Difficulty: Intermediate | Prerequisites: General chemistry acid-base equilibria, pH, pKa basics
Big picture: Solvent extraction is one of the first techniques in organic chemistry where you use acid-base chemistry to physically separate compounds from a mixture. It sits at the intersection of equilibrium, solubility and reactivity. You need a solid grasp of pKa, conjugate acid-base pairs and the difference between strong and weak acids. If you are shaky on those, revisit your gen-chem acid-base notes before reading on.
Solvent extraction separates a mixture of organic compounds by exploiting differences in acid strength. You shake the mixture with aqueous bases of increasing strength (first NaHCO3, then NaOH), and each base pulls a different class of compound into the water layer based on its pKa. Whatever does not react with either base stays in the organic layer as the "neutral" compound.
Solvent extraction (liquid-liquid extraction)
A separation technique in which a solute is transferred from one solvent to another, immiscible solvent by shaking the two layers together. Think of it as using two liquids that refuse to mix to sort compounds by their chemistry.
Partition (distribution)
The equilibrium a solute reaches between two immiscible solvents. In simple terms, each compound "decides" how much of itself ends up in the water layer versus the organic layer, based on its solubility in each.
pKa
The negative log of the acid dissociation constant; a numerical measure of acid strength. Lower pKa = stronger acid. Think of it as a scoreboard: the lower the number, the more readily the compound gives up a proton.
Conjugate base
The species formed when an acid loses a proton. For p-toluic acid (a carboxylic acid), the conjugate base is the p-toluate anion. In simple terms, it is what the acid turns into once it has donated its H+.
Deprotonation
The removal of a proton (H+) from a molecule by a base. This is the key event in every extraction step: a base strips a proton from the acid, converting it into a water-soluble salt.
NaHCO3 (sodium bicarbonate)
A mild base (conjugate acid H2CO3, pKa around 6.4). It is only strong enough to deprotonate acids with pKa below about 6.4, which means it pulls carboxylic acids into the aqueous layer but leaves weaker acids like phenols behind.
NaOH (sodium hydroxide)
A strong base. It deprotonates essentially any organic acid, including phenols (pKa around 10). Used as the second extraction base to remove phenols from the organic layer after NaHCO3 has already removed carboxylic acids.
Aqueous layer
The water-based layer in a two-phase extraction. In this experiment it sits on the bottom (water is denser than tert-butyl methyl ether). This is where ionised, water-soluble salts end up after deprotonation.
Organic layer
The non-aqueous solvent layer, here tert-butyl methyl ether. Neutral, non-ionised compounds remain dissolved here because they are more soluble in the non-polar organic solvent than in water.
The whole technique rests on one idea: if you convert a compound into its ionic (charged) form, it becomes water-soluble and moves from the organic layer into the aqueous layer.
You do this by treating the organic solution with an aqueous base. The base deprotonates the acidic compound, producing a water-soluble salt.
By choosing bases of different strengths, you selectively pull out compounds of different acidity.
Step 1, NaHCO3 wash (mild base): Sodium bicarbonate deprotonates carboxylic acids (pKa around 4 to 5) but not phenols (pKa around 10). The carboxylic acid becomes a carboxylate salt, dissolves in the aqueous layer, and is removed.
In this experiment: p-toluic acid (pKa around 4.2) reacts with NaHCO3 to form sodium p-toluate + CO2 + H2O.
p-tert-Butylphenol (pKa around 10) does not react because it is a weaker acid than carbonic acid (pKa around 6.4).
Step 2, NaOH wash (strong base): Sodium hydroxide deprotonates phenols. The phenol becomes a phenoxide salt, dissolves in the aqueous layer, and is removed.
In this experiment: p-tert-butylphenol + NaOH produces sodium p-tert-butylphenoxide + water.
Step 3, organic layer retained: Acetanilide (the neutral compound) does not have an acidic proton that either base can remove. It stays dissolved in the ether throughout.
A base can only deprotonate an acid that is stronger than the base's own conjugate acid.
NaHCO3 in water is in equilibrium with carbonic acid (H2CO3, pKa around 6.4). Only acids with pKa below about 6.4 are deprotonated by NaHCO3.
NaOH is strong enough to deprotonate any acid with pKa below about 15 to 16, which covers phenols comfortably.
This is why NaHCO3 discriminates between carboxylic acids and phenols, while NaOH does not.
Acetanilide is an amide. Its N-H is very weakly acidic (pKa around 25), far above the reach of NaOH in aqueous solution.
It has no ionisable proton under these conditions, so it never forms a water-soluble salt and remains in the ether.
Reaction 1: p-Toluic acid + NaHCO3
CH3-C6H4-COOH + NaHCO3 --> CH3-C6H4-COO- Na+ + CO2 + H2O
The carboxylic acid is deprotonated. CO2 gas evolves (you see bubbling). The sodium carboxylate salt dissolves in the aqueous layer.
Reaction 2: p-tert-Butylphenol + NaOH
(CH3)3C-C6H4-OH + NaOH --> (CH3)3C-C6H4-O- Na+ + H2O
The phenol is deprotonated by the strong base. The sodium phenoxide dissolves in the aqueous layer.
Non-reaction: p-tert-Butylphenol + NaHCO3
(CH3)3C-C6H4-OH + NaHCO3 --> no reaction
The phenol (pKa around 10) is a weaker acid than carbonic acid (pKa around 6.4), so the equilibrium does not favour deprotonation. This is the basis for separating phenols from carboxylic acids.
Non-reaction: Acetanilide + NaOH
C6H5-NH-COCH3 + NaOH --> no reaction (under these conditions)
The amide N-H (pKa around 25) is far too weakly acidic for aqueous NaOH to deprotonate.
Compound | Approximate pKa | Reacts with NaHCO3? | Reacts with NaOH? |
|---|---|---|---|
p-Toluic acid (carboxylic acid) | 4.2 | Yes | Yes |
Carbonic acid (H2CO3) | 6.4 | Reference point | Yes |
p-tert-Butylphenol (phenol) | 10 | No | Yes |
Water | 15.7 | No | Reference point |
Acetanilide (amide) | ~25 | No | No |
"A lower pKa means the acid is weaker." The opposite is true. Lower pKa = stronger acid. p-Toluic acid (pKa 4.2) is a much stronger acid than p-tert-butylphenol (pKa 10).
"NaOH is selective like NaHCO3." NaOH deprotonates both carboxylic acids and phenols. It is not selective. The selectivity comes from using NaHCO3 first, which removes only the carboxylic acid. If you skip NaHCO3 and go straight to NaOH, you lose the ability to separate those two classes.
"Neutral compounds are extracted last." Neutral compounds are never extracted at all. They remain in the organic layer by default because neither base can ionise them. You recover them by evaporating the solvent.
"The organic layer is always on the bottom." Layer position depends on density. Tert-butyl methyl ether (density around 0.74 g/mL) is less dense than water, so the organic layer floats on top. With denser solvents like dichloromethane, the organic layer sinks to the bottom. Always check.
⚠️ Know the pKa threshold for NaHCO3. You will likely be asked: "Can NaHCO3 deprotonate compound X?" The answer depends on whether the compound's pKa is below about 6.4. Carboxylic acids (pKa 4 to 5): yes. Phenols (pKa 10): no.
⚠️ Be able to predict which layer a compound ends up in after each extraction step. This is a classic exam question: given a mixture and a sequence of washes, state where each compound is at each stage.
⚠️ Understand why the order matters. NaHCO3 before NaOH. If you reverse the order, NaOH removes both the carboxylic acid and the phenol in one step and you cannot separate them from each other.
⚠️ CO2 evolution is a giveaway. When a carboxylic acid reacts with NaHCO3, CO2 is produced. Exam questions may describe bubbling and ask you to explain it.
True or false: NaHCO3 can deprotonate a phenol with pKa 10. Answer: False.
Fill in the blank: A compound must have a pKa below approximately ______ to be deprotonated by NaHCO3. Answer: 6.4.
True or false: Acetanilide moves into the aqueous layer when treated with NaOH. Answer: False.
Fill in the blank: After the NaHCO3 wash, the carboxylic acid is in the ______ layer as a ______ salt. Answer: aqueous; sodium carboxylate.
True or false: If you used only NaOH (no NaHCO3 step), you could still separate a carboxylic acid from a phenol. Answer: False.
Q: A mixture contains benzoic acid (pKa 4.2), 2-naphthol (pKa 9.5) and naphthalene (neutral). Describe how you would separate all three using solvent extraction.
A: Dissolve the mixture in an organic solvent. Extract first with aqueous NaHCO3 to remove benzoic acid (pKa 4.2 < 6.4, so it reacts). Then extract with aqueous NaOH to remove 2-naphthol (pKa 9.5, deprotonated by NaOH but not NaHCO3). Naphthalene remains in the organic layer. Acidify each aqueous extract with HCl to recover the solid compounds.
Q: Explain why p-tert-butylphenol does not react with NaHCO3 but does react with NaOH.
A: p-tert-Butylphenol has a pKa of about 10, which is higher than the pKa of carbonic acid (about 6.4). For NaHCO3 to deprotonate a compound, the compound must be a stronger acid than H2CO3. Since the phenol is weaker, the equilibrium does not favour deprotonation by NaHCO3. NaOH, however, is a strong base that can deprotonate any acid with a pKa up to about 15 to 16, so it readily deprotonates the phenol.
Q: You perform a NaHCO3 extraction and observe vigorous bubbling. What gas is being released, and what does it tell you?
A: The gas is CO2. Its evolution confirms that a carboxylic acid is present in the mixture and is reacting with NaHCO3 to form the carboxylate salt, water and carbon dioxide.
Q: After completing all extraction steps, how do you recover the neutral compound from the organic layer?
A: Dry the organic layer with an anhydrous drying agent (such as Na2SO4), filter or decant, then evaporate the solvent. The neutral compound crystallises or remains as a residue.
Q: A student accidentally performs the NaOH wash before the NaHCO3 wash. What consequence does this have for the separation?
A: NaOH deprotonates both carboxylic acids and phenols. Both compounds move into the same aqueous extract. The student can no longer separate them from each other using this technique, because the selective NaHCO3 step has been bypassed.
This material connects directly to acid-base equilibria from general chemistry. If pKa and Ka calculations feel unfamiliar, go back to those notes first.
Solvent extraction is the conceptual foundation for chromatography (which also separates compounds by differential affinity for two phases) and for understanding drug partitioning across biological membranes.
Melting point determination (covered in the companion lab-technique notes) is used here to assess purity of the recovered solids.
Solvent extraction, liquid-liquid extraction, acid-base extraction, separatory funnel technique, pKa and separation, NaHCO3 extraction, NaOH extraction, carboxylic acid separation, phenol separation, neutral compound isolation, partition coefficient, distribution coefficient, deprotonation, conjugate base, p-toluic acid, p-tert-butylphenol, acetanilide, tert-butyl methyl ether, CHGN-223, organic chemistry lab, Purdue organic chemistry