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WEBSITE DATA FOR CHAPTER 5
TECHNIQUES 2 and 3
Simple and Fractional Semimicroscale Distillation
Technique 2 – Simple Semimicroscale Distillation
2A. Distillation Theory
Distillation techniques often can be used for separating two or more components
on the basis of differences in their vapor pressures. Separation can be
accomplished by taking advantage of the fact that the vapor phase is generally
richer in the more volatile (lower boiling) component of the liquid mixture.
Molecules in a liquid are in constant motion and possess a range of kinetic
energies. Those with higher energies (a larger fraction for the lower boiling
component) moving near the surface have a greater tendency to escape into the
vapor (gas) phase. If a pure liquid (e.g., hexane) is in a closed container,
eventually hexane molecules in the vapor phase will reach equilibrium with
hexane molecules in the liquid phase. The pressure exerted by the hexane vapor
molecules at a given temperature is called the vapor pressure and is represented
by the symbol P°H where the superscript ° indicates a pure component. For any
pure component A, the vapor pressure would be P°A . Suppose a second
component (e.g., toluene) is added to the hexane. The total vapor pressure
(Ptotal) is then the sum of the individual component partial vapor pressures (PH,
PT), where PH is the partial pressure of hexane, and PT is the partial pressure of
toluene as given by DaIton's law.
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Ptotal = PH + PT
or in general
Ptotal = PA + PB + PC + ... + Pn
Assuming that the vapors are ideal, the mole fraction of hexane in the
vapor phase is given by
YH = PH/Ptotal (5.1)
It is important to realize that the vapor pressure (P°A ) and the partial
vapor pressure (PA) are not equivalent, since the presence of a second
component in the liquid system has an effect on the vapor pressure of the first
component. If the solution is ideal, the partial vapor pressure of hexane is given
by Raoult's law
PH = P°H XH (5.2)
where XH is the mole fraction of hexane in the liquid system.
For ideal solutions, Eqs. 5.1 and 5.2 may be combined to obtain the phase
diagram shown in Website figure 5.1W. In this figure and
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Website Fig. 5.1W
Pressure as a function of liquid composition (X) and vapor composition (Y), for
hexane and toluene (temperature held constant at 69 °C).
elsewhere we will drop the subscripts from XH and YH. Here X and Y will
represent the mole fractions of the most volatile component (MVC) (hexane in
this case) in the liquid and vapor phases, respectively.
Website figure 5.1W describes hexane and toluene mixtures at a fixed
temperature. For the region above the X curve, there will be only liquid present.
For the region below the Y curve, there will be only vapor. In the area between
(the sloping, lens-shaped region), liquid and vapor will be present in equilibrium.
This area is the only region of interest to us in examining the distillation process.
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To understand what the phase diagram tells us about the composition of
the liquid and vapor phases, let us imagine that the total pressure of the system
is 500 torr, shown by the horizontal line in new figure 5.2W. At this
Website Fig. 5.2W
Pressure as a function of liquid composition (X) and vapor composition (Y), for
hexane and toluene (temperature held constant at 69 °C).
pressure a liquid of composition X1 will be in equilibrium with a vapor of
composition Y1 . These two points are defined by the intersection of the constant
pressure line with the X and Y curves. It is important to note here that Y will be
greater than X for the equilibrium system. That is, the vapor in equilibrium with
a given liquid will be richer in the more volatile component than in the liquid.
Diagrams such as Website figure 5.2W are not very useful in describing
the distillation process. We need a phase diagram for the mixture at constant
pressure instead of constant temperature. Website figure 5.1W may be
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transformed to the desired diagram if we know how P°H and P°T depend on T.
This information may be supplied by the Clausius–Clapeyron equation or by
appropriate experimental data.
We will obtain a qualitative diagram for temperature as a function of
composition by the following reasoning: (1) The substance having the higher
vapor pressure at a given temperature will have the lower boiling point at a
given pressure. (2) At low temperatures, only the liquid phase will be present,
and at high temperatures only the vapor phase will be present. Thus, the
temperature-composition diagram is shown by Website figure 5.3W.
Website Fig. 5.3W
Temperature as a function of liquid composition (X) and vapor composition (Y).
Note that this figure may also be obtained (qualitatively) by turning Website
figure 5.1W upside down.
Many pairs of liquids do not obey Raoult's law. Often, pairs of liquids
encountered in organic chemistry exhibit a positive deviation from Raoult's law.
The positive deviation means that the pressure above the solution is greater than
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would be predicted by Raoult's law. If this deviation is large, the pressure
composition curve may exhibit a maximum, as shown in Website figure 5.4W.
Here the curves for normal Raoult's law behavior are
Website Fig. 5.4W
Positive deviation from Raoult’s law; L = liquid, V = vapor.
shown as dashed lines for reference. Mixtures in which one of the components is
polar and the other component is at least partly nonpolar often exhibit positive
deviations from Raoult's law.
The temperature–composition diagram for systems showing a positive
deviation from Raoult's law is again obtained by the simple inversion process.
Such a diagram is shown in Website figure 5.5W. In this
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Website Fig. 5.5W
Ethanol-water minimum boiling point phase diagram.
diagram at a temperature of T1 a liquid of composition X1 will be in equilibrium
with a vapor of Y1. At a temperature of Taz, however, the composition of the
liquid and vapor will be the same. This mixture is an azeotropic or constant-boiling
mixture. Water (H2O) and ethanol (CH3CH2OH) form one of the more familiar
azeotropic systems. This mixture exhibits a positive deviation from Raoult's law
and has a minimum boiling azeotrope at 78.1 °C, which consists of 95.6% ethanol
by volume.
2B. Steam Distillation
Another type of distillation and one which depends on materials which exhibit positive
deviations from Raoult's law is the technique of steam distillation.
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Theory
If two substances are immiscible (as the term “steam distillation” implies, one of
the substances will be water), the total vapor pressure (Pt) above the two-phase
liquid mixture is equal to the sum of the vapor pressures of the pure individual
components (Pi) according to Dalton's law.
Pt = P1 + P2 + ... + Pi
In other words, with an immiscible pair of liquids, neither component
lowers the vapor pressure of the other. The two liquids would exert the same
vapor pressure (at a given temperature) even if they were in separate containers.
If a mixture of these two liquids is heated, boiling will occur when the combined
vapor pressures of the liquids equals atmospheric pressure. At this point,
distillation will commence. Condensation of the vapors gives a two-phase
mixture (condensate or distillate) of the organic species and water. The
composition of this distillate is determined by the vapor pressure and molecular
weight of the compounds.
To illustrate this concept let us take an insoluble mixture of
bromobenzene (bp = 156 °C) and water. At 30 °C the vapor pressure of
bromobenzene is 6 torr; water is 32 torr. Therefore, the vapor pressure of the
mixture is the sum of 6 + 32 or 38 torr, at this temperature. At 95 °C, the vapor
pressure of bromobenzene is 120 torr; water is 640 torr. It follows that the sum is
now equal to 760 torr or 1 atm. That is, P°A + P°B = 1 atm.
As a result, the mixture boils and the bromobenzene and water distill
together. A further example is the isolation of cyclohexanone from nonvolatile
reaction byproducts by steam distillation. The normal boiling point of pure
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cyclohexanone is 156 °C and that for water is 100 °C. At about 94.5 °C, the vapor
pressure of cyclohexanone is 112 torr and the vapor pressure of water is 648 torr.
The vapor pressures of cyclohexanone and water add up to 760 torr. Thus the
two compounds steam distill at 94.5 °C.
Several important aspects of steam distillation are summarized below:
1. The mixture boils below the boiling point of either pure component.
In this regard the technique is similar to reduced pressure distillation in that the
liquid distills and condenses at a temperature below its normal boiling point. This
result occurs because compounds that are immiscible in water have very large,
positive deviations from Raoult's law.
2. The boiling point of the mixture will hold constant as long as both
substances are present to saturate the vapor volume.
3. The molar ratio of the two species in the distillate remains constant as
long as aspect 2 holds.
From Dalton's law, the condensate in a steam distillation will consist of
water and the compound in the same molar ratio as the ratio of their vapor
pressures (P°) at the steam distillation temperature. The relationship is
Moles waterMoles of organic species =
P° waterP° organic species
This relationship may be altered to show the weight relationship of the
organic substance to that of water.
Substituting grams per molecular weight (g/MW) for moles we obtain
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g/MW waterg/MW organic species =
P° waterP°organic species
Transposition and rearrangement of this expression leads to
g waterg organic species =
P° water x MW waterP° organic species x MW organic species
Based on this relationship, the weight of water required per weight of
organic species can be calculated. Note that if the vapor pressure of water is
known at the boiling temperature of the mixture that is being steam distilled, the
vapor pressure of the organic substance is 760 torr – P°water.
Example [2BW] of Technique 2B illustrates the application of steam distillation.
Example [2BW]: Isolation of a Natural Product by Steam Distillation: Cinnamaldehyde
From Cinnamon
Common name: cinnamaldehyde
CA number: [14371 - 10 - 9]
CA name as indexed: 2-propenal, 3-phenyl-, (E)-
Purpose. In this example you will extract oil of cinnamon from a native plant source, such as
Cinnamomum zeylanicum, and then purify the principal flavor and odor component of the oil,
cinnamaldehyde. The experiment demonstrates the importance of steam distillation
techniques (at the semimicro level) to the collection of essential oils.
ESSENTIALS OILS.
Let us begin by defining what we mean by the term “metabolite”. The metabolism of an
organism is composed of the biochemical reactions and pathways in that living system. The
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products (most of them organic molecules) derived from this array of molecular
transformations are the metabolites. This vast collection of substances that are generally
referred to as natural products are, in fact, the metabolites of the natural living world.
Natural products are divided into two large families of compounds. Those
metabolites that common to the large majority of all organisms are known as the primary
metabolites. In general, they have well-defined roles in the biochemistry of the system. For
example, the amino acids are the building blocks for protein synthesis in all organisms. The
second great category of natural products is known as the secondary metabolites. Individual
secondary metabolites are far less widely distributed in nature and may be unique to single
species (or even limited to a variety of a particular species). While the biochemical role of
some of these compounds was established early and easily, the majority of these materials
were believed to be of little importance to the functioning of the living system, and their
presence was unexplained until very recently. With the development of chemical
communication theory over the last few decades, however, the vitally important roles of
many of the secondary metabolites in the life cycles of their particular host organisms have
been revealed.
We will now examine a class of secondary metabolites termed essential oils. The
majority of these materials are high-boiling liquids that can be extracted from plant material
via steam distillation techniques. The value of codistilling high-boiling substances was
learned early in the days of alchemy. Because these oils often gave pleasant odors and
flavors, they were considered to be the “essence” of the original plant material. Eventually,
they became known as essentials oils.
These materials were used as flavorings, perfumes, and medicines, and as both
insect repellents and attractants. By the early 1800s, as it became possible to establish the
carbon/hydrogen ratio in organic substances, many of the oils possessing pine-type odors
(the oil of turpentine) were shown to have identical C/H ratios. These materials ultimately
became known as terpenes. The terpenes all have their origin in mevalonic acid, from which
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they utilize, as their building block, a branched five-carbon unit as in isoprene. The terpenes
of the essential oils occur as C10 (monoterpenes), C15 (sesquiterpenes), C20 (diterpenes), C30
(triterpenes), and C40 (tetraterpenes) compounds. Today, this collection of substances
represents a large fraction of the known secondary metabolites, including the steroids.
Terpenes may be polymerized, extending to much higher molecular weights, with between
1000 and 5000 repeating isoprene units (MW = 60,000–350,000) to yield polymers known as
the natural rubbers.
As we have seen, many of the compounds found in the essential oils possess
pleasing properties of taste and odor, and we now know that many of these systems contain
either ketone or aldehyde functional groups. Our senses of taste and smell, however, possess
a wide range of responses to the shape and dimensions of the carbon skeleton supporting
the main functionality that triggers the odor signal. Thus, our sense of odor may involve
simultaneous multiple stimulations by many different molecular species or, as in a number
of cases, the principal response may be to a single component. Since the shape of the odor-
or taste-inducing molecule plays a significant role in the effect, it is not surprising that
chirality can have a dramatic impact on our perception of a particular odor.
One of the classic examples of this type of response is the case of the cyclic ketone
carvone, which contains a single stereocenter (*). The S enantiomer is the principal odor and
flavor component in caraway seed, whereas the R enantiomer gives rise to the odor and
flavor of spearmint!
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What we find pleasant may be offensive to others. A constituent of the oil of lemon
grass is citronell, a C10 unsaturated aldehyde. While we find this compound to have a
pleasant fragrance, it is a potent alarm signal in ants that is shunned by many other insects.
Thus, this terpene aldehyde has been used effectively by both ants and humans as an insect
repellent.
In this example we will isolate the principal component of the oil of cinnamon,
another natually occurring aldehyde, cinnamaldehyde. The oil is first extracted from the
dried parts of the Cinnamomum plant by steam distillation. Although this aromatic aldehyde
is a component of an essential oil, it is not formed from mevalonic acid and is not a terpene.
Cinnamaldehyde is also not an acetogenin nor is it related to usnic acid. The origin of this
fragrant material is shikimic acid, which is part of the plant’s primary metabolism.
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Cinnemaldehyde’s formation fro shikimic acid utilizes one of only two biogenetic
routes in nature that lead to the aromatized benzene ring. The shikimic acid route
contributes to a class of metabolites called the phenylpropanes (Ph—C3), of which
cinnamaldehyde is one of a limited number of simple end products. Another close relative is,
for example, eugenol from oil of cloves.
The pricipal metabolic fate of the phenylpropanes is the formation of lignin
polymers that are the fundamental basis of the structural tissue in all plants. Thus,
cinnamaldehyde itself is a relatively rare example of a primary metabolite which has been
expressed in an essential oil.
DISCUSSION
Steam Distillation. The process of steam distillation can be a valuable technique in the
laboratory for the separation of thermally labile, high-boiling substances from relatively
nonvolatile materials. For steam distillation to be successful, however, the material to be
isolated must be nearly immiscible with water.
Steam distillation is in essence the codistillation (or simultaneous distillation) of two
immiscible liquid phases. By definition, one of these liquid phases is water and the other
phase is usually a mixture of organic substances that have a low solubility in water. Though
steam distillation is widely used as a separation technique for natural products, and
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occasionally for the isolation and/or purification of synthetic products that decompose at
their normal boiling points, it has several limitations. For example, it is not the method of
choice when a dry product is required or if the compound to be isolated reacts with water.
Obvioulsy, steam distillation is not feasible if the compound to be isolated decomposes upon
contact with steam at 100 °C.
Although cinnamaldehyde decomposes at its normal boiling point, it may be
extracted from the plant without degradation by boiling water. Steam distillation, therefore,
is the method of choice for the isolation of this pleasant smelling and tasting aldehyde.
COMPONENT
EXPERIMENTAL PROCEDURE
Estimated time to complete the experiment: 2.5 h.
Components and Equipment. Place 1 g of chopped stick cinnamon (or powder) and 4 mL of
water in a 10-mL round-bottom flask containing a boiling stone. Attach the flask to a
Hickman still head equipped with an air condenser (see equipment diagram, p. 37).
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Distillation Conditions. Place the apparatus on a sand bath maintained at 150–160 °C. Use
an aluminum foil shield or baffle (not shown in drawing) to cover the sand bath. This
procedure will prevent the collection collar of the Hickman still from overheating.
NOTE. The mixture tends to foam during the distillation, so care must be taken to prevent
contamination of the distillate by cinnamon particles, especially if powdered cinnamon is used. If stick
cinnamon is used, first evacuating the flask containing the water and cinnamon for a few minutes and
then returning the system to atmospheric pressure will allow water to fill the pores of the bark and
greatly aids in reducing the foaming problem (See Taber, D. F.; Weiss, A. J. J. Chem. Educ. 1998, 75
633).
Position a thermometer in the throat of the Hickman still to record the distillation
temperature, which should be very close to 100 °C. In early Hickman stills (without sidearm
access) this thermometer makes pipetting of the condensate difficult. It is suggested that in
the latter setup, the thermometer be removed after the initial distillation temperature is
established and recorded.
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Isolation of Cinnamaldehyde. Remove the milky cinnamaldehyde–water (two-phase)
distillate that collects in the collar of the still using a 9-in. Pasteur pipet (or 6-in. in the case of
sidearm stills). Transfer this material to a 12- or 15-mL centrifuge tube. Continue the
distillation for approximately 1 h or until about 5–6 mL of distillate is collected in the
centrifuge tube.
NOTE. Add additional water during the course of the distillation to maintain the original volume in
the flask. Add the water using a 9-in. Pasteur pipet inserted down the neck of the Hickman still, after
first removing the thermometer if it is still in place (the thermometer need not be placed in the still
following the addition).
Extract the combined distillate fractions, which you collected in the centrifuge tube,
with three successive 2-mL portions of methylene chloride (3 x 2 mL). Use the first portion of
methylene chloride to rinse the collection collar of the Hickman still. After each extraction,
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transfer the lower methylene chloride layer (Pasteur filter pipet) to a 25-mL Erlenmeyer
flask.
Dry the combined extracts over anhydrous sodium sulfate.
Purification and Characterization. Transfer the dried methylene chloride solution in at least
two portions, using a Pasteur filter pipet, to a tared 5-mL conical vial. Evaporate the solvent
in a warm sand bath using a slow stream of nitrogen gas. After all the solution has been
transferred and the solvent evaporated, rinse the sodium sulfate with an additional two 0.5-
mL portions of methylene chloride. Transfer the rinses to the same vial and concentrate as
before.
Weigh the flask and calculate the percentage of crude cinnamaldehyde extracted from
the original sample of cinnamon.
Record the infared spectrum and compare it to that reported in the Aldrich Library of
Infared Spectra.
To further characterize the aldehyde, prepare its semicarbazone derivative (see
Chapter 10, Preparation of Derivatives, Aldehydes and Ketones). It has a melting point of
215 °C.
QUESTIONS
5-1W. List several advantages and disadvantages of steam distillation as a method of
purification.
5-2W. Explain why the distillate collected from the steam distillation of cinnamon is cloudy.
5-3W. Calculate the weight of water required to steam distill 500 mg of bromobenzene at
95 °C. The vapor pressure of water at this temperature is 640 torr; that of
bromobenzene is 120 torr.
5-4W. Steam distillation may be used to separate a mixture of p-nitrophenol and
o-nitrophenol. The ortho isomer distills at 93 °C; the para isomer does not. Explain.
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5-5W. A mixture of nitrobenzene and water steam distills at 99 °C. The vapor pressure of
water at this temperature is 733.2 torr. What weight of water is required to steam
distill 300 mg of nitrobenzene.
BIBLIOGRAPHY
Several steam distillation procedures have been reported for isolation of essential oils from
native plant sources:
Bell, C. E.; Clark, A. K.; Taber, D. F.; Rodig, O. D. Organic Chemistry Laboratory: Standard and
Microscale Experiments, 2nd ed., Saunders College: Philadephia, 1996, 153.
Taber, D. F.; Weiss, A. J. J. Chem. Educ. 1998, 75, 633.
Vogel, A. I. Vogel’s Textbook of Practical Organic Chemistry, 5th ed.; Furnis, B. S., et al., Eds.;
Wiley: New York, 1989.
Zubrick, J. W. The Organic Chem Lab Survival Manual, 4th ed.; Wiley: New York, 1997.
TECHNIQUE 3 — Fractional Semimicroscale Distillation
3A. How Spinning Bands Function in Fractional Distillation Columns
To understand how the spinning band improves the performance of the
column, and to see some of the important characteristics of these fractionating
systems, we will analyze them a bit further. Website figure 5.6W shows a very
simple distillation column possessing just one plate.
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Website Fig. 5.6W
Model for fractional distillation with partial reflux.
Overall, however, this system has two theoretical plates.
The compositions of the liquid in the pot and of the infinitesimal amount
of liquid at the first plate are X0 and X1, respectively. Both Y0 and Y1 are
compositions of the vapor in equilibrium with each of the liquids, X0 and X1,
respectively. Since we are talking about the MVC, Y0 > X0, Y1 > X1, and we may
assume that Y1 > Y0 (i.e., we have already established that the vapor will be
enriched in the MVC). The parameter V is the rate at which vapor is transported
upward, L is the rate of downward transport of liquid back to the pot, and D is
the rate at which material is distilled. The parameters L, D, and V are related by
V = L + D. (What goes up, must come down.)
We will first look at the process qualitatively. If D and V are comparable
(L is small), as vapor at composition Y1 is removed, X1 becomes smaller—the
liquid in the first plate becomes less rich in the MVC as the MVC rich vapor is
removed. If, on the other hand, L and V are both large compared with D, the
composition at plate 1 will be maintained at Y0 by the large supply of incoming
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vapor with composition Y0. This relationship may also be seen in the following
quantitative discussion.
The first relationship we have already seen:
V = L + D (5.3)
In addition, if an insignificant amount of material is held up at plate 1, the
moles of MVC going into plate 1 must equal the moles of MVC leaving plate 1;
V . Y0 = L . X1 + D . Y1 (5.4)
We may eliminate V = (L + D) from Eq. 5.4 and rearrange to get
X1 = Y0 – (D/L) . (Y1 – Y0) (5.5)
Now, in Eq. 5.5, Y1 – Y0 will always be greater than 0 under ideal
conditions (since they both represent the MVC). Both D and L are positive.
Therefore, the upper limit for X1 is Y0. This situation would be the case when D/L
= 0 (and nothing is being distilled). In the real world something is being distilled
(D > 0) and X1 will be less than Y0 (and down the line the condensate will be less
rich in the MVC; there will be less separation). You should be able to generalize
this result for one plate for the situations in which we have n plates. When D/L is
small the composition of the first plate is at a maximum; therefore, Y1 is also a
maximum and, furthermore, X2 will be a maximum, and so on.
The purpose of the spinning band is to make D/L as small as possible so
that X1 approaches its upper limit of Y0. The spinning band ensures that the
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optimum separation of a pair of liquids will be achieved. We will have a lot of
vapor going up, a lot of liquid being returned by the spinning action of the spiral
band, and a relatively small amount of material actually being distilled. This
result also implies that if we want to achieve a high degree of separation, the
distillation rate should be low. There is, of course, a compromise between the
low rate of distillation required to obtain maximum separation and a rate that
will allow someone else to use the apparatus and you to get on to other things.
