Answer to question 1.
The answers to this problem are in the tables in the section on Immiscibility.
a) 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.
b) 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.
c) hexanes and methylene chloride: these solvents are miscible and will not form two layers
d) 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.
e) ethanol and water: these solvents are miscible and will not form two layers
f) acetone and toluene: these solvents are miscible and will not form two layers
Answer to question 2.
Diethyl ether and water are called “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.
Answer to question 3.
Hexanes is the better choice because it is a lot less flammable than diethyl ether.
Answer to question 4.
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.
Answer to question 5.
The data can be summarized as follows:
| 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 |
*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.
Answer to question 6.
Five grams of Compound A is dissolved in 90 mL of water. The partition coefficient for Compound A between hexanes and water is 5:
KCompound A(hexanes/water) = 5
a) 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
b) 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 = x g in 30 mL hexanes/5-x g in 90 mL water
5 = 3x/5-x
x = 3.1 g
now, there is 5 - 3.1 g or 1.9 g of the compound in the water, therefore:
K = 5 = x g in 30 mL hexanes/1.9-x g in 90 mL of water
5 = 3x/1.9-x
x = 1.2 g
now, there is 1.9-1.2 g or 0.7 g of the compound in the water, therefore:
K = 5 = x g in 30 mL hexanes/0.7-x g in 90 mL of water
5 = 3x/0.7-x
x = 0.44 g
Now, add the numbers in bold: 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.
Answer to question 7.
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.
Answer to question 8.
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.
Answer to question 9.
a) Could be separated (amine soluble in acid and phenol soluble in strong base).
b) Could be separated (carboxylic acid soluble in weak base and phenol in organic, or then solubilize the phenol in strong base).
c) Could not be separated (both are carboxylic acids).
d) Could not be separated (both are not soluble in either acidic or basic aqueous solutions, so both would remain in the organic layer).
e) Could be separated (the amine would be soluble in acid while the ketone would remain in the organic layer).
Answer to question 10.
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.
Answer to question 11.
Answer to question 12.
Answer to question 13.
a)
b)
Answer to question 14.
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.
Answer to question 15.
The NaHCO3 wash serves to neutralize the acid and to remove water-soluble polar compounds.
Answer to question 16.
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.
