Intuition The whole story in one breathAn alkyl halide is R − X R{-}X R − X X X X electronegative , the C–X bond is polar : carbon is δ + \delta^+ δ + X X X δ − \delta^- δ − δ + \delta^+ δ +
A nucleophile attacks the carbon → substitution (Nu replaces X).
A base grabs a β-hydrogen → elimination (form a C=C, kick out X).
Each comes in two flavours — a one-step version (bimolecular: SN2 / E2) and a two-step version going through a carbocation (unimolecular: SN1 / E1). Everything below is just which one wins, and why .
From alcohols (most common):
R − O H + H X → R − X + H 2 O R{-}OH + HX \rightarrow R{-}X + H_2O R − O H + H X → R − X + H 2 O H I > H B r > H C l HI>HBr>HCl H I > H B r > H C l SOCl 2 \text{SOCl}_2 SOCl 2 R C l + S O 2 + H C l RCl + SO_2 + HCl R C l + S O 2 + H C l P X 3 PX_3 P X 3 P X 5 PX_5 P X 5 From alkenes : Markovnikov addition of H X HX H X H B r HBr H B r anti-Markovnikov .From alkanes : free-radical halogenation R − H + X 2 → h ν R − X R{-}H + X_2 \xrightarrow{h\nu} R{-}X R − H + X 2 h ν R − X C l 2 Cl_2 C l 2 Halide exchange (Finkelstein) : R − C l + N a I → acetone R − I + N a C l ↓ R{-}Cl + NaI \xrightarrow{\text{acetone}} R{-}I + NaCl\downarrow R − C l + N a I acetone R − I + N a C l ↓ N a C l NaCl N a C l
SOCl 2 \text{SOCl}_2 SOCl 2 H C l HCl H C l SOCl 2 \text{SOCl}_2 SOCl 2 O H OH O H − O S O C l -OSOCl − O S O C l S O 2 SO_2 S O 2 H C l HCl H C l as gases , so the reaction is irreversible and the product is easy to purify.
Intuition Two ingredients decide everything(a) Leaving group ability: a good leaving group is a weak base (stable on its own with the electrons). So I − > B r − > C l − ≫ F − I^- > Br^- > Cl^- \gg F^- I − > B r − > C l − ≫ F − (b) Carbocation stability: 3 ∘ > 2 ∘ > 1 ∘ > methyl 3^\circ > 2^\circ > 1^\circ > \text{methyl} 3 ∘ > 2 ∘ > 1 ∘ > methyl + + + make a carbocation (SN1, E1) love 3 ∘ 3^\circ 3 ∘ reach the carbon (SN2) love unhindered 1 ∘ 1^\circ 1 ∘
One concerted step: nucleophile attacks carbon from the side opposite the leaving group; bond making and bond breaking happen together through a 5-coordinate transition state .
Intuition Stereochemistry — Walden inversionBecause Nu comes in opposite to X, the other three bonds flip like an umbrella in a storm. A single stereocentre is inverted (R → S R\to S R → S
Order of reactivity (substrate): methyl > 1 ∘ > 2 ∘ ≫ 3 ∘ \text{methyl} > 1^\circ > 2^\circ \gg 3^\circ methyl > 1 ∘ > 2 ∘ ≫ 3 ∘ steric hindrance ), so 3 ∘ 3^\circ 3 ∘
Favoured by: strong/charged nucleophile, polar aprotic solvent (DMSO, acetone — leaves the nucleophile "naked" and reactive), good leaving group.
Two steps: (1) R − X → R + + X − R{-}X \rightarrow R^+ + X^- R − X → R + + X − (2) R + + Nu → R − Nu R^+ + \text{Nu} \rightarrow R{-}\text{Nu} R + + Nu → R − Nu
Intuition Stereochemistry — racemisationThe carbocation is flat (sp2 ^2 2 . Nu can attack from either face equally → mixture of inversion + retention → racemic (often slight excess of inversion because the leaving group briefly shields one face).
Order of reactivity: 3 ∘ > 2 ∘ > 1 ∘ 3^\circ > 2^\circ > 1^\circ 3 ∘ > 2 ∘ > 1 ∘ Favoured by: polar protic solvent (water, alcohols — they solvate and stabilise the ions), weak/neutral nucleophile, good leaving group.
E2: one step. Strong base removes a β-H while X leaves; rate = k [ R − X ] [ base ] =k[R{-}X][\text{base}] = k [ R − X ] [ base ] anti-periplanar geometry (H and X at 180 ∘ 180^\circ 18 0 ∘ E1: two steps, same carbocation as SN1, then base removes a β-H from R + R^+ R + = k [ R − X ] =k[R{-}X] = k [ R − X ]
Intuition Why anti-periplanar for E2?For the new π \pi π parallel so their electrons overlap into the developing π \pi π anti (180 ∘ 180^\circ 18 0 ∘ why some cyclohexyl halides eliminate slowly — the H must be axial.
Factor
SN2
SN1
E2
E1
Substrate
1 ∘ 1^\circ 1 ∘ 3 ∘ 3^\circ 3 ∘ 2 ∘ , 3 ∘ 2^\circ,3^\circ 2 ∘ , 3 ∘ 3 ∘ 3^\circ 3 ∘
Reagent
strong Nu, weak base
weak/neutral
strong bulky base
weak base
Rate law
k [ R X ] [ N u ] k[RX][Nu] k [ R X ] [ N u ] k [ R X ] k[RX] k [ R X ] k [ R X ] [ B ] k[RX][B] k [ R X ] [ B ] k [ R X ] k[RX] k [ R X ]
Solvent
polar aprotic
polar protic
—
polar protic
Stereo
inversion
racemisation
anti-periplanar
mixture
Worked example Worked decision 3 — 2-bromobutane +t t t
Step: 2 ∘ 2^\circ 2 ∘ strong bulky base + heat. Why? heat and strong base push elimination; bulk pushes Hofmann.Step: E2; bulky base grabs the most accessible β-H.Answer: major = 1-butene (Hofmann) , minor = 2-butene. With small O E t − OEt^- O E t −
Common mistake "More nucleophile → faster, so SN1 speeds up too."Why it feels right: more attacking species usually means faster. Fix: SN1's slow step is ionisation, which the nucleophile isn't part of, so [ Nu ] [\text{Nu}] [ Nu ] not appear in the rate law. Only SN2/E2 are accelerated by reagent concentration.
Common mistake "SN1 gives full inversion like SN2."Why it feels right: both replace X. Fix: SN1 goes through a flat carbocation attacked from both faces → racemisation ; only SN2's backside attack gives clean inversion.
Common mistake "The more stable (Zaitsev) alkene is ALWAYS major."Why it feels right: thermodynamics favours the stable product. Fix: with a bulky base the transition state matters more than product stability — the base can't reach the crowded β-H, so the Hofmann (less substituted) alkene wins.
3 ∘ 3^\circ 3 ∘ Why it feels right: 3 ∘ 3^\circ 3 ∘ Fix: that helps SN1/E1, not SN2. SN2 needs backside access , which 3 ∘ 3^\circ 3 ∘ 3 ∘ 3^\circ 3 ∘
Recall Feynman: explain to a 12-year-oldImagine a kid (the carbon) holding a heavy backpack (the halogen). In the SN2 game, a new friend runs up from behind and shoves the backpack off in one smooth motion — the kid spins around (inversion). In the SN1 game, the kid first drops the backpack on his own (slow), stands there flat-footed, and then a new friend can grab either of his hands. If instead of taking the backpack the friend yanks the kid's belt loop (a nearby hydrogen), the kid's shirt rips open into a new shape — that's elimination , making a double bond. A gentle small friend grabs the easy belt loop; a big clumsy friend can only reach the outer one (Hofmann).
Mnemonic Remember the four"2 for 2, 1 for 1" — SN2 /E2 rate laws are 2nd order (need a partner, one step); SN1 /E1 are 1st order (one lonely substrate ionises). And "Strong Bulky Base = Bye, Hofmann" (bulky base → Hofmann). "Aprotic for SN2, Protic for SN1."
Why is SN1 first order even though two molecules form the product?
What stereochemical outcome proves a reaction went SN2?
When does Hofmann beat Zaitsev?
What is the rate law for SN2 and why? rate = k [ R X ] [ N u ] \text{rate}=k[RX][Nu] rate = k [ R X ] [ N u ] — single concerted step whose transition state contains both partners.
What is the rate law for SN1 and why? rate = k [ R X ] \text{rate}=k[RX] rate = k [ R X ] — the slow ionisation step contains only the substrate, so
[ N u ] [Nu] [ N u ] can't appear.
Stereochemistry of SN2? Backside attack → Walden inversion of configuration.
Stereochemistry of SN1? Planar carbocation attacked from both faces → racemisation .
Substrate order for SN2? methyl > 1° > 2° ≫ 3° (steric hindrance blocks 3°).
Substrate order for SN1? 3° > 2° > 1° (carbocation stability).
Solvent preference: SN1 vs SN2? SN1 polar protic (stabilises ions); SN2 polar aprotic (frees the nucleophile).
Geometry required for E2? H and leaving group anti-periplanar (180° dihedral) for orbital overlap.
Zaitsev rule? Major product is the more substituted (more stable) alkene.
When is the Hofmann (less substituted) alkene major? With bulky bases (e.g. t-BuO⁻) or bulky leaving groups — base can only reach the less hindered β-H.
Why is SOCl₂ preferred over HCl for R-OH → R-Cl? By-products SO₂ and HCl escape as gases → irreversible, clean product.
Finkelstein reaction? R-Cl + NaI in acetone → R-I + NaCl↓ (driven by NaCl precipitation).
What conditions push elimination over substitution? Strong/bulky base, high temperature, hindered substrate.
Leaving group ability order among halides? I⁻ > Br⁻ > Cl⁻ ≫ F⁻ (weaker base = better leaving group).
Carbocations — stability, hyperconjugation, rearrangements Nucleophilicity vs Basicity Markovnikov & anti-Markovnikov addition Alcohols — preparation and reactions Stereochemistry — R/S, optical activity, racemisation Reaction kinetics — molecularity vs order
from alcohols, alkenes, alkanes
irreversible, gas by-products
delta+ carbon electrophilic
Good leaving group is weak base
Carbocation stability 3>2>1
Intuition Hinglish mein samjhoDekho, alkyl halide (R − X R{-}X R − X δ + \delta^+ δ + δ + \delta^+ δ +
Yaad rakhne ka simple funda: SN2 mein nucleophile peeche se (backside) aata hai, isliye structure ulat jaata hai — inversion (Walden inversion). Ye 1° (kam bheed wale) carbon par best chalti hai, kyunki bulky 3° group raasta block kar deta hai. SN1 mein pehle flat carbocation banta hai, dono taraf se attack hota hai, isliye racemic mixture milta hai — aur ye 3° par best hai kyunki tertiary carbocation stable hota hai. Solvent bhi matter karta hai: SN1 ke liye protic (paani, alcohol) jo ions ko stabilise kare; SN2 ke liye aprotic (acetone, DMSO) jo nucleophile ko free aur strong rakhe.
Elimination mein Zaitsev rule kehta hai ki zyada substituted (zyada stable) alkene major banega — ye normal chhote base ke saath. Lekin agar base bulky ho (jaise t-BuOK), to wo bheed wale hydrogen tak pahunch nahi paata, isliye kam substituted alkene major ban jaata hai — isko Hofmann product kehte hain. E2 ke liye ek geometry condition bhi hai: H aur X ko anti-periplanar (180°) hona chahiye taaki orbitals overlap karke naya pi bond bana saken.
Exam tip: pehle substrate dekho (1°/2°/3°), phir reagent (strong/weak, bulky?), phir solvent. Bas in teen cheezon se 80% questions solve ho jaate hain. "2 for 2, 1 for 1" yaad rakho — SN2/E2 second order, SN1/E1 first order.
