Stability Order of Carbocations and Free Radicals (NEET)
Chemistry · Hydrocarbons · NEET
Both carbocations and free radicals follow the same stability order: 3° (tertiary) > 2° (secondary) > 1° (primary) > CH3⁺/•CH3. More alkyl (carbon) groups around the charged or radical carbon means more stability, because alkyl groups push electron density in (inductive effect) and share their C–H bonds with the empty/half-filled orbital (hyperconjugation). Memory hook: "More carbon neighbours = more stable" — count the carbons touching the special carbon.
Carbocation and free radical stability rises as more alkyl groups (carbon neighbours) surround the special carbon: 3° > 2° > 1° > CH3, driven by hyperconjugation and the inductive effect.
Your doubts, answered
How do I decide if a carbocation is 1°, 2° or 3°?
Look only at the carbon that carries the positive charge (the C with the empty orbital). Count how many OTHER carbon atoms are directly bonded to it. 1 carbon neighbour = primary (1°), 2 carbon neighbours = secondary (2°), 3 carbon neighbours = tertiary (3°). Hydrogens do not count, only carbons. Example: in (CH3)3C⁺ the central carbon touches 3 carbons, so it is 3° and very stable.
Why is a tertiary carbocation more stable than a primary one?
Two reasons work together. (1) Inductive effect: alkyl groups (like CH3) push electron density toward the positive carbon, which reduces the positive charge. More alkyl groups = more pushing = more stable. (2) Hyperconjugation: the C–H bonds next to the positive carbon overlap with the empty orbital and spread out the charge. A 3° cation has more neighbouring C–H bonds than a 1° cation, so it is more stable.
Do free radicals follow the same stability order as carbocations?
Yes. The order is the same: 3° > 2° > 1° > •CH3. The reason is also the same — hyperconjugation and the inductive push from alkyl groups stabilise the carbon that has the odd (unpaired) electron, just like they stabilise the positive carbon. So if you learn one order, you know both. This is a favourite NEET shortcut.
What is hyperconjugation in simple words?
Hyperconjugation is when the electrons of a C–H bond that sits next to a positive (or radical) carbon 'leak' into the empty (or half-filled) orbital. It is sometimes called the 'no-bond resonance' or the σ-conjugation effect. More C–H bonds on the neighbouring carbons means more hyperconjugation, and more hyperconjugation means more stability. This is why methyl-rich (branched) cations are the most stable.
Is allyl or benzyl carbocation more stable than tertiary?
Yes, when resonance is possible it beats simple alkyl stabilisation. Allyl (CH2=CH–CH2⁺) and benzyl (C6H5–CH2⁺) cations are stabilised by resonance (the charge spreads over more atoms), so they are often more stable than a normal 3° carbocation. Full order to remember: benzyl ≈ allyl ≥ 3° > 2° > 1° > CH3⁺. NEET usually asks the simple alkyl order first, so master 3° > 2° > 1° > CH3.
⚠️ The NEET trap ✗ Picking a plain secondary carbocation as 'most stable' just because it sits in the middle of a chain. ✓ A secondary carbocation that sits next to a tert-butyl group (many extra C–H bonds) is more stable than a secondary carbocation in a plain chain. Count neighbouring carbons AND the C–H bonds available for hyperconjugation, not just the chain position. 🧠 Stability is decided by the carbons AROUND the special carbon, not by where it is in the chain.
Real NEET questions
NEET 2019 (Odisha) / 2024
The most stable carbocation among the following is: (positive charge shown on the marked carbon)
A · (CH3)3C–C⁺H–CH3 ✓
B · CH3–CH2–C⁺H–CH2–CH3
C · CH3–C⁺H–CH2–CH2–CH3
D · CH3–CH2–C⁺H2
Solution: Option D is a primary carbocation (charged carbon has only 1 carbon neighbour) — least stable, rule it out first. Options A, B and C are all secondary (2 carbon neighbours each). To break the tie, count the C–H bonds available for hyperconjugation next to the positive carbon. In A the positive carbon sits beside a tert-butyl group (CH3)3C–, which provides 9 β-hydrogens plus 3 from the other CH3 = the most hyperconjugation of all four. So A is the most stable carbocation.
NEET 2016 (Phase 1)
The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is:
A · Staggered is less stable than eclipsed, because staggered has torsional strain
B · Eclipsed is more stable than staggered, because eclipsed has no torsional strain
C · Eclipsed is more stable than staggered even though eclipsed has torsional strain
D · Staggered is more stable than eclipsed, because staggered has no torsional strain ✓
Solution: Stability here is decided by torsional strain, the same 'spread things out to be stable' idea. In the staggered form the C–H bonds are as far apart as possible (dihedral angle 60°), so there is no torsional strain and it is the more stable, lower-energy form. The eclipsed form has the bonds lined up, causing repulsion (torsional strain), so it is less stable. Answer D.
Solved Hydrocarbons NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the stability order of carbocations for NEET?
3° > 2° > 1° > CH3⁺ (tertiary most stable, methyl cation least stable). When resonance is possible, benzyl and allyl cations are even more stable than tertiary.
Why are tertiary carbocations the most stable?
Because three alkyl groups surround the positive carbon. They push electron density in (inductive effect) and provide many neighbouring C–H bonds for hyperconjugation, both of which lower the positive charge and stabilise the ion.
Is the free radical stability order the same as carbocations?
Yes, exactly the same: 3° > 2° > 1° > •CH3. Both are stabilised by hyperconjugation and the inductive effect of alkyl groups. Learning one order gives you both.
Which is more stable, allyl carbocation or tertiary carbocation?
The allyl (and benzyl) carbocation is usually more stable because resonance spreads the positive charge over more than one atom, which is a stronger stabilising effect than simple alkyl donation.
How is hyperconjugation different from the inductive effect?
Inductive effect is electron push through sigma bonds by an alkyl group. Hyperconjugation is the overlap of neighbouring C–H bond electrons into the empty (or half-filled) orbital. Both stabilise carbocations and radicals, and more alkyl groups increase both.
