To determine the boiling points of [tex]\( CH_3-O-CH_3 \)[/tex] and [tex]\( CH_3CH_2OH \)[/tex] based on the strength of their intermolecular forces, let's analyze the molecular structure and the types of intermolecular forces present in each compound.
1. Intermolecular Forces:
- [tex]\( CH_3-O-CH_3 \)[/tex] (Dimethyl ether):
- This molecule is an ether. Ethers generally exhibit weaker intermolecular forces such as London dispersion forces and dipole-dipole interactions.
- [tex]\( CH_3CH_2OH \)[/tex] (Ethanol):
- This molecule is an alcohol. Alcohols can form hydrogen bonds, which are significantly stronger than dipole-dipole interactions and London dispersion forces. Hydrogen bonding occurs because the hydrogen atom is bonded to a highly electronegative oxygen atom, making the hydrogen partially positive and enabling it to form strong intermolecular attractions with lone pairs on other oxygen atoms in neighboring molecules.
2. Boiling Point Correlation:
- Stronger intermolecular forces correlate with higher boiling points because more energy (in the form of heat) is required to overcome these forces when transitioning from the liquid phase to the gas phase.
Given the types of intermolecular forces:
- [tex]\( CH_3CH_2OH \)[/tex] with hydrogen bonding will have stronger intermolecular forces than [tex]\( CH_3-O-CH_3 \)[/tex] which primarily relies on weaker London dispersion forces and dipole-dipole interactions.
3. Conclusion:
- [tex]\( CH_3-O-CH_3 \)[/tex] (Dimethyl ether) is expected to have a lower boiling point compared to [tex]\( CH_3CH_2OH \)[/tex] (Ethanol).
Therefore, [tex]\( CH_3-O-CH_3 \)[/tex] is expected to have a lower boiling point compared to [tex]\( CH_3CH_2OH \)[/tex].
