CSIR-NET Named Reactions and Rearrangements — One Principle, Many Names
Most candidates try to memorise named rearrangements as a list of fifty separate facts, and most candidates then mix them up under exam pressure. There is a much smaller thing to learn. The large majority of the rearrangements CSIR-NET asks about are the same reaction — a 1,2-shift of a group, with its bonding electron pair, onto an adjacent electron-deficient atom. Once you know which atom becomes electron deficient and how, the product follows, and the names become labels rather than facts.
The one principle
The electron-deficient atom is created in one of three ways, and that is the entire classification:
| Deficient atom | How it is created | Rearrangements in this family |
|---|---|---|
| Carbon (carbocation) | Loss of water or another leaving group | Pinacol–pinacolone, Wagner–Meerwein, Demjanov, dienone–phenol, retropinacol |
| Nitrogen (nitrene / nitrenium-like) | Loss of a leaving group from N | Hofmann, Curtius, Lossen, Schmidt, Beckmann (N of the oxime) |
| Oxygen (electron-poor O of a peroxide) | O–O bond heterolysis | Baeyer–Villiger, Dakin, Criegee, hydroperoxide (cumene process) |
Retention at the migrating carbon is a single fact that answers a whole class of stereochemistry questions, in Hofmann, Curtius, Schmidt and Baeyer–Villiger alike. If a chiral migrating group is drawn as (R) in the substrate, it is (R) in the product.
The reference table
| Name | Substrate → product | The point examiners test |
|---|---|---|
| Beckmann | Ketoxime → amide (or lactam from a cyclic ketoxime) | The group anti to the departing OH migrates, so E/Z geometry decides the product |
| Hofmann | Primary amide + Br2/OH− → amine, one carbon shorter | Isocyanate intermediate; migration with retention |
| Curtius | Acyl azide → (Δ) isocyanate → amine, one carbon shorter | No oxidant needed; the nitrene-like N is made by losing N2 |
| Lossen | Activated hydroxamic acid → isocyanate → amine | Same family; recognise it by the O-acyl hydroxamate |
| Schmidt | Carboxylic acid + HN3 → amine (one C shorter); ketone + HN3 → amide | Which substrate you started from changes the product class |
| Baeyer–Villiger | Ketone + peracid → ester; cyclic ketone → lactone | Oxygen inserts on the side of the more substituted group |
| Pinacol–pinacolone | 1,2-diol + H+ → ketone | The OH giving the more stable cation leaves first; then the best migrating group shifts |
| Wagner–Meerwein | Carbocation → rearranged carbocation | Skeletal change, usually towards a more stable or less strained cation |
| Wolff (Arndt–Eistert) | α-Diazoketone → ketene → acid, one carbon longer | The one common rearrangement that adds a carbon |
| Favorskii | α-Halo ketone + base → ester or acid | Cyclopropanone (or oxyallyl) intermediate; two different α-halo ketones can give the same product |
| Benzilic acid | 1,2-diketone + OH− → α-hydroxy acid | Aryl migrates to the adjacent carbonyl carbon |
| Fries | Aryl ester + Lewis acid → hydroxyaryl ketone | Low temperature favours the para product, high temperature the ortho |
| Claisen (rearrangement) | Allyl vinyl ether → γ,δ-unsaturated carbonyl; allyl aryl ether → o-allylphenol | [3,3] sigmatropic — do not confuse it with the Claisen condensation |
| Cope / oxy-Cope | 1,5-diene → isomeric 1,5-diene | Also [3,3]; the oxy variant is driven by tautomerisation to a carbonyl |
| Stevens / Sommelet–Hauser | Quaternary ammonium ylide → rearranged amine | [1,2] versus [2,3] pathways from the same kind of ylide |
| Hofmann–Löffler–Freytag | N-haloamine → pyrrolidine | Remote functionalisation by 1,5-hydrogen atom transfer |
Worked example 1 — Beckmann, and why geometry decides everything
Acetophenone oxime, C6H5C(CH3)=N–OH, treated with acid.
Step 1. The OH is protonated and leaves as water, making the nitrogen electron deficient.
Step 2. The group anti (trans) to the departing oxygen migrates from carbon to nitrogen — the migration and the departure are concerted, so only the anti-periplanar group can move.
Step 3. Water adds to the resulting nitrilium ion and tautomerises to the amide.
The two possible answers.
If the OH is anti to phenyl: phenyl migrates to N, giving
C6H5NH–CO–CH3, acetanilide.
If the OH is anti to methyl: methyl migrates, giving
C6H5CO–NH–CH3,
N-methylbenzamide.
Both products are C8H9NO, so a formula check will not separate them — only the oxime geometry does. A question that draws the oxime carefully is telling you it wants this reasoning; one that does not specify geometry cannot have a single correct answer, and saying so is a legitimate response.
Worked example 2 — carbon bookkeeping in the Hofmann rearrangement
Butanamide, CH3CH2CH2CONH2, with Br2 and aqueous NaOH.
Mechanism in four moves: N-bromination → deprotonation to the N-bromoamide anion → loss of bromide with concerted migration of the propyl group from C to N, giving the isocyanate CH3CH2CH2N=C=O → hydrolysis of the isocyanate, losing CO2.
Product: propan-1-amine, CH3CH2CH2NH2.
Count the atoms and check.
Butanamide C4H9NO:
(4 × 12.011) + (9 × 1.008) + 14.007 + 15.999
= 48.044 + 9.072 + 14.007 + 15.999 =
87.12 g mol−1
Propylamine C3H9N:
(3 × 12.011) + (9 × 1.008) + 14.007 = 36.033 + 9.072 + 14.007 =
59.11 g mol−1
Four carbons in, three carbons out — the carbonyl carbon leaves as CO2. That one-carbon loss is shared by Hofmann, Curtius, Lossen and the acid version of Schmidt, and it is the fastest way to eliminate wrong options in a multiple-choice question: count the carbons in each option first.
The mirror image of this idea: the Arndt–Eistert sequence, whose key step is the Wolff rearrangement, adds one carbon to a carboxylic acid. If a question shows a chain growing by one, think Arndt–Eistert; shrinking by one, think the nitrogen family.
Worked example 3 — Baeyer–Villiger regiochemistry
2-Methylcyclohexanone treated with a peracid.
Step 1. The peracid adds to the carbonyl to give the tetrahedral Criegee intermediate.
Step 2. The O–O bond breaks, making the oxygen electron deficient, and one of the two carbon groups migrates from carbon to that oxygen.
Step 3. Which one? The migratory aptitude order usually quoted for this
reaction is
tertiary alkyl > cyclohexyl ≈ secondary alkyl ≈ benzyl > primary alkyl > methyl.
Here the choice is between the secondary carbon bearing the methyl group and the
primary ring carbon on the other side. The more substituted carbon migrates.
Product: the seven-membered lactone in which the new oxygen sits between the carbonyl carbon and the methyl-bearing carbon — 7-methyloxepan-2-one. The methyl-bearing stereocentre is untouched, because migration proceeds with retention.
An honest caution. Migratory aptitude lists differ slightly between textbooks, and the order is not an intrinsic property of a group — it depends on the reaction, the electron demand at the migration terminus and the conformation of the intermediate. In the Baeyer–Villiger the electron-poor oxygen strongly rewards the group best able to stabilise positive charge, which is why the substitution order holds so reliably here. Do not carry the same list unchanged into a pinacol rearrangement.
Worked example 4 — pinacol, and the two-decision rule
Pinacol, (CH3)2C(OH)–C(OH)(CH3)2, with acid.
Decision 1 — which OH leaves? The one giving the more stable carbocation. Both carbons here are equivalent and tertiary, so it does not matter; in an unsymmetrical diol it decides everything.
Decision 2 — which group migrates? A methyl shifts from the adjacent carbon to the cationic centre, giving a new cation stabilised by the neighbouring oxygen lone pair, (CH3)3C–C+(OH)CH3. Loss of a proton gives the ketone.
Product: pinacolone, (CH3)3C–CO–CH3, 3,3-dimethylbutan-2-one.
The trap in unsymmetrical cases. Students apply the migratory-aptitude list first. Do not — decision 1 comes first and often overrides it. In 1,1-diphenylethane-1,2-diol, Ph2C(OH)–CH2OH, the benzylic OH leaves because that cation is far more stable, and then a hydride shifts from the CH2OH carbon, giving Ph2CH–CHO. Phenyl, supposedly the best migrating group, does not migrate at all — because it is on the wrong carbon.
Pericyclic reactions — the selection rules in one table
These are the second large family of "named" processes, and here the answer is a rule, not a memory. All rules below are for the neutral, all-carbon case.
| Class | Electron count | Thermal | Photochemical |
|---|---|---|---|
| Electrocyclic | 4n (e.g. butadiene, 4 e−) | Conrotatory | Disrotatory |
| Electrocyclic | 4n + 2 (e.g. hexatriene, 6 e−) | Disrotatory | Conrotatory |
| Cycloaddition [4+2] | 6 e− | Allowed, suprafacial–suprafacial (Diels–Alder) | Forbidden in that mode |
| Cycloaddition [2+2] | 4 e− | Forbidden suprafacial–suprafacial | Allowed |
| Sigmatropic [1,3]-H | 4 e− | Suprafacial forbidden | Suprafacial allowed |
| Sigmatropic [1,5]-H | 6 e− | Suprafacial allowed — very common | Antarafacial |
| Sigmatropic [1,7]-H | 8 e− | Antarafacial (the vitamin D case) | Suprafacial |
| Sigmatropic [3,3] | 6 e− | Allowed — Cope and Claisen, usually through a chair-like transition state | — |
The Diels–Alder carries two extra rules that are examined separately: the diene must reach the s-cis conformation, and the endo adduct is the kinetic product while the exo is usually the thermodynamic one, so a question specifying low temperature and short time wants endo.
Bond-forming named reactions worth knowing cold
- Aldol, Claisen condensation, Mannich, Michael, Knoevenagel, Robinson annulation — all enolate chemistry. The Robinson annulation is simply a Michael addition followed by an intramolecular aldol condensation.
- Wittig: a stabilised ylide (bearing an ester or ketone) gives the E alkene; a non-stabilised ylide gives the Z alkene. The Horner–Wadsworth–Emmons variant, using a phosphonate, gives E selectively and the water-soluble phosphate by-product is easier to remove.
- Birch reduction: with an electron-donating substituent (anisole, toluene) the substituted carbon is not reduced and you obtain the 1-substituted cyclohexa-1,4-diene; with an electron-withdrawing substituent (benzoic acid) the substituted carbon is reduced, giving the cyclohexa-2,5-diene with the substituent on sp³ carbon. This pair is asked almost as often as any rearrangement.
- Cannizzaro: a non-enolisable aldehyde disproportionates in strong base to the alcohol and the carboxylate; with two different aldehydes and excess formaldehyde it becomes the crossed Cannizzaro.
Mistakes that cost marks
- Claisen condensation vs Claisen rearrangement. Same name, unrelated chemistry — one is enolate acylation, the other a [3,3] sigmatropic shift. Read which one the question means before answering.
- Forgetting the carbon count. Hofmann, Curtius, Lossen and the acid Schmidt all remove one carbon; Arndt–Eistert adds one; Beckmann and Baeyer–Villiger keep the count and insert a heteroatom.
- Drawing inversion at the migrating carbon. These 1,2-shifts are concerted and go with retention. Inversion is an SN2 idea and does not belong here.
- Ignoring oxime geometry in the Beckmann. The group anti to the leaving oxygen migrates. Without the geometry there is no unique product.
- Treating migratory aptitude as a universal constant. It is reaction-dependent, and in the pinacol rearrangement it is subordinate to which cation forms first.
- Applying electrocyclic rules by ring size instead of electron count. Count the π electrons in the open-chain partner, then use 4n or 4n + 2.
- Assuming a thermally forbidden reaction cannot happen at all. "Forbidden" means forbidden as a concerted pericyclic process in that topology; a stepwise radical or ionic route may still operate, and thermal [2+2] reactions of ketenes proceed through an allowed antarafacial component.
Where this appears in the paper
| Exam | Typical task |
|---|---|
| CSIR-NET Chemical Sciences | Predict-the-product items, identify the intermediate, stereochemical outcome of a migration, pericyclic mode assignment, multi-step synthesis sequences |
| GATE Chemistry | Named reactions and their reagents, mechanism-based product prediction |
| IIT-JAM / CUET-PG | Common named reactions, Cannizzaro, aldol, Wittig, basic rearrangements |
| CBSE Class 12 | A smaller set — Cannizzaro, aldol, Hofmann bromamide, Wurtz, Sandmeyer |
CSIR-NET is organised into Part A, Part B and Part C; the number of questions, the marks and the negative-marking rules for each part change between sessions, so read the official notification for yours rather than relying on any secondary summary, this one included.
Build one page per family, not one card per name. Group the rearrangements by which atom becomes electron deficient — carbon, nitrogen, oxygen — and write the carbon-count change beside each. Verifying molecular formulae as you go, as in worked example 2, catches wrong options quickly. There is no named-reaction or mechanism tool in the suite, so this button honestly opens the suite itself rather than pretending a matching calculator exists — the Molar Mass & Composition tool linked below is the one part of this article a calculator can genuinely speed up, since it confirms in seconds whether your proposed product really is the substrate minus CO2 or plus an oxygen.
Open the ABC Chemistry Calculator Suite →Preparing for CSIR-NET, GATE, IIT-JAM or CUET-PG chemistry? ABC Chemistry runs dedicated competitive-exam batches at its coaching centre and fully online for students across India — details at abcchemistry.in.