Question

1. HF boils at 20 °C, and methanol (CH₃OH) boils at 65 °C.  

In which molecule do you expect the hydrogen bonding interactions to be stronger? Why? (2 points)

In which molecule do you expect the (non-hydrogen bonding) dipole-dipole interactions to be stronger? Why? Draw both molecules and sketch the direction of the dipole moment. (4 points)

In which molecule do you expect the London dispersion forces to be stronger? Why? (3 points)

Based on the boiling points given above, which of those three factors is the dominant effect, and how can you tell? (2 points)

           

2. For each of the names below, draw the structure clearly indicated by the name. Then, for each one, indicate whether the name (a) is an acceptable IUPAC name for the structure, (b) is not an acceptable name for the structure, in which case you should give the correct name, or (c) refers to a structure which cannot exist, in which case you should explain why. (10 points)

Answer 1

Hydrogen Bonding is stronger in Metanol as it has electron rich source as Oxygen in it. This causes its boiling point to rise. HF will have lower London dispersion forces. HF is highly polar. CH3OH has induced dipole moment. Dipole moment will be intense in HF as H and F has appreciable difference in its electronegativity. Dipole moment will be H---^F and 3HC----O*H. Hydrogen Bonding is the main factor behind the judgement of boiling point. HF is very stable also. It has high intermolecular bonding. CH3OH will have much weaker London Dispersion forces in Alkyl groups due to induced dipole moment.

Part (c). No names ir structure given.