The crude product was dissolved in an appropriate hot solvent and after slowly cooling the solution, crystals were obtained, which after filtering and drying gave a melting point of 205¡
Trang 1ORGANIC CHEMISTRY TOPICAL:
Separations and Purifications
Test 1
Time: 22 Minutes*
Number of Questions: 17
* 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
Rf
(261)
105
Ha
(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 1–7)
A student attempted to synthesize
N-phenylphthalimide according to the following reaction:
C OH
O
C OH
O
+ H2N
+ 2H 2 O Phthalic acid Analine
N - Phenylphthalimide
O
O
C N C
Reaction l Phthalic acid and aniline were heated to reflux for
one hour Upon cooling, the mixture was dissolved in
ether and placed in a separatory funnel The solution was
then washed with 5% aqueous Na2CO3 followed by 5%
aqueous HCl and allowed to stand over solid anhydrous
Na2SO4 for 15 minutes The ether was then evaporated
and a solid product obtained
The crude product was dissolved in an appropriate
hot solvent and after slowly cooling the solution, crystals
were obtained, which after filtering and drying gave a
melting point of 205¡ to 207¡C The literature value for
the melting point of the expected product is 208¡C
The recrystallized product was analyzed by thin layer
chromatography (TLC) and the chromatogram in Figure 1
was obtained
spotting point
solvent front
Figure 1
A second student carried out the same reaction and obtained the TLC chromatogram shown in Figure 2
spotting point
solvent front
Figure 2
The solubility of the product in several solvents at two different temperatures was also determined These results are presented in Table 1
Table 1
Dichloromethane 1-Propanol Water Toluene
1 In the experiment, which of the following would have
been removed by washing the solution with sodium carbonate?
A Phthalic acid
B N-Phenylphthalimide
C Aniline
D Water
GO ON TO THE NEXT PAGE.
Trang 42 What was the purpose of allowing the washed product
mixture to stand over solid anhydrous sodium sulfate?
A To drive the reaction to completion
B To remove the acidic starting material
C To remove water from the solution
D To increase the purity of the product
3 What conclusion can be drawn from the melting point
data given in the passage?
I The reaction produced a high yield
II The product was fairly pure
III The expected product was obtained
A I only
B II only
C II and III only
D I, II, and III
4 From the solubility data given, which of the four
solvents would be most suitable for recrystallization
of the product?
A Water
B Dichloromethane
C 1-Propanol
D Toluene
5 What is the IUPAC name for aniline?
A Benzenamine
B Cyclohexanamine
C Benzylamine
D Benzoic acid
6 A proton NMR spectrum of the final product was
also taken How would the student be able to confirm
the presence of N-phenylphthalimide from this
spectrum?
A Absorption peaks in the region 10-12 ppm
B Absorption peaks in the region 7-8 ppm
C Absorption peaks in the region 3-4 ppm
D Absorption peaks in the region 1-2 ppm
7 Based on the two TLC chromatograms in Figure 1
and Figure 2, what can be said about the product obtained by the first student compared to that obtained
by the second student?
A The first student obtained a better yield of
product
B The first student obtained a 100% pure product
while the second student obtained a 93% pure product
C The first studentÕs product is more pure than the
second studentÕs
D The first student used a different recrystallizing
solvent
GO ON TO THE NEXT PAGE.
Trang 5Passage II (Questions 8–13)
A student carried out the synthesis of heptanonitrile
by reacting 1-chlorohexane with sodium cyanide (Reaction
1)
1 -Chlorohexane Heptanonitrile
Reaction 1 The student placed sodium cyanide (32g) in a flask
along with a suitable aprotic solvent 1-Chlorohexane
(72g) was then added with continuous stirring and the
resulting mixture was heated for 30 minutes Upon
cooling, distilled water was added and the solution
extracted into ether After washing and drying this layer,
the ether was evaporated, leaving behind a liquid residue
This residue was then distilled under reduced pressure (50
torr) The fraction that boiled between 70¡C and 80¡C was
collected since the desired product has a boiling point of
75¡C at 50 torr The product weighed 53g, and analysis by
gas chromatography (GC) revealed the following
chromatogram:
A
B
8 In the chromatogram shown in Figure 1, the area
under peak A is 5 units and the area under peak B is
180 units If peak B is due to the presence of heptanonitrile, what is the purity of the product?
A 97.0%
B 80.0%
C 5.0%
D 2.7%
9 Based on the information given in the passage, what
would you expect the boiling point of the product to
be at atmospheric pressure?
A 60¡C
B 75¡C
C 80¡C
D 190¡C
1 0 Assuming 100% purity, what is the yield of
heptanonitrile in the experiment?
A 72 111
53 120 5 100
/
B 53 120 5
C 53 111
72 120 5 100
/
D 72 120 5
l 1 How can the fact that heptanonitrile has a longer
retention time in the gas chromatograph be explained?
I It has a higher boiling point
II It is more polar than the other components
in the mixture and thus more attracted to the polar stationary phase of the chromatography column
III It is more soluble in ether during extraction and therefore moves more slowly during the chromatographic separation
IV It is optically active and will move more slowly through the chromatographic separation based on its rotation of plane polarized light
A I and II only
B I, II, and III only
C II and III only
D I, II, III, and IV
GO ON TO THE NEXT PAGE.
Trang 61 2 Reaction 1 most likely proceeds by which of the
following mechanisms?
A Bimolecular elimination
B Bimolecular nucleophilic substitution
C Unimolecular nucleophilic substitution
D Hydrolysis
1 3 In the experiment, why was ether chosen as the
extraction solvent for heptanonitrile?
A The NaCl by-product was not soluble in ether
and would therefore remain in the water layer
while the nitrile would be extracted into the ether
layer
B All nitriles are more soluble in water than in
ether
C Ether and water are completely soluble in each
other
D The sodium salt of the nitrite would remain
dissolved in the water
GO ON TO THE NEXT PAGE.
Trang 7Questions 14 through 17 are
NOT related to a descriptive
passage
1 4 Which of these statements correctly describes the
separation of the compounds below by extraction into
dichloromethane and water?
Acidic Solution Basic Solution
+
CH3(CH2)6C O
OH
CH3(CH2)6C
O Ð
O Caffeine
Octanoic
acid
3
NH
N
N
H 3 C
CH3
N O
CH3
N
CH3
N
H 3 C
A If the solution were acidic, caffeine would be
more soluble in dichloromethane than in water
B If the solution were basic, caffeine would be
more soluble in dichloromethane than in water
C If the solution were basic, octanoic acid would be
more soluble in dichloromethane than in water
D If the solution were acidic, octanoic acid would
dichloromethane
1 5 Which of the following techniques would be most
suitable in the separation and analysis of two
miscible liquids with boiling points of 80¡C and
150¡C?
chromatography
B Recrystallization followed by electrophoresis
C Vacuum distillation followed by electophoresis
D Simple distillation followed by gas
chromatography
1 6 Which of the following factors usually increases the
solubility of a compound in a given solvent?
I Higher temperature
II Similar polarities III Higher molecular weight of the compound
IV Lower density of the solvent
A I, II, III, and IV
B I, II, and IV only
C I, II, and III only
D I and II only
1 7 Separation by thin-layer chromatography is based
primarily on:
A molecular weight.
B relative attraction of the components towards the
mobile and stationary phases
C relative refractive indices of the solvent and the
components
D relative specific rotations of the various
components
END OF TEST
Trang 8ANSWER KEY:
Trang 9Passage I
Sodium carbonate (Na2CO3) is commonly employed in many experiments It is the salt of a weak diprotic acid (carbonic acid H2CO3) and a strong base, formed, for example, by completely reacting carbonic acid with two molar equivalents of NaOH It reacts with acidic compounds by converting them into their corresponding sodium salts:
Na2CO3 + HA → NaHCO3 + NaA
or, in ionic form,
2Na+ + CO32Ð + HA → 2Na+ + HCO3Ð + AÐ
These sodium salts are water soluble and so can be extracted into the aqueous layer Choice A phthalic acid is the only acidic compound out of all of the answer choices This molecule is a dicarboxylic acid and conversion to its corresponding sodium salt involves loss of a proton from each carboxyl functionality resulting in the formation of a -COO−Na+ group As a result, any phthalic acid that contaminates the final product will be dissolved in water and consequently removed
All the other answer choices are wrong since they're not acidic N-phenylphthalimide will not form a sodium salt and
so will not dissolve in the aqueous layer, making choice B incorrect Aniline choice C is basic and so definitely will not form a sodium salt It is in fact treatment with 5% hydrochloric acid, not sodium carbonate, that will result in the removal of any basic contaminants such as aniline Finally, choice D is wrong since it is the addition of anhydrous sodium sulfate that results in the removal of water, not the addition of sodium carbonate
The word ÒanhydrousÓ means Òwithout water,Ó and a substance that is anhydrous, an anhydride, will typically react with water, thus removing it in the process Therefore, even if one is unaware of this experimental procedure, one can still arrive at the correct answer After extracting or washing an organic material in an aqueous solution, the organic layer is wet it contains a small amount of water Several anhydrous salts such as sodium sulfate or calcium chloride can be used as drying agents and so the addition of sodium sulfate will result in removal of water, making choice C the correct response
Choice A is incorrect since the addition of sodium sulfate has nothing to do with driving the reaction to completion The acidic starting material is removed by sodium carbonate not sodium sulfate and so choice B is also wrong Finally, the addition of sodium sulfate has nothing to do with the purity of the product The purity of the product comes into play in when the crude product is dissolved in a hot solvent otherwise known as recrystallization, so choice D is wrong
The melting point of a compound is a physical property of that compound; the expected product must have been obtained since the experimental melting point is quite close to the literature value roman numeral III must be in the correct answer Answer choices A and B can be eliminated The presence of impurities acts to lower the melting point of a substance and also to broaden the range of temperatures over which melting occurs (i.e the melting point is no longer just a point) The melting point of the crystals is 205¡ to 207¡, which indicates the presence of some impurities, but not too much, making roman numeral II correct However, the melting point is not related to the yield of the compound so statement I is wrong Choice C is the correct answer then, since statements II and III are true
This question tests our understanding of the fundamental principles behind recrystallization, but it also incorporates table reading into it In recrystallization, we dissolve an impure product in a solvent at high temperature, then let it cool: as the system cools, the pure product precipitates out, while leaving the impurities behind dissolved in the solvent In choosing
a solvent for recrystallization, then, the solute (product) should be relatively soluble in it at elevated temperatures, and insoluble in it at cooler temperatures The greater the difference between these two solubilities, the greater the yield will be Toluene choice D shows the greatest difference in solubilities, so answer D is correct You can see from the data in the table that the product is much more soluble at 50 degrees Celsius than at 0 degrees Celsius The rest of the answer choices show very little difference between the solubilities in hot and cold solvent In addition, the solute has a low solubility in both choices A water and B dichloromethane, even at high temperature
The accepted IUPAC name for aniline is benzenamine This describes an benzene ring with a primary amine substituent attached to it Choice B is wrong as cyclohexanamine is a six membered carbon ring with an amine substituent on one of its carbons There is no aromatic ring in this compound Benzylamine choice C would be a benzene ring substituted
Trang 10Finally, benzoic acid choice D is a benzene ring with a carboxyl substituent There isn't even an amine functionality (or even an nitrogen atom) in this molecule, so this answer choice is also wrong
cyclohexanamine benzylamine benzoic acid
If you look at the product in Reaction 1, you can see that the only types of proton that are present are aromatic ones There are no protons attached to the nitrogen or carbonyl functionalities, so in the spectrum of the pure product, only this one type of signal should be observed From your knowledge of NMR, you should know that aromatic protons give rise to signals in the region of 7 to 8 parts per million, making choice B the correct response Choices C and D are wrong since these signals are too far upfield to be characteristic of aromatic protons Remember that these protons are deshielded, hence they absorb in the 7-8 ppm region On the other hand, the signal in choice A would be too far downfield and is characteristic
of highly deshielded protons such as carboxylic acid protons
TLC analysis gives a qualitative indication of purity You can tell how many components are in the mixture as they separate out on the chromatogram to give individual spots So, the product obtained by student 1 must be more pure than that obtained by student 2, as the chromatogram in Figure two indicates the presence of an impurity, whereas that in Figure 1 only shows the presence of one compound Although you can see how many compounds are present, you can't determine their amounts, so choice A which talks about yield is wrong The purity of the product cannot be determined quantitatively either,
so choice B is also wrong Finally, the nature of the recrystallization solvent does not determine the nature of the TLC chromatogram, so choice D is also wrong
Passage II
Before we dive into the explanation to this question, let's just briefly discuss the principal behind gas chromatography Basically, the technique is employed so that different components in a mixture can separated The sample to
be analyzed is injected into the GC column and vaporized; an inert carrier gasÑusually helium or nitrogenÑtransports this gaseous sample through the column which contains an adsorbant Depending on the nature of the components, they are attracted to different extents to the adsorbant (also known as the stationary phase) and so are eluted from (swept out from) the column at different times From the chromatogram in Figure 1, it should be evident that there are two components eluted from the column In addition, you should know that the areas under these peaks are proportional to the amount of eluted material In this case, the units of heptanonitrile is 180 the area of peak B divided by the sum of areas A and B which is
185 This figure then has to be multiplied by 100 to get a percentage Without even bothering to carry out the calculation, we should be able to see that 180/185 is really close to 100%, making choice A the correct response
For this question, you need to know that at higher pressures, boiling points are also higher: this is simply a consequence of the definition of boiling Despite our everyday experience telling us that boiling implies a high temperature (100¡C), the actual scientific definition of boiling does not involve any mention of hotness or coldness Boiling occurs when the vapor pressure of the liquid equals the pressure of the environment If this ambient pressure is low (as at high altitudes), then a low vapor pressure would suffice for boiling to occur Conversely, if the ambient pressure is high (as in a pressure cooker), a high vapor pressure is needed for boiling to occur Temperature comes into play because the higher the temperature, the higher the vapor pressure of the liquid The higher the ambient pressure, then, the higher the temperature needs to be in order for the vapor pressure to reach that same valueÑi.e., the higher the boiling point
In this case we have been given the boiling point of the product at 50 Torr If the boiling point of the product at 50 torr is 75°C, then the boiling point at 760 torr or atmospheric pressure would be significantly higher Answer choice D at
190°C is the only one that is much higher than 75°C Choice B is incorrect, because the boiling point is not expected to remain the same when the pressure is increased, and choice A is incorrect as this boiling point is lower not higher Finally, choice C is incorrect as this is not a significant increase in the boiling point As the pressure has drastically increased, you