JAM Organic — The Core Reaction Mechanisms
Organic chemistry in IIT-JAM is mostly product prediction, and product prediction is mostly mechanism. Know which intermediate a reagent generates and the product usually follows with no memorisation. The list of intermediates is short: carbocation, carbanion or enolate, free radical, carbene, and the bridged ions of electrophilic addition. This article works through the mechanisms built on them, plus the two numerical skills that accompany them.
Step zero: read the molecular formula
C = carbons, H = hydrogens, N = nitrogens, X = halogens. Oxygen and sulfur are ignored.
Each degree is one ring or one π bond; a benzene ring is four (three π bonds plus the ring). Before drawing anything, calculate it — it eliminates most wrong options instantly.
Worked example 1.
C₈H₈O: DoU = (2 × 8 + 2 − 8) / 2 = (16 + 2 − 8) / 2 = 10/2 = 5. Four of those are a benzene ring and the fifth is very likely a C=O — consistent with acetophenone.
C₄H₇Cl: DoU = (2 × 4 + 2 − 7 − 1) / 2 = (8 + 2 − 8)/2 = 2/2 = 1 — one ring or one double bond.
C₆H₅NO₂: DoU = (2 × 6 + 2 + 1 − 5) / 2 = (12 + 2 + 1 − 5)/2 = 10/2 = 5 — a benzene ring plus one more π bond, i.e. nitrobenzene. Note that the two oxygens play no part in the count.
1. Electrophilic addition to alkenes
The π bond is the nucleophile. It attacks an electrophile, giving a carbocation or a bridged ion, and a nucleophile then completes the addition. Everything about regio- and stereochemistry falls out of which of those two intermediates forms.
| Reagent | Intermediate | Outcome |
|---|---|---|
| HX | Open carbocation | Markovnikov; rearrangement possible; not stereospecific |
| HBr with peroxides | Free radical | Anti-Markovnikov — HBr only, not HCl or HI |
| Br₂ (in CCl₄) | Bromonium ion | Anti addition |
| Br₂ in water | Bromonium ion | Halohydrin; OH lands on the more substituted carbon |
| BH₃ then H₂O₂/OH⁻ | Concerted four-centre TS | Anti-Markovnikov, syn, no rearrangement |
| Hg(OAc)₂/H₂O then NaBH₄ | Mercurinium ion | Markovnikov, no rearrangement |
| cold dilute KMnO₄ or OsO₄ | Cyclic ester | syn diol |
| O₃ then Zn/H₂O | Ozonide | Cleavage to two carbonyls |
Memorise as a pair: hydroboration–oxidation and oxymercuration–demercuration give opposite regiochemistry and both avoid rearrangement, whereas plain HX does not. An alkene able to rearrange is in the question precisely to test that.
Ozonolysis is also a structure-determination tool — join the two carbonyl carbons of the products with a double bond and you have the starting alkene.
2. Substitution and elimination at saturated carbon
Four pathways compete and the substrate usually settles it: methyl and primary favour SN2, tertiary favours SN1 or E1, and a strong bulky base pushes any of them towards E2. SN2 inverts the centre; SN1 racemises with some excess inversion; E2 needs the hydrogen and the leaving group anti-periplanar.
- Strong nucleophile that is a weak base (I⁻, CN⁻, RS⁻) → substitution.
- Strong, bulky base (t-BuO⁻) → elimination, and the less substituted (Hofmann) alkene.
- Small strong base (EtO⁻) with a secondary or tertiary halide → the more substituted (Zaitsev) alkene.
- Heat always favours elimination, because ΔS is positive when one molecule becomes two.
Polar protic solvents stabilise carbocations and so help SN1; polar aprotic solvents leave anions poorly solvated and therefore more reactive, helping SN2. A dedicated article on the SN1/SN2 comparison is linked below.
3. Carbonyl chemistry — the largest single block
A carbonyl carbon is electrophilic; the α-hydrogen next to it is acidic. Every classical reaction here uses one of those two facts.
Worked example 2 — the aldol reaction, step by step. Ethanal with dilute NaOH.
1. Hydroxide removes an α-hydrogen, giving a resonance-stabilised enolate — the negative
charge is shared between the α-carbon and the oxygen.
2. The enolate carbon attacks the carbonyl carbon of a second molecule of ethanal.
3. The resulting alkoxide takes a proton from water, giving
3-hydroxybutanal — the aldol.
4. On warming, or with acid, it loses water to give but-2-enal, because the new double bond is
conjugated with the carbonyl and that conjugation is the driving force.
The exam variation is the crossed aldol: to get a single product, one partner must have no α-hydrogen (benzaldehyde, formaldehyde) so it can only act as the electrophile.
The rest of the block, compressed:
- Cannizzaro — for aldehydes with no α-hydrogen and concentrated alkali. Hydroxide adds to the carbonyl, then a hydride transfers to a second molecule: one is oxidised to the carboxylate, the other reduced to the alcohol.
- Claisen condensation — the ester enolate attacks a second ester and the alkoxide leaves, giving a β-keto ester.
- Grignard and organolithium — carbanion-like, so they add 1,2 to a carbonyl. They are destroyed by any acidic hydrogen, so a substrate carrying –OH, –NH or –COOH must be protected first.
- Michael addition — stabilised carbanions (malonate, enolates) add 1,4 to an α,β-unsaturated carbonyl.
- Wittig — the phosphorus ylide places the C=C exactly where the C=O was, with no ambiguity about the position.
- Reduction choices — NaBH₄ reduces aldehydes and ketones but not esters or acids; LiAlH₄ reduces essentially all of them. Choosing the wrong one is a common trap.
- Identification tests — Tollens' and Fehling's for aldehydes, iodoform for a methyl ketone or a CH₃CH(OH)– group, 2,4-DNP for any aldehyde or ketone.
4. Aromatic substitution
In electrophilic aromatic substitution the ring attacks the electrophile to give an arenium ion, and a proton is then lost to restore aromaticity. Directing effects follow from whether the substituent can stabilise the positive charge at the ortho and para positions. Activating groups (–OH, –OR, –NH₂, –R) are ortho/para directing; deactivating groups (–NO₂, –CN, –COR, –SO₃H) are meta directing. The halogens are the standard exception: deactivating, yet ortho/para directing. For nucleophilic aromatic substitution, strong electron-withdrawing groups ortho or para to the leaving group allow the addition–elimination route, while a very strong base with no activating group gives the benzyne pathway and therefore a mixture of positions.
Worked example 3 — the numerical that accompanies a mechanism. 5.00 g of benzaldehyde undergoes a crossed Cannizzaro. How much benzyl alcohol is obtained at 78% yield? (M: C₇H₆O = 106.12, C₇H₈O = 108.14 g mol⁻¹)
moles of benzaldehyde = 5.00 / 106.12 = 0.04712 mol
The stoichiometry is 1 : 1 for the alcohol formed from the aldehyde that is reduced, so
theoretical yield = 0.04712 × 108.14 = 5.10 g
Actual = 5.10 × 0.78 = 3.97 g
Sanity check: the product mass is slightly larger than the starting mass per mole because two hydrogens have been added — 108.14 against 106.12 — so a theoretical yield below 5.00 g would have signalled an error.
- Applying the peroxide effect to HCl or HI. Anti-Markovnikov radical addition works for HBr only; the bond energies do not permit the chain for the others.
- Forgetting anti addition with Br₂. The bromonium ion forces the second bromide to attack from the opposite face.
- Allowing a rearrangement in hydroboration or oxymercuration. Neither forms a free carbocation, so neither rearranges.
- Running a Cannizzaro on an aldehyde that has α-hydrogens — it will undergo an aldol reaction instead.
- Using a Grignard reagent on a substrate with an –OH or –NH group without protection.
- Ignoring the DoU. Calculating it first removes most wrong options before you draw a single structure.
- Curved arrows drawn backwards. Arrows start at an electron pair or a bond and point towards the electron-poor atom, never the reverse.
Exam relevance
| Question style | What to check first |
|---|---|
| Predict the major product | Which intermediate the reagent generates |
| Identify a compound from its reactions | DoU, then the positive identification tests |
| Order reactivity or acidity | Stability of the intermediate or of the conjugate base |
| Stereochemistry of the product | syn or anti addition; inversion or racemisation |
| Multiple-select (MSQ) | Each option separately — partial credit rules vary, so read the instructions on the paper |
| Numerical (NAT) | DoU, yield, molar mass, enantiomeric excess |
Take the syllabus, marking scheme and question types for your session from the current official notification.
Do not lose an organic mark to arithmetic. Degree of unsaturation, molar masses for a yield calculation and percentage composition are all quick in the calculator suite — the molar mass tool accepts any formula and the scientific calculator handles the rest, on the same page.
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