Question:hard

Arrange in the correct order of stability (decreasing order) for the following molecules :

Updated On: Apr 2, 2026
  • $(I) > (II) > (III) > (IV)$
  • $\left(IV\right)>\left(III\right)>\left(II\right)\,\approx \left(I\right)$
  • $\left(I\right)>\left(II\right)\,\approx \left(III\right)> \left(IV\right)$
  • $\left(III\right)>\left(I\right)\,\approx \left(II\right)> \left(IV\right)$
Show Solution

The Correct Option is D

Solution and Explanation

To determine the correct order of stability for the given molecules, we need to consider several factors such as resonance, inductive effects, hyperconjugation, and steric hindrance. Let's analyze these factors for each molecule option and derive the order of stability.

  1. Molecule (III): This molecule is highly stable due to extensive resonance. Molecules with resonance typically display higher stability because the electrons are delocalized over several atoms, lowering the overall energy state of the molecule.
  2. Molecules (I) and (II): Both molecules (I) and (II) may have similar stability. Factors like hyperconjugation or the presence of strong electron-donating or withdrawing groups can create a significant impact on stability. If they possess similar characteristics, their stabilities will be comparable.
  3. Molecule (IV): This molecule is the least stable among the options. The lack of resonance or destabilizing steric effects may affect its stability adversely.

Based on this analysis, molecule (III) exhibits the greatest stability due to resonance, followed by molecules (I) and (II), which likely have comparable stability, and finally molecule (IV), which appears to be the least stable due to the absence of stabilizing features.

Therefore, the correct order of stability is option $\left(III\right)>\left(I\right)\,\approx \left(II\right)> \left(IV\right)$.

Was this answer helpful?
0