Organic Chemistry: Some Basic Principles and Techniques • Topic 2 of 3

Electronic Effects & Reaction Mechanisms

Reactions begin with the breaking of a covalent bond. Homolytic fission splits a bond so each atom keeps one electron, giving neutral free radicals ($A−B \rightarrow A^{\bullet}+B^{\bullet}$); it is favoured in non-polar media and by heat or light. Heterolytic fission splits a bond so one atom takes both electrons, giving ions — a carbocation (positive carbon, electron-deficient) or a carbanion (negative carbon, electron-rich).

Nucleophiles and electrophiles

A nucleophile ('nucleus-loving') is electron-rich and donates a lone/bonding pair — $OH^-$, $CN^-$, $NH_3$, $H_2O$. An electrophile ('electron-loving') is electron-deficient and accepts a pair — $H^+$, $NO_2^+$, $BF_3$, $Cl^+$. Curved arrows in a mechanism always show movement from the electron-rich site to the electron-poor site.

Electronic displacement effects

Inductive effect (I): permanent polarisation of a $sigma$-bond by an electronegative atom, transmitted through the chain and falling off with distance. Electron-withdrawing groups ($−NO_2$, $−Cl$) show $−I$; alkyl groups show $+I$ (electron-releasing).
Resonance / mesomeric effect (M): delocalisation of $pi$/lone-pair electrons over a conjugated system, shown by resonance structures and a resonance hybrid. It stabilises ions (e.g. the carboxylate $RCOO^-$) and is stronger than the inductive effect.
Electromeric effect (E): a temporary, complete shift of a $pi$-pair to one atom in the presence of an attacking reagent; it disappears when the reagent is removed.
Hyperconjugation: delocalisation of $sigma(C−H)$ electrons of an alkyl group into an adjacent empty $p$ or $pi$ orbital ('no-bond resonance'). More $\alpha$-hydrogens means more hyperconjugation and greater stability.

Stability of carbocations

Carbocations are stabilised by $+I$ and by hyperconjugation, so the order is $3^\circ > 2^\circ > 1^\circ > CH_3^+$. The same factors make the order of free-radical stability run $3^\circ > 2^\circ > 1^\circ$.

Types of organic reactions

Substitution — one atom/group replaces another (e.g. $CH_4+Cl_2 \rightarrow CH_3Cl+HCl$). Addition — reagent adds across a multiple bond (e.g. $CH_2=CH_2+H_2 \rightarrow CH_3CH_3$). Elimination — a small molecule is removed to form a multiple bond (dehydration of an alcohol). Rearrangement — atoms/groups migrate within a molecule (e.g. $1^\circ \rightarrow 2^\circ$ carbocation shift).

Solved Examples

Worked Example

Classify the fission and product in (i) $Cl_2 \xrightarrow{h\nu} 2Cl^{\bullet}$ and (ii) $CH_3Br \rightarrow CH_3^+ + Br^-$.

Solution

(i) Each Cl keeps one electron → equal split → homolytic fission giving free radicals.
(ii) Br takes both bonding electrons → unequal split → heterolytic fission.
This gives a methyl carbocation $CH_3^+$ and a bromide ion $Br^-$.

Answer: (i) homolytic (free radicals); (ii) heterolytic (carbocation + anion).

Worked Example

Classify each species as nucleophile or electrophile: $OH^-$, $NO_2^+$, $NH_3$, $BF_3$.

Solution

Nucleophiles are electron-rich, donate a pair: $OH^-$ and $NH_3$ (lone pair).
Electrophiles are electron-deficient, accept a pair: $NO_2^+$ (positive) and $BF_3$ (incomplete octet on B).

Answer: Nucleophiles: $OH^-$, $NH_3$; Electrophiles: $NO_2^+$, $BF_3$.

Worked Example

Arrange the carbocations $CH_3^+$, $CH_3CH_2^+$, $(CH_3)_2CH^+$, $(CH_3)_3C^+$ in increasing order of stability.

Solution

Stability rises with $+I$ of alkyl groups and number of $\alpha$-hydrogens (hyperconjugation).
Methyl has none, $1^\circ$ ethyl a little, $2^\circ$ isopropyl more, $3^\circ$ tert-butyl most.
So the order increases from methyl to tertiary.

Answer: $CH_3^+ < CH_3CH_2^+ < (CH_3)_2CH^+ < (CH_3)_3C^+$.

Worked Example

Explain, using resonance, why the carboxylate ion $CH_3COO^-$ is more stable than the ethoxide ion $CH_3CH_2O^-$.

Solution

In $CH_3COO^-$ the negative charge is delocalised over two equivalent oxygen atoms by resonance.
This spreads the charge, lowering energy and giving a stable resonance hybrid.
In ethoxide the charge is localised on one oxygen with no such delocalisation.

Answer: Resonance delocalisation over two oxygens stabilises carboxylate; ethoxide has a localised charge, so it is less stable.

Worked Example

Identify the reaction type: (i) $CH_4 + Cl_2 \xrightarrow{h\nu} CH_3Cl + HCl$; (ii) $CH_2=CH_2 + Br_2 \rightarrow CH_2BrCH_2Br$.

Solution

(i) A H atom of methane is replaced by Cl — one group swaps for another.
That is a substitution reaction (free-radical).
(ii) $Br_2$ adds across the C=C double bond with no atom lost → addition reaction.

Answer: (i) substitution; (ii) addition.

Worked Example

Compare the inductive ($+I$) and hyperconjugative contributions of $−CH_3$ versus $−CCl_3$ on an attached carbocation.

Solution

$−CH_3$ is electron-releasing ($+I$) and has three $\alpha$-C−H bonds for hyperconjugation → it stabilises a carbocation.
$−CCl_3$ has highly electronegative chlorines, so it is electron-withdrawing ($−I$) and has no $\alpha$-C−H for hyperconjugation.
Hence $−CCl_3$ destabilises an adjacent carbocation.

Answer: $−CH_3$ stabilises (+I + hyperconjugation); $−CCl_3$ destabilises (−I, no hyperconjugation).

Key Points

Homolytic fission gives free radicals (equal split); heterolytic fission gives ions — carbocations or carbanions.
Nucleophiles are electron-rich donors ($OH^-$, $CN^-$, $NH_3$); electrophiles are electron-deficient acceptors ($H^+$, $NO_2^+$, $BF_3$).
Inductive (+I/−I) is a permanent $sigma$ effect through bonds; resonance (mesomeric) delocalises $pi$/lone pairs; electromeric is a temporary $pi$ shift; hyperconjugation delocalises $sigma(C−H)$ electrons.
Carbocation and free-radical stability follow $3^\circ > 2^\circ > 1^\circ >$ methyl, due to +I and hyperconjugation.
Organic reactions are substitution, addition, elimination or rearrangement.

Quick Check — 4 MCQs

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Q1.Homolytic fission of a covalent bond produces:

Explanation: Homolytic fission splits the bond equally — each atom keeps one electron, giving two neutral free radicals.

Q2.Which of the following acts as an electrophile?

Explanation: $NO_2^+$ is electron-deficient and accepts an electron pair, so it is an electrophile; the others are nucleophiles.

Q3.The most stable carbocation among the following is:

Explanation: The tertiary carbocation $(CH_3)_3C^+$ is most stabilised by +I and the greatest hyperconjugation.

Q4.Delocalisation of $\sigma(C−H)$ electrons into an adjacent empty orbital is called:

Explanation: Hyperconjugation (no-bond resonance) is the delocalisation of $\sigma$ C−H electrons into an adjacent $p$ or $\pi$ orbital.

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