[YouSheng]

I meet the problem about melting point in exam

Which compound would you expect to have the highest melting point?
A. n-butyl alcohol
B. isobutyl alcohol
C. sec-butyl alcohol
D. tert-butyl alcohol
E. diethyl ether

The answer is D, but I think, unbranched alcohols have large surface to form Van der Waals force with each other, so it should have higher m.p than branched alcohol, but it is not true.
I try to find the reason in internet, some say because of its symmetric.
The concept I can't understand is why symmetric compound give higher m.p but not surface area?
And, is tert-butyl alcohol is more symmetric than n-butyl alcohol?

[OCSaviour]

The melting point is not related to the branching of compounds. That's boiling point. Now, think again.

[Enthalpy]

The melting point is not related to the branching of compounds. That's boiling point. Now, think again.

Yes the melting point relates with branching, quite a lot. But the boiling point, little.

[Babcock_Hall]

Yes, but branching affects melting points for a different reason from the reason it affects boiling point.  With respect to boiling points, molecules lose contact with each other as they enter the gas phase.

[Enthalpy]

Which compound would you expect to have the highest melting point?
A. n-butyl alcohol
B. isobutyl alcohol
C. sec-butyl alcohol
D. tert-butyl alcohol
E. diethyl ether

At least the answer D is correct, that's a good point.

But I can't see how such a question, for which no explanation seems to exist among mankind presently, could possibly be an exam question. Symmetry plays a definite role but is very hard to assess just from the molecule structure. It would make sense, though with big effort, among alkanes. The possible deformations of the molecule (like C-C rotations) seem important too.

The alcohol decides everything here as it makes a hydrogen bond, much stronger than intermolecular forces between alkanes of 4 or 5 carbons. So the melting point would result mainly from possible crystals that offer hydrogen bonds between many molecules.

[Babcock_Hall]

A similar trend occurs in the isomers of C₅H₁₂.  The melting points of n-pentane, 2-methylbutane, and neopentane (2,2-dimethylpropane) are -130, -160, and -17 °C, respectively.

[YouSheng]

Yes, but branching affects melting points for a different reason from the reason it affects boiling point.  With respect to boiling points, molecules lose contact with each other as they enter the gas phase.

I know branching affects boiling point because large surface area provide a high Van der Waals force, but what is the reason for branching affects melting point?

[YouSheng]

Which compound would you expect to have the highest melting point?
A. n-butyl alcohol
B. isobutyl alcohol
C. sec-butyl alcohol
D. tert-butyl alcohol
E. diethyl ether

At least the answer D is correct, that's a good point.

But I can't see how such a question, for which no explanation seems to exist among mankind presently, could possibly be an exam question. Symmetry plays a definite role but is very hard to assess just from the molecule structure. It would make sense, though with big effort, among alkanes. The possible deformations of the molecule (like C-C rotations) seem important too.

The alcohol decides everything here as it makes a hydrogen bond, much stronger than intermolecular forces between alkanes of 4 or 5 carbons. So the melting point would result mainly from possible crystals that offer hydrogen bonds between many molecules.

So there is no theory to explain presently?
Not only alcohols, other organic molecules, such as alkanes, have these properties too.

[Babcock_Hall]

@OP, I would think in terms of the relative freedom of molecules in the liquid vs. the solid state.  Although I think that a qualitative and imperfect explanation is possible, I suspect that Enthalpy is correct in that an explanation that accounted for all of the data would be a difficult undertaking at best.

[OCSaviour]

If we take into account symmetry of a compound, then when the molecule freezes, the molecular with higher symmetry, its groups/atoms will be packed closely, occupying as much free space available. The melting point will be highest for the compound with the highest symmetry, given that they have the same functional group.

[YouSheng]

If we take into account symmetry of a compound, then when the molecule freezes, the molecular with higher symmetry, its groups/atoms will be packed closely, occupying as much free space available. The melting point will be highest for the compound with the highest symmetry, given that they have the same functional group.

Asymmetric compound, such as, pentane, can't packed closely than symmetric compound, such as neopentane?

[Babcock_Hall]

I suspect that both the ability to pack well and the number of degrees of rotational freedom have an effect.

[OCSaviour]

If we take into account symmetry of a compound, then when the molecule freezes, the molecular with higher symmetry, its groups/atoms will be packed closely, occupying as much free space available. The melting point will be highest for the compound with the highest symmetry, given that they have the same functional group.

Asymmetric compound, such as, pentane, can't packed closely than symmetric compound, such as neopentane?

Yes. The tighter the packing, stronger are the intermolecular interactions, higher is the melting point.

[Enthalpy]

Yes. The tighter the packing, stronger are the intermolecular interactions, higher is the melting point.

No. This is only the usual but wrong theory.

[Enthalpy]

A similar trend occurs in the isomers of C₅H₁₂.  The melting points of n-pentane, 2-methylbutane, and neopentane (2,2-dimethylpropane) are -130, -160, and -17 °C, respectively.

Same melting points as yours (from NACA TN No. 1247):

mp
143K   n-pentane
113K   2-methylbutane (mp DEcreases)
257K   2,2-methylbutane (mp INcreases)

So no rule like "more branched has this effect" works. Symmetry seems to favour the solid, but it would depend on a very complicated way on the possibility to stack the molecules. Since they can arrange in any way that favours the solid, with any offset and by alternating the orientations, this is impossible to predict just by looking at the formula nor even the 3D representation of one molecule.

I join an excerpt of a table of melting points and (liquid) densities for nonanes. An "A" in the last column tells that the source is one document (NACA TN No. 1247) written by one team at one time from their own measures on their own products before simulation software existed.

From such tables, I claim that the melting point doesn't relate simply with the degree of branching nor the (liquid) density. Elsewhere I checked that neither the solid density relates consistently with the melting point. But if someone finds a rule usable of just understandable, he would be useful to Mankind. I can provide more data.

I mean it seriously: presently we have no working theory. In the first place, we lack data, even for alkanes. All the explanations and theories that fill the books and courses are wrong. About two (2 = 1+1) groups worldwide conduct research on this topic. There is something useful to do.