Difficulty: Introductory | Prerequisites: Melting Points and Intermolecular Forces study notes (Doc 1), ability to draw substituted benzene rings
This companion set of notes walks through the three compound series used in the CHM 25502 melting point lab. Each series isolates a different structural variable (position of substituent, number of H-bond donors, presence vs absence of an H-bond donor) so you can see how that variable alone drives the melting point. The notes also cover the Mel-Temp apparatus procedure, since the technique is assessed in pre-labs and practical exams.
Three series of substituted aromatic compounds show that melting point depends on the position of the substituent (ortho < meta < para, broadly), the number of available intermolecular H-bond donors, and molecular symmetry. Replacing an -OH or -NH with a methylated version (-OCH₃ or -N(CH₃)₂) removes H-bond donor capacity and typically lowers the melting point.
Positional isomers (ortho, meta, para)
Compounds with the same molecular formula but different positions of a substituent on the benzene ring. Ortho = 2-position (adjacent), meta = 3-position, para = 4-position (opposite). In simple terms, same atoms, different arrangement around the ring.
Nitrophenol
A phenol (benzene + OH) with an -NO₂ group attached to the ring. The series includes 2-, 3- and 4-nitrophenol, all with formula C₆H₅NO₃ and molar mass 139.11.
Nitroanisole (4-nitroanisole)
Similar to nitrophenol, but the -OH is replaced by -OCH₃ (a methoxy group). This removes the hydrogen bond donor. Think of it as "capping" the OH with a methyl, so it can no longer donate an H-bond.
N-methylaniline (nitro-N-methylaniline)
Aniline (benzene + NH₂) with one H on the nitrogen replaced by a methyl group, plus a nitro substituent. The N-H that remains can still act as an H-bond donor, but with reduced capacity compared to a free -NH₂.
N,N-dimethylaniline (3-nitro-N,N-dimethylaniline)
Aniline with both hydrogens on nitrogen replaced by methyl groups. No N-H remains, so no hydrogen bond donation from the nitrogen at all.
Hydroxybenzoic acid
Benzoic acid (benzene + COOH) with an -OH also on the ring. Two H-bond donors are present: the carboxylic acid -OH and the phenolic -OH.
Methoxybenzoic acid (4-methoxybenzoic acid)
Benzoic acid with -OCH₃ on the ring instead of -OH. Only the carboxylic acid -OH can donate H-bonds; the methoxy group cannot.
Compound | Melting Point | Molar Mass | Formula | Key Feature |
|---|---|---|---|---|
2-Nitrophenol | ~45 °C | 139.11 | C₆H₅NO₃ | Intramolecular H-bond (ortho OH...NO₂) |
3-Nitrophenol | 95 °C | 139.11 | C₆H₅NO₃ | Intermolecular H-bonds only |
4-Nitrophenol | ~114 °C | 139.11 | C₆H₅NO₃ | Intermolecular H-bonds + high symmetry |
4-Nitroanisole | 54 °C | 153.14 | C₇H₇NO₃ | No H-bond donor (-OCH₃ replaces -OH) |
What this series teaches:
2-Nitrophenol melts lowest because its ortho arrangement allows an intramolecular H-bond, reducing intermolecular attraction.
4-Nitrophenol melts highest among the three phenols: its para geometry prevents intramolecular bonding, and the symmetric shape packs well.
4-Nitroanisole, despite a higher molar mass, melts much lower than 4-nitrophenol. Replacing -OH with -OCH₃ eliminates the H-bond donor entirely, so only weaker dipole-dipole and dispersion forces hold the lattice together.
Compound | Melting Point | Molar Mass | Formula | Key Feature |
|---|---|---|---|---|
2-Nitro-N-methylaniline | 33-36 °C | 152.15 | C₇H₈N₂O₂ | Intramolecular H-bond (ortho NH...NO₂) |
3-Nitro-N-methylaniline | 64-68 °C | 152.15 | C₇H₈N₂O₂ | Intermolecular H-bonds |
4-Nitro-N-methylaniline | 152 °C | 152.15 | C₇H₈N₂O₂ | Intermolecular H-bonds + high symmetry |
3-Nitro-N,N-dimethylaniline | 60 °C | 166.18 | C₈H₁₀N₂O₂ | No N-H donor (both Hs replaced by CH₃) |
What this series teaches:
The same ortho-intramolecular-bond pattern as Series 1: the 2-isomer melts lowest.
The 4-isomer melts dramatically higher (152 °C vs 64-68 °C for the 3-isomer), showing the combined effect of symmetry and fully available intermolecular H-bonding.
3-Nitro-N,N-dimethylaniline has no N-H at all. Even though it sits at the meta position (no intramolecular bonding possible), its melting point (60 °C) is comparable to the 3-mono-methyl isomer because the loss of the H-bond donor offsets the absence of intramolecular competition.
Compound | Melting Point | Molar Mass | Formula | Key Feature |
|---|---|---|---|---|
2-Hydroxybenzoic acid (salicylic acid) | ~159 °C | 138.12 | C₇H₆O₃ | Intramolecular H-bond (ortho OH...C=O) |
3-Hydroxybenzoic acid | ~201 °C | 138.12 | C₇H₆O₃ | Intermolecular H-bonds |
4-Hydroxybenzoic acid | 213 °C | 138.12 | C₇H₆O₃ | Intermolecular H-bonds + symmetry |
4-Methoxybenzoic acid | 184 °C | 152.15 | C₈H₈O₃ | One fewer H-bond donor (-OCH₃ replaces -OH) |
What this series teaches:
The same positional trend holds: ortho < meta < para for the three hydroxybenzoic acid isomers.
2-Hydroxybenzoic acid (salicylic acid) is the classic example of intramolecular H-bonding lowering the melting point. The ortho -OH can bond internally with the carbonyl of the -COOH.
4-Methoxybenzoic acid melts lower than 4-hydroxybenzoic acid (184 vs 213 °C) despite its higher molar mass, because the -OCH₃ cannot donate a hydrogen bond. The carboxylic acid -OH still donates, which is why the drop is not as dramatic as in Series 1.
In every series, the ortho isomer melts lowest (intramolecular H-bonding).
In every series, the para isomer melts highest among positional isomers (symmetry + intermolecular bonding).
In every series, replacing an -OH or -NH with -OCH₃ or -N(CH₃)₂ removes an H-bond donor and lowers the melting point, even when molar mass increases.
Fill the open end of a capillary tube with 1-2 mm of the solid sample (enough to be visible).
Invert the tube and drop it down a hollow glass tube so it falls onto a hard surface, packing the solid at the closed end.
Repeat until you have a uniform, tightly packed column of solid at the bottom.
Place the capillary tube in the Mel-Temp's channel.
Turn the apparatus on. Confirm the voltage starts at zero.
Set the voltage control to ~40 to heat at roughly 5-10 °C/min.
Watch through the observation window. Note the temperature when the solid first appears wet (onset) and when the last solid disappears (completion). Record this as your rough melting range.
Turn off the apparatus and dispose of the used capillary tube.
Let the apparatus cool to at least 20 °C below your rough melting range.
Load a fresh capillary tube with a new sample.
Insert the tube and heat at a slower rate of ~2 °C/min as you approach the expected melting range.
Record the onset (first crystal appears wet) and completion (last solid converts to liquid). This narrower range is your reported melting point.
Share the data with your group, then dispose of the sample and turn off the apparatus.
Always use a fresh sample for each run. Reheating a melted sample can cause decomposition or polymorphic changes.
The slower heating rate for the precise run is critical: heating too fast gives an artificially high reading because the thermometer lags behind the actual sample temperature.
Record both numbers of the range (onset and completion), not just a single value.
"The 4-nitroanisole should melt higher than 4-nitrophenol because it has a higher molar mass." Molar mass is not the deciding factor here. The -OCH₃ group cannot donate hydrogen bonds, which weakens intermolecular attractions far more than the extra mass strengthens dispersion forces.
"Meta isomers always melt between ortho and para." This is a useful trend but not a universal rule. Other factors (dipole alignment, specific crystal packing) can cause exceptions.
"You can reuse a melted sample for a second run." You should not. The melted and resolidified sample may decompose or crystallise in a different polymorph, giving unreliable results.
"Heating faster gives a more accurate melting point." The opposite. Faster heating causes the thermometer to lag behind the sample temperature, giving a reading that is artificially high.
⚠️ Be ready to rank the melting points of compounds within a series and explain each ranking using intermolecular bonding and symmetry.
⚠️ Know why replacing -OH with -OCH₃ (or -NH with -N(CH₃)₂) lowers the melting point, even when molar mass increases.
⚠️ Be able to describe the Mel-Temp procedure in order, including why you do a rough run first and a slow run second.
⚠️ Understand when and why you record two temperatures (onset and completion) for a melting point range.
⚠️ The cross-series pattern (ortho lowest, para highest, methylation lowers) is a high-probability exam question.
True or false: 4-Nitroanisole melts higher than 4-nitrophenol because it has a larger molar mass. (False. It melts lower because the -OCH₃ cannot donate hydrogen bonds.)
Fill in the blank: In every series, the ______ isomer melts lowest due to intramolecular hydrogen bonding. (ortho / 2-position)
True or false: You should heat as fast as possible during the precise melting point run. (False. A slow rate of ~2 °C/min is needed for accuracy.)
Fill in the blank: The onset of a melting range is recorded when the solid first appears ______. (wet)
True or false: 3-Nitro-N,N-dimethylaniline has no N-H hydrogen bond donor. (True.)
Q: Rank the following in order of increasing melting point and explain: 2-nitrophenol, 3-nitrophenol, 4-nitrophenol.
A: 2-nitrophenol (~45 °C) < 3-nitrophenol (95 °C) < 4-nitrophenol (~114 °C). The ortho isomer forms an intramolecular H-bond that reduces intermolecular forces. The meta isomer hydrogen-bonds intermolecularly but is less symmetric than the para isomer. The para isomer has the strongest intermolecular H-bonding network and the most efficient crystal packing.
Q: 4-Nitrophenol and 4-nitroanisole differ by a single methyl group. Why does that methyl cause a ~60 °C drop in melting point?
A: Replacing the -OH with -OCH₃ removes the hydrogen bond donor. 4-Nitrophenol can donate and accept H-bonds intermolecularly; 4-nitroanisole can only accept (via the O lone pairs and the NO₂). Losing the donor drastically weakens the crystal lattice.
Q: Why is 4-nitro-N-methylaniline's melting point (152 °C) so much higher than 3-nitro-N-methylaniline's (64-68 °C), even though both can form intermolecular H-bonds?
A: Both have an N-H donor, but the para isomer is significantly more symmetric. Better symmetry means tighter crystal packing, more intermolecular contact, and stronger aggregate attractions.
Q: Describe why two melting point runs are performed in the Mel-Temp procedure.
A: The first run (fast heating, ~5-10 °C/min) gives a rough estimate of where the melting range falls. The second run (slow heating, ~2 °C/min) starts 20 °C below the rough range and heats slowly enough for the thermometer to keep up with the sample, producing an accurate, narrow melting range.
Q: A student compares 4-hydroxybenzoic acid (mp 213 °C) with 4-methoxybenzoic acid (mp 184 °C). Both have a carboxylic acid -OH. Why the difference?
A: 4-Hydroxybenzoic acid has two H-bond donors (the carboxylic -OH and the phenolic -OH), while 4-methoxybenzoic acid has only one (the carboxylic -OH). The extra donor in the hydroxy compound creates a denser hydrogen-bonding network in the crystal, raising the melting point.
The structure-property reasoning here carries into every topic where intermolecular forces matter: boiling points, solubility, TLC Rf values, column chromatography elution order and IR spectroscopy (hydrogen-bonded O-H and N-H stretches are broadened). If you can explain why 2-nitrophenol melts lower than 4-nitrophenol, you can apply the same logic to predict relative boiling points or chromatographic behaviour.
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