C/P 10 · Organic laboratory techniques
Separating and Identifying Compounds
Combine extraction, TLC, IR and mass spectrometry to separate and identify a three-component mixture.
Passage
Organic chemists separate and identify compounds by using differences in their properties. In a liquid–liquid extraction, a mixture dissolved in an organic solvent such as diethyl ether is shaken with an aqueous solution. A compound moves into the aqueous layer if it is charged or very polar, and it stays in the ether if it is neutral and nonpolar. Acids and bases can be moved between the layers by changing the pH, because the ionized form of an acid or base is much more soluble in water than its neutral form.
A student has a solution in ether of three compounds: benzoic acid, which is a weak acid (pKa about 4.2), aniline, which is a weak base whose conjugate acid, the anilinium ion, has a pKa of about 4.6, and naphthalene, a neutral hydrocarbon. She shakes the ether solution first with aqueous NaOH (pH about 13), separates the layers, then shakes the remaining ether solution with aqueous HCl (pH about 1), and separates the layers again. She keeps the three liquids separate and labels them: the NaOH layer (Fraction 1), the HCl layer (Fraction 2) and the ether layer that is left (Fraction 3).
Part 1. To identify the compounds in each fraction, the student uses thin-layer chromatography (TLC) on a silica gel plate, which is a polar stationary phase. She spots each of the three pure compounds on the plate, develops it in a nonpolar solvent mixture until the solvent front has moved 8.0 cm from the starting line, and visualizes the spots with ultraviolet light. The retention factor, Rf, is the distance moved by the compound divided by the distance moved by the solvent front. Table 1 shows the distance each compound moved.
Table 1. Distance moved on the TLC plate (solvent front = 8.0 cm).
| Compound | Distance moved by the spot (cm) |
|---|---|
| Naphthalene | 6.4 |
| Aniline | 4.0 |
| Benzoic acid | 2.4 |
Part 2. The student recovers the compound from Fraction 1 by adding concentrated HCl until a white solid forms, filtering the solid and drying it. She records an infrared (IR) spectrum and a mass spectrum. Table 2 shows the main features.
Table 2. Data for the solid isolated from Fraction 1.
| Technique | Observation |
|---|---|
| IR | A very broad absorption from about 2500 to 3300 cm⁻¹ and a strong, sharp absorption near 1700 cm⁻¹ |
| Mass spectrometry | Molecular ion (M⁺) at m/z = 122 |
The student also notes that the exact positions of absorptions in an IR spectrum depend on the physical state of the sample and that a strong absorption near 1700 cm⁻¹ is a typical feature of a C=O bond, while a broad band between 2500 and 3300 cm⁻¹ is typical of an O–H bond that is strongly hydrogen-bonded. In the mass spectrum, the peak at the highest mass usually corresponds to the intact molecule that has lost one electron.
Questions
Question 1 of 6
Which compound is found mostly in Fraction 1, the NaOH layer?
NaOH deprotonates only the acid. The other two are not ionized at pH 13.
The anilinium ion forms in acid, not in a strongly basic solution. At pH 13 aniline is neutral.
A nonpolar neutral hydrocarbon stays in the ether and does not dissolve in water.
Why the answer is what it is
At pH 13, far above the pKa of benzoic acid (4.2), the acid is deprotonated to the benzoate ion, which is charged and water-soluble. Aniline and naphthalene stay neutral and remain in the ether.
Review: Laboratory Techniques and Molecular Identification · Intermolecular Forces · Functional Groups as Reactivity Units · Acids & Bases
Question 2 of 6
Why is aniline found in Fraction 2, the HCl layer?
Protonation, not deprotonation, gives the charged anilinium ion for a base.
Aniline does not react with ether. It moves to water as an ion.
Aniline acts as a base in this experiment, and benzoic acid is the acid.
Why the answer is what it is
The conjugate acid of aniline has a pKa of about 4.6. At pH 1 the base is almost completely protonated to the anilinium ion, which is charged and so is extracted into the aqueous layer.
Review: Laboratory Techniques and Molecular Identification · Intermolecular Forces · Functional Groups as Reactivity Units · Acids & Bases
Question 3 of 6
Why does adding HCl to Fraction 1 cause a white solid to form?
HCl dissolves in water and does not form a solid.
HCl is a strong acid and it protonates the benzoate. It does not deprotonate benzoic acid.
A precipitate forms only if a product is poorly soluble. Here, neutral benzoic acid is poorly soluble in water.
Why the answer is what it is
Lowering the pH below the pKa of benzoic acid converts the benzoate ion back to the neutral acid. The neutral form is much less soluble in water, so it comes out of solution as a solid.
Review: Laboratory Techniques and Molecular Identification · Intermolecular Forces · Functional Groups as Reactivity Units · Acids & Bases
Question 4 of 6
Why does naphthalene travel farthest on the TLC plate?
The most polar compound interacts most strongly with the polar silica and moves the shortest distance.
The movement depends on polarity and interactions with the plate, not simply on mass.
Naphthalene has no O–H or N–H bonds and cannot form hydrogen bonds with the silica gel.
Why the answer is what it is
Silica gel is polar, so polar compounds (benzoic acid, with its carboxylic acid group, and aniline, with its amine) are held back. The nonpolar hydrocarbon naphthalene is held least and moves farthest: Rf = 6.4/8.0 = 0.80.
Review: Laboratory Techniques and Molecular Identification · Intermolecular Forces · Functional Groups as Reactivity Units · Acids & Bases
Question 5 of 6
What functional group is most consistent with the IR spectrum of the solid from Fraction 1?
An alkane has only C–H absorptions. It would not show a strong 1700 cm⁻¹ band or a broad 2500–3300 cm⁻¹ band.
An alcohol would show a broad O–H band, but not a strong C=O absorption near 1700 cm⁻¹.
An amine shows N–H stretching near 3300–3500 cm⁻¹ as sharper bands, with no C=O.
Why the answer is what it is
The combination of a very broad O–H absorption (2500–3300 cm⁻¹) and a strong C=O absorption (near 1700 cm⁻¹) is characteristic of a carboxylic acid.
Review: Laboratory Techniques and Molecular Identification · Intermolecular Forces · Functional Groups as Reactivity Units · Acids & Bases
Question 6 of 6
Benzoic acid has the formula C₇H₆O₂. Is the mass spectrum consistent with the isolated solid being benzoic acid? (C = 12, H = 1, O = 16.)
Adding up the atoms gives 7 × 12 + 6 × 1 + 2 × 16 = 122, not 106.
The molecular ion appears at the mass of the molecule, not at twice that mass.
The mass of the molecular ion depends on the compound. Here it matches the mass of benzoic acid.
Why the answer is what it is
The molecular mass of C₇H₆O₂ is (7 × 12) + (6 × 1) + (2 × 16) = 84 + 6 + 32 = 122. The molecular ion at m/z = 122 matches, supporting the identification.
Review: Laboratory Techniques and Molecular Identification · Intermolecular Forces · Functional Groups as Reactivity Units · Acids & Bases