Question

1. Define and compare elemental, inductive and resonance effects.

2. Define and compare various intermolecular forces.

3. Illustrate how intermolecular forces effect physical properties.

4. Draw and name all isomers of C5H12.

5. Define and discuss kinetic vs thermodynamic control of a reaction.

Answer 1

Ans-(1) Inductive effects occur through the sigma bond network.
Resonance effects occur through the pi bond network.

(2) Ionic bonds > H bonding > Van der Waals dipole-dipole interactions > Van der Waals dispersion forces.

1. Ionic forces are interactions between charged atoms or molecules. Positively charged ions, such as Na(+) , Li(+), and Ca(2+), are termed cations. Negatively charged ions, such as Cl(–), Br(–), HO(–) are called anions (I always got this straight through remembering that the “N” in “Anion” stood for “Negative”) The attractive forces between oppositely charged ions is described by Coulomb’s Law, in which the force increases with charge and as decreases as the distance between these ions is increased. The highly polarized (charged) nature of ionic molecules is reflected in their high melting points (NaCl has a melting point of 801 °C) as well as in their high water solubility (for the alkali metal salts, anyway; metals that form multiple charges like to leave residues on your bathtub)

2. Hydrogen bonding occurs in molecules containing the highly electronegative elements F, O, or N directly bound to hydrogen. Since H has an electronegativity of 2.2 (compare to 0.9 for Na and 0.8 for K) these bonds are not as polarized as purely ionic bonds and possess some covalent character. However, the bond to hydrogen will still be polarized and possess a dipole.

The dipole of one molecule can align with the dipole from another molecule, leading to an attractive interaction that we call hydrogen bonding. Owing to rapid molecular motion in solution, these bonds are transient (short-lived) but have significant bond strengths ranging from (9 kJ/mol (2 kcal/mol) (for NH) to about 30 kJ/mol (7 kcal) and higher for HF. As you might expect, the strength of the bond increases as the electronegativity of the group bound to hydrogen is increased.

So in a sense, HO, and NH are “sticky” – molecules containing these functional groups will tend to have higher boiling points than you would expect based on their molecular weight.

3. Van Der Waals Dipole-dipole interactions.

Other groups beside hydrogen can be involved in polar covalent bonding with strongly electronegative atoms. For instance, each of these molecules contains a dipole:

These dipoles can interact with each other in an attractive fashion, which will also increase the boiling point. However since the electronegativity difference between carbon (electronegativity = 2.5) and the electronegative atom (such as oxygen or nitrogen) is not as large as it is for hydrogen (electronegativity = 2.2), the polar interaction is not as strong. So on average these forces tend to be weaker than in hydrogen bonding.

4. Van der waals Dispersion forces (London forces)

The weakest intermolecular forces of all are called dispersion forces or London forces. These represent the attraction between instantaneous dipoles in a molecule. Think about an atom like argon. It’s an inert gas, right? But if you cool it to –186 °C, you can actually condense it into liquid argon. The fact that it forms a liquid it means that something is holding it together. That “something” is dispersion forces. Think about the electrons in the valence shell. On average, they’re evenly dispersed. But at any given instant, there might be a mismatch between how many electrons are on one side and how many are on the other, which can lead to aninstantaneous difference in charge.

It’s a little like basketball. On average, every player is covered one-on-one, for an even distribution of players. But at any given moment, you might have a double-team situation where the distribution of players is “lumpy” (it also means that somebody is open). In the valence shell, this “lumpiness” creates dipoles, and it’s these dipoles which are responsible for intermolecular attraction.

The polarizability is the term we use to describe how readily atoms can form these instantaneous dipoles. Polarizability increases with atomic size. That’s why the boiling point of argon (–186 °C) is so much higher than the boiling point of helium (–272 °C). By the same analogy, the boiling point of iodine, (I-I, 184 °C) is much higher than the boiling point of fluorine (F-F, –188°C).

For hydrocarbons and other non-polar molecules which lack strong dipoles, these dispersion forces are really the only attractive forces between molecules. Since the dipoles are weak and transient, they depend on contact between molecules – which means that the forces increase with surface area. A small molecule like methane has very weak intermolecular forces, and has a low boiling point. However, as molecular weight increases, boiling point also goes up. That’s because the surface over which these forces can operate has increased. Therefore, dispersion forces increase with increasing molecular weight. Individually, each interaction isn’t worth much, but if collectively, these forces can be extremely significant. How can a gecko lizard walk on walls? Look at its feet.

[Determining trends for hydrocarbons can get a little bit tricky depending on the exact structure – symmetry also plays a role in boiling points and melting points. I talked about this in detailpreviously.]

Bottom line:

1. The intermolecular forces increase with increasing polarization of bonds.
2. Strength of forces (and therefore impact on boiling points) is ionic > hydrogen bonding > dipole dipole > dispersion
3. Boiling point increases with molecular weight, and with surface area

(3)

)

(4)

• When temperature is low, there is enough energy to form the 1,2-product – and that’s it. The product ratio is determined by the reaction rate (i.e. the height of the transition state D). Such a reaction is said to be under kinetic control.
• When temperature is high, there is enough energy to form both the 1,2- and the 1,4-products. Furthermore, the reaction to form products is reversible (i.e. there is enough energy to go from E → D and thence back to C. In this case the ratio of products is determined by the relative thermodynamic stabilities (i.e. the height of E). Such a reaction is said to be under thermodynanic control.