Question

Exercise 2 Separation of a Mixture Based on Acid-Base Properties

One purpose of this exercise is to learn how to use a separatory funnel to extract a single component away from other compounds in solution. To do so, we will apply the principles of solubility and acid-base behavior you’re seeing in class. One of the compounds is neutral in the acid-base sense. It has no ability to either donate or accept a proton from an aqueous solution, and will remain uncharged at all times. One of the three is very weakly acidic, a phenol, and the third is more strongly acidic, a carboxylic acid. The neutral compound will not respond to changes in pH but the other two will. The three compounds are listed below, with their structures.

Hypothesis: We should be able to completely separate a three-component mixture, one compound at a time, based on solubility properties related to the compound’s pKa, at a specific pH.

Why is this process so important that it’s the second in a series of lab skills that we teach you? One reason is that it applies principles you’ve learned in class to a real-world situation. Acid-base chemistry is one of the chief concepts in an organic chemistry course.

Another reason is that the outcome of a laboratory synthesis in organic chemistry is messy, and there needs to be a way to isolate and clean up the desired product. Organic synthesis, the “art of making new organic compounds,” is not like your experiences in the kitchen. There, for example, if you decide to make a cake, you end up with the cake, and a few dishes to wash. A normal consequence of doing an organic reaction on the other hand, is the fact that the desired products usually form within a messy mixture of side-products and unreacted starting compounds, which must be removed so that the product can be isolated, purified, and its identity confirmed. The sequence of steps that accomplishes this, once the reaction is complete, is called the workup. Exactly what the organic chemist decides to do in the workup depends on the component’s physical and chemical properties — usually meaning the polarities and acid-base character of the products, by-products, and reactants, catalysts, and solvents that may be present. As you saw in class and in Ch 2, entire classes of compounds can be sent into, or made to precipitate out of solution by the choice of solvent or a change in pH. This can be very useful. Extraction is a workup procedure.

Key ideas

Since each of the three compounds in our mixture has more than five carbons, we expect that none will be water-soluble. In fact, since all three have a dipole, we predict that if we choose a moderately non-polar solvent that also has a dipole, all three may dissolve in it. (“Like dissolves like.”)

Diethyl ether is such a solvent (“ether” and ethyl ether are synonyms), and once all three are dissolved in it, we can combine the ether solution with a water solution of a weak base (NaHCO3). Ether and water do not dissolve appreciably in each other, and will separate into two layers. The ether layer floats on the water because it is less dense than water.

When these two layers are shaken together, there will be just enough contact of the base (dissolved in the water) and the 1-naphthoic acid, (dissolved in the ether) to cause the acid and base to react to form a water-soluble salt. What we will have done is to cause one component of the mixture to become soluble in water (the naphthoic acid now in its ionic form) and at the same time, because it has an ionic charge, insoluble in the ether.

If we drain off the water and set it aside, and extract the ether solution again, with a fresh portion of the bicarbonate solution, we can remove more of the naphthoicacid from the ether. This is a statistically controlled process. Each time the ether solution is shaken with the aqueous base solution, a specific percentage of the acid is converted to its water-soluble ionic form. Thus if you extract only one time, you will leave a large amount of unchanged acid behind in the ether. Two extractions are better, and three are usually recommended. Each time, the new extract (here, the water) is added to the previous extract. That is, we combine all the same extracts into one flask. In a written procedure, you would see the phrase, “The aqueous phases were combined…” or “the ether phases were combined and…” How can we now separate the remaining two components? Let’s review equilibrium, equilibrium constants, and pKa.

• A good “working definition” of pKa states that when the pKa of the acid (or base) is equal to the pH of the solution it’s dissolved in, then 50% of the compound is in its ionic form (and is therefore water-soluble), and 50% of the compound is in its neutral form, (and is therefore water-insoluble if the compound has more than 5 carbons). For example, the pKa of 2-naphthol is 9.5. If you place a sample of this compound in a very basic water solution at pH 11.5, now two pH units to the “base side” of the pKa, there is over one hundred times more of the 2-naphthol in the ionized (and therefore water-soluble!) form than in the non-ionized form, because there are so many hydroxide ions present that can remove the proton from the hydroxyl-group. Remember that both pKa and pH are base-10 logarithmic scales. A change of 1 pH unit is a 10-fold change in concentration. However, if you add some acid and begin to neutralize this basic solution, let’s say to a pH of 9.5, then only 50% of the molecules are still ionized (and water-soluble.) (What’s true about the other 50%?) If you continue adding acid, until the solution is at a pH of 7.5, now over one hundred times fewer of the 2-naphthol molecules are ionized than before. What would you see happening in that solution?

If we started to add base again to the solution just described, until the pH of the aqueous phase was now 13.5, what would you see? And why?

Study the diagram below for a “visual” explanation that students often find helpful.

What can be easily lost in this process is the concept that these numbers represent statistical, equilibrium-controlled amounts. When it looks like the 2-naphthol is precipitating nicely at pH 7.5, how much of this compound is still ionized and therefore still in solution? Out of one gram of 2-naphthol, what amount is still dissolved at pH 7.5?

Using the same kind of example, this time for 1-naphthoic acid, the diagram would look like this.

The pKa of 2-naphthol is 9.5, and the pKa of 1-naphthoic acid is 3.7.

Our advice to you is to test both the NaOH and the sodium bicarbonate solutions with pH paper before you start. Write the observation in your lab notebook. Think about how much of each of the compounds would be soluble or insoluble at those two pH regions.

Because the pKa of the phenol is higher than that of the carboxylic acid, it is a much weaker acid than a carboxylic acid and it requires a much stronger base for solubility. Therefore if we treat the ether solution with an aqueous NaOH solution, a strong base, we should be able to cause the following “solubility switch” to function:

Now we’ve caused the 2-naphthol to become water soluble (because most of it is in its ionic form at this pH) but insoluble in the ether (because its ionic charge makes it far too polar to dissolve in a non-polar solvent). Again we need to repeat the extraction using a fresh portion of the sodium hydroxide solution, and combine the water extracts.

After all of these extractions, what remains in the ether should be only one compound, the less polar and neutral 2-methoxynaphthalene. (Is it really the only compound that could be left in the ether?) To purify this 2-methoxynaphthalene/ether solution, we must wash the ether solution of its impurities, and then remove traces of water and other water-soluble materials, which we discard. The process is identical, and we will use a saturated NaClsolution this time, but we will discard the water layer instead of saving it. (It’s just the wash water, not the extract, see?) Washing an ether solution with a very concentrated salt solution in order to “pull” water from the ether is another example of osmotic processes at work.

Do NOT discard any of the three products isolated today! Save all of them. The three products from this Exercise will be used over the next 2-3 weeks.

Procedure, step 1. Obtain 1.20 g of the mixture. Assume that it contains equal amounts (by weight) of each of the three components. Dissolve it in 40 mL of ether, and pour the solution into your separatory funnel.

Now add 25 mL of 5% NaHCO3 solution, a weak base, to the ether solution in theseparatory funnel, and extract (as demonstrated), draining the lower, aqueous layer in a labeled 125 mL Erlenmeyer. Repeat with a fresh 25 mL portion of the weak base, pouring this into the separatory funnel containing the ether. Extract again. Combine the water extracts in the same 125 L flask. You now have a flask containing water, sodium bicarbonate, and which of the three compounds from the mixture? Write this in your notebook! Your separatory funnel should still contain the ether solution, now minus one of the original three compounds.

Based on the appearance and the Rf values from your reference card done at the start of the exercise, identify each spot. The spot shows the presence of that substance in the solution. A single spot visible to UV light and iodine vapor indicates that there is only one compound present. This is what we hope to see aswe remove one compound at a time from the mixture.

To bring back the solid naphthoic acid, take your erlenmeyer flask containing the aqueous bicarbonate extracts to the HOOD, add concentrated HCl in small portions to the combined water extracts, stirring with a glass rod, until the solution is distinctly acidic. Watch for fizzing and foaming! Then chill the slurry in an ice-water bath. (Why chill it? What does temperature have to do with solubility? Why insist on a glass stirring rod?)

Set up a filtration system as demonstrated using a filter flask connected to the two lengths of black rubber tubing joined with a plastic check-valve (arrow pointing toward the sink), filter adapter, and the small white, porcelain Hirsch funnel with a filter paper disk. Be sure to clamp the flask to a ring stand for support. (If you have a lot of solid, use the larger Büchner funnel instead.) Wet the filter paper, start the suction, and pour the chilled 1-naphthoic acid slurry onto the filter paper, rinsing with a little ice-cold water that has a drop of HCl in it. (Why the HCl? What does pH have to do with 1-naphthoic acid’s solubility in water?)

Let air flow through the crystals. Gently lift out the damp filter paper and solid with a bent spatula as demonstrated, place it on a watch glass, and then set this aside in your lab drawer to finish drying until next week. You cannot determine the mass until the solid is completely dry.

Wastes: Pour the water from the filter flask, neutralized with sodium bicarbonate,into the bottle labeled “Contaminated Aqueous Waste,” in the hood.

Procedure, step 2. Now add 25 mL of 5% NaOH solution, the strong base, to the ether in the separatory funnel and extract. In the same way that you did for Part 1, repeat with a fresh 25 mL portion of the strong base, combining and saving the aqueous layers in a second labeled erlenmeyer. Again, leave the ether solution in your stoppered separatory funnel.

Don’t forget to make a new TLC card showing the mixture, the water phase, and the remaining ether phase after the Part 2 extraction. It should resemble the drawing on the previous page. Draw an exact facsimile of its final appearance into your lab notebook.

To bring back the solid 2-naphthol, take your erlenmeyer flask containing the aqueous sodium hydroxide extracts to the HOOD as before, acidify with HCl as you did in the first part, chill, and filter. No fizzing will be seen. (Why? What gas was given off before? Would you expect to see that when using NaOH?) The solid will be a light tan color. The solid will be damp and cannot be weighed yet. Save this on another watch glass in your drawer for next week.

Wastes: Pour the water from the filter flask, neutralized with sodium bicarbonate,into the bottle labeled “Contaminated Aqueous Waste,” in the hood.

Procedure, step 3. Finally, wash the ether layer that remains in your separatoryfunnel by adding a 25 mL portion of a saturated NaCl solution, mixing well and separating as before, and make another TLC card as before, drawing the facsimile in your notebook. This time however, you discard the aqueous layer, in the “contaminated aqueous wastes” container. Carefully transfer the ether to a clean 50 or 125 mL erlenmeyer flask and add small amounts of anhydrous sodium sulfate as demonstrated, until the added solid no longer appears clumped or “chunky.” Wait about 10-15 minutes (patience!), then decant (transferring the ether but none of the solid) into a pre-weighed, 50 mL or 100 mL round-bottom flask, rinsing the drying agent left behind with a few mL of clean ether to complete the transfer, and recover the solid 2-methoxynaphthalene by evaporating the ether on a rotary evaporator. You will be shown how to use this apparatus during the lab time. When your solid appears, you can stop the machine, remove the flask, and carefully wipe it dry. Record the dry weight of flask-plus-compound and subtract the weight of the empty flask to obtain the mass of the compound. Record data in your notebook! Transfer this solid to a labeled vial and save it for next week. (Hint: Use a spatula bent into a curve at the end, and scrape the solid out onto a piece of weighing paper. Set your vial on another piece of weighing paper and transfer. If you spill any, it can be captured on the paper. No losses!)

What consequence would there have been if you had extracted our original three compounds with the NaOH first, then with the bicarbonate?

Why did we use concentrated HCl used instead of dilute HCl when you recovered your compounds? After all, the final pH would have been exactly the same either way.

Answer 1

I hope the questions are in bold and not somewhere in text given above.

The extraction is done with bicarbonate first and then with NaOH. But if you are doing it in reverse order, then extraction will not be proper. The three compounds mentions in the text are 2-methoxy naphthalene, 2-naphthol and naphthoic acid. Naphthol is a weak acid, naphthoic acid is strong acidic and naphthalene is neutral. NaOH is a strong base and bicarbonate is a weak base. Naphthol being a weak acid will only solubilise in strong base. Naphthoic acid is a strong acid, so it will be soluble in both strong and weak base, i.e. NaOH and NaHCO3. If you add aqueous NaOH solution to the mixture, then both naphthoic acid and naphthol will go in water layer and the naphthalene will go in organic layer, which is ether here. If you separate these layers and add NaHCO3 solution to the ether layer, you will not be left with anything which will be soluble in NaHCO3 solution. So, extraction will be difficult. On the other side, If you add NaHCO3 initially to the mixture of compounds, then naphthoic acid will go in aqueous layer and naphthalene will go in ether layer. Then when NaOH will be added, naphthol will go in this layer. Thus, all the compounds are in different layers, and easy to recover.

If concentrated HCl is used instead of dilute HCl, even if the pH is kept same to recover the compounds, then there are more chances of side reactions during recovery, which is undesirable. And by adding concentrated HCl, there is increased evolution of heat. When NaHCO3 is neutralized with HCl, CO2 is evolved. If HCl is concentrated, then there will be lot of CO2 evolved and we have to handle the system with a lot of care during the recovery. These problems are not there while using dilute HCl.