M I N I S C A L E P R O C E D U R E
Preparation Sign in at www.cengage.com/login to answer Pre-Lab Exercises, access videos, and read the MSDSs for the chemicals used or produced in this pro- cedure. Read or review Sections 2.2, 2.4, 2.9, 2.13, 2.16, 2.19, 2.21, 2.29, and 4.4.
Apparatus A 125-mL separatory funnel, a 125-mL round-bottom flask, and appara- tus for steam distillation using an internal steam source, simple distillation, and flameless heating.
Setting Up Determine the weight of 2.5 mL of lemon grass oil, and then place it and 60 mL of water in the round-bottom flask. Equip this flask for steam distillation as shown in Fig. 2.44, except replace the steam inlet tube with a stopper; use the heating source suggested by your instructor. Have your instructor check your assembled apparatus before continuing the experiment.
Distillation Heat the liquid to boiling and adjust the heat source so that the distilla- tion proceeds as rapidly as possible; avoid applying excess heat to the stillpot, especially when the volume of water in the flask has been reduced below about 25 mL. Continue the distillation until oil droplets no longer appear in the distillate, which should occur after about 30 mL of distillate has been collected. Allow the dis- tillate to cool to room temperature or below, using an ice-water bath if necessary.夹
Isolation Transfer the cooled distillate to a separatory funnel and extract it sequen- tially with two 15-mL portions of diethyl ether. The funnel should be vented frequently to avoid the buildup of pressure. Transfer the organic layer from the separatory funnel to a 50-mL Erlenmeyer flask, and add about 0.5–1.0 g of anhy- drous calcium chloride. Allow the ethereal solution to remain in contact with the drying agent until the organic layer is dry, as evidenced by its being completely clear. If the experiment is stopped at this point, loosely stopper the flask and store it in a hood; never leave flasks containing diethyl ether in your locker drawer.夹
Chapter 4■ LiquidsDistillation and Boiling Points 149
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Decant the dried organic solution into a tared 125-mL round-bottom flask.
Equip the flask for simple distillation and evaporate the solvent using simple distil- lation. Alternatively, one of the techniques in Section 2.29 as directed by your instructor may be used to concentrate the solution. After the ether is completely removed, the pot residue is crude citral.
Analysis Determine the weight of citral isolated and calculate the percentage recov- ery of citral based on the weight of the original sample of lemon grass oil. If instructed to do so, save or submit a sample of citral for GLC or GC-MS analysis (Sec. 6.4). You may be asked to perform one or more of the following tests on citral.
1.Test for a carbon-carbon double bond. Perform the tests for unsaturation described in Section 25.8.
2.Test for an aldehyde. Perform the chromic acid test outlined in Section 25.7D.
3.Analyze the citral using gas chromatography to assess the purity of the product.
W R A P P I N G I T U P
Flush the aqueous solution remaining in the distillation flask down the drain. Do the same with the aqueous steam distillate once you have completed the extraction. Pour the solution from the test for unsaturation into the container for halogen-containing liquids. Neutralize the solution for the chromic acid test and then pour it into the con- tainer for hazardous heavy metals.
E X E R C I S E S
1.Why was a steam distillation rather than a simple distillation performed in the isolation of citral from lemon grass oil?
2.What type of product is expected from the reaction of citral with Br2/CH2Cl2? With chromic acid?
3.Provide structures for the semicarbazone and the 2,4-dinitrophenylhydrazone of citral.
4.Why does the citral float on the surface of the aqueous distillate rather than sinking to the bottom?
5.Explain why the substitution of 1-propanol, bp 97 °C (760 torr), for water in a steam distillation would not work.
6.Suppose that you are to steam-distill a sample of a natural product whose vapor pressure at 100 °C is known to be half that of citral. What consequence would this have on the amount of distillate required per mole of the natural product present?
7. Both geranial (1) and vitamin A (3) are members of the class of natural products called terpenes. This group of compounds has the common characteristic of being biosynthesized by linkage of the appropriate number of five-carbon units having the skeletal structure shown below. Determine the number of such units present in each of these terpenes and indicate the bonds linking the various individual units.
C C C C
C 150 Experimental Organic Chemistry■ Gilbert and Martin
Chapter 4■ LiquidsDistillation and Boiling Points 151 8.In the reduction of nitrobenzene to aniline (Eq. 4.12), the product is readily iso-
lated from the reaction mixture by steam distillation.
a. From this information, what do you know about the miscibility of aniline in water?
b. At approximately what temperature will aniline and water codistil?
c. What is the normal boiling point of aniline?
9.Using Raoult’s and Dalton’s laws, explain how the boiling point of a binary mixture depends on the miscibility of the two liquids.
H I S T O R I C A L H I G H L I G H T
Reducing Automobile Emissions
NH2 NO2 [H]
(reduction)
From a smog alert in Los Angeles, CA, to an “Ozone Action Day” in Austin, TX, it is clear that the air we breathe is under attack. The “aggressors” in this attack include volcanoes, plants, and trees. Volcanoes spew tons of noxious gases and particulate matter into the atmosphere during an eruption, and trees and plants create the haze that makes the Blue Ridge Mountains appear “blue.” This haze results from photochemical reactions between atmospheric oxygen and the volatile hydrocarbons that are produced by and emit- ted from the flora that cover these beautiful moun- tains. Humans are also important contributors to air pollution through activities such as power generation and manufacturing processes. Primary sources of the volatile chemicals that contribute to degradation of air quality are the automobiles we drive, the buses we ride, and the trucks we depend on in many ways.
These chemicals include carbon monoxide and car- bon dioxide, oxides of nitrogen and sulfur, abbreviated as NOxand SOx, respectively, and hydrocarbons, HC.
Unfortunately, we do not presently have a com- plete picture of how these emitted substances inter- act with one another in the atmosphere to cause environmental problems, but some broad generaliza- tions are possible. For example, carbon dioxide is judged to be the most significant “greenhouse gas”
contributing to global warming. Carbon monoxide depletes hydroxyl radicals, HO, in the atmosphere, and this may lead to formation of ozone under certain atmospheric conditions. Although stratospheric ozone is beneficial because it blocks harmful radiation from
penetrating the atmosphere—the existence of an ozone “hole” over the Earth’s poles has received wide publicity and is a problem of great environmen- tal concern—the presence of ozone at lower altitudes is not. Ozone reacts with carbon-carbon double bonds present in materials like rubber tires, thereby shortening their useful life. It also reacts with plant and animal tissues, including your skin, to cause health-related problems. The nitrogen oxides are formed by oxidation of atmospheric nitrogen at the high temperatures and pressures attending the com- bustion of gasoline and diesel fuel in air; they are pri- marily comprised of nitric oxide, NO. This oxide is converted to nitrogen dioxide, NO2, by atmospheric oxygen and sunlight, which, in turn, reacts with hydrocarbons to form ozone, O3, or with water to form nitrate, NO3. Both O3and NO2are very environ- mentally damaging. Nitrate in the form of nitric acid contributes to acid rain, which has a major negative impact on natural habitats such as lakes and forests.
Similarly, sulfur oxides, formed from oxidation of sul- fur-containing components in fuels, may ultimately be converted to sulfuric acid, which is another con- tributor to acid rain.
In many urban areas, one-third to one-half or even more of the NOx and HC pollutants are pro- duced by motor vehicles having internal combustion engines. Indeed, on a national basis, some 30% of all smog-forming emissions are produced by auto- mobiles, buses, and trucks, so they are obvious tar- gets for addressing the matter of air pollution. It is (4.12)
(Continued)
152 Experimental Organic Chemistry■ Gilbert and Martin
H I S T O R I C A L H I G H L I G H T Reducing Automobile Emissions (Continued) known that most pollutants are produced during
the period when the engine is warming to its normal operating temperatures. This is because only the lower-molecular-weight hydrocarbons in gasoline are efficiently burned (oxidized) at lower engine tem- peratures; their higher-molar-mass relatives may not burn at all or are only partially oxidized to carbon dioxide and water. So a key to decreasing the emis- sion of pollutants is to make the burning of gasoline more efficient in a cold engine.
The desired increased efficiency in burning can be effected through the technology illustrated on page 148. The gasoline in the regular tank is pumped into a chamber, in which the heat of the engine distills the gasoline. The lower-boiling hydrocarbons are selectively vaporized, condensed, and then stored in a special tank. With appropriate control devices, this more volatile “light” gasoline can be fed into the engine when the engine is cold; once it warms up, the “regu- lar” gas is used, and the “light” gasoline in the spe- cial tank is replenished, ready for use the next time a
“cold-start” is needed.
This simple system is remarkably effective in reducing emissions of pollutants: Emissions of hydro- carbon are decreased by some 50% and that of par- tially oxidized hydrocarbons by over 80%. Although the system has not yet been commercialized, Ford Motor Company is in the process of attempting to bring this technology to the marketplace, and other car, truck, and bus manufacturers are sure to follow if it works.
Relationship of Historical Highlight to Experiments
The device illustrated in the figure depends on distil- lation to effect separation of the lower-molecular- weight components of gasoline. This is a fractional distillation, in effect, one that potentially has impor- tant implications for lessening air pollution. As you may know, chemists have long used this technique to separate and purify volatile liquids having differing boiling points.
Engine
Fuel pump
Heated gasoline
Regular fuel tank Heated
vapor
3 1
2
4
(1) Vapor separator that uses the heat from the running engine to vaporize “regular” gasoline; (2) condenser in which vapors are condensed; (3) alternate fuel tank where the distillate, or “light”
gasoline, is stored; (4) valve controlled by vehicle’s computer, which, based on the temperature of the engine, delivers either
“light” or “regular” gasoline to the fuel pump and then to the engine.
Extraction
Extraction is a technique commonly used in organic chemistry to separate a material you want from those you do not. While the term extraction may be unfamiliar to you, the process is actually something you commonly perform. For example, many of you probably start the day, especially after a long night of studying, with an extraction when you brew a pot of coffee or tea. By heating coffee grounds or tea leaves with hot water, you extract the caffeine, together with other water-soluble compounds such as dark-colored tannins, from the solid material. You can then drink the liquid, which is certainly more enjoyable than eating coffee grounds or tea leaves, to ingest the caffeine and benefit from its stimulating effect. Similarly, when you make a soup, the largely aqueous liquid portion contains numerous organic and inorganic compounds that have been extracted from spices, vegetables, fish, or meat, and these give your culinary creation its distinctive flavor. In the procedures found in this chapter, you will have an opportunity to develop your existing experimen- tal skills further by isolating organic compounds using different types of extractions.
5.1 I N T R O D U C T I O N
The desired compound from a reaction is frequently part of a mixture, and its isolation in pure form can be a significant experimental challenge. Two of the more common methods for separating and purifying organic liquids and solids are recrystallization and distillation; these procedures are discussed in Chapters 3 and 4, respectively. Two other important techniques available for these purposes are extraction and chromatography. As you will see here and in Chapter 6, both of these methods involve partitioning of compounds between two immiscible phases. This process is termed phase distribution and can result in separation of compounds if they distribute differently between the two phases.
Distribution of solutes between phases is the result of partitioning or adsorption phenomena. Partitioning involves the difference in solubilities of a substance in two immiscible solvents—in other words, selective dissolution. Adsorption, on the other hand, is based on the selective attraction of a substance in a liquid or gaseous mixture to the surface of a solid phase. The various chromatographic techniques depend on both of these processes, whereas the process of extraction relies only on partitioning.
Extraction involves selectively removing one or more components of a solid, liquid, or gaseous mixture into a separate phase. The substance being extracted will partition between the two immiscible phases that are in contact, and the ratio of its distribution between the phases will depend on the relative solubility of the solute in each phase.
153
C H A P T E R 5
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5.2 T H E O R Y O F E X T R A C T I O N
Liquid-liquid extractionis one of the most common methods for removing an organic compound from a mixture. This process is used by chemists not only in the isolation of natural products but also in the isolation and purification of products from most chemical reactions.
The technique involves distributing a solute, A, between two immiscible liquids, Sx, the extracting phase, and So, the original phase. The immiscible liquids normally encountered in the organic laboratory are water and some organic solvent, such as diethyl ether, (C2H5)2O, or dichloromethane, CH2Cl2. At a given temperature, the amount of A, in g/mL, in each phase is expressed quantitatively in terms of a constant, K, commonly called the partition coefficient (Eq. 5.1). Strictly speaking, the volumes of solution should be used in the definition of [A], but if the solutions are dilute, only slight errors result if volumes of solvent are used. Furthermore, a close approximation of the partition coefficient K may be obtained by simply dividing the solubility of A in the extracting solvent Sxby the solubility of A in the original solvent So.
(5.1) The process of liquid-liquid extraction can be considered a competition between two immiscible liquids for solute A, with solute A distributing or parti- tioning between these two liquids when it is in contact with both of them. The mathematical expression of Equation 5.1 shows that at equilibrium, the ratio of con- centrations of A in the two phases will always be constant.
Equation 5.1 leads to several important predictions. For example, if K > 1, solute A will be mainly in the extracting solvent, Sx, so long as the volume, Vx, of this solvent is at least equal to the volume, Vo, of the original solvent So; the amount of solute remaining in So will depend on the value of K. By recasting Equation 5.1 into Equation 5.2, it is easily seen that if the volumes of Sxand Soare equal, the value of K is simply the ratio of the grams of A in each solvent. Because the product of the ratios Ax/Aoand Vo/Vxmust be constant for a given solvent pair, increasing the volume of extracting solvent Sxwill result in a net increase in the amount of solute A in Sx.
(5.2) There are practical, economic, and environmental reasons, however, that limit the quantities of organic solvents that can realistically be used in extractions. The question thus arises as to whether it is better to perform a single extraction with all of the solvent, or to perform several extractions with smaller volumes. For example, if an organic compound is dissolved in 10 mL of water, and only 30 mL of diethyl ether is used, will three extractions with 10-mL portions of ether provide better recovery of solute than a single one with 30 mL?
Applying Equation 5.2 would provide the answer, but the process is tedious for multiple extractions. This equation can be generalized, however, to accommodate multiple extractions in terms of the fraction, FA, of solute A remaining in the original solvent Soafter n extractions, using a constant volume of an immiscible solvent Sx for each extraction. Thus, the amount of solute remaining in Soafter one extraction with Sxis obtained by rearranging Equation 5.3 to Equation 5.4. The fraction, FA, of A still in the original solvent is obtained through Equation 5.5, in which Ciand Cfare the initial and final concentrations, respectively. For a second extraction,
K ⫽grams of A in Sx
grams of A in So ⫻ mL of So mL of Sx ⫽ Ax
Ao ⫻Vo Vx K ⫽[A] in Sx
[A] in So 154 Experimental Organic Chemistry■ Gilbert and Martin
Chapter 5■ Extraction 155 Equations 5.6–5.8 result. Equation 5.8 can be generalized to Equation 5.9 when n extractions are performed.
(5.3) where Ao⫽amount (grams) of solute in Sobefore extraction
A1⫽(Ao– Ax)⫽amount (grams) of solute in Soafter extraction Voand Vx⫽volume (mL) of original and extracting solvents, respectively
(5.4)
(5.5) where Ci⫽Ao/Soand Cf⫽A1/So, respectively
(5.6)
(5.7)
(5.8)
(5.9) Performing calculations according to Equation 5.9 for the case of K⫽5, we obtain FA⫽1/16 for a single extraction with 30 mL of diethyl ether, and 1/216 for three extractions with 10-mL portions. This means that 6.3% of A remains in the aqueous phase when one extraction is performed, whereas the value drops to only 0.5% in the case of three successive extractions with the same total volume of solvent. The amount of A that could be isolated is increased by some 6% with multiple extractions. Setting K at 2 gives corresponding values for FAof 1/7 and 1/27, respectively, which translates to a 10% increase in the quantity of A that is removed from the aqueous layer with multiple extractions.
From these calculations, it is clear that multiple extractions become increas- ingly important as the value of K decreases. The improved recovery of solute from multiple extractions, though, must be balanced with the practical consideration that the relatively small increase in recovery may not justify the additional time and solvent required to perform multiple extractions unless the product is of great value. Importantly, the partition coefficients K are generally large for the extrac- tions required in the procedures provided in this textbook, so only one or two extractions are usually required.
Selection of the appropriate extracting solvent is obviously a key for successfully using this technique to isolate and purify compounds, and important guidelines for making the correct choice are summarized here.
1.The extracting solvent must not react in a chemically irreversible way with the components of the mixture.
FA = Cf
Ci = a Vo KVx + Vobn FA¿ =
Cf
Ci = a Vo KVx + Vob2 K = Aoa Vo
KVx + Vob2 K = aA1 - A2
Vx b aVo A2b FA =
Cf Ci =
Vo KVx + Vo A1 =
(AoVo) KVx + Vo K ⫽ aAo - A1
Vx b aVo A1b
2.The extracting solvent must be immiscible, or nearly so, with the original solution.
3.The extracting solvent must selectively remove the desired component from the solution being extracted. That is, the partition coefficient K of the component being removed must be high, while the partition coefficients of all other components should be low.
4.The extracting solvent should be readily separable from the solute. Use of a volatile solvent facilitates its removal by simple distillation or one of the other techniques outlined in Section 2.29.
5.3 B A S E A N D A C I D E X T R A C T I O N S
The discussion of Section 5.2 focuses on partitioning one substance between two immiscible solvents. Now consider what happens if a mixture of two or more compounds is present in a given volume of solvent Soand an extraction using a solvent Sxis performed. If the partition coefficient K of one component, A, is signifi- cantly greater than 1.0 and if those of other components are significantly less than 1.0, the majority of A will be in Sx, whereas most of the other compounds will remain in S0. Physical separation of the two solvents will give a partial separation, and thus purification, of the solute A from the other components of the mixture.
Solutes differing significantly in polarity should have very different coeffi- cients K for partitioning between nonpolar and polar solvents. For example, con- sider the distribution of two organic compounds, the first neutral and nonpolar, and the second ionic and polar, between a nonpolar solvent and a polar solvent. If a solution of these compounds in the nonpolar solvent is shaken with the polar solvent, the neutral compound will preferentially partition into the nonpolar phase, whereas the polar constituent will preferentially partition into the polar phase.
Separating the two phases effects a separation of the two solutes.
Carboxylic acids and phenols (1 and 3, Eqs. 5.10 and 5.11, respectively) are two classes of organic compounds containing functional groups that are polar and hydrophilic (water-loving). Unless they contain fewer than about six carbon atoms, such compounds are generally either insoluble or only slightly soluble in water because of the hydrophobic (water-avoiding) properties of the carbon- containing portion, R or Ar, of the molecule. They are soluble in common organic solvents like dichloromethane or diethyl ether that have at least modest polarity (Table 3.1, p. 96). Consequently, carboxylic acids or phenols dissolved in diethyl ether, for example, will largely remain in that phase when the solution is extracted with water.
R C O O
H
B–Na+(aq) R C O– O
B Na+ + H +
1 2
A carboxylic acid Water insoluble Kwater/org < 1
A base (as sodium salt)
A carboxylate (as sodium salt) Water soluble Kwater/org > 1 156 Experimental Organic Chemistry■ Gilbert and Martin
(5.10)