecular ion This molecular ion is identical to the original neutral molecule except that it is missing one electron, making it positively charged The molecular ion is usually unstable and will itself fragment to give other positively charged fragments These positively charged fragments are separated
on the basis of their charge to mass ratios When these fragments are accelerated by an electric field and passed through a magnetic field, they will be deflected by different amounts Ions with different masses strike the electric multiplier at different times; their appearance are recorded on a chart The instrument can be calibrated to give the molecular weight and approximate number of fragments
Since a methyl group is 15 atomic mass units, the loss of a 15 amu fragment from the molecular ion generally indicates that a methyl group must have been present in the molecule Therefore, choice C is correct An ethyl group (choice B) would have been indicated by a peak at 29 amu less than the molecular weight Choice A is incorrect since the mass spectrum indicates only the masses of atoms and groups and not bonding relationships Answer D is false because the fragmentation of molecules does provide much useful information regarding the structure of the molecule Again, the correct answer is choice C
2 The correct answer for question 2 is choice A The multiplicity, or shape, of the signal is a result of the
protons on adjacent atoms The signal from protons on a particular carbon is split by protons on adjacent carbons due to spin-spin coupling The information that is provided by the splitting of the proton NMR signals, along with the chemical shift and peak area data, accounts for most of the usefulness of proton NMR to identify compounds In this question, all signals in the NMR spectrum of both compounds are singlets This implies that none of the hydrogen atoms in the molecules of either product A or B have any other hydrogen atoms on adjacent atoms For choice B to be correct, there would
be only one type of hydrogen atom in each of products A and B and there would be only one peak in the entire NMR spectrum This is not the case as both product A and B have two separate but unsplit proton signals Choice C is also incorrect Because this is proton NMR, no information about carbon atoms are obtained directly and so choice C is incorrect Choice D is incorrect because it is simply another way of phrasing choice B Again, choice A is correct
location on the NMR spectrum of the more common types of protons These are called the chemical shifts of protons The shifts are commonly expressed on the delta scale as parts per million or ppm values You should have some of the more common chemical shifts committed to memory; a list of them can be found in your organic chemistry home study book You should certainly know that aldehyde and carboxylic acid protons appear far downfield (downfield is to the left in the spectrum): the carboxylic acid proton appears between 10.5 and 12 and the aldehyde proton appears between 9 and 10 The MCAT has included these chemical shifts more than once You should also know that if electronegative elements are close to the proton of interest, the proton will be deshielded, meaning that its signal will appear further downfield A useful mnemonic is: deshield = downfield = left Anyway, vinyl protons appear at around 5-6 ppm and so answer A is correct Aromatic protons usually give a signal between 6 and 8.5 ppm (choice B), an aldehyde proton comes at 9-10 ppm (choice C), and those which are part of an alcohol group come between 3 and 5 ppm (choice D) Answers B, C, and D are wrong, and again, the correct answer is A Again, if you had trouble with this question, consult your organic home study book
4 The correct choice for question 4 is C In order to answer this question, you need to remember the IR
absorbance values of some major functional groups The functional groups that are most readily identified by IR are the carbonyl group, with a sharp absorption between 1700 to 1750 cm–1, the hydroxyl group, showing a strong broad absorption band between 2500 and 3600 cm–1, the exact region depends upon what type of O–H linkage is in question: alcohol, carboxylic acid, or phenol Keep in mind that all you have to know is that a hydroxyl group will show up as a strong broad band in this region Another absorption that you should be familiar with is that of a C–O single bond stretch This appears
as a strong sharp absorption in the region of 1000 to 1300 cm–1 This type of absorption is seen for alcohols, ethers, and esters With this in mind, an absorption at 1200 cm–1 will most likely correspond to the stretching of a carbon/oxygen single bond This makes choice C the correct response Choices A and B are incorrect as the absorbance values are incorrect for these functional groups Choice D is also wrong: carbon-carbon single bonds show only very weak IR absorbances if they show up at all Again, choice C is the correct answer
there is some form of double bond present, so you could have narrowed the answer choices down to C and D The fact that it peaks below 200 nm suggests that there is an isolated double bond, rather than a conjugated system Don't worry if you didn't know this though; you can identify the products without the UV data In fact, the NMR spectrum is probably the most useful in identifying the products You can see that product A has two types of hydrogen atoms because there are two signals The ratio of these two peak areas is given as 1 to 3 It is very important to understand that the ratio of peak areas only reveals the ratio of the number of hydrogens in the compound, not the actual numbers of hydrogens in the compound The ratio provided by the NMR data for product A matches only the first compound shown in choice D, which has two types
Trang 10second answer choices are possible The NMR spectrum of product B has two signals, implying two types of hydrogen atoms, and the ratio of these hydrogens is 1 to 3 The second compound of choice A would have three types of hydrogen and
is thus incorrect The second compound in choice B would have two types of hydrogen but they are in a 4 to 4 ratio and this can therefore be eliminated The second compound in choice C has two types of hydrogen but the ratio is 4 to 6 and is incorrect The second compound in choice D has two types of hydrogen in the correct ratio, 3 to 9 or 1 to 3 Thus choices
A, B, and C could not be correct and choice D is the right answer
elimination reaction because the original compound has no double bonds You would also expect an elimination product because the reaction in the passage is between a tertiary alkyl halide and a strong base In addition, you are told that product B contains oxygen and no chlorine, so it must have lost the chlorine and gained an oxygen containing group Because chlorine
is a good leaving group, product B is most likely the result of a substitution reaction This would be logical since elimination is almost always accompanied by the formation of some substitution product
You could have also derived the answer from the experimental data in the passage, but you really should be know that the reaction presented in this passage is a model reaction for elimination and nucleophilic substitution Sodium ethoxide
is a very strong base and would cause elimination to produce the alkene The ethoxide ion can also act as a nucleophile, replacing the chloride The other answers are wrong for a variety of reasons, of which I will mention the most obvious Choice B involves hydrolysis, or reaction with water Water could not be present with sodium ethoxide, which would react with any water present Choice C mentions photolysis, or reaction with light, which is not given as a reaction condition, so this is also wrong Choice D is also wrong because there are no dienes in the starting materials which could undergo a [4 + 2] cycloaddition A cycloaddition reaction is one in which another 2 unsaturated molecules combine to form a cyclic
molecule Cycloaddition reactions are further classified according to the number of pi electrons involved in the transition state A [4+2] reaction means that 6 pi electrons are involved in the transition state: 4 pi electrons from one reactant and 2
from another Again, A is the correct response
Passage II (Questions 7–12)
NMR, which indicates it is aromatic We can therefore eliminate choice A, since cyclohexyl ethyl ketone contains no aromatic ring It can also be rejected because its molecular weight would be 140 atomic mass units, not 134 amu as indicated
by the MS for student 1's compound An ethyl group is indicated by the presence of a triplet and a quartet in the NMR, which student 1 has The molecular weight is indeed 134 amu with fragments peaks as expected from such a compound The fragments are the phenyl ring at 77 amu, the ethyl carbonyl fragment at 57 amu, and the phenyl carbonyl fragment at 105 amu The IR peak of student 1's compound at 1710 cm–1 also indicates the presence of a C=O group, so this also rules out choice D is this does not possess a carbonyl group Choice C can be eliminated because it would have shown a peak at 9.5 ppm in the NMR due to the aldehyde hydrogen, and the compound in choice D would have shown a peak at 3400 cm–1 in the
IR spectrum So the correct choice for this question is choice B
amu indicated by the parent peak Subtracting the weight of the fragment recorded (57 amu) from the total weight gives a difference of 77 amu Of the groups listed, only the phenyl group C6H5 - has a weight of 77 amu This gives choice B as the correct answer The other groups have weights as follows: carbonyl - 28 amu, ethyl - 29 amu, methyl - 15 amu Again the correct answer is choice B
coupling of the hydrogen with hydrogens on an adjacent carbon A single signal is split into n+1 peaks, where n is the number of adjacent hydrogens Because the signal is split into a quartet or four signals, the correct answer must be A Choice B is incorrect because the signal that would be produced by four adjacent hydrogens in question would be a quintet or 5 peaks Aromatic protons would absorb in the 6 to 8.5 ppm region, so choice C is incorrect Lastly, the area of the signal is irrelevant to this question In addition, the signal at 2.0 ppm would be 2.5 times smaller than the aromatic signal so choice
D is not right Again, the correct answer is A
1 0 The correct answer for question 10 is choice C Answer A is wrong because this process does not occur
in IR spectroscopy but is the basis for NMR spectroscopy The process of promoting electrons from a molecular orbital to
an antibonding orbital mentioned in B is the basis for UV spectroscopy IR spectroscopy, in contrast is a result of polar covalent bonds absorbing IR light and vibrating with more energy The area around 2900 cm–1 is characteristic of the C–H bond and so answer C is correct IR light usually does not have enough energy to break covalent bonds, so choice D is incorrect Again, answer choice C is the correct response
The only aldehyde in the list is 3-phenylpropionaldehyde, so this must be the compound that student 2 had Since it has 2 alpha hydrogen atoms adjacent, the aldehyde hydrogen would be a triplet All of the other answer choices do not absorb in the region 9 to 10 ppm, so again, choice D is the correct answer
1 2 The correct answer here is choice C The dominant absorbance at 1700-1740 cm–1 in the IR spectrum is