Understanding Reaction Intermediates The Key to Unlocking Chemical Pathways Part 1

Reaction Intermediates

Reactive Intermediates Part— I

Carbocations · Carbanions · Rearrangements · Neighbouring Group Participation  |  IIT-JAM · GATE · CSIR-NET · BITSAT · TGT · PGT

1. What Are Reactive Intermediates?

When a bond breaks during an organic reaction, it doesn't always give you the final product directly. There's usually a short-lived species sitting in the middle — something unstable, highly reactive, and gone almost as soon as it forms. That's a reactive intermediate.

Their instability is exactly what makes them reactive. The moment they appear, they want to transform into something more stable — and that drive determines the products you get. Trapping or identifying these intermediates has been one of chemistry's most powerful tools for understanding reaction mechanisms.

This chapter focuses on two of the most important ones: carbocations (positive charge on carbon) and carbanions (negative charge on carbon). Both come from heterolytic bond cleavage — where both electrons of a bond go to one of the two atoms involved.

⚡ Exam Quick Recall: Heterolytic cleavage → one species gets both electrons. The one that keeps the electrons = carbanion. The one that loses both = carbocation.

2. Structure of Carbocations

A carbocation forms when carbon loses its bonding electrons to become positively charged. The most common route is heterolytic cleavage of haloalkanes:

(CH₃)₃C–Br  →  (CH₃)₃C⁺  +  Br⁻

They also form from alkenes and alkynes (by proton or electrophile addition), from alcohols (via protonation then water loss), and from amines (via diazotisation followed by N₂ elimination).

The structural facts are crisp. A carbocation carbon is:

  • Trivalent — bonded to three atoms or groups only
  • sp² hybridised — three hybrid orbitals in a plane, one vacant unhybridised pz orbital perpendicular to that plane
  • Trigonal planar — bond angles of 120°
  • Electron-deficient (sextet) — only 6 electrons around carbon, not 8
  • Acts as a Lewis acid — will accept electrons from any available source

Raman, IR, and NMR spectroscopic data on simple alkyl cations all confirm this planar geometry. The methyl cation has three σ bonds formed by sp²–s overlap; in the tert-butyl cation, each sp²-hybridised orbital of the central carbon overlaps with sp³ orbitals of the three methyl groups.

⚡ Exam Tip: In MCQs asking about carbocation hybridisation — always sp², not sp³. The empty p orbital is above and below the plane, making it reactive toward nucleophiles.

Once formed, a carbocation can do three things: react with a nucleophile, lose a proton to form an alkene, or rearrange to a more stable carbocation. The path it takes depends on what's available and which cation is more stable.


3. Stability of Carbocations

Carbocations are classified as methyl, primary (1°), secondary (2°), or tertiary (3°) based on how many carbon atoms directly attach to the positively charged carbon. The stability follows a clear trend driven by three effects.

3.1 Stability by Inductive Effect (+I Effect)

Alkyl groups donate electrons toward the carbocation through the sigma framework — this is the positive inductive effect (+I). More alkyl groups around C⁺ means more electron density is pushed toward it, reducing the intensity of the positive charge and stabilising the species.

Stability:   (CH₃)₃C⁺ > (CH₃)₂CH⁺ > CH₃CH₂⁺ > CH₃⁺
               tert-butyl    isopropyl     ethyl     methyl

Electron-withdrawing groups (–I effect) attached to C⁺ do the opposite — they pull what little electron density remains away from the cationic carbon, making it even more deficient and less stable.

3.2 Stability by Hyperconjugation (No-Bond Resonance)

Hyperconjugation is sometimes called no-bond resonance because it involves σ-bond electrons delocalising into the empty p orbital. The C–H sigma bonds of adjacent methyl groups partially overlap with the vacant p orbital of C⁺, reducing the electron deficiency.

The rule is simple: count the C–H bonds on groups directly adjacent to C⁺.

CarbocationAdjacent C–H bondsContributing structures
Methyl (CH₃⁺)00 (no hyperconjugation)
Ethyl (CH₃–CH₂⁺)33
Isopropyl [(CH₃)₂CH⁺]66
tert-Butyl [(CH₃)₃C⁺]99 (most stable by hyperconj.)

More contributing structures = better delocalisation = greater stability. Both inductive effect and hyperconjugation give the same order — tert > sec > pri > methyl.

3.3 Stability by Resonance

This is the most powerful stabilising force. When an unsaturated group or a lone pair is present adjacent to C⁺, the positive charge delocalises over multiple atoms via pi-orbital overlap. The more resonance structures, the more stable the ion.

Allyl cation (CH₂=CH–CH₂⁺): The positive charge is shared between two carbons via two resonance structures. Technically a primary carbocation — but far more stable than ethyl or propyl due to resonance.

Benzyl cation (C₆H₅–CH₂⁺): The benzene ring's six π electrons delocalise into the vacant p orbital of CH₂⁺. Four resonance structures spread the positive charge across the ring and the benzylic carbon. More stable than allyl because more contributing structures exist.

Diphenylmethyl > Benzyl; Triphenylmethyl (trityl) > Diphenylmethyl. Trityl has nine contributing resonance structures — it's stable enough to be isolated as a salt.

Heteroatom assistance: If O, N, or S sits adjacent to C⁺, its lone pair donates directly into the empty orbital. This is stronger than even allylic resonance — the oxocarbenium or iminium ion formed is particularly stable.

Aromatic carbocations: The tropylium ion (cycloheptatrienyl cation, C₇H₇⁺) is the gold standard. All seven carbons are sp², the ring is planar, six π electrons satisfy Hückel's 4n+2 rule (n=1), and the cation is fully aromatic. It's the most stable carbocation known.

Stability order of conjugated cyclic cations:
Tropylium (C₇H₇⁺) > Cyclopropenylium (C₃H₃⁺) > Cyclopentadienyl cation (C₅H₅⁺)

The cyclopentadienyl cation has 4 π electrons — antiaromatic by Hückel's rule — and is therefore quite unstable. Don't confuse it with the cyclopentadienyl anion, which has 6 π electrons and is aromatic.

⚡ Exam Trap: Resonance stabilisation > inductive stabilisation, always. An allyl carbocation (1°) beats an isopropyl carbocation (2°) in terms of stability when resonance is active.

4. Neighbouring Group Participation (NGP) & Anchimeric Assistance

Sometimes a group that isn't directly at the reaction site — sitting one carbon away — jumps in and assists the leaving group's departure. It temporarily forms a bond with the reaction centre, stabilises an intermediate, and then hands off to the external nucleophile. This is neighbouring group participation (NGP), and the rate enhancement it causes is called anchimeric assistance.

Classic NGP groups include: COO⁻, OH, OR, NR₂, NHCOR, SH, SR, halogens (I > Br > Cl).

Why is it important?

NGP changes both the rate and the stereochemistry of a reaction. In a simple SN2, you'd expect inversion. In a simple SN1, you'd get a racemic mixture. When NGP operates, you often get retention of configuration — a hallmark clue in exam questions.

Classic Example: Hydrolysis of 2-Bromopropanoic acid

When (R)-2-bromopropanoic acid is treated with AgOH, only (S)-lactic acid (with overall retention of configuration) forms. At first glance that's shocking — SN2 should invert, SN1 should racemise.

The explanation: the carboxylate oxygen attacks the adjacent carbon bearing Br in an intramolecular SN2 step (causing one inversion). This forms a strained α-lactone intermediate. External OH⁻ then attacks in a second SN2 step (causing a second inversion). Two inversions = net retention. That's NGP in action.

⚡ Exam Rule: If a question says "retention of configuration" with a nucleophilic substitution at a chiral centre, think NGP immediately. It involves two sequential SN2 inversions.

NGP is reduced when: (a) the carbocation is already very stable on its own, (b) a very good nucleophile is present in solution, or (c) the leaving group is exceptionally good (leaves before the NGP group gets a chance to assist).


5. Non-Classical Carbocations (Bridged Carbocations)

Classical carbocations have the positive charge localised on one carbon or delocalised by resonance with a pi bond in the allylic position — and they can always be drawn with a single Lewis structure. Non-classical carbocations are different. Their positive charge is delocalised by a C–C or C–H sigma bond, or by a pi bond that is not in the allylic position. Three atoms share two electrons — no single Lewis structure can draw them correctly.

The term "non-classical" was introduced by J.D. Roberts for the cyclopropylcarbinyl cation, which he proposed exists as a tricyclobutonium structure. Saul Winstein extended it to explain the dramatic reactivity difference he observed in norbornyl systems.

The Norbornyl Case (C–C Single Bond NGP)

The exo-2-norbornyl brosylate undergoes acetolysis 350 times faster than the endo isomer. Both give only exo-norbornyl acetate, and the exo product is completely racemic. Winstein's explanation: the C1–C6 bond assists the departure of the exo brosylate in a concerted fashion, forming a symmetrical bridged non-classical carbocation (A). The acetate ion can then attack either C1 or C2 with equal probability — giving the 50:50 racemic product.

The endo isomer reacts by SN1 then slowly converts to the bridged ion, explaining the slight inversion (not complete racemisation) and the much slower rate.

C=C Double Bond as NGP

Acetolysis of tosylate I proceeded 10¹¹ times faster than tosylate II because the double bond in tosylate I is geometrically locked in the right position for backside assistance, forming a bridged non-classical carbocation with retention of configuration.

Aromatic Ring as NGP (Phenonium Ion)

Acetolysis of L-threo-3-phenyl-2-butyltosylate gives 96% threo acetate as a racemic mixture — explainable only if the phenyl ring forms a cyclic phenonium ion intermediate. Without NGP, you'd get the erythro product.

Methyl Group as NGP

Solvolysis of neopentyl tosylate produces a rearranged product via a bridgehead carbocation. Isotope labelling studies confirm methyl participates as the neighbouring group.

⚡ Key Distinction: Classical carbocation = single Lewis structure possible. Non-classical = no single Lewis structure works; positive charge delocalised by sigma bond or non-allylic pi bond.

6. General Reactions of Carbocations

Four things a carbocation can do once formed:

  1. React with a nucleophile — Lewis acid + Lewis base → substituted product. e.g., tert-butyl cation + Cl⁻ → tert-butyl chloride.
  2. Lose a proton (deprotonation) — from an adjacent carbon → alkene forms. e.g., (CH₃)₃C⁺ → (CH₃)₂C=CH₂ + H⁺.
  3. Rearrange — 1,2-hydride shift or 1,2-alkyl shift → new (more stable) carbocation.
  4. Addition to an unsaturated compound — cation adds to a double bond, generating a new cation at another carbon. This is the basis of Friedel–Crafts alkylation and cationic polymerisation.

In Friedel–Crafts reactions, an alkyl halide generates a carbocation in the presence of AlCl₃. This electrophile adds to the aromatic ring, forming an arenium ion (Wheland intermediate), which then loses a proton to restore aromaticity.


7. Rearrangement Reactions of Carbocations

Rearrangements happen because carbocations chase stability. If a 1,2-shift — moving a hydrogen (hydride shift) or alkyl/aryl group with its electron pair — produces a more stable cation, it will happen. The migrating group always moves with its bonding electron pair to the adjacent positively charged carbon.

The energy diagram tells the story: starting material (A) → initial carbocation B → possible bridged/non-classical transition state C → rearranged carbocation D (lower energy than B) → product E. D is more stable than B, that's the driving force.

7.1 Wagner–Meerwein Rearrangement

Originally defined for acid-catalysed rearrangements of alcohols, the term now broadly covers any carbocation rearrangement involving a 1,2-shift. The driving force is forming a more stable carbocation, with steric relief as a bonus.

Classic example — Neopentyl alcohol + HCl:

Neopentyl alcohol [(CH₃)₃C–CH₂OH] reacts with HCl. The expected product would be neopentyl chloride. What actually forms is 2-chloro-2-methylbutane (tert-amyl chloride).

(CH₃)₃C–CH₂OH + HCl → (CH₃)₃C–CH₂⁺ (primary) → CH₃–C⁺(CH₃)–CH₂CH₃ (tertiary) → tert-amyl chloride

Reaction Mechanism

Step by step: protonation of OH, then loss of water gives a primary carbocation. A methyl group migrates (with its electrons) from the adjacent quaternary carbon to the primary C⁺ — 1,2-alkyl shift. The tertiary carbocation formed then captures Cl⁻.

Wagner–Meerwein rearrangements are abundant in terpene chemistry. The conversion of isoborneol to camphene is a textbook example. α-Pinene reacting with HCl gives bornyl chloride via a concerted rearrangement — the gem-dimethyl bridge migrates simultaneously as Cl⁻ approaches from the opposite face, making the reaction fully stereospecific.

Ring expansion also falls under this umbrella. 2-Cyclobutylpropan-2-ol in acid gives 1,2-dimethylcyclopentene — the four-membered ring expands to five via a methylene group shift, relieving angle and torsional strain even though this moves from a 3° to a 2° carbocation (strain relief overrides the stability argument here).

The Tiffeneau–Demjanov rearrangement is a specific application used to convert cyclic ketones into ring-expanded homologues. A cyanohydrin is made, the amine is diazotised, and N₂ loss generates a primary cation that immediately rearranges via ring expansion.

Other Examples:

The Nametkin rearrangement is the variant where a methyl group specifically migrates in terpene rearrangements.

Another Good Example:

7.2 Pinacol–Pinacolone Rearrangement

This rearrangement converts a vicinal diol (1,2-diol) to a ketone under acidic conditions. Its name comes from the classic transformation of pinacol (2,3-dimethyl-2,3-butanediol) → pinacolone (3,3-dimethyl-2-butanone).

Mechanism (four clear steps):

  1. Protonation of one –OH group by acid
  2. Loss of water → tertiary carbocation
  3. 1,2-methyl shift → new oxocarbenium ion (more stable because the positive charge on carbon is adjacent to oxygen's lone pair)
  4. Deprotonation of oxonium → ketone

The intermediate carbocation was confirmed by isotopic labelling experiments using H₂¹⁸O — incorporation of ¹⁸O into the recovered diol proved the cation is genuinely free (solvated) and not a concerted process.

Relative Stability Controls Which OH Protonates

When the two carbons bearing –OH have different substituents (unsymmetrical diol), the more stable carbocation forms preferentially. In 2-methyl-1,1-diphenylpropane-1,2-diol, protonation at C-2 gives the benzyl cation (C-1 bears two phenyl groups) — far more stable than protonating C-1 and forming a secondary cation. Therefore only the methyl-migrated ketone (iii) forms, not the phenyl-migrated one.

Now Let's Understand How It's Works Suppose Two Possible Carbocations Are Forming (a) and (b)

It is clear that structure (a) is more stable than (b) because the positive charge on the benzylic carbon is delocalized over both benzene rings. Now Structure (a) Will Give Ketone (iii)

Migratory Aptitude

When there's a choice of which group migrates, the more electron-rich group migrates faster because it better stabilises the partial positive charge at the bridged transition state.

Migratory aptitude:   p-CH₃OC₆H₄ > p-CH₃C₆H₄ > C₆H₅ > CH₃ > H

Why phenyl over methyl? In the bridged non-classical intermediate, a phenyl group delocalises the positive charge through the ring (like a phenonium ion), while a methyl group can only donate through hyperconjugation — much weaker. A p-methoxy phenyl does it even better.

Pinacol rearrangement can also cause ring expansion in cyclic diols. 1,2-Diols of cyclopentane derivatives give cyclohexanone products. The mechanism is identical — carbocation formation, then the ring carbon migrates.

Deamination reactions of 1,2-amino alcohols behave similarly (the diazonium ion provides the driving force for carbocation formation). The conversion of 1,1-diphenyl-2-aminopropan-1-ol with HNO₂ gives a pinacolone-type product with full concerted stereospecificity.

1,2-Bromohydrins also undergo pinacol-type rearrangements with Ag⁺ ions. 2-Bromo-1,1,2-trimethylpropan-1-ol gives 3,3-dimethylbutan-2-one when heated with Ag⁺ — the silver coordinates to Br, assists ionisation, and the methyl group migrates.

7.3 Benzil–Benzilic Acid Rearrangement

This is the base-promoted rearrangement of an aromatic diketone (benzil) to a salt of an α-hydroxy acid (benzilic acid). Aliphatic diketones give poor yields because aldol condensation competes.

Mechanism:

  1. OH⁻ adds to one carbonyl of benzil → tetrahedral alkoxide anion
  2. Phenyl group migrates (as phenyl anion with its electrons) to the adjacent carbonyl carbon → new C–C bond, expelling O⁻
  3. Proton transfer → benzilic acid anion

The migration terminus here carries only a partial positive character (carbonyl carbon), unlike pinacol where it has a full positive charge. That's the key structural distinction between the two rearrangements.

⚡ Exam Comparison:
Pinacol: acid-catalysed, vicinal diol → ketone, migration to a full C⁺.
Benzilic acid: base-catalysed, aromatic diketone → α-hydroxy acid, migration to a partial C⁺ (carbonyl carbon).

8. Carbanions — Generation, Structure & Stability

A carbanion is the opposite of a carbocation — it's a carbon bearing a negative charge, with a full octet (8 electrons). Like carbocations, they arise from heterolytic bond cleavage, but this time carbon retains both electrons.

Generation of Carbanions

  • Base abstracts a proton from a C–H bond: carbonyl compounds, nitro compounds, and active methylene compounds are common substrates. NaNH₂, NaOEt, LDA are typical bases.
  • Decarboxylation: –COO⁻ → carbanion + CO₂ (important in malonic ester synthesis).
  • Nucleophile addition to a double bond: Nu⁻ adds to C=C → carbanion on the adjacent carbon.
  • Metal + alkyl halide: Mg with RX gives Grignard reagent (R–MgX). Li, Na, Zn, Hg also work. Carbon gets a partial negative charge in the C–metal bond — this reversal of polarity is called umpolung.
  • Ylides and dithianes: treatment of phosphonium salts or dithianes with alkyl lithium generates stabilised carbanions.

Structure of Carbanions

A simple alkyl carbanion prefers pyramidal (sp³) geometry — the lone pair sits in an sp³ orbital at the apex of a tetrahedron. This is lower energy than a planar sp² arrangement because sp³ orbitals hold electrons closer to the nucleus than pure p orbitals.

However, carbanions that can delocalise the lone pair (benzyl, allyl, adjacent to C=O, C≡N, etc.) prefer planar (sp²) geometry, maximising orbital overlap with the π system.

Simple carbanions undergo rapid inversion of configuration (like a flipping umbrella) — the energy barrier is only about 2 kcal/mol for the methyl carbanion. For CF₃⁻, it's about 120 kcal/mol because the three electronegative F atoms stabilise the sp³ arrangement so strongly.

Stability of Carbanions

Four key factors control carbanion stability — and they're essentially the inverse of carbocation stability trends.

1. s-Character of the Carbanion Carbon

More s-character = electrons held closer to nucleus = more stable anion.

Stability:   sp (alkynyl) > sp² (vinyl, aryl) > sp³ (alkyl)
pKₐ order:   RC≡C–H (~25) < H₂C=CH₂ (~37) < CH₄ (~44)

2. Inductive Effect

Electron-withdrawing groups (F, Cl, CN, C=O, NO₂) stabilise carbanions by pulling electron density away. The alkyl substitution effect is the reverse of carbocations:

Stability:   CH₃⁻ > 1° > 2° > 3°   (opposite to carbocations!)

3. Resonance / Conjugation

Conjugation with C=O, C≡N, NO₂, or aromatic rings delocalises the negative charge, dramatically increasing stability. The effect order:

NO₂ > RCO > COOR > SO₂ > CN ≈ CONH₂ > Halogens > H > R

Benzyl anion is more stable than allyl anion because it has more contributing resonance structures. Triphenylmethyl anion (pKₐ of triphenylmethane = 33) is stable enough to exist in solution with sodamide in liquid ammonia.

4. Aromatisation

Cyclopentadienyl anion (C₅H₅⁻) has 6 π electrons — it's aromatic (4n+2, n=1). Its pKₐ is ~16, making cyclopentadiene one of the most acidic hydrocarbons. Compare with a simple alkene (pKₐ ~37). Aromaticity gained upon deprotonation is a massive stabilising force.

⚡ Remember: Carbanion stability order is exactly OPPOSITE to carbocations. More alkyl groups = less stable carbanion (because alkyl groups donate electrons, making the already electron-rich carbon worse). Cyclopentadienyl anion is aromatic and very stable; cyclopentadienyl cation is antiaromatic and very unstable.
PropertyCarbocationCarbanion
Charge on CPositive (+)Negative (–)
Electrons around C6 (sextet)8 (octet)
Hybridisationsp²sp³ (usually)
GeometryTrigonal planarPyramidal (usually)
Acts asElectrophile / Lewis acidNucleophile / Lewis base
Alkyl group effectStabilises (more alkyl = more stable)Destabilises (more alkyl = less stable)
Stability by substitution3° > 2° > 1° > methylmethyl > 1° > 2° > 3°

9. Reactions of Carbanions

Carbanions are strong nucleophiles and strong bases. Their main reactions:

  • Displacement (SN2): carbanion attacks an electrophilic carbon. Malonate and acetoacetate alkylations are textbook examples — the carbanion from diethyl malonate attacks an alkyl halide to give a mono-alkylated product.
  • Elimination (E1cb): carbanion forms first, then the adjacent leaving group departs. β-Phenylethyl bromide with base → styrene via carbanion intermediate.
  • Condensation: aldol, Claisen, and Dieckmann condensations all involve carbanion intermediates attacking carbonyl groups.
  • Michael Addition: carbanion adds to the β-carbon of an α,β-unsaturated carbonyl compound.

Rearrangement Reactions of Carbanions

Carbanions rearrange far less readily than carbocations. The key difference: in carbocation rearrangements, the migrating group moves with a pair of electrons; in carbanion rearrangements, the group migrates without a pair of electrons. True 1,2-carbon-to-carbon alkyl shifts are essentially unknown for carbanions.

What does occur are 1,2-shifts from N or S to the carbanion carbon:

Favorskii Rearrangement

α-Halo ketones react with base (NaOMe, NaOH) to give carboxylic acid derivatives with a different carbon skeleton. A cyclopropanone intermediate forms — the base abstracts the α-proton on the side opposite the halogen, and the resulting carbanion/enolate displaces the halide in an intramolecular reaction. Hydroxide then opens the strained ring at the less hindered side.

Reaction Mechanism :

Ring contraction is observed with cyclic α-halo ketones. 2-Chlorocyclobutanone → cyclopropanecarboxylic acid is the classic example.

For ketones without an α-hydrogen on the side opposite the halide, a "semibenzilic" pathway operates — nucleophile adds to carbonyl, halide departs, with a 1,2-shift.

Examples:

[2,3]-Wittig Rearrangement

Allyl ethers deprotonated with BuLi at –78°C give homoallylic alcohols via a [2,3]-sigmatropic rearrangement. A five-membered cyclic transition state forms; the process is thermally allowed and stereospecific.

Stevens Rearrangement

Quaternary ammonium salts with an electron-withdrawing group α to N undergo base-catalysed rearrangement to tertiary amines. NaOH abstracts the α-proton → ylide forms → benzyl (or allylic) group migrates from N to C without its electrons. Configuration at the migrating centre is retained. The priority rule: benzyl migrates before methyl.


10. Ambident Ions and Their Reactions

An ambident nucleophile (from Latin: ambi = both, dent = tooth) is a species with two distinct nucleophilic sites — it can attack from either end. The classic example is the enolate ion, where both oxygen and carbon carry negative charge:

R–CH=C(–O⁻)–R'  ⟷  R–C⁻H–C(=O)–R'
   O-attack site           C-attack site

Other common ambident ions: CN⁻ (attacks via C → nitrile; via N → isocyanide), SCN⁻ (S → thiocyanate; N → isothiocyanate), NO₂⁻ (O → nitrite ester; N → nitroalkane), and phenoxide ion (O → ether; C → C-alkylated product).

What Controls Which End Attacks?

HSAB principle: Hard acids (like carbocations in SN1) prefer hard bases (O⁻ in enolate = harder). Soft acids (carbon in SN2, less polarised) prefer soft bases (C⁻ in enolate = softer). So: SN1 conditions → O-alkylation; SN2 conditions → C-alkylation.

Solvent effects: In protic solvents, the more electronegative atom (O, N) is hydrogen-bonded and solvated, making it less available — C-attack predominates. In polar aprotic solvents (DMSO, DMF), the hard atom becomes free again — O-attack increases. The reaction of β-naphthoxide with benzyl bromide gives 95% O-alkylation in DMSO but 85% C-alkylation in CF₃CH₂OH — a dramatic illustration of solvent control.

Percentage of N- and C-alkylated Products in Different Solvents

Examples:

Steric effects: When the O-site is sterically blocked (e.g., 2,6-dimethylphenoxide), C-alkylation wins because the oxygen is flanked by two methyl groups.

Counter ion effects: K⁺ counter ion with an enolate → reaction stays SN2 → N-alkylation or C-alkylation. Ag⁺ counter ion → coordinates with halide leaving group → reaction becomes SN1 → O-alkylation.

Some Named Reactions of Ambident Ions

  • Nef Reaction: nitronate anion (from RCH₂NO₂ + base) → reacts at C with soft electrophiles; at O with hard electrophiles like MeOSO₂F.
  • Kolbe Nitrile Synthesis: alkyl halide + KCN → nitrile (C-attack); with AgCN → isocyanide (N-attack). Silver activates SN1, favouring the harder N-attack... wait — that seems counter-intuitive until you remember isocyanides bond via C: Ag⁺ makes the carbon electrophile harder, favouring N-attack in isocyanide formation.
  • Active Methylene Alkylation (Acetylacetone): with CH₃OTs (soft, SN2) → 97% C-alkylation; with CH₃I (harder) → 3% C-alkylation, 97% O-alkylation. The leaving group hardness reverses the outcome.
  • Alkylation Order:


11. Solved MCQs

Q1. A carbocation behaves like a Lewis acid because it:

  • a) Is electron deficient and can accept electrons
  • b) Can donate a proton during an acid-base reaction
  • c) Is trivalent and positively charged
  • d) Can form a bond with a compound having negative charge
✅ Answer: (a) — A Lewis acid is an electron pair acceptor. The empty p orbital of a carbocation makes it a perfect electron pair acceptor.

Q2. The hybridisation of carbon in a carbocation is:

  • a) sp³
  • b) sp²
  • c) sp
  • d) dsp²
✅ Answer: (b) sp² — trigonal planar, with one empty p orbital perpendicular to the plane.

Q3. Which carbocation is most stable?

  • a) CH₃⁺
  • b) CH₃CH₂⁺
  • c) (CH₃)₂CH⁺
  • d) (CH₃)₃C⁺
✅ Answer: (d) — tert-Butyl cation is most stable due to maximum inductive (+I) effect and 9 hyperconjugating C–H bonds.

Q4. The tropylium ion (C₇H₇⁺) is highly stable because it is:

  • a) A tertiary carbocation
  • b) Antiaromatic
  • c) Aromatic — follows Hückel's rule with 6 π electrons
  • d) Stabilised only by inductive effect
✅ Answer: (c) — All 7 carbons are sp², the ring is planar, 6 π electrons satisfy 4n+2 (n=1). Fully aromatic.

Q5. Which of the following is a non-classical carbocation?

  • a) tert-Butyl cation
  • b) Benzyl cation
  • c) Norbornyl cation
  • d) Allyl cation
✅ Answer: (c) — The 2-norbornyl cation has a bridged structure with the C1–C6 bond delocalising the positive charge — cannot be drawn with a single Lewis structure.

Q6. In the hydrolysis of (R)-2-bromopropanoic acid with AgOH, the product has:

  • a) Inverted configuration (SN2)
  • b) Racemised configuration (SN1)
  • c) Retained configuration (two SN2 inversions via NGP)
  • d) No reaction
✅ Answer: (c) — NGP by the carboxylate group causes two sequential inversions → net retention.

Q7. The migratory aptitude in pinacol rearrangement follows which order?

  • a) H > CH₃ > C₆H₅ > p-CH₃OC₆H₄
  • b) p-CH₃OC₆H₄ > p-CH₃C₆H₄ > C₆H₅ > CH₃ > H
  • c) CH₃ > C₆H₅ > H
  • d) All groups migrate equally
✅ Answer: (b) — More electron-rich groups migrate faster because they better stabilise the bridged transition state via electron donation.

Q8. Favorskii rearrangement involves:

  • a) Migration of a group as a cation
  • b) Formation of a cyclopropanone intermediate via carbanion
  • c) Acid-catalysed rearrangement of diols
  • d) Rearrangement of aromatic diketones with base
✅ Answer: (b) — An α-haloketone under base gives a carbanion that displaces the halide intramolecularly to form a cyclopropanone, which then opens.

Q9. In the reaction of CN⁻ with a primary alkyl halide (SN2), the product formed is:

  • a) Isocyanide (C-bonded)
  • b) Nitrile (C-bonded)
  • c) Both in equal amounts
  • d) Nitrile only with Ag⁺
✅ Answer: (b) — In SN2 conditions, CN⁻ is the softer nucleophile (C attacks) → nitrile. With Ag⁺ (SN1 conditions, harder electrophile), N-attack gives isocyanide.

Q10. Cyclopentadienyl anion is more stable than cycloheptatrienyl anion because:

  • a) It has more resonance structures
  • b) It is aromatic with 6 π electrons; the heptatrienyl anion is antiaromatic with 8 π electrons
  • c) Cyclopentadienyl anion is a carbocation
  • d) It has sp³ hybridised carbon
✅ Answer: (b) — C₅H₅⁻ has 6 π electrons (aromatic, 4n+2, n=1). C₇H₇⁻ has 8 π electrons (antiaromatic, 4n, n=2).

12. Practice Questions (25–30 Qs for Exam Drill)

Work through these before looking at any solutions. They're modelled on IIT-JAM, GATE, CSIR-NET, and TGT/PGT paper patterns.

Q1. Arrange the following in decreasing order of carbocation stability: allyl, benzyl, tert-butyl, methyl. Justify your order.
Q2. The tert-butyl carbocation has how many hyperconjugating structures? Write them out briefly.
Q3. What product is expected when neopentyl alcohol reacts with HBr? Explain the mechanism, identifying the type of carbocation rearrangement.
Q4. Draw all resonance structures of the benzyl cation. Why is triphenylmethyl cation more stable than benzyl cation?
Q5. Explain why exo-2-norbornyl brosylate reacts 350 times faster in acetolysis than the endo isomer. What intermediate is involved?
Q6. In the Favorskii rearrangement of 2-chlorocyclohexanone with NaOMe, what is the final product? Draw the cyclopropanone intermediate.
Q7. Compare the stability of the following carbanions: CH₃⁻, (CH₃)₂CH⁻, (CH₃)₃C⁻, CF₃⁻. Explain the role of the inductive effect.
Q8. State the Hückel rule and apply it to determine whether the following are aromatic, antiaromatic, or nonaromatic: cyclopentadienyl cation, cyclopentadienyl anion, cyclopropenyl cation, cyclooctatetraene.
Q9. What is "anchimeric assistance"? Give one example showing how it leads to retention of configuration.
Q10. In the pinacol rearrangement of 2-methyl-1,1-diphenylpropane-1,2-diol, which ketone forms preferentially? Explain why in terms of relative carbocation stability.
Q11. What is the difference between a classical and non-classical carbocation? Give a structural criterion to identify each.
Q12. Write the complete mechanism for the Wagner–Meerwein rearrangement of 3,3-dimethylbut-1-ene with HCl.
Q13. Why does the allyl carbanion prefer C-protonation (thermodynamic) over O-protonation (kinetic) upon reprotonation of an enolate? Connect this to HSAB theory.
Q14. The pKₐ of cyclopentadiene is ~16 while that of cycloheptatriene is ~36. Explain this difference.
Q15. In the Grignard reagent CH₃MgBr, which atom bears the partial negative charge? What organic chemistry concept describes this polarity reversal?
Q16. What is the role of the solvent in determining the site of attack by an ambident nucleophile? Distinguish between the effects of protic and polar aprotic solvents.
Q17. Explain the Stevens rearrangement. What types of groups migrate, and what is the stereochemical outcome at the migration origin?
Q18. The [2,3]-Wittig rearrangement is described as a sigmatropic reaction. What does this mean, and what is the key structural requirement in the substrate?
Q19. Show how isotope labelling with H₂¹⁸O provides evidence for a carbocation intermediate in the pinacol rearrangement.
Q20. Acetolysis of tosylate I (containing a fixed double bond) proceeds 10¹¹ times faster than tosylate II. What type of non-classical carbocation is involved, and why does the geometry of the substrate matter so much?
Q21. Draw and explain the resonance structures of the methoxymethyl cation (CH₃OCH₂⁺). Why is it more stable than the ethyl cation?
Q22. In the Benzil–Benzilic acid rearrangement, what nucleophile initiates the reaction? Write the complete mechanism with all intermediates.
Q23. Consider two carbocations: (a) cyclopropenyl cation (C₃H₃⁺) and (b) cyclopentadienyl cation (C₅H₅⁺). Which is more stable, and why? Draw all resonance structures for both.
Q24. What does "migratory aptitude" mean in the context of pinacol rearrangement? Arrange the following in order: H, CH₃, C₆H₅, p-MeOC₆H₄.
Q25. The Tiffeneau–Demjanov rearrangement converts a cyclic ketone into a homologue with a larger ring. Write the step-by-step mechanism starting from cyclohexanone.
Q26. Explain why the carbanion stability order (methyl > 1° > 2° > 3°) is opposite to the carbocation stability order. Use the inductive effect in your explanation.
Q27. In the reaction of 2,6-dimethylphenoxide with an alkyl iodide, C-alkylation is preferred over O-alkylation. Explain why.
Q28. A quaternary ammonium salt (PhCOCH₂⁺N(CH₃)₂CH₂Ph Br⁻) reacts with NaOH. Identify the type of rearrangement and write the product.
Q29. Describe the E1cb mechanism of elimination. How does it differ from E2? Give one example involving a carbanion intermediate.
Q30. In the reaction of acetylacetone's enolate with CH₃OTs vs CH₃I, different O/C alkylation ratios are observed. Explain this using HSAB and the nature of the leaving group.

End of Chapter Notes — Reactive Intermediates I  |  Good luck with your exam prep!

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