Reaction Mechanisms — Concepts, Formulas & Examples

A reaction mechanism is the step-by-step description of how bonds break and form during a reaction. CBSE Class 12 and NEET test specific mechanisms — SN1, SN2, E1, E2, electrophilic addition and electrophilic aromatic substitution.

Core Concepts

Curved arrows

In a mechanism, curved arrows show movement of electron pairs. The tail is where the electrons come from, the head is where they go. Single-headed arrows (fish hooks) show single electron movement (radicals).

Rules for drawing curved arrows:

Arrow starts from an electron-rich source (lone pair, bond, negative charge)
Arrow ends at an electron-poor destination (positive charge, electronegative atom, empty orbital)
Full arrows = two electrons (heterolytic). Half arrows = one electron (homolytic/radical)
Never draw an arrow from a positive charge — positive charges lack electrons

Bond breaking

Homolytic fission: Bond breaks evenly — one electron goes to each fragment. Produces free radicals. Occurs in non-polar bonds or with UV light/heat. Example: Cl $_2$ → 2Cl $\cdot$ (under UV).

Heterolytic fission: Bond breaks unevenly — both electrons go to one fragment. Produces a cation and an anion. Occurs in polar bonds in the presence of polar solvents. Example: $(CH_3)_3C-Br \rightarrow (CH_3)_3C^+ + Br^-$

Reactive intermediates

Intermediate	Structure	Stability order	Found in
Carbocation (R $^+$ )	3 bonds, empty p orbital	3° > 2° > 1° > CH $_3^+$	SN1, E1, electrophilic addition
Carbanion (R $^-$ )	3 bonds + lone pair	CH $_3^-$ > 1° > 2° > 3°	E1cb, Grignard
Free radical (R $\cdot$ )	3 bonds + 1 unpaired $e^-$	3° > 2° > 1° > CH $_3\cdot$	Halogenation, polymerisation
Carbene (:CR $_2$ )	2 bonds + lone pair	Singlet/triplet	Reimer-Tiemann, cyclopropane formation

SN2 mechanism

Concerted — bond making and breaking happen simultaneously. Nucleophile attacks from the opposite side of the leaving group (backside attack). Inversion of configuration. Rate depends on both nucleophile and substrate concentration.

$\text{Rate} = k[\text{substrate}][\text{nucleophile}]$ (second order)

Characteristics of SN2:

One-step mechanism (no intermediate)
Backside attack → Walden inversion (configuration flips)
Favoured by: strong nucleophile, primary substrate, polar aprotic solvent (DMSO, DMF, acetone)
Hindered by: bulky substrate (steric hindrance), weak nucleophile
Reactivity order: CH $_3$ X > 1° > 2° >> 3° (not observed)

SN1 mechanism

Two-step. Leaving group leaves first, forming a carbocation. Nucleophile then attacks. Rate depends only on substrate. Racemisation (nucleophile attacks from either side).

$\text{Rate} = k[\text{substrate}]$ (first order)

Characteristics of SN1:

Two-step: (1) ionisation → carbocation, (2) nucleophilic attack
Planar carbocation → attack from both sides → racemisation (mixture of R and S)
Favoured by: stable carbocation (3° > 2°), polar protic solvent (water, alcohols), weak nucleophile
Reactivity order: 3° > 2° > 1° >> CH $_3$ (opposite to SN2)

How to decide SN1 vs SN2: Primary substrate + strong nucleophile → SN2. Tertiary substrate + weak nucleophile → SN1. Secondary can go either way — the nucleophile and solvent determine the path.

E2 mechanism

Concerted elimination. Base removes a beta hydrogen, electrons flow to form the double bond, leaving group leaves. Trans/anti-periplanar geometry required.

$\text{Rate} = k[\text{substrate}][\text{base}]$ (second order)

Key features:

One-step, no intermediate
Requires anti-periplanar (trans) arrangement of H and leaving group
Strong, bulky bases (like tert-butoxide) favour E2 over SN2
Follows Zaitsev’s rule — more substituted alkene is major product

E1 mechanism

Two-step: ionisation to carbocation, then loss of a proton from the carbocation. Same first step as SN1, so E1 and SN1 often compete.

SN1 vs SN2 vs E1 vs E2 — decision guide

Factor	SN2	SN1	E2	E1
Substrate	Primary, methyl	Tertiary	Any (bulky base helps)	Tertiary
Nucleophile/Base	Strong Nu	Weak Nu	Strong, bulky base	Weak base
Solvent	Polar aprotic	Polar protic	—	Polar protic
Rate law	2nd order	1st order	2nd order	1st order
Intermediate	None	Carbocation	None	Carbocation
Stereochemistry	Inversion	Racemisation	Anti-periplanar	—

Electrophilic addition to alkenes

Alkene is the nucleophile. Electrophile (H $^+$ from HX) adds to one carbon, forming a carbocation. X $^-$ then attacks the carbocation. Markovnikov rule follows from the more stable carbocation.

H $^+$ adds to the less substituted carbon (gives the more stable carbocation on the other carbon). $\text{CH}_3\text{CH=CH}_2 + \text{H}^+ \rightarrow \text{CH}_3\overset{+}{\text{C}}\text{HCH}_3 \text{ (2° carbocation --- stable)}$

\text{CH}_3\overset{+}{\text{C}}\text{HCH}_3 + \text{Br}^- \rightarrow \text{CH}_3\text{CHBrCH}_3 \text{ (Markovnikov product)}

Anti-Markovnikov addition: With HBr in the presence of peroxides (radical mechanism), the Br adds to the less substituted carbon. This is called the peroxide effect or Kharasch effect. It works only with HBr (not HCl or HI).

Electrophilic aromatic substitution

Electrophile attacks the aromatic ring, forming a cyclohexadienyl cation (arenium ion). A proton is then lost to restore aromaticity. Benzene is preserved at the end.

General mechanism:

The electrophile (E $^+$ ) attacks the pi cloud of benzene, forming a sigma complex (arenium ion). Aromaticity is temporarily lost.

A base removes a proton from the carbon bearing E, restoring the aromatic ring. The overall result: one H on benzene is replaced by E.

Examples: Nitration (E $^+$ = NO $_2^+$ ), halogenation (E $^+$ = Cl $^+$ or Br $^+$ ), Friedel-Crafts alkylation (E $^+$ = R $^+$ ), Friedel-Crafts acylation (E $^+$ = RCO $^+$ ), sulphonation (E $^+$ = SO $_3$ ).

Worked Examples

Nucleophile attacks from opposite side of leaving group. The other three groups flip like an umbrella in the wind. Configuration at the chiral carbon is inverted — R becomes S and vice versa.

Tertiary > secondary > primary > methyl. Alkyl groups donate electron density (+I effect and hyperconjugation), stabilising positive charge. Determines SN1/E1 preference.

If benzene underwent addition (like alkenes), it would lose aromaticity — the stabilisation energy of 150 kJ/mol would be sacrificed. In substitution, the aromatic ring is restored in the final step. Thermodynamics strongly favours substitution.

Common Mistakes

Drawing arrows from positive charge. Arrows always start from electrons, not from positive charges.

Confusing SN1 and SN2 rates. SN1 is first order in substrate; SN2 is first order in both.

Writing that E2 has a carbocation intermediate. E1 does; E2 is concerted.

Forgetting Walden inversion in SN2. If the starting material is R, the product is S. Students often draw the product with the same configuration.

Applying the peroxide effect to HCl and HI. It works only with HBr. HCl has too strong a bond for radical formation, and HI adds the normal Markovnikov way even with peroxides.

Exam Weightage and Revision

JEE Main 2024 tested SN1 vs SN2 prediction. NEET 2023 asked about Markovnikov addition. CBSE boards ask mechanism questions as three-mark short answers. Mechanisms are the backbone of organic chemistry — understanding them makes every reaction predictable.

When a question gives a scenario, identify the core mechanism first, then match it to the concepts above. Most wrong answers come from reading the scenario too quickly.

Learn three mechanism types — SN2, SN1/E1, electrophilic addition. Most NEET mechanism questions use one of these.

Practice Questions

Q1. Predict whether the following reaction proceeds by SN1 or SN2: (CH $_3$ ) $_3$ CBr + H $_2$ O → ?

SN1. The substrate is tertiary (forms a stable 3° carbocation). The nucleophile (H $_2$ O) is weak. The solvent (water) is polar protic, which stabilises the carbocation. All factors point to SN1. Product: (CH $_3$ ) $_3$ COH (tert-butanol) with racemisation if the substrate is chiral.

Q2. Why does SN2 show inversion of configuration?

The nucleophile attacks from the backside — the side opposite to the leaving group. As the new bond forms and the old bond breaks, the three remaining groups flip to the opposite side (like an umbrella inverting in wind). This converts R configuration to S and vice versa. This stereochemical inversion is called Walden inversion.

Q3. What is the product of HBr addition to propene in the presence of peroxides?

1-bromopropane (anti-Markovnikov). The peroxide generates Br $\cdot$ radicals. Br $\cdot$ adds to the less substituted carbon (CH $_2$ ), forming a more stable 2° radical on the other carbon. This radical then abstracts H from HBr. The result is anti-Markovnikov addition: CH $_3$ CH $_2$ CH $_2$ Br.

Q4. Why does benzene prefer substitution over addition?

Benzene has aromatic stabilisation energy of about 150 kJ/mol. Addition would destroy the aromatic pi system, costing this energy. Substitution restores aromaticity in the final step — only one H is replaced by the electrophile, and the aromatic ring remains intact. Thermodynamics strongly favours the pathway that preserves aromaticity.

Q5. Arrange in order of SN1 reactivity: CH $_3$ Br, (CH $_3$ ) $_2$ CHBr, (CH $_3$ ) $_3$ CBr.

SN1 reactivity follows carbocation stability: 3° > 2° > 1°. (CH $_3$ ) $_3$ CBr > (CH $_3$ ) $_2$ CHBr > CH $_3$ Br. The tertiary substrate ionises most easily because it forms the most stable carbocation.

FAQs

What is the difference between mechanism and rate law? A mechanism describes the actual sequence of bond-breaking and bond-forming steps. A rate law is an experimentally determined mathematical relationship between rate and concentrations. The rate law must be consistent with the mechanism, but multiple mechanisms can sometimes give the same rate law.

Can SN1 and E1 occur simultaneously? Yes. Both SN1 and E1 share the same first step (carbocation formation). Once the carbocation forms, it can either be attacked by a nucleophile (SN1) or lose a proton (E1). Higher temperatures favour elimination (E1) because it has a higher activation energy.

What determines whether a reaction follows Markovnikov or anti-Markovnikov? The stability of the intermediate. In ionic addition (no peroxides), H $^+$ adds first, and the carbocation forms on the more substituted carbon (more stable) — this gives Markovnikov product. In radical addition (with peroxides), Br $\cdot$ adds first, and the radical forms on the more substituted carbon (more stable) — this gives anti-Markovnikov product.

Mechanisms are chemistry’s way of explaining. Once you can draw arrows, you understand what is actually happening in a reaction — not just memorise the outcome.