Organic Compounds Containing Halogens: Chemistry | JEE Main
The correct order of melting points of dichlorobenzenes is
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Hint 1 of 2
What primary physical factor determines the unusually high melting point of para-substituted benzenes compared to ortho and meta isomers?
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Step-by-step solutionView
Correct answer
The melting point order is p-dichlorobenzene > o-dichlorobenzene > m-dichlorobenzene because the high symmetry of the para-isomer allows more efficient crystal lattice packing.
Option analysis
Why each option works or fails
A ·
Believing that m-dichlorobenzene has a higher melting point than o-dichlorobenzene while still identifying p-dichlorobenzene as the highest. Between the ortho and meta isomers, ortho-dichlorobenzene has a significantly higher dipole moment and slightly better packing, giving it a higher melting point (-17 °C) than meta-dichlorobenzene (-24 °C).
B ·
None. This is the correct choice. Para-dichlorobenzene packs most efficiently in a solid lattice due to its centrosymmetric structure, resulting in a much higher melting point (53.5 °C), followed by ortho (-17 °C) and meta (-24 °C).
C ·
Assuming that permanent dipole-dipole attractions govern solid-state melting points, which predicts ortho > meta > para because ortho has the highest dipole moment and para has zero net dipole. Melting points depend predominantly on crystal lattice stability and molecular packing symmetry rather than net dipole moments; para-dichlorobenzene packs best despite having zero dipole moment.
D ·
Confusing boiling point trends (where ortho > para > meta or ortho ≈ para > meta) with melting point trends, or placing the most symmetric isomer at the lowest melting point. Remember that molecular symmetry dramatically increases melting point by allowing tighter lattice packing, making the para isomer by far the highest melting of the three.
Reviewed route
Solution
StepWorking
01concept
Melting point depends mainly on crystal lattice packing and molecular symmetry.
The para-isomer is highly symmetrical. It fits much better into the crystal lattice. Therefore, it has a significantly higher melting point than the ortho- and meta-isomers.
Between the ortho- and meta-isomers, ortho-dichlorobenzene has a higher dipole moment. It also packs slightly better than the meta-isomer.
This makes the overall melting point order: para > ortho > meta.
02option_verdict
The correct order is para-dichlorobenzene > ortho-dichlorobenzene > meta-dichlorobenzene.
The symmetrical para-isomer has the highest melting point of 323 K. The ortho-isomer follows with a melting point of 256 K. The meta-isomer has the lowest melting point at 249 K.
03option_verdict
Ranks ortho > meta > para or misses that para has the highest melting point due to superior crystal packing efficiency.
04option_verdict
This places the ortho isomer with the highest dipole moment at the highest melting point (ortho > para > meta). It ignores that crystal lattice symmetry dominates melting point rather than molecular dipole moment.
05option_verdict
Places meta isomer above ortho or para, which contradicts both symmetry packing and intermolecular forces.
✓discriminator
Symmetry dictates melting point in aromatic isomers: para always has the highest melting point due to compact crystal packing. This immediately isolates para > ortho > meta.
Hints that build this answer step by step
What primary physical factor determines the unusually high melting point of para-substituted benzenes compared to ortho and meta isomers?
Symmetry and close packing efficiency in the crystal lattice
Comparing ortho-dichlorobenzene and meta-dichlorobenzene, which isomer has the higher melting point?
Why does para have a much higher melting point than ortho and meta, while boiling points are nearly identical?
Boiling point involves breaking intermolecular forces in a disordered liquid phase where dipole moments dominate (ortho > para ~ meta). Melting point requires breaking a rigid crystal lattice, where molecular symmetry allows para to pack much more closely and stably.