GATE Organic — The Reaction Mechanisms That Repeat
Organic chemistry looks endless when you memorise reactions one at a time. It becomes finite the moment you sort it by mechanism, because only a handful of elementary steps exist: a nucleophile attacks an electrophile, a proton moves, a group leaves, a bond migrates to an electron-poor atom, or electrons move round a ring in one concerted step. This guide organises the mechanisms GATE Chemistry keeps returning to, with the reasoning you are expected to show.
1. Substitution at saturated carbon
The whole of SN1 versus SN2 sits on two ideas: how stable the carbocation would be, and how crowded the back face is. Rate laws follow from the mechanism, not the other way round.
SN1: rate = k[RX] — carbocation intermediate, racemisation with excess inversion
Two refinements GATE likes. SNi (ROH + SOCl₂ without a base) gives retention through a front-side ion pair. Neighbouring group participation — an adjacent ester, sulfide, amine or π system attacking first — gives a rate acceleration that looks impossible for the substrate, plus net retention, because two inversions cancel. "Unexpectedly fast, retention of configuration" in a question means NGP.
Worked example 1 — reading a rate law. For a primary halide, k = 3.0 × 10⁻⁴ dm³ mol⁻¹ s⁻¹, [RX] = 0.050 mol dm⁻³, [Nu⁻] = 0.10 mol dm⁻³.
rate = k[RX][Nu⁻] = 3.0 × 10⁻⁴ × 0.050 × 0.10
= 3.0 × 10⁻⁴ × 5.0 × 10⁻³ = 1.5 × 10⁻⁶ mol dm⁻³ s⁻¹
Now double only [Nu⁻]: rate becomes 3.0 × 10⁻⁶ mol dm⁻³ s⁻¹, exactly twice. For a genuine SN1 substrate the same change would leave the rate unaltered. The units of k are the giveaway before you calculate anything: dm³ mol⁻¹ s⁻¹ means second order overall.
Worked example 2 — how much inversion? An optically pure substrate whose product would have [α] = +52.0° if formed with complete inversion is solvolysed. The isolated product shows [α] = +24.0° under identical conditions. What fraction went through each path?
Enantiomeric excess = 24.0 / 52.0 = 0.4615, i.e. 46.15% ee.
Let i = % inverted and r = % retained. Then i + r = 100 and
i − r = 46.15.
Adding: 2i = 146.15 → i = 73.08%, so r = 26.92%.
Interpretation: 2 × 26.92 = 53.85% of the material is racemic (an SN1 free-ion pathway) and the remaining 46.15% is cleanly inverted (SN2 or a tight ion pair). Check: 53.85 + 46.15 = 100 ✓.
2. Elimination, and the competition with substitution
E2 is concerted and needs the H and the leaving group anti-periplanar; that single requirement decides the answer in most conformationally locked questions. E1 shares the carbocation of SN1, so the two always appear together. E1cb operates when the β-hydrogen is acidic (β to a carbonyl or nitro group) and the leaving group is poor.
- Small, strong base (EtO⁻) → Zaitsev, the more substituted alkene.
- Bulky base (t-BuO⁻) or a bulky/charged leaving group (–NR₃⁺, –SR₂⁺) → Hofmann, the less substituted alkene.
- Heat favours elimination over substitution, because ΔS is positive for elimination.
In rigid cyclohexanes the H and the leaving group must both be axial. A substrate that cannot reach that arrangement eliminates towards the other side, or not at all — the standard menthyl/neomenthyl trap.
3. Carbocation rearrangements
Any mechanism with a cationic intermediate must be checked for a 1,2-shift: a hydride or alkyl group migrates with its bonding pair to the adjacent empty p orbital, but only if the new cation is more stable. Stability: benzylic ≈ allylic > 3° > 2° > 1° > methyl.
Worked example 3 — pinacol → pinacolone, step by step.
1. Protonate one –OH of 2,3-dimethylbutane-2,3-diol.
2. Lose water → a tertiary carbocation on C2.
3. A methyl group on C3 migrates to C2 with its electron pair. C3 now carries the
remaining –OH and the positive charge.
4. That cation is stabilised by the oxygen lone pair — it is a protonated ketone.
5. Lose H⁺ → 3,3-dimethylbutan-2-one (pinacolone).
The exam point is step 3: the driving force is not that the methyl "wants" to move, but that an oxygen-stabilised cation is far lower in energy than a plain tertiary one. Where the two carbons differ, migratory aptitude generally runs aryl > hydride ≈ alkyl, and electron-donating substituents on the migrating aryl ring accelerate it — but the stereoelectronic requirement (anti-periplanar alignment) can override the intrinsic order, so always draw the conformation before choosing.
4. Named rearrangements to an electron-poor atom
Group them by which atom the group migrates to, and the list stops being scary.
| Migration terminus | Rearrangement | What decides the product |
|---|---|---|
| Electron-poor carbon | Pinacol, Wagner–Meerwein | Which OH ionises; anti-periplanar group migrates |
| Electron-poor nitrogen | Beckmann (oximes) | The group anti to the leaving –OH migrates |
| Electron-poor nitrogen | Hofmann, Curtius, Schmidt, Lossen | Amine with one carbon fewer; isocyanate intermediate |
| Electron-poor oxygen | Baeyer–Villiger | Higher migratory aptitude group ends up on O |
| Carbene / carbenoid | Wolff (Arndt–Eistert) | Ketene intermediate; one-carbon homologation |
For Baeyer–Villiger oxidation the aptitude order is tertiary alkyl > cyclohexyl ≈ secondary alkyl ≈ benzyl ≈ phenyl > primary alkyl > methyl, so an unsymmetrical methyl ketone gives the acetate ester. In the Beckmann rearrangement the geometry of the oxime fixes the amide, not the size of the group — which is why oxime geometry is always specified in the question.
5. Carbonyl chemistry: addition, then a decision
Almost every classical condensation is "nucleophile adds to C=O, then something leaves or a proton shifts".
- Aldol — enolate adds to another carbonyl; heat or acid dehydrates to the enone. Crossed aldols need one partner with no α-H.
- Claisen condensation — enolate of an ester attacks another ester; the alkoxide leaves. Needs a full equivalent of base because the β-keto ester product is deprotonated.
- Cannizzaro — only for aldehydes with no α-H; hydride transfer gives one alcohol and one carboxylate.
- Michael addition — soft, stabilised nucleophile adds 1,4 to an enone; hard nucleophiles (RLi, RMgX) add 1,2.
- Wittig — the ylide fixes the alkene position with certainty; stabilised ylides tend to give E, non-stabilised ylides Z.
6. Aromatic substitution and pericyclic selection rules
In electrophilic aromatic substitution the intermediate is the arenium ion, and directing effects follow from whether the substituent can stabilise the positive charge at the ortho/para positions. Halogens are the standard exception: deactivating by induction, ortho/para directing by resonance. For nucleophilic aromatic substitution, choose between addition–elimination (needs strong electron-withdrawing groups ortho/para to the leaving group) and the benzyne route (very strong base, no activating group, mixture of positions).
Pericyclic reactions are pure bookkeeping once you count electrons:
Photochemical: the opposite in each case
Cycloaddition, thermal: [4+2] allowed suprafacial–suprafacial; [2+2] needs light
Sigmatropic: thermal [1,5]-H allowed suprafacially; [1,3]-H is not
- Skipping the rearrangement check. If your mechanism has a secondary carbocation next to a tertiary carbon, the examiner expects the shift.
- Quoting a rate law you did not derive. Order comes from the rate-determining step; do not assume "bimolecular" from the balanced equation.
- Ignoring anti-periplanarity in E2 and in migrations. Draw the Newman or the chair before choosing a product.
- Using Beckmann migratory aptitude. Beckmann is decided by geometry; Baeyer–Villiger is decided by aptitude. Swapping the two is a very common loss.
- Attempting a Cannizzaro on an aldehyde with α-hydrogens — it will aldolise instead.
- Curved arrows drawn from positive to negative. Arrows always start at an electron pair or a bond and point at the electron-poor atom.
How these appear in the paper
| Question type | What is really being tested |
|---|---|
| Predict the major product | Correct intermediate, then the lowest-energy path from it |
| Identify the intermediate | Carbocation, carbanion, carbene, nitrene, benzyne or arenium ion |
| Order the rates | Substrate structure, leaving group, solvent, base strength and bulk |
| Stereochemical outcome | Inversion, retention, racemisation, syn vs anti addition |
| Numerical (NAT) | Rate from a rate law, ee or optical purity, degree of unsaturation |
Take the syllabus and marking scheme for your session from the current official notification.
Do the numerical part without slips. Organic NAT questions still come down to arithmetic — a rate from a rate law, an enantiomeric excess, a percentage yield, a molar mass for a mass-to-mole step. The calculator suite has the scientific calculator, molar mass and mass ↔ mole tools together on one page, so you can check those steps in seconds.
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