Questions: Which compound has the higher boiling point? (CH3)2CHCOCH3 (CH3)3CCH2OH

Transcript text: Part 2 of 3 Which compound has the higher boiling point? $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOCH}_{3}$ $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}$

Solution

Solution Steps

Step 1: Identify the Compounds

The two compounds given are:

$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCOCH}_{3}$, which is 2-butanone (also known as methyl ethyl ketone).
$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{OH}$, which is tert-butyl alcohol.

Step 2: Analyze Intermolecular Forces

The boiling point of a compound is influenced by the strength of its intermolecular forces. The main types of intermolecular forces are:

Van der Waals (London dispersion) forces: Present in all molecules.
Dipole-dipole interactions: Present in polar molecules.
Hydrogen bonding: A strong type of dipole-dipole interaction occurring when hydrogen is bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine.

2-butanone is a ketone, which means it has dipole-dipole interactions due to the polar carbonyl group (C=O). However, it does not have hydrogen bonding.

Tert-butyl alcohol, on the other hand, has an -OH group, which allows for hydrogen bonding. Hydrogen bonding is generally stronger than dipole-dipole interactions and significantly increases the boiling point.

Step 3: Compare Boiling Points

Given that tert-butyl alcohol can form hydrogen bonds, it will have stronger intermolecular forces compared to 2-butanone, which only has dipole-dipole interactions and van der Waals forces. Therefore, tert-butyl alcohol is expected to have a higher boiling point.