[VISUAL] Based on the information in the table above, which of the compounds has the highest boiling — States of Matter Chemistry Question

STEPS:

1. Analyze the chemical structures and molecular weights of the given compounds:
* Butanal ($\text{C}_4\text{H}_8\text{O}$, $72.10\text{ g/mol}$): An aldehyde containing a polar carbonyl ($\text{C=O}$) group. The hydrogen atoms are only bonded to carbon atoms (there are no $\text{O–H}$, $\text{N–H}$, or $\text{F–H}$ bonds).
* Pentane ($\text{C}_5\text{H}_{12}$, $72.15\text{ g/mol}$): A nonpolar hydrocarbon containing only carbon-carbon and carbon-hydrogen single bonds.
* Propanoic acid ($\text{C}_3\text{H}_6\text{O}_2$, $74.08\text{ g/mol}$): A carboxylic acid containing both a carbonyl ($\text{C=O}$) group and a hydroxyl ($\text{–OH}$) group, which features a hydrogen atom covalently bonded directly to a highly electronegative oxygen atom.
* *Note:* All three compounds have extremely similar molar masses ($\sim 72\text{–}74\text{ g/mol}$). This means their London dispersion forces (LDFs), which depend on electron cloud size and polarizability, are highly comparable. This allows us to focus almost entirely on differences in their primary polar intermolecular forces (IMFs).

2. Identify the types of intermolecular forces (IMFs) present in each compound:
* Pentane: Because it is nonpolar, it exhibits only weak London dispersion forces (LDFs).
* Butanal: Because it has a polar carbonyl group, it exhibits both LDFs and dipole-dipole interactions. However, it cannot form intermolecular hydrogen bonds because it does not have a hydrogen atom bonded directly to a highly electronegative atom (N, O, or F).
* Propanoic acid: Because it has an $\text{–OH}$ group, its molecules can engage in highly stable intermolecular hydrogen bonding in addition to exhibiting dipole-dipole interactions and LDFs.

3. Relate intermolecular forces to boiling points:
* The boiling point of a molecular liquid is a measure of the thermal energy required to overcome the attractive intermolecular forces holding the molecules together in the liquid phase.
* Stronger IMFs require more thermal energy to break, resulting in a higher boiling point.
* Hydrogen bonding is the strongest type of polar intermolecular attraction among these substances. Since propanoic acid can form these strong intermolecular hydrogen bonds, it will have the strongest overall cohesive forces and therefore the highest boiling point.

4. Combine the findings to select the correct option:
* Propanoic acid has the highest boiling point because it is the only compound among the three that can form intermolecular hydrogen bonds. This corresponds to Option D.

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WHY_OTHERS_WRONG:

• Option A is incorrect: Although butanal contains an oxygen atom, its hydrogen atoms are only bonded to carbon. For intermolecular hydrogen bonding to occur, a hydrogen atom must be covalently bonded to $\text{N}$, $\text{O}$, or $\text{F}$. Therefore, butanal cannot form intermolecular hydrogen bonds and has a lower boiling point than propanoic acid.
• Option B is incorrect: While pentane has a slightly longer carbon chain skeleton (5 carbons) than propanoic acid (3 carbons) and butanal (4 carbons), its lack of any polar groups makes it completely nonpolar. The weak LDFs in pentane are far easier to overcome than the strong hydrogen bonds in propanoic acid, giving pentane a much lower boiling point.
• Option C is incorrect: Having more $\text{C–H}$ bonds does not increase a compound's boiling point in a way that overrides polar intermolecular forces. Because $\text{C–H}$ bonds are essentially nonpolar, they only contribute to weak LDFs, which are insufficient to match the strength of hydrogen bonding in propanoic acid.