Case: During organic reactions, a covalent bond between two carbon atoms or a carbon atom and some other atom is broken and a new bond is formed. The fission of bonds may be homolytic or heterolytic forming free radicals, carbocations, carbanions etc. This depends upon the displa

Question

Case: During organic reactions, a covalent bond between two carbon atoms or a carbon atom and some other atom is broken and a new bond is formed. The fission of bonds may be homolytic or heterolytic forming free radicals, carbocations, carbanions etc. This depends upon the displacement of electron in a bond and is governed by inductive effect, electromeric effect, resonance effect and hyperconjugation. Questions: (i) Give correct order of relative stability of the following carbanions: CH3(-), CH3CH2(-), CH2Cl(-), CHCl2(-), CCl3(-) (all as carbanions, negative charge on the indicated carbon). [1 mark] (ii) What happens when CH3Cl undergoes homolytic fission? [1 mark] (iii) Do (CH3)3C+ and (CH3)3C- have the same structure? Explain. [2 marks]

**OR**

In place of part (iii) only: Which of the two Ph3C+ or (CH3)3C+ is more stable and why? [2 marks]

Accepted Answer

> > [!TLDR] > Chlorine's electron-withdrawing (-I) effect stabilises a carbanion by dispersing its negative charge, so more Cl atoms means more stability; CH3Cl's homolytic fission gives a methyl radical and a chlorine radical; and the tert-butyl cation and carbanion have different geometries (planar vs pyramidal) because a carbanion carries an extra lone pair the cation doesn't have. > > **(i) Relative stability of carbanions:** > > $$CH_3^-,\ CH_3CH_2^-,\ CH_2Cl^-,\ CHCl_2^-,\ CCl_3^-$$ > > Chlorine is electronegative and exerts an electron-withdrawing **inductive (-I) effect**, which pulls electron density away from the carbanion's negative centre, spreading out (delocalising) the charge and stabilising the carbanion. More Cl atoms means a stronger cumulative -I effect and greater stability. > > The methyl group (in $CH_3CH_2^-$, effectively an ethyl carbanion with an extra $CH_3$ attached) is electron-DONATING (+I effect), which intensifies (concentrates) the negative charge, destabilising the carbanion relative to plain $CH_3^-$. > > Order of INCREASING stability (least stable to most stable): > $$CH_3CH_2^- < CH_3^- < CH_2Cl^- < CHCl_2^- < CCl_3^-$$ > > (As Cl atoms increase from 0 to 3, the -I effect and hence stability keeps increasing.) > > **(ii) Homolytic fission of CH$_3$Cl:** > > In homolytic fission, the C-Cl bond breaks symmetrically, with each fragment retaining one electron of the shared bonding pair, generating two neutral free radicals: > > $$CH_3-Cl \xrightarrow{h u ext{ or heat}} CH_3^{\bullet} + Cl^{\bullet}$$ > > i.e., a methyl free radical ($CH_3^\bullet$) and a chlorine free radical/atom ($Cl^\bullet$) are formed. > > **(iii) Do $(CH_3)_3C^+$ and $(CH_3)_3C^-$ have the same structure?** > > **No**, they have different structures. > …

Intermediate	How It Forms	Key Feature	Example Reaction
Carbocation	A carbon atom loses a bonding electron pair (heterolytic cleavage)	Carbon with only 6 valence electrons (positive charge, $sp^2$ hybridised, planar)	$S_N1$ substitution, electrophilic addition to alkenes
Carbanion	A carbon atom gains a bonding electron pair	Carbon with 8 valence electrons plus an extra lone pair (negative charge, $sp^3$ hybridised, pyramidal)	$S_N2$ substitution (as a nucleophile), base-catalysed reactions
Free Radical	A bond breaks symmetrically (homolytic cleavage)	Carbon with 7 valence electrons (unpaired electron, neutral, highly reactive)	Free-radical halogenation of alkanes, polymerisation

Concept understanding — Reaction Intermediates

Reaction Intermediates: The Transient Guests of a Chemical Reaction

The Intuition: Why Do Intermediates Exist?

The Precise Statement

The Three Classic Examples

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