The four competing pathways
When an alkyl halide $\ce{R-X}$ meets a species with a lone pair, that species can play two roles. As a nucleophile it attacks the electron-deficient carbon bearing the halogen and replaces the leaving group — this is substitution. As a base it pulls a hydrogen from the carbon adjacent to the halogen, expelling the halide and forging a double bond — this is elimination. NCERT classifies the reactions of haloalkanes precisely along these lines: nucleophilic substitution, elimination, and reaction with metals.
Each role splits further by molecularity, giving four named mechanisms. Substitution is either bimolecular (SN2) or unimolecular (SN1); elimination is either bimolecular (E2) or unimolecular (E1). The unimolecular pair share a rate-determining ionisation to a carbocation; the bimolecular pair are concerted, single-step processes whose rate depends on both reagents.
| Pathway | Role of reagent | Rate law | Intermediate | Generic outcome |
|---|---|---|---|---|
| SN2 | Nucleophile | Rate = k[RX][Nu] | None (concerted) | Substitution, inversion of configuration |
| SN1 | Nucleophile | Rate = k[RX] | Carbocation | Substitution, racemisation |
| E2 | Base | Rate = k[RX][Base] | None (concerted) | Alkene, β-elimination |
| E1 | Base | Rate = k[RX] | Carbocation | Alkene, β-elimination |
The full mechanistic detail of each route lives on its own page; here the goal is the higher skill of prediction — given a substrate and a reagent, naming which of these four will dominate.
Why there is a fork at all
NCERT puts it memorably: "A chemical reaction is the result of competition; it is a race that is won by the fastest runner. A collection of molecules tend to do, by and large, what is easiest for them." An alkyl halide bearing a β-hydrogen, when met by a base or nucleophile, has two competing routes — substitution and elimination — and which one is taken depends on the nature of the alkyl halide, the strength and size of the base or nucleophile, and the reaction conditions.
The same hydroxide ion illustrates the duality. Attacking carbon, it substitutes; abstracting a β-proton, it eliminates. The carbon-halogen bond is polarised — carbon carries a partial positive charge, halogen a partial negative charge — so the electron-rich reagent is drawn to that carbon, but the β-hydrogens it could remove are only one bond away. The product distribution therefore reports on a kinetic contest, not on which product is more stable in isolation.
"Major product" means kinetically favoured, not most stable overall
Students often assume the alkene "wins" because it has a new π bond, or that the alcohol "wins" because substitution is simpler. Neither is a rule. The major product is whatever forms fastest under the stated conditions. Change the solvent from water to ethanol, or raise the temperature, and the same halide can flip its major product entirely.
Read the reagent and conditions first. Only then decide between substitution and elimination.
Factor 1 — Substrate class (methyl, 1°, 2°, 3°)
The degree of substitution at the carbon bearing the halogen is the first filter, because it controls which mechanisms are even available. NCERT records two opposing reactivity orders: SN2 falls as branching rises (methyl > 1° > 2° > 3°) because bulky groups block the back-side approach of the nucleophile, while SN1 rises with branching (3° > 2° > 1°) because the carbocation grows more stable. Carbocation stability also gates E1, so the two unimolecular routes share the same substrate preference.
| Substrate | β-H present? | Dominant tendency | Reasoning (NCERT) |
|---|---|---|---|
Methyl, CH3X | No | SN2 only | Three small H atoms; no β-H so elimination impossible |
| Primary (1°) | Usually | SN2 (E2 only with bulky strong base) | Unhindered carbon favours back-side attack |
| Secondary (2°) | Usually | SN2 or E2 — the reagent decides | NCERT: 2° halide gives SN2 or elimination by base strength/size |
| Tertiary (3°) | Usually | SN1 or E1/E2 — never SN2 | Bulky groups block SN2; stable 3° carbocation enables ionisation |
NCERT compresses this into a usable summary: a primary alkyl halide prefers SN2; a secondary halide goes SN2 or elimination depending on the strength of the base/nucleophile; a tertiary halide goes SN1 or elimination depending on the stability of the carbocation or the more substituted alkene. Notice that the secondary and tertiary rows are decided not by the substrate alone but by the reagent — which is Factor 2.
A methyl halide cannot eliminate
Elimination is β-elimination: it needs a hydrogen on the carbon next to the one carrying the halogen. A methyl halide $\ce{CH3X}$ has no β-carbon, so it can never give an alkene. With any reagent it does substitution. Watch for this in "which compound cannot undergo elimination" framings.
No β-hydrogen → no elimination, regardless of reagent or temperature.
Factor 2 — Nucleophile versus base
Every reagent has both nucleophilic character (willingness to donate a pair to carbon) and basic character (willingness to grab a proton). The balance, not the absolute strength, decides the pathway. NCERT states the steric logic directly: a bulkier nucleophile prefers to act as a base and abstract a proton rather than approach a tetravalent carbon, and vice versa.
| Reagent character | Example | Drives | Why |
|---|---|---|---|
| Strong, small nucleophile (weak base) | $\ce{I-}$, $\ce{CN-}$, $\ce{RS-}$ | SN2 | Reaches carbon easily; little tendency to grab a proton |
| Strong, bulky base | $\ce{(CH3)3CO-}$ (tert-butoxide) | E2 | Too hindered to reach carbon; abstracts accessible β-H |
| Strong base that is also a good Nu | $\ce{HO-}$, $\ce{C2H5O-}$ | Both — solvent and substrate tip it | Acts as Nu in water; more as base in ethanol / on 3° halide |
| Weak nucleophile (neutral, weak base) | $\ce{H2O}$, $\ce{ROH}$ | SN1 / E1 | Cannot force a concerted step; waits for carbocation |
Two reagents make excellent diagnostic anchors. Cyanide and iodide are textbook SN2 reagents — small, polarisable, weakly basic — so they substitute even on secondary halides. tert-Butoxide is the textbook E2 base — strongly basic but so bulky it cannot squeeze onto carbon — so it eliminates even where a smaller base might substitute. Between these extremes sits hydroxide and ethoxide, strong bases that are also competent nucleophiles, which is exactly why their behaviour swings with the solvent.
The back-side approach that makes SN2 so sensitive to bulk is worked out step by step in the SN2 mechanism page.
Factor 3 — Temperature and solvent
Two physical conditions can override a substrate's default leaning. Temperature is the simpler of the two. Elimination breaks more bonds and creates more independent particles in its transition state than substitution, so it carries a larger, more favourable entropy of activation. As temperature rises, the entropy term in the free energy of activation grows in importance and elimination is increasingly favoured. In practice, heating a borderline reaction increases the fraction of alkene — which is why dehydrohalogenation with alcoholic KOH is run hot.
Solvent acts in two ways. Polar protic solvents (water, alcohols, acetic acid) stabilise ions through hydrogen bonding and so encourage the ionisation step of the unimolecular routes; NCERT notes that SN1 reactions are generally carried out in polar protic solvents. The same solvent that supplies a proton to solvate the departing halide thereby speeds carbocation formation. A protic solvent also cages a small anionic nucleophile, blunting its nucleophilic strength while leaving its basic strength comparatively intact — nudging a reagent like ethoxide toward elimination. Less polar or aprotic media leave bimolecular pathways (SN2, E2) to compete on their own merits.
A decision framework
The three factors collapse into a short sequence of questions. Read the substrate, then read the reagent, then check the conditions. The flowchart below is the working version of NCERT's three-line summary.
Aqueous KOH versus alcoholic KOH
No single contrast captures the framework better than the one NCERT poses as Exercise 6.20: treatment of alkyl chlorides with aqueous KOH gives alcohols, but with alcoholic KOH the major products are alkenes. The substrate and the formula of the reagent are identical; only the medium differs, and that alone flips the pathway.
In water, KOH is fully dissociated and the hydroxide ion is heavily solvated by hydrogen bonding. A solvated $\ce{HO-}$ behaves as a nucleophile, attacks the carbon and substitutes:
$$\ce{C2H5Cl ->[\text{aq. KOH}] C2H5OH + Cl^-}$$
In ethanol, a significant fraction of the reagent exists as the ethoxide ion, and the less-solvated species is a stronger base than nucleophile. It abstracts a β-hydrogen, and hydrogen halide is eliminated — β-elimination, also called dehydrohalogenation — to give the alkene:
$$\ce{C2H5Cl ->[\text{alc. KOH},\ \Delta] CH2=CH2 + KCl + H2O}$$
When elimination can give more than one alkene, Saytzeff's (Zaitsev's) rule selects the major product — the more highly substituted, more stable alkene, the one with the greater number of alkyl groups on the doubly bonded carbons. NCERT's standing example is 2-bromopentane, which gives pent-2-ene as the major product:
$$\ce{CH3CHBrCH2CH2CH3 ->[\text{alc. KOH}] CH3CH=CHCH2CH3\ (\text{major}) + CH2=CHCH2CH2CH3}$$
Worked predictions
CH3CH2CH2Br with (a) KCN in ethanol, (b) tert-butoxide, hot. Name the dominant pathway.
(a) A primary halide with a strong, small, weakly basic nucleophile ($\ce{CN-}$) is the textbook SN2 case → $\ce{CH3CH2CH2CN}$ (butanenitrile). (b) The same primary halide with a strong but very bulky base at high temperature cannot accommodate back-side attack; the base abstracts a β-H, giving E2 → $\ce{CH3CH=CH2}$ (propene). The substrate is fixed; the reagent and temperature flip the outcome.
2-Bromobutane with (a) aqueous KOH, (b) alcoholic KOH, Δ.
A secondary halide can go either way, so the medium decides. (a) Aqueous → solvated $\ce{HO-}$ acts as nucleophile, SN2 substitution → butan-2-ol, $\ce{CH3CH(OH)CH2CH3}$. (b) Alcoholic and hot → $\ce{EtO-}$ acts as base, E2 elimination. By Saytzeff, the major alkene is but-2-ene over but-1-ene: $\ce{CH3CH=CHCH3}$.
2-Bromo-2-methylpropane (tert-butyl bromide) warmed in aqueous ethanol with no added strong base.
A tertiary halide cannot do SN2. In a polar protic medium with only a weak nucleophile present, it ionises to the stable tertiary carbocation (rate-determining), then partitions between capture by solvent (SN1 → 2-methylpropan-2-ol) and loss of a β-proton (E1 → 2-methylprop-1-ene). Warming raises the elimination share. NCERT's rule for 3° halides — SN1 or elimination depending on carbocation stability and the more substituted alkene — is exactly this fork.
Aryl and vinyl halides do neither SN2 nor E2 here
The whole framework above assumes an sp³ C–X bond. In haloarenes the halogen sits on an sp² carbon, the C–X bond has partial double-bond character from resonance, and the phenyl cation is unstable — so NCERT rules out the SN1 route and notes haloarenes are extremely unreactive toward ordinary nucleophilic substitution. Do not apply the alkyl-halide decision tree to chlorobenzene.
Check hybridisation first: this framework is for alkyl (sp³) halides only.
Choosing the pathway in one screen
- Four routes: SN2 and E2 are concerted and bimolecular; SN1 and E1 share a rate-determining carbocation.
- Substrate gate: methyl/1° → SN2; 2° → reagent decides; 3° → never SN2, instead SN1/E1/E2. No β-H → no elimination.
- Reagent decides: strong small nucleophile → SN2; strong bulky base → E2; weak nucleophile in protic solvent → SN1/E1.
- Conditions: high temperature favours elimination; polar protic solvents favour the unimolecular (carbocation) routes.
- The NEET anchor: aqueous KOH → alcohol (substitution); alcoholic KOH, hot → alkene (elimination), with Saytzeff picking the more substituted alkene.