Key Fact
Hyperconjugation
Stabilising interaction between σ C-H bonds and adjacent unsaturated π system or carbocation. More α-H atoms → more hyperconjugation → more stable. Tertiary carbocation > secondary > primary.
-- NCERT, p. 28Free Radicals Carbocations
8 MCQs1 revision card9-step worked example
Source: NCERT Alcohols, Phenols and EthersPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The trap that costs marks: when NEET gives you four carbocations and asks you to rank their stability, a common error is writing the order backwards — placing methyl or primary carbocations as most stable. The correct order is 3° > 2° > 1° > methyl, and missing this loses you 5 marks (4 + 1 negative).
What are carbocations and free radicals?
A carbocation is a carbon atom bearing a positive charge with only six electrons in its valence shell — it is electron-deficient. A free radical is a species with an unpaired electron on carbon. Both are reactive intermediates; they form during bond-breaking and exist briefly before reacting further (NCERT Class 11 Chemistry, Chapter 9, page 28).
Carbocation stability — why 3° beats 1°
Stability depends on how well the positive charge is dispersed. Two effects matter:
-
Hyperconjugation: Adjacent C–H bonds (α-hydrogens) donate electron density into the empty p-orbital of the carbocation. A tertiary carbocation has 3 alkyl groups, each contributing α-H bonds. More α-H bonds → more hyperconjugative stabilisation → more stable. A methyl carbocation (CH₃⁺) has zero α-H bonds — no hyperconjugation at all.
-
Inductive effect (+I): Alkyl groups are electron-donating through σ-bonds, partially neutralising the positive charge. More alkyl groups → greater +I effect.
Both effects point the same way: 3° > 2° > 1° > CH₃⁺ (NCERT Class 11 Chemistry, Chapter 9, page 30).
Resonance-stabilised carbocations
Allyl (CH₂=CH–CH₂⁺) and benzyl (C₆H₅–CH₂⁺) carbocations are primary by substitution count, but resonance delocalises the charge across π-electrons. This makes them significantly more stable than a typical primary carbocation — comparable to or exceeding secondary carbocations in stability.
Free radical stability follows the same order (3° > 2° > 1° > methyl) for the same reasons: hyperconjugation stabilises the species with the unpaired electron.
Watch out: NEET 2025 tested carbocation stability ranking directly. If you see a ranking question, write out the order explicitly before scanning the options — don't reverse it under pressure.
Practice MCQs
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which of the following is a carbocation?
MCQ 2Easy RecallPractice
How many valence electrons does the carbon atom in a carbocation possess?
MCQ 3Easy RecallPractice
Free radical stability follows the order:
MCQ 4Direct ApplicationPractice
Arrange the following carbocations in decreasing order of stability: (I) (CH₃)₃C⁺ (II) (CH₃)₂CH⁺ (III) CH₃CH₂⁺ (IV) CH₃⁺
MCQ 5Direct ApplicationPractice
The allyl carbocation (CH₂=CH–CH₂⁺) is more stable than a simple primary carbocation because of:
MCQ 6Direct ApplicationPractice
Which carbocation is most stable? (I) CH₃⁺ (II) CH₃CH₂⁺ (III) (CH₃)₂CH⁺ (IV) C₆H₅CH₂⁺ (benzyl)
MCQ 7Concept TrapPractice
A tertiary carbocation is more stable than a primary carbocation. Which of the following best explains this observation?
MCQ 8CalculationPractice
Consider the following carbocations: (I) (CH₃)₃C⁺ (II) CH₂=CH–CH₂⁺ (III) (CH₃)₂CH⁺ (IV) C₆H₅CH₂⁺ The correct stability order is:
Quick recall before you leave
Worked Example
- 1
Given
Four carbocations of different types: (I) tertiary, (II) allyl (primary with adjacent C=C), (III) primary, (IV) benzyl (primary with adjacent aromatic ring).
- 2
Required
Decreasing order of stability.
- 3
Concept
Carbocation stability is governed by (a) hyperconjugation — more α-C–H bonds donate more electron density, and (b) resonance — charge delocalisation through π-systems (allyl, benzyl). Resonance with an aromatic ring (benzyl) provides more extensive delocalisation than resonance with a single C=C (allyl).
- 4
Formula / Rule
3° > 2° > 1° > CH₃⁺ (hyperconjugation + inductive) Benzyl ≈ allyl >> simple 1° (resonance elevates these) Benzyl > allyl (aromatic ring has more resonance contributors)
- 5
Substitution / Classification
- (I) (CH₃)₃C⁺: tertiary — 9 α-H bonds, strong hyperconjugation, strong +I - (IV) C₆H₅CH₂⁺: benzyl — primary by substitution, but aromatic resonance delocalises charge over several ring carbons - (II) CH₂=CH–CH₂⁺: allyl — primary by substitution, but resonance across one C=C (two contributing structures) - (III) CH₃CH₂⁺: simple primary — only 3 α-H bonds, weak hyperconjugation, no resonance
- 6
Reasoning
Benzyl (IV) tops the list — aromatic resonance provides the most extensive delocalisation. Tertiary (I) is next — 9 α-H bonds give very strong hyperconjugation that exceeds allyl's limited two-carbon resonance. Allyl (II) is still well above simple primary. Primary (III) has only modest hyperconjugation and no resonance.
- 7
Final answer
**IV > I > II > III** — i.e., C₆H₅CH₂⁺ > (CH₃)₃C⁺ > CH₂=CH–CH₂⁺ > CH₃CH₂⁺ Note: No numerical constants or calculations are involved in this ranking problem — it is a qualitative comparison of electronic effects.
- 8
Common trap
Ignoring resonance and ranking purely by degree of substitution would place (I) first and both (IV) and (II) below it. NEET distractors exploit exactly this: an option showing I > IV > II > III tempts students who consider only hyperconjugation and forget resonance stabilisation.
- 9
Similar NEET-style question
"Among CH₃⁺, (CH₃)₂CH⁺, C₆H₅CH₂⁺, and (CH₃)₃C⁺, which is the most stable carbocation?" — the answer is benzyl, C₆H₅CH₂⁺, because aromatic resonance dominates over hyperconjugation alone. ---
Before solving, remember these
Key Fact
Reaction intermediates
Carbocations (e⁻ deficient, sp²): stability tertiary > secondary > primary > methyl. Carbanions (e⁻ rich, sp³): stability methyl > primary > secondary > tertiary. Free radicals (one unpaired e⁻): stability tertiary > secondary > primary.
-- NCERT, p. 30Formulas
Carbocation stability order
Stability from hyperconjugation (more α-H) and inductive donation (alkyl groups). Resonance can elevate primary cations.
| Symbol | Quantity | SI Unit |
|---|---|---|
| stability | relative | - |
Valid when
- Gas phase or aprotic solvent
- Compare similar reaction conditions
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student writes 1° > 2° > 3° (linear with substitution count, but inverted) or treats methyl as more stable.
When it triggers
Question gives multiple carbocations and asks for stability ranking.
How to avoid
Stability: 3° > 2° > 1° > methyl. Hyperconjugation (more α-H = more stable). Resonance can elevate (allyl, benzyl > 1°).
Category: Similar Terms
Student treats inductive (through sigma bonds, decreases with distance) like resonance (through pi system, often dominant).
When it triggers
Comparison of substituent effects (acidity, basicity, dipole moment).
How to avoid
Inductive: through-bond, weakens with distance, only sigma. Resonance: through-pi-system, often more powerful, requires conjugation.
Category: Similar Terms
Student numbers carbon chain from wrong end, giving higher locants to substituents than necessary.
When it triggers
IUPAC naming question with multiple substituents.
How to avoid
Choose end that gives the LOWEST set of locants for all substituents (compare set, not first-encountered). Functional group has priority for lowest locant.
Category: Similar Terms
Student conflates optical (chirality, R/S) with geometrical (cis/trans). They're different stereoisomerism types.
When it triggers
Question about isomerism of a specific compound.
How to avoid
Optical isomerism requires chiral center (sp³ with 4 different groups). Geometrical isomerism requires restricted rotation (C=C with two different groups on each carbon).
Root cause: concept gap
Correction
Stability follows hyperconjugation: more α-H → more stable. Order: 3° > 2° > 1° > methyl. Allyl/benzyl resonance-stabilised.
Root cause: concept gap
Correction
Hyperconjugation requires α-C-H bond. tert-Butyl carbocation: 9 α-H → very stable. No-α-H carbocation: no hyperconjugation.
Root cause: concept gap
Correction
Inductive: through sigma, weakens with distance, weak. Resonance: through pi, often dominant for activated systems.
Root cause: rushed under time pressure
Correction
Try numbering from BOTH ends; pick the one giving the lowest SET of locants. Functional group has priority for lowest locant.
Root cause: concept gap
Correction
Optical: chirality (sp³ with 4 different groups). Geometrical: restricted rotation (C=C). Different molecular features required.
Past Year Questions
11 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
Which one of the following compounds can exist as cis-trans isomers?
11, 2-Dimethylcyclohexane
2Pent-1-ene
32-Methylhex-2-ene
41, 1-Dimethylcyclopropane
NTA Answer: Option 1(final)
NEET 2025
1II > III > I
2II > I > III
3I > II > III
4III > II > I
NTA Answer: Option 2(final)
NEET 2025
1A, C and D only
2B and E only
3A and C only
4D and E only
NTA Answer: Option 3(final)
NEET 2024Revised key
A compound with a molecular formula of C H has two tertiary carbons. Its IUPAC name is : 6 14
1n-hexane
22-methylpentane
32,3-dimethylbutane
42,2-dimethylbutane
NTA Answer: Option 3(revised_final)
NEET 2024Revised key
1Three resonance structures can be drawn for ozone
2BF has non-zero dipole moment 3
3Dipole moment of NF is greater than that of NH 3 3
4Three canonical forms can be drawn for CO2− ion 3
NTA Answer: Option 4(revised_final)
NEET 2024Revised key
1propylamine
2butylamine
3butanamide
4–bromobutanoic acid
NTA Answer: Option 1(revised_final)
NEET 2023
1H = U + n g RT
2H – U = –nRT
3H + U = nR
4H = U – n RT g
NTA Answer: Option 1(final)
NEET 2022
The incorrect statement regarding chirality is
1A racemic mixture shows zero optical rotation
2S 1 reaction yields 1 : 1 mixture of both enantiomers N
3The product obtained by S 2 reaction of haloalkane having chirality at the reactive site shows inversion N of configuration
4Enantiomers are superimposable mirror images on each other
NTA Answer: Option 4(final)
NEET 2022
10.2303
21.3818
30.9212
40.4606
NTA Answer: Option 3(final)
NEET 2022
The correct IUPAC name of the following compound is
16-bromo-4-methyl-2-chlorohexan-4-ol
21-bromo-5-chloro-4-methylhexan-3-ol
36-bromo-2-chloro-4-methylhexan-4-ol
41-bromo-4-methyl-5-chlorohexan-3-ol
NTA Answer: Option 2(final)
NEET 2021
1(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
2(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
3(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
4(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
NTA Answer: Option 1(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Order carbocations by stability. 3° > 2° > 1° > methyl. Allyl/benzyl extra stabilised by resonance.
InterpretationEasy
Common distractors
ignores resonance
Compares only by alkyl substitution
Sources
NCERT refs: Class 11 Chemistry Chapter 9, p.28 | Class 11 Chemistry Chapter 9, p.30
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