Q.Which of the following compounds will have the highest melting point and why?

Concept: Melting Point and Molecular Symmetry — For isomeric aromatic compounds, the melting point is highest for the most symmetrical isomer because symmetry allows tighter packing in the crystal lattice, increasing lattice energy.

Reasoning:

1. All three are isomers of $\mathrm{C_8H_8Cl_2}$ with the two methyl groups fixed at positions 1 and 4 (para to each other). The only difference is the placement of the two chlorine atoms.
2. Compound (II) has the two Cl atoms at positions 2 and 5 — this places the molecule in a fully symmetrical arrangement: a centre of symmetry and a twofold axis. Such high symmetry leads to efficient molecular packing in the solid state. …

Why symmetry controls melting point

Melting is the process where a crystal lattice breaks down into a disordered liquid. The stronger the intermolecular forces and the better the molecules fit together in the solid, the more energy (heat) is needed to melt them. For non-polar or weakly polar molecules like these dichloro-dimethylbenzenes, the dominant forces are London dispersion forces. These forces depend on surface contact — the more closely and extensively molecules can nestle together, the stronger the overall attraction.

Symmetry is the key. A highly symmetrical molecule packs into a crystal with fewer voids and more uniform contacts. Think of stacking identical bricks versus stacking oddly shaped stones: the bricks form a denser, more stable pile. The same principle applies here.

All three compounds are isomers with the same molecular formula, so their electron counts and polarizabilities are similar. The difference in melting point comes almost entirely from packing efficiency.

Step-by-step analysis

1. Identify the molecular structures

All three have two methyl groups fixed at positions 1 and 4 (para to each other). The only variation is where the two chlorine atoms sit on the remaining four ring carbons (positions 2, 3, 5, 6).

• (I) 2,3-dichloro-1,4-dimethylbenzene: Cl atoms on adjacent carbons (2 and 3). This puts both Cl atoms on the same side of the ring — a crowded arrangement (though the molecule still keeps one mirror plane, between the two Cl atoms).
• (II) 2,5-dichloro-1,4-dimethylbenzene: Cl atoms at positions 2 and 5, which are para to each other. The molecule has a centre of symmetry: a 180° rotation about the centre gives the same arrangement. This is the most symmetrical isomer.
• (III) 2,6-dichloro-1,4-dimethylbenzene: Cl atoms at positions 2 and 6, which are meta to each other and flank one of the methyl groups. The molecule has a plane of symmetry (through the 1-4 axis) but no centre of symmetry.

2. Rank the symmetry

• (II) has a centre of symmetry (inversion centre) — clearly the most symmetrical of the three.
• (III) has one mirror plane (through the C1–C4 axis, mapping the two flanking Cl atoms onto each other) but no centre of symmetry. …

Concept: Symmetry & Melting Point in Aromatic Compounds

For isomeric aromatic compounds, higher molecular symmetry leads to better packing in the crystal lattice, which increases the melting point. This is because symmetrical molecules fit together more efficiently, requiring more energy (higher temperature) to overcome the intermolecular forces.

Method: Symmetry Analysis for Melting Point Comparison

Steps:

1. Draw the structures of all three isomers with the given substitution pattern (both $\mathrm{CH_3}$ groups at positions 1 and 4).

2. Identify the symmetry elements in each molecule:

   • Look for planes of symmetry, axes of rotation, and centre of symmetry.
   • More symmetry elements → better crystal packing → higher melting point.

3. Rank the isomers from most symmetrical to least symmetrical.

4. Conclude — the most symmetrical isomer will have the highest melting point.

Applying the method:

Compound (II) — 2,5-dichloro-1,4-dimethylbenzene:

• The two Cl atoms are para to each other.
• The molecule has a centre of symmetry (inversion centre) and a two-fold rotation axis perpendicular to the ring through its centre. Apart from the molecular plane itself (which every planar molecule has), it has no other mirror plane — any candidate plane perpendicular to the ring would map a Cl onto a CH₃ or an H.
• This is the most symmetrical structure — it packs most efficiently in the crystal.

Compound (I) — 2,3-dichloro-1,4-dimethylbenzene:

• Cl atoms on adjacent carbons — no centre of symmetry. …

🧠 The Core Concept

Melting point in isomeric aromatic compounds depends on symmetry and packing efficiency in the crystal lattice.

• Higher symmetry → better packing → stronger intermolecular forces (mostly van der Waals) → higher melting point.
• Lower symmetry → poorer packing → lower melting point.

For these three isomers, all have the same molecular formula and the same number of substituents — only the positions of the two Cl atoms differ.

✗ Common Mistake #1: Assuming "more polar" means higher melting point

Why it's wrong:

Polarity (dipole moment) is not the main factor here. All three have C–Cl bonds, but the net dipole varies. Students often think a larger dipole means stronger intermolecular forces → higher melting point.

But melting point depends more on how well molecules pack in the solid state, not just dipole strength.

Example in this problem:

• Compound (II) is fully symmetrical (Cl atoms para to each other, CH₃ groups para to each other). → Net dipole is zero (or very small). → Yet it has the highest melting point because it packs perfectly.

How to avoid:

✓ Focus on symmetry and crystal packing, not just polarity.

✓ Remember: symmetrical molecules often have higher melting points even if nonpolar.

✗ Common Mistake #2: Thinking "more steric hindrance" always lowers melting point

Why it's wrong:

Steric hindrance can affect packing, but symmetry can override it.

• Compound (I) has two Cl atoms on adjacent carbons — a crowded arrangement with only one mirror plane and no centre of symmetry. → Packs less efficiently than (II) → melts below (II).
• Compound (III) has Cl atoms flanking one CH₃ — also only one mirror plane, no centre. → Also melts below (II). Since (I) and (III) have comparable symmetry (one plane each), symmetry alone cannot rank them against each other — don't assert a specific (I)-vs-(III) order.

How to avoid:

✓ Compare the overall molecular shape — not just local crowding.

✓ Draw the molecules and check for mirror planes or rotational symmetry.

✗ Common Mistake #3: Ignoring the "para-dimethyl" backbone

Why it's wrong:

All three have the two CH₃ groups fixed at positions 1 and 4 (para to each other).

Students sometimes forget this and treat the compounds as if the methyl groups can move — but they are fixed.

This means the only variable is the Cl positions.

How to avoid:

✓ Always note the fixed substituents first.

✓ Then focus on the variable part — here, the Cl atoms.

✓ Correct Answer & Reasoning

Highest melting point: Compound (II) — 2,5-dichloro-1,4-dimethylbenzene.

Why:

• It has a center of symmetry and a mirror plane — highest symmetry.
• Molecules pack very efficiently in the crystal lattice. …