Question
Arrange the following carbocations in increasing order of stability:
Explain the role of hyperconjugation in determining this order. (JEE Main 2024)
Solution — Step by Step
Count the carbon atoms directly attached to the positively charged carbon:
- → methyl (0° — no α-carbons)
- → primary (1°)
- → secondary (2°)
- → tertiary (3°)
The degree directly tells us how many α-carbon–hydrogen bonds are available for hyperconjugation.
Hyperconjugation works through the overlap of a C–H σ-bond at the α-carbon with the empty p-orbital on the carbocation. More α-H atoms = more hyperconjugative structures = greater electron delocalization = lower energy = more stable.
| Carbocation | α-H atoms | No. of hyperconjugative structures |
|---|---|---|
| 0 | 0 | |
| 3 | 3 | |
| 6 | 6 | |
| 9 | 9 |
For the ethyl carbocation , one hyperconjugative (no-bond resonance) structure looks like:
The positive charge spreads from the carbocation carbon onto the hydrogen. This delocalization lowers the overall charge density on any single atom, stabilising the species.
More hyperconjugative structures → more resonance stabilisation → more stable carbocation.
Final answer: methyl < primary < secondary < tertiary
Why This Works
Hyperconjugation is essentially no-bond resonance. The electrons in a C–H σ-bond at the α-position partially “donate” into the adjacent empty p-orbital of the carbocation. This is the same principle as resonance — spreading charge over more atoms always lowers energy.
The tertiary carbocation has three methyl groups, giving 9 α-H atoms and therefore 9 hyperconjugative structures. The positive charge is smeared across all nine hydrogens plus the central carbon — that’s 10 atoms sharing the burden.
Compare this to the methyl cation , which has no α-carbons at all. There is nothing to hyperconjugate with. The full positive charge sits on one lonely carbon atom, making it the most reactive and least stable.
Quick rule for JEE: number of hyperconjugative structures = number of α-H atoms. For a 3° carbocation with three —CH₃ groups, that’s 3 × 3 = 9 structures. This number comes up directly in MCQs.
Alternative Method — Inductive Effect Argument
Even without drawing hyperconjugative structures, we can arrive at the same order using the inductive effect.
Alkyl groups are electron-donating through σ-bonds (+I effect). Each methyl group pushes electron density toward the positively charged carbon, reducing the charge and stabilising the cation.
- → three methyl groups donating → maximum +I stabilisation
- → no alkyl groups → zero stabilisation
Both arguments (hyperconjugation and inductive effect) give the same order because more substitution means both more α-H atoms and more alkyl +I donors. In JEE problems, hyperconjugation is the preferred explanation when α-H count is specifically mentioned.
Common Mistake
Students often confuse the number of alkyl groups with the number of α-H atoms. A 3° carbocation has 3 alkyl groups but 9 α-H atoms — the hyperconjugative structures come from the H atoms, not from counting the methyl groups themselves. If a question asks “how many hyperconjugative structures does have?”, the answer is 9, not 3.
Getting this wrong costs marks in JEE Main when the question gives you numerical options like 3, 6, 9, and 12.