Q.Which is the correct increasing order of boiling points of the following compounds?
1-Iodobutane, 1-Bromobutane, 1-Chlorobutane, Butane

Concept: Boiling Point Trends – Boiling point depends on molecular mass (London forces) and polarity (dipole-dipole interactions). For alkyl halides, mass dominates; for butane vs. haloalkanes, the absence of a dipole and much lower mass makes butane the lowest.

Reasoning:

1. Butane ($C_4H_{10}$, ~58 g/mol) has only weak London forces and no dipole → lowest boiling point. …

Boiling point trends in organic compounds are governed by two main factors: the strength of intermolecular forces and the molecular mass. For a series of similar compounds, like the 1-halobutanes, the key force is London dispersion (van der Waals) forces. These forces increase with molecular size and surface area — larger, heavier molecules have more electrons and a larger electron cloud, making them more polarizable. Stronger dispersion forces mean more energy is needed to separate molecules, hence a higher boiling point.

Butane is a simple alkane with no polar C–X bond. It experiences only weak London forces. The haloalkanes, on the other hand, have a polar C–X bond, which adds dipole-dipole interactions. However, for the 1-halobutane series, the difference in dipole moments is small — the dominant effect is the increasing size and polarizability of the halogen as we go from Cl to Br to I.

Let’s work through the reasoning step by step.

1. Identify the compounds and their molecular masses.

   • Butane ($\ce{C4H10}$): $M \approx 58\ \text{g/mol}$
   • 1-Chlorobutane ($\ce{C4H9Cl}$): $M \approx 92.5\ \text{g/mol}$
   • 1-Bromobutane ($\ce{C4H9Br}$): $M \approx 137\ \text{g/mol}$
   • 1-Iodobutane ($\ce{C4H9I}$): $M \approx 184\ \text{g/mol}$

   Mass increases steadily from butane to 1-iodobutane.

2. Consider the intermolecular forces at play.

   All four molecules experience London dispersion forces. The haloalkanes also have dipole-dipole interactions because the C–X bond is polar. But the dipole moment of C–Cl is about 1.9 D, C–Br about 1.8 D, and C–I about 1.6 D — these are quite similar. The real difference comes from polarizability: iodine is much larger and more polarizable than chlorine, so the dispersion forces in 1-iodobutane are significantly stronger than in 1-chlorobutane.

3. Rank the boiling points.

   • Butane has the lowest mass and no dipole — it will have the lowest boiling point ($-0.5\ ^\circ\text{C}$).
   • 1-Chlorobutane boils at about $78\ ^\circ\text{C}$.
   • 1-Bromobutane boils at about $101\ ^\circ\text{C}$.
   • 1-Iodobutane boils at about $130\ ^\circ\text{C}$.

   So the order is: Butane < 1-Chlorobutane < 1-Bromobutane < 1-Iodobutane. …

Concept: Boiling Point Trends in Haloalkanes

The boiling point of haloalkanes depends on two competing factors:

• Molecular mass (higher mass → higher boiling point)
• Intermolecular forces (stronger forces → higher boiling point)

For alkyl halides with the same carbon chain, the dominant factor is molecular mass/polarisability: all four compounds rely mainly on London dispersion forces, and although the three 1-halobutanes are actually polar (C–X dipole moments ≈ 1.6–1.9 D), their dipole moments are so similar that polarity does not change the ordering — the growing dispersion forces from Cl → Br → I do.

Method: Trend Analysis Based on Molecular Mass

Step 1: List the compounds with their molecular masses

Step 2: Apply the rule …

Common Mistakes & How to Avoid Them

Mistake 1: Confusing the trend of boiling points with bond strength or reactivity

The error: Students often think that since C–I bond is weakest (most reactive in substitution reactions), 1-iodobutane should have the lowest boiling point. They incorrectly apply the "stronger bond = higher boiling point" logic.

Why it's wrong: Boiling point depends on intermolecular forces (van der Waals forces), not bond strength within the molecule. Larger, more polarisable molecules have stronger London dispersion forces.

How to avoid: Remember:

• Boiling point ∝ molecular size and surface area (for similar shapes)
• For alkyl halides: larger halogen → more electrons → stronger dispersion forces → higher boiling point
• Bond strength is irrelevant here

Mistake 2: Thinking polarity is the dominant factor

The error: Students argue that C–Cl bond is most polar, so 1-chlorobutane should have the highest boiling point due to dipole-dipole interactions.

Why it's wrong: While dipole-dipole forces matter, dispersion forces dominate for larger molecules. The increase in dispersion forces from Cl → Br → I outweighs the decrease in polarity.

How to avoid: Compare the total number of electrons:

• Butane: 34 electrons
• 1-Chlorobutane: 50 electrons
• 1-Bromobutane: 68 electrons
• 1-Iodobutane: 86 electrons

More electrons → stronger temporary dipoles → higher boiling point.

Mistake 3: Forgetting that butane has no dipole

The error: Students place butane somewhere in the middle, thinking it has "some" polarity.

Why it's wrong: Butane is non-polar (only C–C and C–H bonds). It has only weak dispersion forces, while all haloalkanes have additional dipole-dipole interactions. …