Reaction Mechanisms, Part 1 · 10 sections ·
about 26,308 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM
Organic chemistry looks like thousands of unrelated reactions until you see that almost all of them are the same handful of moves, repeated. This part teaches those moves. It starts with the curly arrow — what it actually means, and the rules that make an arrow legal — and builds up to reading a reaction energy profile and predicting which of two products a reaction will give and why. By the end you should be able to look at an unfamiliar reaction and make a sensible guess at its mechanism, which is exactly what Part C of the exam asks you to do. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.
The 10 sections in Part 1
1The curly arrow — what it means and what makes one legal Free below
2Finding the nucleophile and the electrophile — from orbitals, not from memory
3The electronic toolkit — induction, resonance, hyperconjugation and sterics
4Reaction coordinate diagrams
5Intermediates versus transition states
6The rate-determining step, and the Hammond postulate
7Kinetic versus thermodynamic control
8Acidity and basicity — the master variable
9Hard and soft acids and bases in organic reactivity
10The map — what the remaining eight parts do
The curly arrow — what it means and what makes one legal
Free extract
Section A.1 of Part 1,
reproduced in full from the book — figures and all. No sign-in, no paywall on this section.
Beginner layer
The notation is the subject. A student who draws arrows correctly is already doing mechanism; a student who draws them decoratively is memorising pictures.
Sir Robert Robinson introduced the curved arrow in the 1920s, and it remains the most compact piece of notation in chemistry. It is also routinely misunderstood, because it looks like it is describing motion — something sliding from here to there — and students then draw it from whatever seems to be moving.
Curly arrow (full-headed, also called double-barbed): the movement of one pair of electrons. Its tail is placed on the pair that moves — a lone pair, or a bond. Its head is placed where that pair ends up — between two atoms if it becomes a bond, or on an atom if it becomes a lone pair.
Fishhook arrow (half-headed, also called single-barbed): the movement of a single electron. Used only for radical and single-electron-transfer mechanisms, which are Part 3 and parts of Part 8. Mixing the two notations in one scheme is a guaranteed lost mark. Note that ‘double-headed arrow’ is reserved for something else entirely — the straight, two-ended resonance arrow ↔ — so do not use it for the curly arrow.
Read the definition again and notice what it does not say. It does not say the arrow starts at an atom. It does not say the arrow starts at a charge. It does not say the arrow shows an atom moving. The arrow is about a specific pair of electrons, and you should be able to point at that pair in the structure you have drawn.
Every curly arrow is a claim about two electrons. Arrow 1 starts on the lone pair of oxygen and ends between O and C, because that pair becomes the new C–O bond. Arrow 2 starts on the C–Br bond, not on Br, because it is the bonding pair that moves; it ends on Br, which keeps them as a lone pair and leaves with a negative charge. Two arrows, two pairs, and the total charge is −1 before and after. Count charge on every mechanism you draw: it catches more errors than any other single check.
The rules that make an arrow legal
Five rules cover essentially everything. Learn them as a checklist and run through them the first fifty times you draw a mechanism; after that they become automatic.
Rule 1 — Start on electrons. The tail sits on a lone pair or on a bond. Never on a positive charge, never on an empty orbital, never on an atom symbol with nothing to give. Rule 2 — Point at where the pair ends up. The head lands between two atoms if the pair becomes a bond, or on an atom if it becomes a lone pair. The destination must be able to take it: an atom with a vacancy, or one that will simultaneously release a pair of its own. Rule 3 — Respect the octet. No second-period atom (C, N, O, F) may end a step with more than eight valence electrons. If your arrow would give carbon five bonds, another arrow must leave that carbon in the same step. Rule 4 — Conserve charge. The total charge on the left equals the total charge on the right, every step, without exception. Rule 5 — Conserve atoms. Curly arrows move electrons. If an atom has appeared or vanished between two structures, you have made an arithmetic error, not a chemical discovery.
Rule 3 is the one that does real work, so it deserves a sentence of its own. Carbon has four valence orbitals and no more. If a nucleophile is going to bond to a carbon that already has four bonds, one of those four bonds must break in the same step. That single constraint is why substitution at saturated carbon is a two-arrow process, why SN2 goes with inversion, and why SN1 needs a carbocation first. All of Part 4 is downstream of Rule 3.
Learn these four and you have the grammar. Nearly every step in the next eight parts is one of them, or two of them happening at once. (a) is addition to an electron-poor centre with a vacancy — BF₃, a carbocation, a metal. (b) is substitution. (c) is electrophilic addition to a π bond. (d) is deprotonation. What changes from reaction to reaction is which orbitals play the parts, not the moves themselves.
The errors, and how to catch them yourself
Arrows start where the electrons are, not where the charge is. The two drawings describe the same overall change, so an examiner marking only products would not separate them — but the left-hand one is a claim that electrons flow out of an empty orbital, which is not a thing that happens. Get into the habit of asking, for every arrow you draw: which specific pair of electrons is this? If you cannot name it — that lone pair, this π bond, that C–H σ bond — the arrow is wrong.
⚠ Common mistakes & exam traps
Drawing the arrow from a positive charge. A cation’s plus sign marks an absence of electrons. Nothing can flow out of it. The arrow must come from the nucleophile’s lone pair and point at the cationic centre. This is the single most common arrow error in scripts, and examiners look for it.
Arrows that break the octet. Attacking a neutral, four-coordinate carbon with a single arrow gives a five-bonded carbon. Either the leaving group departs in the same step (two arrows, SN2) or it left in a previous step (SN1). Similarly, protonating a carbonyl oxygen and then drawing the nucleophile onto a still-doubly-bonded carbon without breaking the π bond gives carbon five bonds.
Starting the arrow on the atom instead of on the bond. When a leaving group departs, the pair that moves is the C–LG bonding pair. Drawing the arrow from a lone pair already on the leaving group says something different and wrong: it would make the leaving group attack, not leave.
One arrow where two are needed in deprotonation. Removing a proton needs the base’s lone pair to attack H and the C–H (or O–H) bonding pair to collapse onto carbon (or oxygen). Drawing only the first arrow leaves hydrogen with four electrons.
Arrows pointing from the electrophile to the nucleophile. Electrons flow from electron-rich to electron-poor. If your arrow starts at the δ+ carbon of a carbonyl and points at the incoming amine, you have drawn the reaction backwards.
Treating an arrow as a statement about timing. Two arrows drawn in the same step assert that the two changes are part of one elementary event; they do not assert that the two bonds are half-made and half-broken to the same extent. That question — how far each bond has gone at the transition state — is a real and separate one, and it is what Sections A.6 and Part 5 are about.
Charge that does not balance. If you begin with a neutral molecule and an anion and end with two neutral species, an arrow is missing or an extra one has crept in. Check the charge on every step. It takes three seconds and catches most errors.
Push the arrows for the reaction of ammonia with a proton, and then explain why the product cannot react with a second proton. Easy
Locate the electrons. Nitrogen in NH₃ has three bonding pairs and one lone pair. That lone pair is the only thing available to move.
Locate the acceptor. H⁺ is a bare proton: a 1s orbital with no electrons at all. It is the perfect acceptor — empty, low-lying and small.
One arrow. Tail on the nitrogen lone pair; head between N and H. No other arrow is needed, because nothing has to break: the proton had no bonds to lose.
Check the rules. Nitrogen ends with four bonds and no lone pair — eight electrons, octet intact (Rule 3 satisfied). Charge: 0 + (+1) = +1 on the left, NH₄⁺ is +1 on the right (Rule 4 satisfied).
Why not a second proton. NH₄⁺ has no lone pair left. There are no electrons whose tail an arrow could sit on, so no legal arrow can be drawn. This is not a matter of charge repulsion being unfavourable — it is that the move cannot be written at all. Ammonium is not a base under any conditions.
A student draws the base-mediated formation of an enolate from acetone with a single arrow, from the C–H bond to the oxygen. What is wrong, and what is the correct arrow set? Medium
Count what has to happen. Three separate electron pairs move: the base picks up the proton, the C–H pair moves onto carbon, and — if you are drawing the delocalised enolate directly — that carbon pair pushes into the C=O π system while the π pair moves onto oxygen.
The student’s single arrow fails Rule 5 in spirit and Rule 3 in fact. A C–H bond pair arriving on oxygen leaves the proton stranded with no bond and no electrons, and it never says what removed it. The hydrogen has simply been abandoned.
Correct set, drawn as a single concerted step. Arrow 1: lone pair on the base → H. Arrow 2: the C–H bonding pair → into the C–C bond region (forming the new π bond). Arrow 3: the C=O π pair → onto oxygen.
Alternative, equally correct. Draw only arrows 1 and 2, giving the localised carbanion, then a separate resonance arrow set converting it to the O-anion. Two resonance forms of one species — not two steps of a mechanism. Use the double-headed straight resonance arrow, not the reaction arrow, or you have claimed something false.
Check. Charge: base (−1) + ketone (0) = −1; conjugate acid (0) + enolate (−1) = −1. Octets: carbon ends with three bonds and a lone pair or with a π bond — eight electrons either way. Oxygen ends with one bond and three lone pairs — eight.
Advanced / reference layer
What the arrow is a cartoon of, what it deliberately leaves out, and the three situations in which the notation genuinely fails.
The beginner layer treats the curly arrow as bookkeeping. It is bookkeeping, but it is bookkeeping for something, and knowing what makes the difference between using the notation and understanding it.
A chemical reaction is a continuous deformation of one electron-density distribution into another along a path on a potential-energy surface. Nothing anywhere in that description is discrete, and electrons are not localised in pairs that travel from place to place. What the curly arrow encodes is the dominant orbital interaction driving that deformation: a filled donor orbital on one fragment mixing with an empty acceptor orbital on the other, so that in the product the pair which occupied the donor now occupies a bonding combination that did not previously exist.
A curly arrow is a two-orbital interaction drawn as a cartoon. A filled donor orbital (the HOMO — a lone pair, a π bond, sometimes a σ bond) overlaps an empty acceptor orbital (the LUMO — an empty p orbital, a π*, a σ*). The pair drops into the new bonding combination; the antibonding combination goes up and stays empty; net stabilisation ∆E. Second-order perturbation theory says ∆E grows as the square of the overlap and falls as the energy gap ∆ε widens — which is why a high-lying HOMO and a low-lying LUMO make a fast pair, and why this one picture underlies nucleophilicity, HSAB and frontier-orbital selectivity alike.
Two things follow immediately from that expression, and both are used constantly in the rest of this book. Stabilisation grows with the square of the overlap — so geometry matters enormously, and a donor orbital pointing the wrong way donates nothing. And stabilisation grows as the energy gap shrinks — so a high-energy HOMO and a low-energy LUMO react fast even when neither carries much charge. Section A.2 turns this into a working method; Section A.9 turns it into HSAB.
What the arrow does not say
Three silences in the notation are responsible for most of the confusion in the literature, and being explicit about them is worth a great deal.
It does not fix the extent of bond formation at the transition state. Two arrows in one step assert concertedness — one elementary event, one transition state — and nothing more. An SN2 reaction with a very good leaving group and a weak nucleophile has a transition state in which C–LG is nearly broken and C–Nu barely formed. It is still drawn with the same two arrows. The extent of each bond change is described by the position of the transition state (A.6) and by More O’Ferrall–Jencks analysis, not by the arrows.
It does not identify the rate-determining step. A mechanism with four arrow sets does not tell you which of the four steps is slow. That comes from the energy profile (A.4–A.6) and from kinetics (Part 9).
It is not unique. Many reactions can be drawn with more than one legal arrow set that gives the same product — most commonly, a proton transfer written before rather than after a nucleophilic attack. Choosing between them is a matter of evidence (often a pH–rate profile or an isotope effect), not of drawing.
Where the notation genuinely fails
Situation
Why arrows struggle
What is used instead
Pericyclic reactions
The arrows can be drawn, and they give the right product — but they are arbitrary: the same Diels–Alder can be drawn with arrows going clockwise or anticlockwise, and neither direction means anything physical. The arrows also fail to predict the crucial facts — allowedness, stereospecificity, the photochemical reversal.
Orbital symmetry: Woodward–Hoffmann correlation diagrams, FMO analysis, or the aromatic-transition-state (Hückel/Möbius) count. Part 6 develops this.
Radical and SET processes
Pairs do not move; single electrons do. Two-electron notation is simply the wrong notation.
Fishhook (half-headed) arrows, and spin-density arguments. Part 3.
Hypervalent and three-centre bonding
A 3-centre–4-electron bond in a phosphorane or a trihalide is not constructible from localised two-centre pairs, so no arrow set describes the bonding correctly.
Delocalised MO description; the arrows are then used only as a bookkeeping convenience across the step.
Metal-catalysed steps
Oxidative addition, migratory insertion and reductive elimination involve d orbitals and formal oxidation-state changes at the metal; curly arrows are at best an analogy.
Elementary-step formalism and electron counting — the coordination-chemistry and organometallic treatment.
Very short-lived ‘intermediates’
If a putative intermediate would live less than one bond vibration (≈10⁻¹³ s) it is not a chemical species, so a mechanism drawn through it is misleading even though every arrow is legal.
The lifetime criterion and enforced-concerted analysis (A.5, and Part 4 on borderline solvolysis).
None of these is a reason to abandon curly arrows — they remain the most efficient reasoning tool in organic chemistry. They are a reason to know that the notation is a model with a domain of validity, which is exactly the kind of thing a Part C question likes to probe.
Note: A useful discipline when you meet an unfamiliar reaction: draw the arrows, then ask “could I have drawn a different legal set?” If yes, you have found a genuine mechanistic question, and somewhere in the literature there is an experiment that settled it. That is how most of Part 9 came to exist.
Med
Ozone is drawn with a formal positive charge on the central oxygen and a formal negative charge on one terminal oxygen. A student, wanting to show ozone attacking an alkene, draws an arrow starting at the positive charge on the central oxygen. Diagnose the error and state which arrow is correct for the first step of ozonolysis.
Show solution
The error is Rule 1: an arrow cannot start at a formal charge. Worse, the central oxygen of ozone is the electron-poor end — it bears the positive formal charge precisely because it has three bonds and only one lone pair. It is an acceptor, not a donor. For the first step of ozonolysis (a 1,3-dipolar cycloaddition), the correct arrow set has the alkene π pair attacking the electrophilic terminal oxygen — the one drawn doubly bonded in the O⁻–O⁺=O form — the ozone π pair shifting onto the central oxygen, and the negatively charged terminal oxygen’s lone pair forming the second new C–O bond — three arrows in a closed cycle, which is the signature of a concerted cycloaddition. Note the general lesson: formal charge is a bookkeeping label, not a reliable guide to where the electrons actually are. In ozone the terminal oxygens carry the electron density and the central oxygen does not, exactly opposite to what a quick glance at the formal charges suggests.
Med
Draw the complete arrow set for the acid-catalysed hydration of a ketone, R₂C=O + H₂O → R₂C(OH)₂, and count the arrows.
Show solution
Step 1 (protonation, 1 arrow). A lone pair on the carbonyl oxygen → the proton of H₃O⁺. Simultaneously (a second arrow if you draw H₃O⁺ explicitly) the O–H bonding pair of H₃O⁺ → onto its oxygen, releasing water. This gives the protonated ketone, which is best drawn as the resonance hybrid with substantial positive charge on carbon. Step 2 (nucleophilic addition, 2 arrows). A lone pair on water → the carbonyl carbon; the C=O π pair → onto oxygen. Two arrows are compulsory here: one arrow alone would give carbon five bonds and break Rule 3. Product is the protonated hydrate, an oxocarbenium-derived cation bearing +OH₂. Step 3 (deprotonation, 2 arrows). A lone pair on a water molecule → one of the O–H protons; that O–H bonding pair → onto oxygen. Gives the neutral hydrate and regenerates H₃O⁺. Count: six arrows across three steps — 2 + 2 + 2 (five, if you write the proton as ‘H⁺’ and do not draw the acid explicitly, which removes the second arrow of step 1). Charge is +1 throughout, and the catalyst is regenerated — both checks that the scheme is complete. The general pattern protonate, add, deprotonate is the backbone of most of Part 7.
Hard
Explain, in orbital terms, why a curly arrow drawn from the lone pair of chloride to a tertiary carbocation is a sensible mechanistic statement, whereas one drawn from the lone pair of chloride to a tertiary alkane is not — even though both would ‘form a C–Cl bond’.
Show solution
The question is what the acceptor orbital is, and how it lies in energy. Carbocation. The acceptor is a genuinely empty 2p orbital on an sp²-hybridised carbon. It is non-bonding, so it lies low — well below the σ* manifold — giving a small energy gap to the chloride lone pair, and it is sterically exposed on both faces. Large overlap integral, small ∆ε, so ∆Estab is large. The interaction is strongly stabilising and no bond need break. Legal by every rule, and fast — typically diffusion-limited. Alkane. There is no empty orbital of comparable energy. The lowest available acceptor is a σ*(C–C) or σ*(C–H) orbital, which lies very high (that is what makes alkanes unreactive), so ∆ε is large and ∆Estab is negligible. Worse, donating into σ* means breaking the corresponding σ bond, so a second arrow is compulsory — and the leaving group would have to be a carbanion or a hydride, both catastrophically bad leaving groups. The step is not forbidden by the notation; it is forbidden by the energetics, and the notation quietly shows you why the moment you are made to write the second arrow. This is the general test: if the acceptor orbital is σ*, ask what leaves, and ask whether that species can survive.
Read the rest of Part 1
The remaining 9 sections of this part —
Finding the nucleophile and the electrophile — from orbitals, not from memory, The electronic toolkit — induction, resonance, hyperconjugation and sterics, Reaction coordinate diagrams… — and all nine parts of
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