Q.Arrange the following compounds in increasing order of dipole moment.
$CH_3CH_2CH_3$, $CH_3CH_2NH_2$, $CH_3CH_2OH$

Concept: Dipole moment depends on both the magnitude of the bond dipole and the molecular geometry. For small molecules with similar shapes, the polarity of the functional group determines the order.

Reasoning:

1. $CH_3CH_2CH_3$ (propane) is a nonpolar hydrocarbon. The C–H and C–C bonds have negligible dipole moments, so the net dipole moment is essentially zero.
2. $CH_3CH_2NH_2$ (ethylamine) has a polar N–H bond. Nitrogen is less electronegative than oxygen, so the N–H dipole is moderate. The lone pair on nitrogen also contributes, but the overall dipole is smaller than that of an alcohol. …

Why dipole moment trends matter

Dipole moment ($\mu$) is a measure of the net polarity of a molecule. It depends on two things: the magnitude of individual bond dipoles (how electronegative the atoms are) and the molecular geometry (whether those dipoles cancel or reinforce each other). For small organic molecules, the key is to compare the polarity of functional groups and the symmetry of the carbon skeleton.

Here, all three compounds have a three-carbon chain. The difference lies in the functional group at one end: a methyl group ($-CH_3$), an amino group ($-NH_2$), or a hydroxyl group ($-OH$). The carbon skeleton is nearly identical, so the trend is driven by the polarity of the C–X bond (where X = H, N, or O) and the presence of lone pairs.

Step-by-step reasoning

1. Propane ($CH_3CH_2CH_3$)

   Propane is a hydrocarbon with only C–C and C–H bonds. The electronegativity difference between carbon (2.55) and hydrogen (2.20) is small — about 0.35 on the Pauling scale. Each C–H bond has a tiny dipole, but the molecule is nearly symmetric. The central carbon has two methyl groups attached, and the overall shape is such that the small bond dipoles largely cancel. The net dipole moment of propane is very close to zero — typically around 0.08 D. For all practical purposes, it is nonpolar.

2. Ethylamine ($CH_3CH_2NH_2$)

   The amino group introduces a polar C–N bond (electronegativity difference ~0.5) and a polar N–H bond (difference ~0.8). Nitrogen also has a lone pair, which contributes to the dipole. However, the molecule is not perfectly symmetric: the nitrogen atom is at one end, and the lone pair points away from the carbon chain. The net dipole moment of ethylamine is about 1.2–1.3 D. This is moderate — higher than propane, but lower than ethanol. …

Method: Polarity & Intermolecular Forces Analysis

This method uses the dipole moment as a measure of molecular polarity — the greater the charge separation and the more asymmetric the molecule, the higher the dipole moment.

Step 1: Identify the functional group and bond polarity in each compound

Step 2: Compare electronegativity difference

• O–H bond: Electronegativity difference ≈ 1.4 → largest dipole
• N–H bond: Difference ≈ 0.9 → moderate dipole
• C–H bond: Difference ≈ 0.4 → negligible dipole

Step 3: Consider molecular symmetry

• $CH_3CH_2CH_3$ is symmetric → dipoles cancel → dipole moment ≈ 0 D …

🧠 The Core Concept First

Dipole moment ($\mu$) depends on:

• Magnitude of charge separation (polarity of bonds)
• Molecular geometry (vector sum of bond dipoles)

For the three compounds:

• $CH_3CH_2CH_3$ (propane) — nonpolar C–C and C–H bonds → $\mu \approx 0$
• $CH_3CH_2NH_2$ (ethylamine) — polar N–H bonds, lone pair on N → moderate $\mu$
• $CH_3CH_2OH$ (ethanol) — very polar O–H bond, high electronegativity difference → largest $\mu$

Correct order:

$CH_3CH_2CH_3 < CH_3CH_2NH_2 < CH_3CH_2OH$

✗ Common Mistake #1: Confusing dipole moment with boiling point

What students do:

They think “more polar = higher boiling point” and then assume the same order applies to dipole moment.

Why it’s wrong:

Boiling point depends on intermolecular forces (H-bonding, van der Waals), not just dipole moment. Ethanol has strong H-bonding, but its dipole moment is not the only factor.

How to avoid:

• Remember: dipole moment is a molecular property (intrinsic), boiling point is a bulk property.
• For dipole moment, focus on bond polarity + geometry, not H-bonding strength.

✗ Common Mistake #2: Ignoring the lone pair contribution in amines

What students do:

They think $NH_2$ is less polar than $OH$, so they put ethylamine’s dipole moment close to propane.

Why it’s wrong:

The lone pair on nitrogen creates a significant dipole moment — it’s not just the N–H bonds. The vector sum of N–H dipoles plus lone pair gives a net dipole that is substantial.

How to avoid:

• Always account for lone pairs when calculating dipole moment.
• In amines, the lone pair points away from the molecule, adding to the dipole.

✗ Common Mistake #3: Forgetting that propane has zero net dipole

What students do:

They think “C–H bonds are slightly polar, so propane has a small dipole.”

Why it’s wrong: …