Section 4. Answers to Exercises
Experimental Organic Chemistry: A Miniscale and Microscale Approach
SECTION 4. ANSWERS TO EXERCISES
CH 3
SOLIDS. RECRYSTALLIZATION AND MELTING POINTS
Sec. 3.2
Recrystallization
1.
Page references for each term are: a. page 94; b. page 95; c. page 95; d. page 95; e. page 98; f. page 95; g.
pages 99; h. page 98; i. page 97; j. page 100; k. page 100; l. page 100; m. page 95; n. page 95.
2.
Ethyl ethanoate and 2-propanone.
3.
Selections b and c are appropriate reasons to perform a recrystallization.
4.
(1) Select a solvent or mixture of solvents in which the solid is insoluble when cold but soluble when heated.
Cool the hot solution in which the solid is soluble to determine whether crystals reform. (2) Dissolve the
impure solid in a minimum volume of the selected solvent or mixture of solvents. (3) If the solution is
colorless, subject it to hot filtration if solid impurities are present in the solution. If the solution is colored,
treat it with decolorizing carbon before performing the hot filtration after addition of a filter-aid. (4) Let the
hot filtrate cool slowly to room temperature. This step allows for the formation of crystals from the solution.
(5) Subject the solution containing the crystals to vacuum filtration. Remove as much solvent as possible and
add a small portion of cold solvent to the crystals on the filter to wash them. (6) Isolate the crystals and allow
them to dry to remove residual amounts of solvent.
5.
Crystals may form in the funnel, thus making the filtration more difficult and reducing the amount of
crystalline product that will be recovered.
6.
a. The filter flask will contain the filtrate after filtration has been effected.
b. The filter trap serves to prevent water from backing up from a water aspirator, if one is used for the
vacuum source, and to provide a place where the vacuum may be released.
c. The Büchner funnel is used to collect the desired crystalline product.
7.
Pressure changes in the system may allow water to enter the flask, contaminating the filtrate.
8.
(1) Select a solvent or mixture of solvents in which the solid is insoluble when cold but soluble when heated.
Cool the hot solution in which the solid is soluble to determine whether crystals reform. (2) Dissolve the
impure solid in a minimum volume of the selected solvent or mixture of solvents. (3) If the solution is
colorless, subject it to hot filtration if solid impurities are present in the solution. If the solution is colored,
treat it with decolorizing carbon before performing the hot filtration after addition of a filter-aid. (4) Let the
hot filtrate cool slowly to room temperature. This step allows for the formation of crystals from the solution.
(5) Subject the solution containing the crystals to vacuum or Craig tube filtration. Remove as much solvent
as possible and add a small portion of cold solvent to the crystals to wash them. If a Craig tube is being used,
a second centrifugation will be required. (6) Isolate the crystals and allow them to dry to remove residual
amounts of solvent.
9.
The packing in the tip of the filter-tip pipet lessens the possibility that the hot solvent will accidentally squirt
from the pipet during the transfer.
10.
Crystals may form in the pipets, making the filtration more difficult and reducing the amount of crystalline
product that will be recovered.
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Section 4. Answers to Exercises
Experimental Organic Chemistry: A Miniscale and Microscale Approach
11.
Such particles may pass between the ground-glass shoulder of the Craig tube and the Teflon component.
12.
It may be decolorized by adding decolorizing carbon to the hot solution containing the dissolved solute,
followed by hot filtration to remove the carbon.
13.
They are removed by hot filtration of the solution containing the solute dissolved in the solvent.
14.
When a hot solution has been treated with decolorizing carbon and filtered, the carbon may not be completely
removed, as evidenced by the presence of a dark tint in the filtrate. A small amount of filter-aid may be
added to the filtrate, the solution heated again just to boiling and refiltered. The carbon is trapped in the
filter-aid and thus removed. Alternatively, the filter-aid may be added to the decolorized solution prior to the
initial gravity filtration.
15.
a. The substance being purified must be insoluble or nearly insoluble in cold solvent.
b. The compound being purified should be soluble in the hot solvent but insoluble or nearly so when cooled.
c. The boiling point of the solvent should be low enough that it can be removed easily from the crystals in the
final drying step.
d. The boiling point of the solution should be lower than the melting point of the solid being purified.
e. The solvent should not react chemically with the substance being purified.
16.
Polar solutes will be more soluble in polar solvents than non-polar solvents, because similar intermolecular
forces interact between the molecules. For the same reason, non-polar solutes will be more soluble in nonpolar solvents than polar solvents. In other words, the solute will dissolve in the solvent most “like” the
solute.
17.
Apply the criterion of solvent selection stating that the solid should be soluble in hot solvent and nearly
insoluble in cold solvent and that the compound should crystallize as completely as possible when the
solution is cooled. For solid A, cyclohexane appears to be the solvent of choice. For solid B, none of the
solvents listed meet the criteria so use of a mixed solvent for the recrystallization must be considered.
Because the solid is soluble in water and ethanol at room temperature but insoluble in hot and cold
dichloromethane, petroleum ether, and toluene, the mixed solvent must be one of the solvents in which the
compound is soluble at room temperature and one of the solvents in which it is insoluble. Water is not
miscible with dichloromethane, petroleum ether, and toluene, whereas ethanol is. Thus, a possible mixed
solvent involves ethanol and either dichloromethane, petroleum ether, or toluene. The actual choice would
have to be determined experimentally.
18.
The mixture would be heated with water until the benzoic acid dissolved. The resulting heterogeneous
mixture could be decanted or subjected to hot gravity filtration to separate the sand from the solution. Upon
cooling, pure benzoic acid should separate from the solution.
19.
According to the Merck Index, the solubility of benzoic acid is 68 g/L at 95 °C, corresponding to a solubility
of 0.068 g/mL. The volume of water required to dissolve 1 g of benzoic acid at 95 °C is computed as follows:
1 g/(0.068 g/mL) = 14.7 mL.
20.
The calculations depend on the experimental data obtained, so the following only illustrates the procedure for
doing the calculations. (1) For acetanilide, assume that the original sample weighed 0.5 g and that 15 mL of
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Section 4. Answers to Exercises
Experimental Organic Chemistry: A Miniscale and Microscale Approach
hot water was used for its dissolution. The solubility in water of acetanilide at 0 °C is given as 0.53 g/100
mL, or 0.0053 g/mL of water. Its solubility in 15 mL of water would then be (0.0053 g/mL)(15 mL) = 0.08 g;
this represents the amount of acetanilide remaining in solution after crystallization is complete. (2) For
benzoic acid, assume that the original sample also weighed 0.5 g and that 11 mL of hot water was used for
dissolution. Using the same approach as with acetanilide, the amount of benzoic acid remaining dissolved at
0 °C after crystallization is complete is 0.0022 g.
21.
Ethanol (bp 78 °C) is more volatile than 1-octanol (bp 196 °C), so ethanol is more easily removed from the
crystals during the drying process. Hexane (bp 69 °C) has a higher boiling point than pentane (bp 36 °C), and
thus can be heated to a higher temperature during the dissolution process. The advantage here is that less
solvent is required for dissolution. Water (bp 100 °C) has a higher boiling point than methanol (bp 65 °C), so
water can be heated to a higher temperature during the dissolution process and less solvent will be required.
Other advantages of using water versus methanol are that water is much less expensive and is nonflammable.
22. a. Toluene is the best choice of solvents, because the sample is insoluble in the cold solvent, yet soluble in the
hot solvent. This should allow both solute and impurity to dissolve when the solvent is heated to boiling;
upon cooling, the impurities should remain solvated, while the solute precipitates from solution. Filtration
then yields the pure solute, leaving the impurities behind in the mother liquor.
b. Ethanol would not be a suitable recrystallization solvent because the solute is soluble even in the cold
solvent; i.e. the solute would never precipitate from solution.
Petroleum ether would not be a suitable solvent because the solid sample would never dissolve in the
solvent, even upon heating; thereby, leaving the impurities encapsulated in the solute.
c. Water:
H2 O
Ethanol:
CH3CH2OH
Toluene:
C6H5CH3
Petroleum Ether: Primarily a mixture of hexane isomers, C6H14, identified by their boiling range: 35–60
°C.
2-Butanone:
CH3COCH2CH3
Acetic Acid:
CH3CO2H
d. Most polar solvent:
Least polar solvent:
23.
Water
Petroleum ether
a. This decreases the amount of solid that will be recovered because too much of the compound will remain
in solution on cooling.
b. The solid may contain adsorbed solvent that contains impurities, which may produce isolated crystals that
are impure.
c. If this occurs, some of the crystals will be dissolved, leading to a decreased yield of pure solid.
d. Using a large excess of decolorizing carbon results in a decreased yield of crystals because some of the
desired compound will be adsorbed on the carbon.
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Section 4. Answers to Exercises
Experimental Organic Chemistry: A Miniscale and Microscale Approach
e. Crystals obtained in this manner will consist of the desired compound along with the impurities that
solidified with the desired compound.
f. This will cause very small crystals to form, and the likelihood that impurities will be adsorbed on their
surface is increased.
24.
Concentrating the solution also concentrates the impurities in it.
Upon cooling the concentrate, co-
crystallization of the impurities along with the desired compound is more likely.
25.
The dissolution of a solid depends on the surface of the solid that is in contact with the solvent. Large crystals
have smaller surface areas than small ones and thus will dissolve more slowly.
26.
Operating at the temperature extremes allows one to work at the solubility extremes of the desired material in
the chosen solvent. At the temperature high (the boiling point of the solvent), the solute dissolves thereby
also releasing the impurity, which also dissolves in the solution. At the temperature low (usually 0 °C--the
temperature of an ice-water bath), when precipitation is desired, the solubility of the desired material is at a
minimum and crystallization of pure material result is possible, while leaving the impurity dissolved in
solution.
27.
The purpose of scratching the flask is to induce crystallization. This works by providing a rough surface on
which the first crystals may form. Another common way to induce crystallization is by adding a seed crystal
(pure solute crystal from some other source) to the cold solution; this provides a crystal surface onto which
other crystals will readily form.
28.
Yes. The mass attributed to the impurity will invariably be lost during the recrystallization, because this
material is removed.
29.
The same solvent used in the recrystallization should be used to rinse the filter cake. The solvent should be
cooled to 0 °C prior to use to avoid loss of material due to solubility of the solute in the warmer solvent.
30.
If the boiling point of the solvent is higher than the melting point of the solute, the latter will melt rather than
dissolve when the solvent is heated to boiling. Furthermore, high boiling solvents are difficult to remove
completely from the filter cake because they will not evaporate readily, and thus the product will be
contaminated with traces of solvent.
31.
The impurities may be found in the mother liquor (filtrate) at the end of the recrystallization process.
32.
1. Combine the impure stilbene and a minimum amount (∼2 mL) of ethanol in an Erlenmeyer flask and heat
the solution to boiling.
2. As soon as boiling is observed, add just enough additional ethanol to make the solution homogeneous,
while still boiling.
3. If some insoluble material persists, perform a hot filtration.
4. Allow the solution to cool to ambient temperature, then place the Erlenmeyer flask into an ice water bath.
5. If crystals do not form, scratch the flask or add a seed crystal.
6. Once crystals have formed, vacuum-filter the solution and rinse the filter cake with cold ethanol.
7. Allow the crystals to air dry on the filter paper, and then scrape the crystals from the filter paper and allow
further drying.
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Section 4. Answers to Exercises
33.
Experimental Organic Chemistry: A Miniscale and Microscale Approach
Choose one of the available solvents in which the crystals were soluble and dissolve the crystals in a small
amount of this solvent. Then slowly add one of the available solvents in which the crystals were insoluble,
but one that is also miscible with the first solvent chosen, until a cloudy appearance results. This creates a
binary solvent system with properties of both solvents. See if this binary solvent system fits the criteria for a
good recrystallization solvent system. For example, varying ratios of ethyl acetate and petroleum ether often
serve as good binary solvent systems. Usually, some trial and error is required to find an appropriate binary
solvent system.
34.
a. The poison will adsorb strongly onto the surface of the activated charcoal.
b. Normal excretion will remove the charcoal containing the poison from the individual’s body.
Sec. 3.3
1.
Physical Constants: Melting Points
a. Heating the sample too fast, sample size, state of subdivision of the sample, use of an uncalibrated
thermometer.
b. Sample not packed tightly in the capillary tube, use of an uncalibrated thermometer.
c. Impure sample, heating the sample too fast.
2.
Page references for each term are: a. page 115; b. page 115; c. page 115; d. page 114; e. page 114.
3.
When a solid melts, the intermolecular forces (London forces, hydrogen bonding, ionic bonding) that hold the
molecules together in a tight lattice are disrupted and the molecules acquire greater kinetic energy and
experience weaker intermolecular interactions. When a solid dissolves, the intermolecular forces holding the
molecules together are broken and replaced by intermolecular interactions between molecules of solid and
molecules of solvent.
4.
a. Since a melting point is an experimental measurement—the temperature when the solid starts to melt until
the temperature when it is completely melted—it is observed as a range of value and must be reported as
such, rather than a single discrete point.
b. Yes.
5.
A eutectic mixture exhibits a sharp melting point range, much like a pure substance; thus, such a mixture is
indistinguishable from a pure substance when making a melting point determination. Encountering a eutectic
mixture would be a rare occurrence because the mixture is eutectic at only one particular composition unique
to that particular binary mixture. All other compositions of that mixture would exhibit the typical profile of
an impure solid—depressed melting point over a wide range.
6.
The melting point would be depressed (lower temperature than the pure substances) and melt over a range of
temperatures (not a sharp melting point).
7.
It is undesirable because paper is a solid material that does not melt. Its presence in the sample makes it
difficult to determine the temperature at which the sample completely melts.
8.
Seal the top of the capillary tube in a flame so that the vapors cannot escape.
9.
a. Instead of turning from solid to liquid, these solids will remain solid or become an amorphous solid, while
changing in color, typically to brown or black, which is the result of thermal decomposition.
b. For example:
D-asparagine
[H2NCOCH2CH(NH2)CO2H] mp 235 °C (dec).
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Section 4. Answers to Exercises
10.
Experimental Organic Chemistry: A Miniscale and Microscale Approach
a. False. This statement is usually true, but the addition of an impurity occasionally will cause the melting
point to increase.
b. False. This statement is also usually true, but if the substance happens to be a eutectic mixture, it would
exhibit a sharp melting point.
c. False. The statement would be true if the melting point was neither lowered nor raised, since the addition
of an impurity occasionally causes the melting point to increase.
d. True.
11.
Combine a small amount of the pure sample with either benzoic acid or 2-naphthol, mix them thoroughly, and
determine the melting point of the mixture. If the mixture melting point is the same as either pure sample, the
identity of the sample has been determined. If the mixture melting point is lowered and broad, the pure
sample is not the compound with which it was mixed.
12.
The Thiele micro melting-point technique is not sufficiently accurate to permit one to determine the precise
temperature at which a compound melts. This technique always produces a melting point range.
13.
a. The melting points of pure and pure R are about 160 °C and 180 °C, respectively.
b. The composition of the eutectic mixture is about 67% Q and 33% R, and the melting point is about 80 °C.
c. This mixture will not melt at 120 °C or at 75 °C, but it will melt at 160 °C.
d. At 120 °C, the melting point-composition curve is intersected at two different points, corresponding to the
following approximate compositions: 45 mol % Q and 55 mol % R and 80 mol % Q and 20 mol % R.
From the melting point information alone, it is not possible to differentiate between these two possible sets
of compositions. However, one could, for example, add a little pure Q to the mixture, and determine the
melting point. If the it decreases, the composition of the original mixture would have been 45 mol % Q
and 55 mol % R and if it increases, the composition of the original mixture would have been 80 mol % Q
and 20 mol % R.
14.
When properly determined with a pure solid, a melting point range of no more than 0.5–1.0 °C should be
observed, but few compounds exhibit such a narrow melting range.
a. This compound is relatively pure.
b. This compound is very impure as judged by the broad melting point.
c. The purity of this compound cannot be assessed because impurities are introduced as the compound
decomposes, possibly lowering the melting point.
d. This compound is probably very pure.
15.
The recrystallization may have accidentally resulted in formation of the eutectic mixture, which would have a
sharp melting point. The reason a lower temperature was not reported for the solid prior to recrystallization
would be that too little of the solid melted at the eutectic temperature to be visible.
16.
Mix the unknown sample with salicylic acid in a 50:50 ratio and determine the melting point range. Repeat
this process separately with benzanilide and triphenylmethanol. Out of the three mixtures, the one exhibiting
a sharp melting point will be that in which the unknown and the additive are identical.
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