- #1
least squares
- 4
- 0
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. -______________-