Extraction Technique Quiz
1) 1) Consider the following solvent pairs. If mixed together, which pairs would form two layers? If they form two layers, which solvent would be on top?
- Hexanes and water
- Water and methylene chloride
- Hexanes and methylene chloride
- Methanol and hexanes
- Ethanol and water
- Acetone and toluene
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The answers to this problem are in the tables in the section on Immiscibility.
- Hexanes and water: these are immiscible so will form two layers; hexanes (d=0.68) will be on top and water (d=1.0) will be on the bottom.
- Water and methylene chloride: these are immiscible so will form two layers; water (d=1) will be on the top and methylene chloride (d=1.33) will be on the bottom.
- Hexanes and methylene chloride: these solvents are miscible and will not form two layers.
- Methanol and hexanes: these are immiscible so will form two layers; hexanes (d=.68) will be on the top and methanol (d=.79) will be on the bottom.
- Ethanol and water: these solvents are miscible and will not form two layers.
- Acetone and toluene: these solvents are miscible and will not form two layers.
2) You have 100 mL of water and add 4 mL of ether to it. You notice only one layer. Then, you add 6 mL more and see a small layer of a clear liquid on top of the water. Explain what is happening.
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Diethyl ether and water are immiscible, however, the term immiscible is not an exact one. Ether is soluble in water to a small extent. As noted in the table in the middle of the miscibility section, 5.6 mL of ether is soluble in 100 mL of water. So, if only 4 mL of ether is added to 100 mL of water, it will dissolve in the water. Once over 5.6 mL ether per 100 mL water is added, it will form a separate layer. Since ether is less dense than water, the ether will form a layer on top of the water.
3) You want to extract a compound from water, and have determined that the compound is equally soluble in both hexanes and diethyl ether. Which solvent would be the better choice for the extraction?
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Hexanes is the better choice because it is a lot less flammable than diethyl ether, and because its solubility in water is lower.
4) The distribution coefficients for a compound are as follows:
- Kcyclohexane/water = 1.5
- Kpentane/water = 11.2
- Kdiethyl ether/water = 5.1
You have an aqueous mixture of the compound. Which solvent system, cyclohexane/water, pentane/water, or diethyl ether/water, would give the most efficient extraction into the organic solvent?
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Pentane/water would be the most efficient extraction system, since it has the highest partition coefficient. This means that for the three solvents cyclohexane, pentane, and diethyl ether, an extraction with pentane would result in the highest amount of the compound in the organic phase.
5) One gram of a compound requires the following quantities of solvent to dissolve: 47 mL of water, 8.1 mL of chloroform, 370 mL of diethyl ether, or 86 mL of benzene. Calculate the solubility of the compound in these four solvents (as g/100 mL). Estimate the partition coefficient of the compound between chloroform and water, ethyl ether and water, and benzene and water. Which solvent would you choose to extract the compound from an aqueous solution?
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The data is summarized in the table below. K is calculated from K = Corganic solvent/Cwater. Chloroform would be the solvent of choice to extract the compound from an aqueous solution because the partition coefficient for this pair of solvents is the highest.
| Solvent |
Given |
g/100 mL |
K |
| Water |
1 g/47 mL |
2.1 |
- |
| Chloroform |
1 g/8.1 mL |
12.3 |
5.9 |
| Diethyl ether |
1 g/370 mL |
.27 |
.13 |
| Benzene |
1 g/86 mL |
1.2 |
.57 |
6) Five grams of Compound A is dissolved in 90 mL of water. The distribution coefficient for Compound A between hexanes and water is 5 (in other words, KCompound A(hexanes/water) = 5.)
- How much of Compound A will be in the hexanes if you extract it from the water one time with 90 mL of hexanes?
- How much of Compound A will be in the hexanes if you extract it from the water with three sequential extractions using 30 mL of hexanes each time, and then combine the hexanes extracts?
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- How much of Compound A will be in the hexanes if you extract it from the water one time with 90 mL of hexanes?
K = 5 = x g in 90 mL/5-x g in 90 mL
5 = x/5-x (90 mL cancels)
x = 4.2 g
- How much of Compound A will be in the hexanes if you extract it from the water with three sequential extractions using 30 mL of hexanes each time, and then combine the hexanes extracts?
K = 5 = x1 g in 30 mL hexanes/5-x1 g in 90 mL water
5 = 3x1/5-x1
x1 = 3.1 g
Now, there is 5 - 3.1 g or 1.9 g of the compound in the water, therefore:
K = 5 = x2 g in 30 mL hexanes/1.9-x2 g in 90 mL of water
5 = 3x2/1.9-x2
x2 = 1.2 g
Now, there is 1.9-1.2 g or 0.7 g of the compound in the water, therefore:
K = 5 = x3 g in 30 mL hexanes/0.7-x3 g in 90 mL of water
5 = 3x3/0.7-x3
x3 = 0.44 g
Now, add x1, x2 and x3: 3.1 + 1.2 + 0.44 g, and you get 4.74 g. Compare this value with the 4.2 g obtained with one extraction with 90 mL of hexanes (a, above), and you see why the efficiency of extraction is improved if you split the organic extraction solvent up into several portions and do multiple extractions.
7) You carefully purified phenanthrene from a very long, complicated procedure. It took days. It is very important for you to hand the purified phenanthrene in to your TA in one hour to get a good grade. Oops! You accidentally put the phenanthrene into a vial which you thought was clean, but instead had a lot of NaCl clinging to the sides. What can you do to quickly purify the phenanthrene again?
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Phenanthrene is very soluble in organic solvents, like hexanes, while sodium chloride is soluble in water. You could purify the phenanthrene by adding water and hexanes, shaking, and allowing the layers to separate. Separate and save the organic layer containing the phenanthrene, throw away the aqueous layer containing the sodium chloride. Dry the organic layer to re-isolate the phenanthrene.
8) Show the product of the following reactions. If there is no reaction, write "NR".
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Note on i) - under the right conditions, HCl will react with C6H5OH to give C6H5Cl. However, under the conditions for extraction, it will not react to form an ionic species.
9) Which of the following compounds could be separated by acid-base extraction?
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- Could be separated (amine soluble in acid and phenol soluble in strong base).
- Could be separated (carboxylic acid soluble in weak base and phenol in organic, or then solubilize the phenol in strong base).
- Could not be separated (both are carboxylic acids).
- Could not be separated (both are not soluble in either acidic or basic aqueous solutions, so both would remain in the organic layer).
- Could be separated (the amine would be soluble in acid while the ketone would remain in the organic layer).
10) If you test an unknown organic compound and find it to be soluble in 5% aqueous NaHCO3, what functional group does it probably contain?
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It is most likely a carboxylic acid. Phenols will not be water-soluble unless a strong base is used (10% NaOH), and amines are only water-soluble if acid is used. Most other organic compounds are not water soluble.
11) You are going to extract a compound into diethyl ether from water, using two successive extractions. Make a detailed step-wise summary of all operations in the procedure, paying attention to the location of the solute at each step and the position of each layer in the separatory funnel. Start with step 1 as "Put the separatory funnel in the ring on the ringstand."
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- Put the sep funnel in the ring on the ringstand.
- Put the aqueous solution containing the compound you want into the sep funnel.
- Add ether and shake.
- Allow the layers to separate; the ether will be on top and most of your compound will be in it, the aqueous layer is on the bottom.
- Drain off the lower aqueous layer and save it in a flask.
- Pour the ether layer from the sep funnel into a flask and save it.
- Pour the aqueous layer back into the sep funnel.
- Add a fresh portion of ether to the sep funnel.
- Shake and then allow the layers to separate; the ether will be on top and most of your compound will be in it, the aqueous layer is on the bottom.
- Drain off the lower aqueous layer; you do not need this layer anymore since most of the compound has been extracted into the ether.
- Pour the ether layer from the sep funnel into the same flask as before (step 6) - this is where your compound is, in the ether.
12) Repeat the previous problem using methylene chloride as the extraction solvent.
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- Put the sep funnel in the ring on the ringstand.
- Put the aqueous solution containing the compound you want into the sep funnel.
- Add methylene chloride and shake.
- Allow the layers to separate; the methylene chloride will be on botom and most of your compound will be in it, the aqueous layer is on the top.
- Drain the lower methylene chloride layer and save it in a flask.
- Leave the aqueous layer in the sep funnel and now add a fresh portion of methylene chloride to it.
- Shake and then allow the layers to separate; the methylene chloride will be on bottom and most of your compound will be in it, the aqueous layer is on the top.
- Drain the lower methylene chloride layer into the same flask as before (step 5) - this is where your compound is, in the methylene chloride.
- You do not need the aqueous layer in the sep funnel anymore since most of the compound has been extracted into the methylene chloride.
13) Outline in a flowchart how you would separate the following compounds:
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14) A reaction work-up for an aqueous reaction mixture calls for extraction first with diethyl ether and then a wash with saturated aqueous sodium chloride. What is the purpose of the saturated sodium chloride wash?
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Saturated sodium chloride serves two functions in an extraction procedure. It pulls water from the organic layer to dry it, and it also helps force the organic compound into the organic layer. To some extent, the saturated sodium chloride will also pull inorganics from the organic layer into the aqueous layer.
15) You have prepared an organic compound by reacting it with strong aqueous acid. The work-up calls for extraction into methylene chloride, then a wash with 5% NaHCO3. What is the purpose of the NaHCO3 wash?
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The NaHCO3 wash serves to neutralize the acid and to remove water-soluble polar compounds.
16) What can you do if you do not know which layer is which in an extraction procedure?
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Drop a small amount of water into the neck of the separatory funnel. Watch it carefully: if it remains in the upper layer, that layer is the aqueous layer. If it sinks to the bottom of the upper layer to the interface between the two liquids, the bottom layer is the water layer. If this does not work, remove a small amount of the top layer from the separatory funnel and place it in a test tube. Add a small amount of water to the test tube, mix it and allow it to settle: if you now see two layers, the top layer in the separatory funnel is the organic layer. If you see only one layer, the top layer in the separatory funnel is the aqueous layer.
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