Two Isomers, One Distillable: Separating 2- and 4-Nitrophenol
A second-year laboratory class nitrates phenol and obtains 2.1 g of a dark mixture that TLC shows to be two products. The demonstrator insists on steam distillation rather than column chromatography, because the column is booked by another group.
Steam distillation gives a pale yellow distillate from which product A is recovered as a low-melting solid (mp 44–46 °C). The residue left in the distillation flask contains product B (mp 113–115 °C). Analytical TLC of the crude mixture on silica, eluting with toluene–ethyl acetate (9:1), gives two spots: A at Rf 0.68 and B at Rf 0.24.
Both isomers have the molecular formula C₆H₅NO₃ (Mr 139.1). Both dissolve in 2 mol dm⁻³ NaOH and are recovered on acidification. A classmate argues that because the two compounds have the same molar mass and the same functional groups, they should have almost identical physical properties, so the differences must come from an impurity.
1. Assign each spot and each melting point to the ortho or para isomer, and justify the assignments using structure and intermolecular forces.
Read the worked answer
A = 2-nitrophenol (ortho); B = 4-nitrophenol (para).
The deciding factor is whether the hydrogen bond is intramolecular or intermolecular.
2-Nitrophenol (ortho):
- The –OH and –NO₂ groups are on adjacent carbons, so they can form a six-membered intramolecular hydrogen bond.
- The –OH is therefore "used up" internally: it cannot hydrogen-bond to neighbouring molecules.
- Fewer intermolecular hydrogen bonds → weaker lattice → low melting point (44–46 °C) and high vapour pressure → steam volatile.
- On silica, fewer free –OH groups are available to hydrogen-bond to surface silanol groups, so it is less strongly retained and moves further up the plate → Rf 0.68.
4-Nitrophenol (para):
- The groups are too far apart (across the ring) to bond intramolecularly, so every –OH and –NO₂ must hydrogen-bond between molecules.
- Extensive intermolecular hydrogen bonding → strong crystal packing → high melting point (113–115 °C), negligible vapour pressure at 100 °C → not steam volatile.
- Strong hydrogen bonding to silica silanols → strong retention → Rf 0.24.
The Rf order and the melting-point order are therefore two independent pieces of evidence pointing the same way, and they agree with which compound distilled.
2. Evaluate the classmate's claim that identical molar mass and identical functional groups should give identical properties, and explain the NaOH solubility and the acidification step.
Read the worked answer
The claim is false because position, not just identity of groups, controls intermolecular forces.
- Molar mass affects only the dispersion (London) contribution, which is the same for both isomers here. The dominant term is hydrogen bonding, and hydrogen bonding is geometry-dependent.
- In the ortho isomer the same two groups that hydrogen-bond in the para isomer instead bond to each other. The functional groups are identical; the connectivity of the hydrogen-bond network is not.
- A ~70 °C melting-point difference with identical Mr is entirely normal for positional isomers (similarly, 2-hydroxybenzoic acid melts far below the 4-isomer).
Why an impurity is a poor explanation: an impurity would depress and broaden a melting point, not raise it, and would normally give an extra spot or a streak on the TLC plate. The plate shows exactly two sharp spots, and the two melting points are both sharp, so an impurity does not fit the observations.
NaOH and acidification: the phenolic –OH has pKa ≈ 7.2, so both isomers are deprotonated by 2 mol dm⁻³ NaOH to the phenolate anion, which is ionic and water-soluble. Adding strong acid reprotonates the phenoxide, and the neutral nitrophenol precipitates. Note that the nitro group is not affected - it is the phenol that is being titrated by the base, and this behaviour is common to both isomers, so it cannot be used to tell them apart.
3. The demonstrator chose steam distillation over column chromatography for a 2.1 g batch. Justify that choice chemically, and state two limitations of the method that the class should record.
Read the worked answer
Justification. Steam distillation separates on the basis of vapour pressure at about 100 °C, and the two isomers differ enormously there: the ortho compound has a vapour pressure of the order of kilopascals at 100 °C (no intermolecular hydrogen bonding to break), while the para compound's is negligible. Co-distillation with steam therefore carries over essentially pure A while B stays in the flask.
It is also the safer choice: nitroaromatic compounds can decompose or deflagrate if overheated in a flask, and steam distillation keeps the mixture near 100 °C rather than raising it to the boiling point of the mixture under reduced pressure or at atmospheric pressure. It needs no adsorbent, no solvent, and no column, and is cheap at gram scale.
Limitations to record:
- The distillate is very dilute - most of its mass is water. Recovery requires solvent extraction (e.g. diethyl ether), drying over a drying agent such as anhydrous MgSO₄, and then evaporation, each step losing material and adding time.
- Incomplete transfer. Distillation must be monitored (TLC of the residue) until no more A comes over; some A always remains, so the recovery is not quantitative, and the residue is a mixture enriched in B rather than pure B.
- B is not purified at all by this method - the residue still contains tarry by-products and needs a separate recrystallisation.
- Safety: 2-nitrophenol is toxic and is absorbed through skin, so gloves and a fume cupboard are required even though the temperature is modest.
Judgement: steam distillation is the right call for a 2 g teaching batch where the ortho isomer is the target, but it is not a general separation - for larger scale or for purifying the para isomer, chromatography or recrystallisation would be needed.