Q.Arrange the following compounds in increasing order of their boiling points.
$CH_3CHO$, $CH_3CH_2OH$, $CH_3OCH_3$, $CH_3CH_2CH_3$

The key idea is that boiling point depends on the strength of intermolecular forces: hydrogen bonding > dipole-dipole > weak London forces.

Step 1: Identify the dominant force in each compound.

• $CH_3CH_2OH$ (ethanol) has an –OH group → strong hydrogen bonding.
• $CH_3CHO$ (acetaldehyde) has a polar C=O bond → dipole-dipole interactions.
• $CH_3OCH_3$ (dimethyl ether) is polar but cannot hydrogen-bond (no O–H) → dipole-dipole, weaker than aldehyde.
• $CH_3CH_2CH_3$ (propane) is nonpolar → only weak London dispersion forces. …

Why boiling point trends matter

Boiling point is a measure of how much energy you need to overcome the intermolecular forces holding molecules together in the liquid phase. The stronger those forces, the higher the boiling point. For small organic molecules, three types of forces dominate:

• London dispersion forces — present in all molecules; increase with molecular size and surface area.
• Dipole-dipole interactions — present in polar molecules; stronger than dispersion forces of similar magnitude.
• Hydrogen bonding — a special, very strong dipole-dipole interaction when H is bonded to N, O, or F.

The trick is to identify which of these forces each compound can experience, then compare them fairly — keeping molecular mass roughly similar so we aren't comparing apples to oranges.

Here, all four compounds have 2–3 carbon atoms, so molecular masses are close (around 44–60 g/mol). That means differences in boiling point come almost entirely from the type and strength of intermolecular forces, not from size.

Step-by-step reasoning

1. Identify the intermolecular forces each molecule can have

   • $CH_3CH_2CH_3$ (propane): A nonpolar hydrocarbon. Only London dispersion forces. No dipole, no hydrogen bonding.
   • $CH_3OCH_3$ (dimethyl ether): A polar molecule (C–O bond has a dipole, and the molecule is bent at oxygen). It has dipole-dipole interactions, but no O–H bond — so no hydrogen bonding. Dispersion forces are similar to propane (same number of electrons, roughly).
   • $CH_3CHO$ (acetaldehyde): A polar molecule with a carbonyl group (C=O). It has a strong dipole. Importantly, it has no O–H or N–H bond, so it cannot form hydrogen bonds with itself. However, the carbonyl oxygen can act as a hydrogen bond acceptor if a donor is present — but here, there is no donor. So only dipole-dipole and dispersion forces.
   • $CH_3CH_2OH$ (ethanol): Has an O–H group. This allows strong hydrogen bonding between molecules (each ethanol can form two hydrogen bonds: one via the H, one via the lone pairs on O). This is the strongest intermolecular force among the four.
   Watch out

   A common mistake is to think that any molecule containing oxygen can hydrogen-bond. That's false. Hydrogen bonding requires a hydrogen atom bonded directly to N, O, or F. Acetaldehyde ($CH_3CHO$) has an oxygen, but the hydrogen atoms are all bonded to carbon — so no hydrogen bonding.

2. Rank by strength of dominant intermolecular force

   • Hydrogen bonding (ethanol) > dipole-dipole (acetaldehyde and dimethyl ether) > only London dispersion (propane).

   So ethanol should have the highest boiling point, and propane the lowest.

3. Compare the two dipole-dipole molecules: acetaldehyde vs. dimethyl ether

   Both have similar molecular mass (44 vs. 46 g/mol). Which has stronger dipole-dipole interactions? …

Method: Intermolecular Forces Analysis

This method ranks boiling points by identifying and comparing the strength of intermolecular forces in each molecule. Stronger forces → higher boiling point.

Step 1: Identify the type of intermolecular forces present

Step 2: Rank by force strength

General order of strength:

Hydrogen bonding > dipole-dipole > London dispersion

• $CH_3CH_2CH_3$ — only London forces → lowest boiling point
• $CH_3OCH_3$ — dipole-dipole, but no H-bonding
• $CH_3CHO$ — stronger dipole than ether (C=O is more polar than C–O–C) …

Mistake 1: Forgetting to check intermolecular forces first

What students do wrong:

They guess based on molecular size alone (e.g., “propane is bigger than ethanol, so it boils higher”).

Why it’s wrong:

Boiling point depends on the strength of intermolecular forces, not just size.

• $CH_3CH_2OH$ (ethanol) has hydrogen bonding — the strongest force here.
• $CH_3CHO$ (acetaldehyde) has dipole-dipole forces.
• $CH_3OCH_3$ (dimethyl ether) also has dipole-dipole but weaker than aldehyde.
• $CH_3CH_2CH_3$ (propane) has only weak London dispersion forces.

How to avoid:

Always list the type of intermolecular force for each compound before comparing.

Mistake 2: Confusing boiling point with melting point trends

What students do wrong:

They apply symmetry rules (which matter more for melting points) to boiling points.

Why it’s wrong:

Boiling point depends on how strongly molecules stick together in liquid, not on crystal packing. Symmetry helps in solids, not in liquids.

How to avoid:

For boiling points, focus on polarity, hydrogen bonding, and molar mass — not molecular shape symmetry.

Mistake 3: Ignoring relative strength of dipole-dipole forces

What students do wrong:

They treat all dipole-dipole forces as equal.

Why it’s wrong:

$CH_3CHO$ has a stronger dipole (due to the C=O bond) than $CH_3OCH_3$ (C–O–C has a smaller dipole moment). So acetaldehyde boils higher than dimethyl ether.

How to avoid:

Compare dipole moments or at least the polarity of functional groups — carbonyl > ether.

Mistake 4: Overlooking hydrogen bonding in alcohols

What students do wrong:

They rank ethanol below acetaldehyde because “acetaldehyde has a double bond.”

Why it’s wrong:

Hydrogen bonding (O–H···O) is much stronger than any dipole-dipole force. Ethanol will always have the highest boiling point among these four.

How to avoid:

Remember: Alcohols > Aldehydes/Ketones > Ethers > Alkanes (for similar molar mass).

Mistake 5: Not checking molar mass when forces are similar

What students do wrong:

They rank ether and alkane incorrectly when forces are weak.

Why it’s wrong: …