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JEE Mains Chemistry · Organic Chemistry - Some Basic Principles and Techniques

Reaction Intermediates, Bond Fission and Reagents

A covalent bond breaks evenly into two free radicals or unevenly into a carbocation and an anion; how stable these short-lived species are, and whether a reagent gives or takes an electron pair, decides how a reaction runs.

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Why this matters

Twenty-four PYQs, four of them asking for a number, and one from 2026. Sixteen are about carbocations: their shape, their order of stability or of hydride affinity, how many hyperconjugating hydrogens hold one up, and the shifts and ring expansions that move the charge. Eight cover bond fission and the other species: which fission gives ions, free radicals and carbanions ranked by stability, nucleophiles and electrophilic centres counted, the radical from benzoyl peroxide, and a Grignard reagent destroyed by an O–H group.

Concept 1 of 2: Carbocation stability, hydride affinity and rearrangement

A carbocation is a carbon with only six electrons: sp², flat, with an empty p orbital, and hungry for electrons. Anything that feeds electron density into that empty orbital steadies it: alkyl groups through hyperconjugation and the +I effect, and a neighbouring π system or lone pair through resonance. The more stable the cation, the less energy it releases when it finally takes a hydride ion.

Definition

  • Shape: sp², trigonal planar, with the empty p orbital at right angles to the plane. It is an electrophile.
  • Alkyl order: 3∘>2∘>1∘>CH3+3^\circ > 2^\circ > 1^\circ > \mathrm{CH_3^+}, matching the α-hydrogens: 9 in tert-butyl, 6 in isopropyl, 3 in ethyl, none in methyl.
  • Resonance outweighs hyperconjugation: benzylic and allylic cations are stabilised, and each extra phenyl ring helps more, Ph3C+>Ph2CH+>PhCH2+\mathrm{Ph_3C^+ > Ph_2CH^+ > PhCH_2^+}.
  • The tropylium ion, C7H7+\mathrm{C_7H_7^+}, is aromatic (6 π electrons in a planar ring) and very stable. Cyclopropyl groups on the cationic carbon also stabilise it strongly: tricyclopropylmethyl is an exceptionally stable cation.
  • A lone-pair donor that can reach the positive carbon (−OCH3\mathrm{-OCH_3}, −NR2\mathrm{-NR_2}) stabilises it by +R: a para −OCH3\mathrm{-OCH_3} on a benzyl cation helps, a meta one cannot reach the charge. A −NO2\mathrm{-NO_2} group destabilises it.
  • Vinyl CH2=CH+\mathrm{CH_2{=}CH^+} and ethynyl cations are very unstable: the charge sits on a carbon with more s-character.
  • Hydride affinity is the energy released when a cation captures H−\mathrm{H^-}. The more stable the cation, the LOWER its hydride affinity.
  • Rearrangement: a 1,2-hydride or 1,2-methyl shift turns a cation into a more stable one. A cation next to a cyclobutane ring expands it to a cyclopentane ring to relieve strain, and ring growth stops at six.

Hyperconjugation count and alkyl order

hyperconjugating H=number of α-H3∘>2∘>1∘>CH3+\text{hyperconjugating H} = \text{number of } \alpha\text{-H} \qquad 3^\circ > 2^\circ > 1^\circ > \mathrm{CH_3^+}

Worked example

Arrange (CH3)2CH+\mathrm{(CH_3)_2CH^+}, CH3CH2+\mathrm{CH_3CH_2^+}, (C6H5)2CH+\mathrm{(C_6H_5)_2CH^+} and CH3+\mathrm{CH_3^+} in decreasing order of stability, and then in decreasing order of hydride affinity.
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The same idea in a real exam question:

JEE Mains · 2023 · 10 April 2023 · Q129Moderate

Example 1 · Organic Chemistry - Some Basic Principles and Techniques · Reaction Intermediates, Bond Fission and Reagents

The decreasing order of hydride affinity for following carbocations is: (A) (B) (C) (D) Choose the correct answer from the options given below:

More stable means LOWER hydride affinity

Hydride affinity measures how eagerly a cation grabs H−\mathrm{H^-}. A stable cation is not eager, so the order of hydride affinity is the reverse of the order of stability.

A meta donor cannot reach the charge

On a benzyl cation, the positive charge spreads only to the ortho and para ring carbons. An −OCH3\mathrm{-OCH_3} at the meta position cannot donate its lone pair to the charge, so it helps far less than a para one.

The vinyl cation is not allylic

In CH2=CH+\mathrm{CH_2{=}CH^+} the charge sits ON the double-bond carbon, which cannot spread it. In the allyl cation, CH2=CH−CH2+\mathrm{CH_2{=}CH{-}CH_2^+}, it sits next to the double bond and is shared by resonance.

The methyl cation has no hyperconjugation

Hyperconjugation needs a C–H bond on the carbon NEXT to the cationic carbon. CH3+\mathrm{CH_3^+} has no such carbon, so it has no hyperconjugating hydrogen at all.
Drill 15 more on carbocation stability, hydride affinity and rearrangement

Concept 2 of 2: Bond fission, free radicals, carbanions and reagent types

A covalent bond breaks in one of two ways. If each atom keeps one electron (homolysis), two free radicals form, usually with heat, light or a peroxide. If one atom keeps both electrons (heterolysis), ions form, and the reactions that follow are called ionic or polar reactions. A reagent then either brings an electron pair (a nucleophile) or looks for one (an electrophile).

Definition

  • Homolysis → free radicals → free-radical reactions. Heterolysis → a carbocation and an anion, or a carbanion and a cation → ionic reactions.
  • Free radicals: allylic, benzylic and propargylic radicals are the most stable (resonance); then 3∘>2∘>1∘>CH3∙3^\circ > 2^\circ > 1^\circ > \mathrm{CH_3^{\bullet}}. A radical on an sp² or sp carbon (vinyl, ethynyl) is the least stable.
  • Carbanions: sp³, pyramidal, with a lone pair; they are nucleophiles and bases. The alkyl order is reversed: CH3−>1∘>2∘>3∘\mathrm{CH_3^- > 1^\circ > 2^\circ > 3^\circ}. −R groups (C=O, NO2\mathrm{NO_2}, CN) stabilise them, and the aromatic cyclopentadienyl anion (6 π electrons) is very stable, while the antiaromatic cyclopropenyl anion is very unstable.
  • Nucleophiles give an electron pair, from a lone pair or a π bond: OH−\mathrm{OH^-}, CN−\mathrm{CN^-}, NH3\mathrm{NH_3}, H2O\mathrm{H_2O}, RSH, R2S\mathrm{R_2S}, C=C. Electrophiles accept a pair: H+\mathrm{H^+}, carbocations, BF3\mathrm{BF_3}, AlCl3\mathrm{AlCl_3}.
  • Electrophilic centres in a molecule: the carbon of C=O, the carbon of C≡N, and the β-carbon of a C=C conjugated with C=O.
  • Benzoyl peroxide gives benzoyloxy radicals, which lose CO2\mathrm{CO_2}: (C6H5COO)2→2 C6H5COO∙→2 C6H5∙+2 CO2\mathrm{(C_6H_5COO)_2 \to 2\,C_6H_5COO^{\bullet} \to 2\,C_6H_5^{\bullet} + 2\,CO_2}.
  • A Grignard reagent, RMgX, reacts as a carbanion. Any O–H or N–H group protonates it: with an alcohol it gives the alkane RH.
SpeciesFormed byCarbon: hybridisation and shapeElectrons on the carbonBehaves as
CarbocationHeterolysis; carbon loses the pairsp², trigonal planar6 (a sextet)Electrophile
CarbanionHeterolysis; carbon keeps the pairsp³, pyramidal8, with one lone pairNucleophile and base
Free radicalHomolysissp², nearly planar7, with one unpaired electronNeutral, very reactive; starts chain reactions
Radical and carbocation stability follow the same alkyl order; carbanion stability runs the other way.
All three are short-lived intermediates; none is isolated in an ordinary reaction.
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The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q43Moderate

Example 2 · Organic Chemistry - Some Basic Principles and Techniques · Reaction Intermediates, Bond Fission and Reagents

Ionic reactions with organic compounds proceed through: (A) Homolytic bond cleavage (B) Heterolytic bond cleavage (C) Free radical formation (D) Primary free radical (E) Secondary free radical Choose the correct answer from the options given below:

Carbanion stability runs opposite to carbocation stability

Alkyl groups push electrons towards the carbon. That helps a positive carbon and hurts a negative one, so the methyl carbanion is the most stable simple carbanion and the tertiary one the least.

Ionic reactions come from heterolysis

Homolysis gives neutral radicals, which lead to free-radical reactions. Only heterolysis gives the ions that ionic reactions need.

A π bond can make a nucleophile

Ethene has no lone pair and no charge, yet its π electrons attack electrophiles such as H+\mathrm{H^+} and Br+\mathrm{Br^+}. Count alkenes when a question asks for nucleophiles.
Drill 7 more on bond fission, free radicals, carbanions and reagent types

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

  • Carbocation stability, hydride affinity and rearrangement

    Hyperconjugation count and alkyl order

    hyperconjugating H=number of α-H3∘>2∘>1∘>CH3+\text{hyperconjugating H} = \text{number of } \alpha\text{-H} \qquad 3^\circ > 2^\circ > 1^\circ > \mathrm{CH_3^+}

Reference tables (1)

Bond fission, free radicals, carbanions and reagent types3 rows
SpeciesFormed byCarbon: hybridisation and shapeElectrons on the carbonBehaves as
CarbocationHeterolysis; carbon loses the pairsp², trigonal planar6 (a sextet)Electrophile
CarbanionHeterolysis; carbon keeps the pairsp³, pyramidal8, with one lone pairNucleophile and base
Free radicalHomolysissp², nearly planar7, with one unpaired electronNeutral, very reactive; starts chain reactions
Radical and carbocation stability follow the same alkyl order; carbanion stability runs the other way.
All three are short-lived intermediates; none is isolated in an ordinary reaction.

Watch out for (7)

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