Question

the four compounds below have similar molar masses. rank the pure liquids of the four compounds in order of increasing boiling point (i.e. 1 = lowest boiling point and 4 = highest boiling point). c(ch3)4 ch3ch2och2ch3 ch3ch2ch2ch2oh ch3ch2ch2ch2ch3

Explanation:

Step1: Analyze intermolecular forces

Inter - molecular forces affect boiling points. C($\mathrm{CH_3}$)$_4$ is a non - polar molecule with only London dispersion forces. $\mathrm{CH_3CH_2OCH_2CH_3}$ is polar, having dipole - dipole forces in addition to London dispersion forces. $\mathrm{CH_3CH_2CH_2CH_2OH}$ has hydrogen bonding (due to the $-\mathrm{OH}$ group), dipole - dipole forces, and London dispersion forces. $\mathrm{CH_3CH_2CH_2CH_2CH_3}$ is non - polar with only London dispersion forces.

Step2: Compare London dispersion forces

Among non - polar molecules, the molecule with more surface area has stronger London dispersion forces. $\mathrm{CH_3CH_2CH_2CH_2CH_3}$ has a more extended structure compared to C($\mathrm{CH_3}$)$_4$, so it has stronger London dispersion forces.

Step3: Rank boiling points

Hydrogen bonding is the strongest intermolecular force among these. So, $\mathrm{CH_3CH_2CH_2CH_2OH}$ has the highest boiling point (rank 4). Among the non - polar molecules, $\mathrm{CH_3CH_2CH_2CH_2CH_3}$ has stronger London dispersion forces than C($\mathrm{CH_3}$)$_4$, so $\mathrm{CH_3CH_2CH_2CH_2CH_3}$ has a higher boiling point than C($\mathrm{CH_3}$)$_4$ (rank 3 for $\mathrm{CH_3CH_2CH_2CH_2CH_3}$ and rank 1 for C($\mathrm{CH_3}$)$_4$). $\mathrm{CH_3CH_2OCH_2CH_3}$ has dipole - dipole forces and ranks 2.

Answer:

C($\mathrm{CH_3}$)$_4$: 1
$\mathrm{CH_3CH_2OCH_2CH_3}$: 2
$\mathrm{CH_3CH_2CH_2CH_2CH_3}$: 3
$\mathrm{CH_3CH_2CH_2CH_2OH}$: 4