Introduction to General Organic Chemistry

Q: What is the hybridisation and geometry of a methyl carbocation?

The methyl carbocation CH3+ is sp2-hybridised with a planar trigonal geometry and H-C-H bond angles of about 120 degrees. There is an empty unhybridised p-orbital perpendicular to the plane of the three C-H bonds, which is why the cation is a strong electrophile.

Q: How are free radicals detected experimentally?

Free radicals have an unpaired electron, so they are paramagnetic. The technique used is electron spin resonance (ESR), also called electron paramagnetic resonance (EPR). Radicals can also be trapped chemically using spin traps, or their presence inferred indirectly from product distributions and the effect of radical initiators or inhibitors on the reaction.

Q: Why can carbocations not be formed at bridgeheads?

A carbocation must be planar (sp2) because it has an empty p-orbital perpendicular to a trigonal plane. A bridgehead carbon in a bicyclic system like norbornyl or adamantyl is locked in a rigid, roughly tetrahedral geometry and physically cannot become planar. So bridgehead carbocations are extremely difficult to form. Bridgehead radicals and carbanions, which do not need to be planar, form much more easily.

Q: What are reactive intermediates and how are they different from synthetic intermediates?

Reactive intermediates like carbocations, carbanions, free radicals, carbenes, nitrenes, and benzyne are short-lived species (10^-6 s to a few seconds) that appear during a reaction but cannot be isolated. Synthetic intermediates are stable compounds that are actually prepared and purified in the lab, and then used as starting materials for the next step of a multi-step synthesis.

Q: Why is a Grignard reagent considered a carbanion source?

In a Grignard reagent R-Mg-X, magnesium is far less electronegative than carbon, so the C-Mg bond is highly polar with a significant delta minus on carbon. The alkyl group therefore behaves like a carbanion: it attacks electrophilic carbons in aldehydes, ketones, esters, and CO2. This is why Grignards form new carbon-carbon bonds so readily.

Q: What is the difference between an electrophile and a nucleophile?

An electrophile is an electron-poor species that accepts an electron pair (it is a Lewis acid, either positive or neutral electron-deficient, like H+, NO2+, AlCl3, or a carbocation). A nucleophile is an electron-rich species that donates an electron pair (it is a Lewis base, either negative or neutral with a lone pair, like OH-, CN-, NH3, or a carbanion). In every ionic reaction, a nucleophile attacks an electrophile.

ChemistrySome Basic Principles of Organic ChemistryFor JEE aspirants

General Organic Chemistry (GOC) is the study of how covalent bonds break and which short-lived species drive organic reactions. Every organic reaction starts with a bond cleavage, either homolytic (each atom keeps one electron, forming free radicals) or heterolytic (one atom takes both electrons, forming a carbocation and a carbanion). The three key reactive intermediates in JEE and NEET organic chemistry are carbocations, carbanions, and free radicals. Their structure, hybridization, and stability trends decide the mechanism and product of nearly every organic transformation.

Key Ideas - Quick Reference

Homolytic cleavage: $A : B \to A^{∙} + B^{∙}$ (each atom gets one electron, free radicals formed, favoured in non-polar solvents and UV light).
Heterolytic cleavage: $A : B \to A^{+} + B^{-}$ or $A^{-} + B^{+}$ (both electrons go to one atom, ions formed, favoured in polar solvents).
Carbocation $R_{3} C^{+}$ : sp $^{2}$ , planar, 120°, empty p-orbital. Stability: $3^{\circ} > 2^{\circ} > 1^{\circ} > C H_{3}^{+}$ ; benzyl and allyl are extra-stable by resonance.
Carbanion $R_{3} C^{-}$ : sp $^{3}$ , pyramidal, lone pair at one apex. Stability: $C H_{3}^{-} > 1^{\circ} > 2^{\circ} > 3^{\circ}$ (reverse of carbocation).
Free radical $R_{3} C^{∙}$ : sp $^{2}$ , planar, unpaired electron in p-orbital. Stability parallels carbocation: benzyl $>$ allyl $> 3^{\circ} > 2^{\circ} > 1^{\circ} > C H_{3}^{∙} >$ vinyl.
Electron-releasing groups (alkyl, $- O C H_{3}$ , $- N H_{2}$ ) stabilise carbocations; electron-withdrawing groups ( $- N O_{2}$ , $- C N$ , $- C = O$ ) stabilise carbanions.

1. Bond Cleavage - the Starting Point of Every Organic Reaction

Organic reactions proceed through the formation of reactive intermediates, and these intermediates are generated when a covalent bond in the reactant breaks. Depending on how the shared pair of electrons is divided between the two atoms, bond cleavage is of two types: homolytic and heterolytic.

The three intermediates you must know are:

Free radicals like $C ∙ H_{3}$ (methyl radical)
Carbocations like $C \oplus H_{3}$ (methyl cation)
Carbanions like $C ⊖ H_{3}$ (methyl carbanion)

1.1 Homolytic (Symmetrical) Cleavage

In homolytic cleavage, the two electrons shared in the covalent bond are split equally: one electron goes to each of the two bonded atoms. The species formed are called free radicals, and this movement is denoted by half-headed (fish-hook) arrows.

A - B heat / UV A^{∙} + B^{∙}

Example: $C H_{3} - C l h ν C H_{3}^{∙} + C l^{∙}$

Key features of free radicals produced this way:

They are electrically neutral (no positive or negative charge).
They are extremely reactive because of the unpaired electron.
Homolysis is favoured in non-polar solvents, gas phase, and by UV light or heat.

Geometry change: The methyl radical $C ∙ H_{3}$ is sp $^{2}$ -hybridised with three $σ$ $C - H$ bonds and a singly-occupied p-orbital, HCH angle 120°, and all three $C - H$ bonds coplanar. So when a methyl radical is formed by homolytic cleavage of $C H_{3} - X$ , the carbon changes geometry from tetrahedral to planar and rehybridises from sp $^{3}$ to sp $^{2}$ .

Figure 1: Homolytic cleavage of

A - B

. Each atom retains one electron and forms an electrically neutral free radical.

1.2 Heterolytic (Unsymmetrical) Cleavage

When a covalent bond between two atoms A and B breaks so that both electrons of the shared pair are taken away by one atom, the mode is called heterolytic cleavage. This is indicated by a full-headed curved arrow that denotes a two-electron shift.

A : B \to A^{+} + B^{: -} or A : B \to A^{: -} + B^{+}

The more electronegative atom keeps the electron pair and becomes the anion; the other becomes the cation.

Heterolysis produces charged species (a cation and an anion).
It usually occurs in polar covalent bonds and is favoured by polar solvents (water, ethanol) which stabilise the ions.
When a carbocation is formed, the carbon changes from sp $^{3}$ to sp $^{2}$ , just like in homolysis but with a positive charge and an empty p-orbital instead of an unpaired electron.

Figure 2: Heterolytic cleavage of a polar

A - B

bond. Both bonding electrons go to the more electronegative atom, producing charged ions.

1.3 Homolytic vs Heterolytic - Side-by-Side

Feature	Homolytic Cleavage	Heterolytic Cleavage
Electron distribution	One electron to each atom	Both electrons to one atom
Species formed	Free radicals (neutral)	Cation and anion (charged)
Arrow notation	Half-headed (fish-hook)	Full-headed curved arrow
Favoured by	Non-polar solvents, UV, heat, peroxides	Polar solvents, polar bonds
Typical bond	Non-polar (e.g. $C l - C l$ , $C - H$ )	Polar (e.g. $C - B r$ , $C - O$ )

1.4 Curly Arrow Notation - Reading Mechanisms

Every organic mechanism uses curly arrows to show where electrons move. Learning to read them is the single most important skill in organic chemistry.

Figure 3: Curly arrow notation. A fish-hook (half-headed) arrow tracks a single electron; a full-headed arrow tracks a pair. The arrow always points from where the electrons start to where they end up.

Golden rule: An arrow shows electron movement, not atom movement. It starts from a bond or a lone pair (the electron source) and ends at an atom or an empty orbital (the electron sink).

2. Reaction Intermediates - an Overview

Most organic reactions occur through the involvement of certain chemical species which are generally short-lived (10 $^{- 6}$ seconds to a few seconds) and highly reactive, and hence cannot be isolated. These are called reactive intermediates.

Because they are so unstable, reactive intermediates are detected by spectroscopic methods (ESR, NMR), trapped chemically, or their presence is confirmed by indirect evidence such as product distribution and kinetics.

Do not confuse reactive intermediates with synthetic intermediates. Synthetic intermediates are stable compounds that are actually prepared, isolated, and purified for use as starting materials in a synthesis. Reactive intermediates cannot be stored in a bottle.

3. Carbocations (earlier called Carbonium Ions)

Carbocations are the key intermediates in a huge number of organic reactions, particularly in nucleophilic substitution ( $S_{N} 1$ ) and electrophilic addition reactions of alkenes.

3.1 Structure of Carbocations

In a carbocation, the positively charged carbon atom is bonded to three other atoms and has no non-bonding electrons. It is sp $^{2}$ -hybridised with a planar structure and bond angles of about 120°. There is a vacant unhybridised p-orbital lying perpendicular to the plane of the three $C - H$ (or $C - R$ ) bonds.

Figure 4: Structure of methyl carbocation

C + H_{3}

. The central carbon is sp

^{2}

hybridised with all three C-H bonds at 120° in one plane (shown by wedge-and-dash notation). The empty p-orbital lies perpendicular to this sp

^{2}

plane.

3.2 Stability of Carbocations

There is a marked increase in carbocation stability with additional alkyl substitution. This is why the rate of $H X$ addition to alkenes decreases in the order:

(C H_{3})_{2} C = C H_{2} > C H_{3} - C H = C H_{2} > C H_{2} = C H_{2}

The reason is the stability of the carbocation formed in the rate-determining step. Alkyl groups are electron-releasing (+I effect). Three alkyl groups release more electron density than two, and two more than one, so the positive charge is dispersed more effectively. The final order is:

Stability of carbocations:

3^{\circ} > 2^{\circ} > 1^{\circ} > C + H_{3}

Additional stabilisation from resonance and heteroatoms:

Allyl and benzyl cations are extra-stable because the positive charge is delocalised over two atoms (allyl) or the ring (benzyl) by resonance.
Substituent effects on benzyl cation: electron-donating groups like $p$ -methoxy ( $- O C H_{3}$ ) and $p$ -amino ( $- N H_{2}$ ) stabilise it by 14 and 26 kcal/mol respectively, while electron-withdrawing groups like $p$ -nitro ( $- N O_{2}$ ) destabilise it by about 20 kcal/mol.
Heteroatoms with lone pairs next to the cationic centre (like $C H_{3} - O - C + H_{2}$ ) donate their electrons into the empty p-orbital, giving very stable "oxocarbenium" cations that can even be isolated as solids.
Cyclopropylmethyl cations are even more stable than benzyl cations because the bent orbitals of the cyclopropyl ring conjugate with the empty p-orbital.
Bridgehead carbocations are very hard to form because the carbon cannot become planar. This is direct evidence that carbocations must be planar.
Vinyl cations $(C H_{2} = C + H)$ are unusually unstable because they lack effective resonance and hyperconjugation.

Rule of thumb: Stability

\propto

number of hyperconjugated structures

\propto

number of

α

-hydrogens on the cationic carbon. This explains the alkyl-substitution stability order.

Figure 5: Resonance stabilisation of the allyl cation. The positive charge is shared between the two terminal carbons, giving the ion much greater stability than a plain primary cation.

Figure 6: Resonance stabilisation of the benzyl cation. The positive charge delocalises from the benzylic carbon into the ortho and para positions of the aromatic ring, producing exceptional stability.

Figure 7: Hyperconjugation in

(C H_{3})_{3} C^{+}

. A

σ

C - H

bond on an

α

-carbon aligned with the empty p-orbital donates electron density into it. With three methyl groups (nine

α

-hydrogens), tert-butyl cation is far more stabilised than any 2° or 1° cation.

Figure 8: Stability order of alkyl carbocations. Greater alkyl substitution disperses the positive charge and increases stability.

4. Carbanions

A carbanion is a chemical species that bears a negative charge on carbon and possesses eight electrons in its valence shell. Carbanions are produced by heterolytic cleavage of a covalent bond in which the shared pair of electrons is retained by the carbon atom.

4.1 Structure of Carbanions

A carbanion carries an unshared pair of electrons and thus behaves as a base and a nucleophile. The best description is that the central carbon is sp $^{3}$ -hybridised with the lone pair occupying one apex of the tetrahedron. So carbanions have pyramidal structures similar to those of amines like $N H_{3}$ .

They rapidly interconvert between two pyramidal forms through an umbrella-flip (just like the nitrogen inversion in amines). This is why chiral carbanions almost always give a racemic mixture instead of retention of configuration.

Figure 9: Structure of methyl carbanion

C - H_{3}

. The carbon is sp

^{3}

hybridised with a lone pair at one apex of the tetrahedron, producing a pyramidal geometry similar to ammonia.

4.2 Stability of Carbanions

Because a carbanion has a lone pair and a negative charge, anything that removes electron density stabilises it, and anything that adds electron density destabilises it. This is the exact opposite of carbocations.

Electron-attracting groups like $- C N$ , $- N O_{2}$ , and $> C = O$ increase stability by pulling the negative charge away.
Electron-releasing groups like $- C H_{3}$ decrease stability by pushing more electron density onto the already negative carbon.
Benzyl carbanion is stabilised by resonance because the lone pair delocalises into the ring.

Stability of carbanions:

C - H_{3} > 1^{\circ} > 2^{\circ} > 3^{\circ}

4.3 Generation and Properties

The best-known source of carbanion character is the Grignard reagent ( $R - M g - X$ ) and other organometallic compounds where carbon is bonded to a less electronegative metal such as lithium, sodium, potassium, or zinc. The $C - M$ bond is highly polar with carbon carrying a partial negative charge, so it acts as a source from which the organic group can be transferred with its electrons.

Properties: Carbanions are nucleophilic and basic. Since the negative charge sits on carbon (which is less electronegative than nitrogen), a carbanion is a stronger base and better nucleophile than the corresponding amine. In fact, alkyl carbanions are basic enough to remove a proton from ammonia.

5. Free Radicals

A free radical is a species that has one or more unpaired electrons. In ordinary molecules all electrons are paired and the total magnetic moment is zero, but radicals have a net magnetic moment and are therefore paramagnetic. Free radicals are usually detected by electron spin resonance (ESR), also called electron paramagnetic resonance (EPR).

5.1 Structure of Free Radicals

Simple alkyl radicals have a planar (trigonal) structure, i.e. they are sp $^{2}$ -hybridised with the odd electron sitting in the unhybridised p-orbital. A pyramidal sp $^{3}$ structure is also possible for some radicals, so free radicals are not always strictly planar.

Figure 10: Structure of methyl free radical

C ∙ H_{3}

. The central sp

^{2}

carbon is planar with all three C-H bonds at 120° (shown by wedge-and-dash notation). The single unpaired electron sits in the p-orbital perpendicular to that plane.

Evidence for planarity: When a free radical is generated at a chiral centre, the resulting product is racemic (optical activity is lost). A planar radical is attacked at either face with equal probability, giving both enantiomers in equal amounts. On the other hand, radicals can be generated at bridgeheads (unlike carbocations), which shows a pyramidal geometry is also possible.

5.2 Stability of Free Radicals

Just like carbocations, the stability of alkyl free radicals follows the order tertiary $>$ secondary $>$ primary $>$ methyl, and this is again explained on the basis of hyperconjugation. The stabilising effect in allylic and benzylic radicals arises from resonance with the adjacent $π$ -system.

Ease of formation / stability of alkyl free radicals:
benzyl

>

allyl

> 3^{\circ} > 2^{\circ} > 1^{\circ} > C ∙ H_{3} >

vinyl

Bond-dissociation energies confirm this: about 19 kcal/mol less energy is needed to form a benzyl radical from toluene than to form a methyl radical from methane. Triphenylmethyl-type radicals are exceptionally stable, largely because steric hindrance prevents dimerisation, not just resonance.

5.3 Free Radicals in Action - Chlorination of Methane

The classic example of a free-radical reaction is the chlorination of methane, $C H_{4} + C l_{2} h ν C H_{3} C l + H C l$ . It proceeds in three stages: initiation (radicals are born), propagation (radicals do the work in a chain), and termination (radicals combine and disappear).

Step 1: Initiation. A $C l_{2}$ molecule absorbs a UV photon and the $C l - C l$ bond breaks homolytically to give two chlorine radicals.

Figure 11: Initiation step. UV light (

h ν

) homolytically cleaves the

C l - C l

bond to produce two chlorine free radicals

C l^{∙}

Step 2: Propagation. Each new radical reacts to make yet another radical, so the chain keeps going. A chlorine radical abstracts a hydrogen from methane, producing $H C l$ and a methyl radical. The methyl radical then attacks $C l_{2}$ , producing $C H_{3} C l$ and a new chlorine radical, which starts the cycle again.

Figure 12: Propagation. A chlorine radical abstracts a hydrogen from

C H_{4}

to give

H C l

and a methyl radical; the methyl radical then reacts with

C l_{2}

to give

C H_{3} C l

and regenerates a chlorine radical, sustaining the chain.

Termination steps (

C l^{∙} + C l^{∙} \to C l_{2}

C H_{3}^{∙} + C l^{∙} \to C H_{3} C l

C H_{3}^{∙} + C H_{3}^{∙} \to C H_{3} C H_{3}

) remove radicals from the pot and finally stop the chain. A single initiation event can trigger thousands of propagation cycles, which is why radical reactions are so fast.

6. Electrophiles and Nucleophiles

Every ionic organic reaction is essentially a handshake between an electron-poor species (electrophile) and an electron-rich species (nucleophile). Recognising which is which lets you predict the direction of arrow-pushing and the product of the reaction.

6.1 Electrophiles ("electron loving")

An electrophile is any species that accepts a pair of electrons during a chemical reaction. Electrophiles are either positively charged or electron-deficient at some atom. They are the same as Lewis acids.

Positive electrophiles: $H^{+}$ , $N O_{2}^{+}$ , $R - C + H_{2}$ , $B r^{+}$ , $R_{3} C +$ (all carbocations).
Neutral electrophiles: $B F_{3}$ , $A l C l_{3}$ , $F e B r_{3}$ , $S O_{3}$ , $> C = O$ (carbonyl carbon), $R - X$ (the alkyl carbon).

6.2 Nucleophiles ("nucleus loving")

A nucleophile is any species that donates a pair of electrons during a chemical reaction. Nucleophiles are either negatively charged or carry a lone pair that they can share. They are the same as Lewis bases.

Negative nucleophiles: $O H^{-}$ , $C N^{-}$ , $C l^{-}$ , $R O^{-}$ (alkoxide), $R - C - H_{2}$ (all carbanions).
Neutral nucleophiles (with lone pair): $H_{2} O$ , $N H_{3}$ , $R O H$ , $R_{2} N H$ , $R_{2} S$ .

Figure 13: Nucleophile-electrophile reaction. Hydroxide (

O H^{-}

) uses its lone pair to attack the

δ^{+}

carbon of

C H_{3} B r

; simultaneously, the

C - B r

bond electrons leave as

B r^{-}

. This two-arrow, single-step mechanism is

S_{N} 2

substitution.

6.3 Quick Comparison

Feature	Electrophile	Nucleophile
Meaning	Electron loving	Nucleus loving
Electrons	Accepts a pair	Donates a pair
Charge	Positive or neutral electron-deficient	Negative or neutral with lone pair
Lewis role	Lewis acid	Lewis base
Example	$H^{+}$ , $N O_{2}^{+}$ , $A l C l_{3}$ , carbocation	$O H^{-}$ , $C N^{-}$ , $N H_{3}$ , carbanion
In reactions	Attacked by nucleophile	Attacks electrophile

7. Types of Organic Reactions

Every organic reaction, no matter how complex, belongs to one of these four broad categories. Recognising the type helps you write the right mechanism and predict the product.

Figure 14: The four types of organic reactions. Substitution and elimination compete on the same substrate; addition is characteristic of

π

-bonds; rearrangement converts a less-stable intermediate to a more-stable one.

7.1 Substitution

One atom or group in a molecule is replaced by another. Classic example: $C H_{3} B r + O H^{-} \to C H_{3} O H + B r^{-}$ . Sub-types: nucleophilic ( $S_{N} 1$ , $S_{N} 2$ ), electrophilic (aromatic $S_{E} A r$ ), and free-radical substitution.

7.2 Addition

Two atoms or groups add across a multiple bond ( $π$ bond broken, two $σ$ bonds made). Classic example: $C H_{2} = C H_{2} + H B r \to C H_{3} C H_{2} B r$ . Sub-types: electrophilic (alkenes, alkynes), nucleophilic (carbonyls), and free-radical addition.

7.3 Elimination

Two atoms or groups leave adjacent carbons and a $π$ bond forms between them. Classic example: $C H_{3} C H B r C H_{3} a l c . K O H C H_{3} C H = C H_{2} + H B r$ . Sub-types: E1, E2. Elimination is the mirror image of addition.

7.4 Rearrangement

Atoms in a molecule reorganise to give a more stable isomer. Commonly seen when a carbocation intermediate can shift a hydride ( $H^{-}$ ) or alkyl group to itself from an adjacent carbon to convert a less-stable cation (e.g. 1° or 2°) into a more stable one (2° or 3°). This is why $S_{N} 1$ reactions sometimes give unexpected products.

8. All Three Intermediates - At a Glance

Feature	Carbocation $R_{3} C^{+}$	Carbanion $R_{3} C^{-}$	Free radical $R_{3} C^{∙}$
Electrons on C	6 (electron-deficient)	8 (with lone pair)	7 (unpaired)
Charge	Positive	Negative	Neutral
Hybridisation	sp $^{2}$	sp $^{3}$ (mostly)	sp $^{2}$ (mostly)
Geometry	Planar, 120°	Pyramidal, ~107°	Planar (trigonal)
Extra feature	Empty p-orbital	Lone pair at apex	Odd electron in p-orbital
Behaviour	Electrophile / Lewis acid	Nucleophile / Lewis base	Neutral, reactive
Formed by	Heterolysis (loses pair)	Heterolysis (keeps pair)	Homolysis
Stability order	$3^{\circ} > 2^{\circ} > 1^{\circ}$	$1^{\circ} > 2^{\circ} > 3^{\circ}$	$3^{\circ} > 2^{\circ} > 1^{\circ}$
Detected by	NMR, kinetics	NMR, trapping	ESR (paramagnetic)

Solved Examples

Solved Example 1

The greater the s-character in an orbital, the ______ is its energy.
(A) Greater (B) Lower (C) Both (D) None

Solution:

The 2s orbital is closer to the nucleus and lower in energy than the 2p orbital. An sp orbital (50% s) is closer to the nucleus than sp $^{2}$ (33% s) or sp $^{3}$ (25% s). So more s-character means the electrons are held closer to the nucleus and the orbital is lower in energy.

Answer: (B) Lower.

Solved Example 2

Which of the following carbocations is most stable?
(A)

C H_{3} - C + H - C H_{3}

(isopropyl)
(B)

(C H_{3})_{3} C +

(tert-butyl)
(C)

P h - C + H_{2}

(benzyl)
(D)

C H_{2} = C H - C + H_{2}

(allyl)

Solution:

Compare the stabilisation available to each cation:

(A) 2° carbocation, only two alkyl groups (+I and hyperconjugation).
(B) 3° carbocation, three alkyl groups (strong hyperconjugation, nine $α$ -hydrogens).
(C) Benzyl cation, resonance-stabilised over the ring (four canonical forms).
(D) Allyl cation, resonance-stabilised over two atoms.

The benzyl cation is stabilised by resonance over the aromatic ring plus a small $α$ -hyperconjugation contribution. This is generally the most stable among the four.

Answer: (C) Benzyl cation.

Solved Example 3

Which

C - C l

bond is weaker?
(a)

P h - C H_{2} - C l

(b)

P h - C H_{2} - C H_{2} - C l

Solution:

Bond-dissociation energy of $C - C l$ depends on the stability of the resulting carbocation (or radical) after cleavage.

(a) $P h - C H_{2} - C l$ breaks to give the benzyl cation $P h - C + H_{2}$ , which is resonance-stabilised over the ring.
(b) $P h - C H_{2} - C H_{2} - C l$ breaks to give a primary cation $P h - C H_{2} - C + H_{2}$ that is not directly conjugated with the ring.

Since the benzyl cation is far more stable, the $C - C l$ bond in (a) is weaker and easier to break.

Answer: (a) $P h - C H_{2} - C l$ has the weaker $C - C l$ bond.

Solved Example 4

Arrange the following in the increasing order of

C - B r

bond energy:
(a)

C H_{3} - C H_{2} - C H_{2} - B r

(1° propyl)
(b)

(C H_{3})_{3} C - B r

(3° tert-butyl bromide)
(c)

C H_{3} (C H_{2})_{4} - B r

(1° pentyl)
(d)

C H_{2} = C H - B r

(vinyl bromide)
(e)

P h - C H_{2} - B r

(benzyl bromide)

Solution:

Weaker $C - B r$ bond $\Leftrightarrow$ more stable resulting cation.

(d) Vinyl cation is the least stable (sp carbon, no hyperconjugation), so $C - B r$ is strongest.
(b) Tert-butyl cation is 3° and highly stable, so $C - B r$ is weak.
(e) Benzyl cation is resonance-stabilised, so $C - B r$ is weakest.
(a) and (c) are both primary alkyl; roughly equal, but (a) is slightly less stabilised.

Increasing bond energy: (e) $<$ (b) $<$ (a) $\approx$ (c) $<$ (d)

Solved Example 5

Which

C - I

bond is weaker and why?
(i)

C H_{3} - O - C H_{2} - I

(ii)

C H_{3} - C H_{2} - I

Solution:

When (i) breaks heterolytically, it gives the cation $C H_{3} - O - C + H_{2}$ . The oxygen lone pair donates into the empty p-orbital of carbon and creates an extra $π$ -bond, forming a very stable oxocarbenium ion.

Cation (ii) is just a primary ethyl cation with no such stabilisation.

So the $C - I$ bond in (i) is weaker because the resulting cation is far more stable.

Solved Example 6

Identify the more stable carbanion in each pair:
(a)

C H_{3} - C H_{2} - C - H_{2}

and

C H_{3} - C - H - C H_{3}

(b)

C - H_{3}

and

C H_{3} - C - H - C H_{3}

(c)

P h - C - H_{2}

and

P h - C H_{2} - C - H_{2}

Solution:

(a) The propyl carbanion $C H_{3} - C H_{2} - C - H_{2}$ is 1°; the isopropyl carbanion is 2°. Fewer alkyl groups pushing electron density means less destabilisation, so $C H_{3} - C H_{2} - C - H_{2}$ (1°) is more stable (inductive effect argument).
(b) Methyl carbanion $C - H_{3}$ has no +I alkyl groups; the isopropyl carbanion has two. So $C - H_{3}$ is more stable.
(c) Benzyl carbanion $P h - C - H_{2}$ has its lone pair directly conjugated with the ring (resonance-stabilised). The homobenzyl carbanion $P h - C H_{2} - C - H_{2}$ is not conjugated. So $P h - C - H_{2}$ is more stable.

Solved Example 7

Alkenes undergo electrophilic addition and benzene undergoes electrophilic substitution, even though both proceed through a carbocation intermediate. Explain.

Solution:

In alkenes, the $π$ -electrons are localised in the double bond. When an electrophile attacks, the resulting carbocation is a plain (usually secondary or tertiary) cation. The nucleophile then adds to this cation, giving an addition product and destroying the double bond permanently.

In benzene, the $π$ -electrons are delocalised over the ring (aromatic system, 6 $π$ -electrons). Electrophile attack gives a resonance-stabilised arenium ion (sigma complex). Restoring aromaticity is a huge driving force, worth about 36 kcal/mol of resonance energy. So instead of the nucleophile adding, a proton is lost and the ring becomes aromatic again, giving a substitution product.

Solved Example 8

Which of the following statements is correct?
(A) Allyl carbonium ion (

C H_{2} = C H - C + H_{2}

) is more stable than propyl carbonium ion.
(B) Propyl carbonium ion is more stable than allyl carbonium ion.
(C) Both are equally stable.
(D) None of these.

Solution:

The allyl cation is stabilised by resonance: the positive charge is delocalised over two carbon atoms via the adjacent double bond.

$C H_{2} = C H - C + H_{2} ⟷ C + H_{2} - C H = C H_{2}$

The propyl cation ( $C H_{3} - C H_{2} - C + H_{2}$ ) is a plain primary cation with only weak inductive stabilisation and no resonance. Hence the allyl cation is far more stable.

Answer: (A).

Common Mistakes to Avoid

Watch out

Reversing the carbanion stability order. Carbanions follow $C H_{3}^{-} > 1^{\circ} > 2^{\circ} > 3^{\circ}$ , the opposite of carbocations. Alkyl groups are electron-releasing, which destabilises a negatively charged carbon.
Confusing homolytic and heterolytic arrows. Fish-hook (half-headed) arrows show one electron moving (homolytic). Full-headed curved arrows show two electrons moving together (heterolytic). Mixing them up is a common exam mark loss.
Assuming vinyl cations are stable. Vinyl cations ( $C H_{2} = C + H$ ) are actually less stable than $C + H_{3}$ . The sp-hybridised cation carbon is not helped by resonance or hyperconjugation.
Forgetting that carbocations must be planar. A bridgehead carbocation (like at the bridgehead of norbornane) cannot become planar and is nearly impossible to form. This is often tested.
Treating radicals and carbocations as identical. They have the same alkyl-substitution stability order and both are sp $^{2}$ , but radicals are neutral and paramagnetic, while carbocations are positive and diamagnetic.
Missing the sp $^{3}$ nature of carbanions. Carbocations and radicals are sp $^{2}$ , but carbanions are sp $^{3}$ (pyramidal, with a lone pair). Do not assume all three intermediates share the same geometry.

Frequently Asked Questions

Q1. What is the difference between homolytic and heterolytic bond cleavage?

In homolytic cleavage, the two shared electrons split equally, giving each atom one electron and forming two neutral free radicals. In heterolytic cleavage, both electrons go to one atom, giving a cation and an anion. Homolysis is favoured by non-polar solvents, UV light, and heat; heterolysis needs polar bonds and polar solvents.

Q2. Why is the stability order of carbanions opposite to that of carbocations?

A carbocation is electron-deficient, so electron-releasing alkyl groups (+I effect) stabilise it, making 3° $>$ 2° $>$ 1° $>$ methyl. A carbanion is already electron-rich, so alkyl groups pushing more electron density onto it destabilise it. Fewer alkyl groups (methyl carbanion) means less destabilisation, giving methyl $>$ 1° $>$ 2° $>$ 3°.

Q3. What is the hybridisation and geometry of a methyl carbocation?

The methyl carbocation $C + H_{3}$ is sp $^{2}$ -hybridised with a planar (trigonal) geometry and $H - C - H$ bond angles of about 120°. There is an empty unhybridised p-orbital perpendicular to the plane of the three $C - H$ bonds, which is why the cation is a strong electrophile.

Q4. Why is the benzyl carbocation exceptionally stable?

The empty p-orbital on the benzylic carbon is in conjugation with the aromatic ring's $π$ -system. This lets the positive charge delocalise over the ortho and para positions of the ring, giving four resonance structures. This spread of charge dramatically lowers the energy of the cation. Electron-donating groups on the ring (like $- O C H_{3}$ , $- N H_{2}$ ) stabilise it further.

Q5. How are free radicals detected experimentally?

Free radicals have an unpaired electron, so they are paramagnetic. The technique used is electron spin resonance (ESR), also called electron paramagnetic resonance (EPR). Radicals can also be trapped chemically using spin traps, or their presence inferred indirectly from product distributions and the effect of radical initiators or inhibitors on the reaction.

Q6. Why can carbocations not be formed at bridgeheads?

A carbocation must be planar (sp $^{2}$ ) because it has an empty p-orbital perpendicular to a trigonal plane. A bridgehead carbon in a bicyclic system (like norbornyl or adamantyl) is locked in a rigid, roughly tetrahedral geometry and physically cannot become planar. So bridgehead carbocations are extremely difficult to form. Bridgehead radicals and carbanions, which do not need to be planar, form much more easily.

Q7. What are reactive intermediates and how are they different from synthetic intermediates?

Reactive intermediates (carbocations, carbanions, free radicals, carbenes, nitrenes, benzyne) are short-lived species (10 $^{- 6}$ s to a few seconds) that appear during a reaction but cannot be isolated. Synthetic intermediates are stable compounds that are actually prepared and purified in the lab, and then used as starting materials for the next step of a multi-step synthesis.

Q8. Why is a Grignard reagent considered a carbanion source?

In a Grignard reagent $R - M g - X$ , magnesium is far less electronegative than carbon, so the $C - M g$ bond is highly polar with a significant $δ^{-}$ on carbon. The alkyl group therefore behaves like a carbanion: it attacks electrophilic carbons in aldehydes, ketones, esters, and $C O_{2}$ . This is why Grignards form new $C - C$ bonds so readily.

Q9. What is the role of s-character in the energy of an orbital?

The s orbital is spherically symmetric and its electrons are held closest to the nucleus. Greater s-character in a hybrid orbital (sp $>$ sp $^{2}$ $>$ sp $^{3}$ ) means the electrons are on average closer to the nucleus, so the orbital is lower in energy, more electronegative, and forms shorter, stronger bonds. This explains why sp-hybridised carbons in alkynes are more acidic than sp $^{3}$ carbons in alkanes.

Q10. What is the difference between an electrophile and a nucleophile?

An electrophile is an electron-poor species that accepts an electron pair (it is a Lewis acid, either positive or neutral electron-deficient, like $H^{+}$ , $N O_{2}^{+}$ , $A l C l_{3}$ , or a carbocation). A nucleophile is an electron-rich species that donates an electron pair (it is a Lewis base, either negative or neutral with a lone pair, like $O H^{-}$ , $C N^{-}$ , $N H_{3}$ , or a carbanion). In every ionic reaction, a nucleophile attacks an electrophile.

Q11. What are the four main types of organic reactions?

The four types are substitution (one group replaces another, e.g. $C H_{3} B r + O H^{-} \to C H_{3} O H$ ), addition (two groups add across a $π$ bond, e.g. $C H_{2} = C H_{2} + H B r \to C H_{3} C H_{2} B r$ ), elimination (two groups leave adjacent carbons and a $π$ bond forms, e.g. dehydrohalogenation of alkyl halides), and rearrangement (atoms reorganise to give a more stable isomer, commonly seen when carbocation intermediates shift hydrides or alkyl groups to become more stable).

Previous year questions on Introduction to General Organic Chemistry

2 questions from past papers, each with a step-by-step solution.

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