Qualitatively assessing melting/boiling points.

[least squares]

Homework Statement

Problem #1: List in order of increasing boiling points, n-pentane, neopentane, 1-chloropentane, and octane.
Problem 2: List in order of increasing melting points, cyclohexane, n-hexane, trans-oct-3-ene, cis-oct-3-ene, acetanilide, and aniline.

Homework Equations

Intermolecular forces, symmetry, molecular weight.

The Attempt at a Solution

Problem #1:
I immediately sort the molecules based on the strength of their intermolecular forces. This leaves me with 2 groups: the nonpolar, and the polar molecules.
This leaves me with n-pentane, neopentane, and octane. Of these three, octane will have the highest b.p. because it has the largest molecular mass.
Between n-pentane and neopentane, n-pentane will have more significant London dispersion forces due to its linear nature (which allows for a greater degree of intermolecular interaction and opportunity for polarization compared to the compact and spherical neopentane).
Pentyl chloride will have the highest boiling point because it has a net dipole moment. This enables the molecules to participate in dipole-dipole interactions as a bulk.

However, in the answer key, I don't understand why octane is listed as having a higher b.p. than 1-chloropentane. I understand that larger molecules will tend to have higher boiling points, but how are we to qualitatively assess that increased mass of octane will "overpower" the dipole interactions of the alkyl chloride?

Problem 2:
I take the same approach, I immediately separate the molecule into nonpolar and polar molecules.
Of the 6, acetanilide and aniline will have the highest melting points as they are polar. Between the two, I reasoned that acetanilide will have a higher melting point because it has a polar carbonyl group in addition to the h-bonding provided by the nitrogen bonded to a hydrogen. This stands in contrast to the mere nitrogen bonded to two hydrogens in the aniline. I have to admit, my reasoning here is fuzzy. As it stands, I'm comparing carbonyl & N-H (acetanilide) to N-H & N-H (aniline). How do we know that dipole + hydrogen bonding is better than hydrogen bonding + hydrogen bonding? I know that hydrogen bonding is a stronger intermolecular interaction than dipole-dipole or dipole-hydrogen bond.
Of the remaining four molecules, I know that the two alkenes, by virtue of being longer, will have higher melting points. Between the cis and trans conformation, the trans conformation will have a higher melting point because, again, its relatively linear shape allows for a greater degree of interaction and therefore greater London forces compared to the cis conformation.
WHAT confuses is the cyclohexane vs. n-hexane argument. I am tempted to argue as I had previously done, that n-hexane has a higher melting point because its linear shape allows for a greater degree of interaction than cyclohexane. I'm imagining the n-hexane as a bulk, which are easier to "pack" tightly together because they are linear.
HOWEVER, the answer key states that between cyclohexane & n-hexane, neopentane and n-pentane, cyclohexane and neopentane have higher melting points because of their "symmetry".
Colour me lost. I have utterly no clue why neopentane has a higher melting point than n-pentane, but lower boiling point than n-pentane. -______________-

 

[janhaa]

However, in the answer key, I don't understand why octane is listed as having a higher b.p. than 1-chloropentane. I understand that larger molecules will tend to have higher boiling points, but how are we to qualitatively assess that increased mass of octane will "overpower" the dipole interactions of the alkyl chloride?

In addition to a larger molar mass, octane also has a larger surface area.
So its possible to establish multiple intermolecular forces.

 

[least squares]

Thanks, janhaa, for the response. I hadn't considered that!

That actually leads me to one of my important questions. The larger surface area argument works in favor of n-alkanes over its branched counterparts when comparing which boiling points will be higher.

Why, then, does this argument fall apart when we consider melting points? Apparently the 'symmetry' of cyclohexane renders its melting point higher than n-hexane. I have no idea what that even MEANS.

 

[least squares]

Dudes and dudettes, I are so confuse.

Is the oxygen in H20 really sp3 hybridized?... I know it has bent molecular geometry, and tetrahedral electronic geometry. However, it is only bonded to two other atoms. Wouldn't it only need two hybridized orbitals (sp) in order to bond with both H atoms. The other two lone pairs can be housed in the empty p orbitals.

 

[least squares]

I suppose I'll accept the idea that an atom needs to hybridize as well to store lone pairs. The only instance where it does not hybridize an additional p is when that p is unavailable due to its involvement as a pi bond bro.

 

[epenguin]

Apparently the 'symmetry' of cyclohexane renders its melting point higher than n-hexane. I have no idea what that even MEANS.

They have nearly the same molecular weight. In fact that of n-hexane is sllghtly higher than that of cycohexane, so if that were all that mattered n-hexane should melt at a very slightly higher temperature than cyclohexane.

Now consider in however vague a way a crystal structure of each. If the cyclohexane molecule arrives it has to fit into the stucture, arriving in the right orientation with respect to the crystal lattice, which it sometimes will. Now change the orientation of the arriving molecule by 360/6 = 60° . It looks exactly the same as before and will fit in just the same. Same for 120°, 180°, 240° and 300° - six orientations. When n-hexane fits into a lattice there is only one way to rotate it so it still fits in. so forming the crystal is more favoured for cyclo- than for n- hexane - there are more ways to do it.

Now I should say I am stretching a forum rule against spoonfeeding to explain like that. You should use this to check back where you have missed it in a textbook you must have or have access to and where it is surely explained better than you can expect anyone to do off the cuff here, and so learn to use the textbook better.