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Molecular Stability

Molecular Stability

3 patterns

Aromaticity, resonance stabilization, ring strain, and carbocation stability. These determine which molecules persist and which decompose. You will encounter this when predicting whether a compound will survive reaction conditions or fall apart.

Incorrect
1,3-Cyclohexadiene (non-aromatic)

Correct
Benzene (aromatic)
Why it's wrong

1,3-Cyclohexadiene lacks the full cyclic delocalization that gives benzene its aromatic stability. Without aromaticity, the molecule does not benefit from the ~150 kJ/mol resonance stabilization energy.

Why it's correct

Benzene is significantly more stable than 1,3-cyclohexadiene due to aromatic stabilization. Its six pi electrons are fully delocalized across the ring, giving it roughly 150 kJ/mol of extra resonance energy compared to what you would expect from isolated double bonds.

Wikipedia: Aromaticity
Incorrect
Cyclopropane (ring strain)

Correct
Cyclohexane (no strain)
Why it's wrong

Cyclopropane is highly strained because its 60-degree bond angles deviate enormously from the ideal tetrahedral angle of 109.5 degrees. This gives it about 115 kJ/mol of ring strain energy.

Why it's correct

Cyclohexane is nearly strain-free because its chair conformation allows bond angles close to the ideal tetrahedral angle of 109.5 degrees. It has essentially zero ring strain.

Wikipedia: Ring strain
Incorrect
Methyl cation (primary)

Correct
Tert-butyl cation (tertiary)
Why it's wrong

The methyl cation has no adjacent alkyl groups to stabilize its positive charge through hyperconjugation. This makes it one of the least stable simple carbocations and very difficult to observe experimentally.

Why it's correct

Tertiary carbocations are more stable than primary ones because the three adjacent alkyl groups donate electron density through hyperconjugation and inductive effects, spreading the positive charge over a larger volume.

Wikipedia: Carbocation stability