Q.Out of o- and p-dibromobenzene which one has higher melting point and why?

Concept: Melting Point and Molecular Symmetry — Melting point depends on molecular symmetry and packing efficiency in the solid state, not just molecular mass.

Reasoning:

1. Both $o$-dibromobenzene and $p$-dibromobenzene have the same molecular formula $\ce{C6H4Br2}$ and similar molar mass.
2. $p$-Dibromobenzene is a symmetrical molecule — the two bromine atoms are opposite each other. This symmetry allows the molecules to pack more efficiently in the crystal lattice, leading to stronger intermolecular forces (higher lattice energy). …

Why Symmetry Matters for Melting Points

Melting point is not about how strongly molecules interact in general — it’s about how well they pack together in the solid state. A molecule that fits neatly into a crystal lattice (like bricks in a wall) will have strong intermolecular forces holding the crystal together. A molecule that is bent or asymmetrical creates gaps and weak points in the lattice, so less energy is needed to break it apart.

For disubstituted benzenes, the relative positions of the two bromine atoms dramatically change the molecular shape and symmetry.

Step-by-Step Reasoning

1. Identify the molecular structures

• o-Dibromobenzene (1,2-dibromobenzene): The two bromine atoms are adjacent on the benzene ring. This creates a molecule with a bent, asymmetrical shape — the bromines are on one side, making the molecule polar and lopsided.
• p-Dibromobenzene (1,4-dibromobenzene): The two bromine atoms are directly opposite each other. This gives a highly symmetrical, rod-like shape — the molecule has a centre of symmetry and is non-polar overall.

2. Consider crystal packing efficiency

In the solid state, molecules arrange themselves to minimise empty space. p-Dibromobenzene, being symmetrical and flat, can stack neatly like tiles. Its molecules fit into a regular, repeating lattice with strong intermolecular forces (primarily London dispersion forces, which are significant for large, polarisable bromine atoms) acting uniformly in all directions.

o-Dibromobenzene, with its bent shape, cannot pack as tightly. The protruding bromine atoms create steric hindrance, forcing molecules apart and leaving voids in the crystal. This looser packing means weaker overall intermolecular forces holding the crystal together.

3. Compare the actual melting points

The experimental data confirms this reasoning:

Concept: Symmetry & Packing in Molecular Crystals

The melting point of an organic compound depends on how efficiently its molecules pack in the solid state. Higher symmetry → better packing → stronger intermolecular forces (especially van der Waals forces) → higher melting point.

Method: Symmetry-Based Packing Analysis

Step 1: Identify molecular symmetry

• p-Dibromobenzene (1,4-dibromobenzene): The two Br atoms are opposite each other. The molecule is highly symmetric (planar, with a centre of symmetry).
• o-Dibromobenzene (1,2-dibromobenzene): The two Br atoms are adjacent. The molecule is less symmetric (no centre of symmetry, dipole moment present).

Step 2: Predict crystal packing efficiency

• p-isomer: Symmetrical shape allows molecules to stack neatly in the crystal lattice, like bricks. This maximises intermolecular contact (especially between Br atoms and the aromatic rings).
• o-isomer: Asymmetrical shape creates gaps in the lattice, reducing packing efficiency. …

This is a classic trap in organic chemistry, especially for JEE, NEET, and board exams. Let’s break it down.

The Correct Answer

p-Dibromobenzene has a higher melting point than o-dibromobenzene.

Why?

• p-Dibromobenzene is symmetrical.
• Symmetry allows the molecules to pack more efficiently in the crystal lattice.
• Stronger intermolecular forces (especially dipole-induced dipole and London dispersion forces) require more energy to overcome → higher melting point.
• o-Dibromobenzene is less symmetrical, so it packs less efficiently → lower melting point.

Common Mistakes Students Make

1. ✗ Thinking “ortho has stronger dipole → higher melting point”

Why it’s wrong:

• Yes, o-dibromobenzene has a net dipole moment (the two Br atoms are on adjacent carbons).
• But melting point depends more on crystal packing efficiency than on dipole strength alone.
• p-dibromobenzene has zero net dipole, but its symmetry allows much better packing → stronger overall lattice energy.

How to avoid:

• Remember: For melting point, symmetry > dipole moment in most cases.
• Visualise the molecules: p-isomers are like flat, symmetric tiles that stack neatly; o-isomers are bent and leave gaps.

2. ✗ Confusing melting point with boiling point

Why it’s wrong:

• For boiling point, o-dibromobenzene does have a higher value than p-dibromobenzene (because of stronger dipole-dipole interactions in the liquid state).
• Many students memorise “ortho > para for boiling point” and wrongly apply it to melting point.

How to avoid:

• Boiling point → depends on intermolecular forces in liquid (dipole moment matters more).
• Melting point → depends on crystal packing in solid (symmetry matters more).
• Make a clear mental table:

3. ✗ Forgetting that p-isomers are more symmetric

Why it’s wrong:

• Some students think “para” means “opposite” but don’t connect that to molecular shape.
• They might even think both are equally symmetric.

How to avoid:

• Draw the structures:
  • p-dibromobenzene: Br atoms at 1 and 4 positions → molecule has a centre of symmetry.
  • o-dibromobenzene: Br atoms at 1 and 2 → no centre of symmetry, molecule is bent.
• Symmetry → better packing → higher melting point.

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