Organic Mechanisms: Curly Arrows, Electrophiles, Nucleophiles
Draw curly arrows and identify electrophiles and nucleophiles in organic mechanisms.
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Mechanism questions reward a chemically consistent electron story. This lesson connects the prescribed reaction language, bond fission, reactive species and electronic effects so that curly arrows become consequences of structure rather than decorative marks.
Use it beside Representations and Nomenclature and return to the Organic Chemistry hub to apply the toolkit to each functional-group family.
Definitions (Must Know)
A. Core reaction language
- Addition: two species combine and no atom group is replaced.
- Substitution: one atom or group is replaced by another.
- Elimination: atoms or groups are removed to form a multiple bond.
- Condensation: two molecules join with loss of a small molecule.
- Hydrolysis: a bond is broken by reaction with water.
- Oxidation / reduction: in organic chemistry, track oxygen gain/hydrogen loss or oxygen loss/hydrogen gain, while using oxidation states when needed.
B. Bond fission and intermediates
- Homolytic fission: each atom takes one bonding electron, producing radicals.
- Heterolytic fission: one atom takes both bonding electrons, producing ions.
- Free radical: a species with an unpaired electron.
- Carbocation: an organic ion with a positively charged carbon centre.
C. Reagent roles
- Electrophile: an electron-pair acceptor; a Lewis acid.
- Nucleophile: an electron-pair donor; a Lewis base.
Primary, secondary and tertiary describe how many carbon groups are attached to the carbon centre being classified.
Key Ideas (What Earns Marks)
| Feature | What it tells you |
|---|---|
| lone pair, negative charge or pi bond | possible electron-pair source |
| positive charge or δ + centre | possible electron-pair destination |
| full-headed curly arrow | movement of an electron pair |
| single-headed arrow | movement of one electron in a radical mechanism |
| polar bond | possible heterolytic cleavage and ionic reaction |
| UV light | likely homolytic cleavage and radicals |
| stable carbocation | makes an SN1 or carbocation pathway more plausible |
| crowded reaction centre | can slow a one-step backside attack |
A curly arrow starts at the electron pair that moves: a lone pair, a negative charge-bearing atom or a bond. It ends at the atom receiving that pair or at the bond being formed. When a bond breaks heterolytically, the arrow starts on that bond.
Detailed Explanations
A. Electronic and steric effects
- The inductive effect transmits electron donation or withdrawal through sigma bonds and changes charge stability.
- The mesomeric effect arises when a lone pair, charge or pi bond is delocalised across adjacent p orbitals.
- A steric effect occurs when bulky groups impede approach to a reaction centre.
Use these effects to explain evidence; do not cite them as unexplained labels. For example, a more substituted carbocation is stabilised by electron-releasing alkyl groups, whereas crowding around a carbon centre hinders SN2 attack.
B. Mechanism-selection workflow
- Identify the functional group and the bond that changes.
- Assign the reagent’s role: radical source, electrophile, nucleophile, acid, base, oxidant or reductant.
- Use the solvent, temperature, catalyst and kinetic information.
- Choose the reaction family and likely intermediate.
- Draw every required electron movement.
- audit atoms, charge, lone pairs and catalyst regeneration.
C. One toolkit, several mechanisms
- Free-radical substitution: single-headed arrows; initiation, propagation and termination.
- Electrophilic addition: the alkene pi bond attacks an electrophile.
- Electrophilic substitution: the arene attacks an electrophile, then restores aromaticity.
- Nucleophilic substitution: a nucleophile attacks an electron-poor carbon as the leaving group departs, in one step or through a carbocation.
- Nucleophilic addition: a nucleophile attacks the carbonyl carbon and the pi pair moves to oxygen.
The arrow does not mean “this species attacks”; it identifies exactly which electrons move and where.
Worked Examples
Modelled example 1
Use the nitrogen lone pair to form a bond
Problem
Study the worked solution
Locate an available electron pair
Method
Identify the lone pair on nitrogen.Reason
This pair is available to form a new covalent bond.Working
Start the first full-headed arrow at the nitrogen lone pair and end it at the H atom of H-Cl.Apply the definition
Method
Classify ammonia as a nucleophile.Reason
A nucleophile donates an electron pair to an electron-deficient centre.Working
Start the second arrow at the H-Cl bond and end it at Cl. Draw NH₄ + with four N–H bonds and separate Cl⁻. Total charge is 0 = (+1) + (-1).
Guided practice 2
Classify the Proton
Problem
Classify by electron-pair role
Hints
Hint 1: electron state
Hint 2: definition
View solution step by step
Identify electron deficiency
Method
Recognise that H⁺ has no electron pair to donate.Reason
It can form a bond by receiving an electron pair from another species.Working
H⁺: electron-pair acceptor.State the classification
Method
Call H⁺ an electrophile.Reason
Electrophiles accept electron pairs.Working
H⁺ is an electrophile because it accepts an electron pair.
Guided practice 3
Transfer the arrow rules to oxygen
Problem
Construct before revealing
Sketch the species and arrows on paper. For each arrow, record its electron source and destination; then count O–H bonds and the total charge.
Hints
Hint 1: new bond
Hint 2: bond breaking
View solution step by step
Move both pairs
Method
Draw O lone pair → H, and H–Cl bond → Cl.Reason
The first pair forms the new O–H bond; the second breaks H–Cl heterolytically.Working
Both arrows have full heads. Neither starts at the positive end of H–Cl.Rebuild the products
Method
Draw water with two O–H bonds and a separate chloride ion.Reason
Oxygen gains one bond and becomes neutral; chlorine receives the departing bond pair and is negative.Working
OH⁻ + HCl → H₂O + Cl⁻. The charge remains -1 and every atom is retained.
Common misconception 4
Attack Site in Bromoethane
Learner claim
Track bond polarity and attraction
View solution step by step
Assign the bond polarity
Method
Label carbon δ + and bromine δ-.Reason
Bromine draws bonding electron density toward itself.Working
C\delta⁺-Br\delta⁻.Choose the attack site
Method
Attack the δ + carbon with the nucleophile.Reason
The nucleophile donates an electron pair to the electron-deficient centre.Working
Nucleophile → carbon bonded to Br.
Challenge 5
Fission of Chlorine
Bond-fission transfer
Read products and infer electron movement
Hints
Hint 1: product symbol
Hint 2: split
View solution step by step
Infer the electron split
Method
Give one bonding electron to each chlorine atom.Reason
Equal sharing produces two neutral radicals, each with one unpaired electron.Working
Cl-Cl → Cl. + .Cl.Name fission and condition
Method
State homolytic fission initiated by UV light.Reason
Photochemical energy breaks the halogen bond and starts a radical chain.Working
Cl₂ → [UV] 2Cl.: homolytic fission.
Common Mistakes
- Starting an arrow at H⁺, a carbocation or a δ + atom.
- Drawing the arrowhead towards a lone pair that is donating electrons.
- Using full-headed arrows for radical-chain steps.
- Omitting the arrow from a breaking bond to the atom that receives the pair.
- Drawing an intermediate without its charge, or exceeding a second-period atom’s octet.
- Claiming that tertiary always means SN1 or primary always means SN2 without considering solvent, nucleophile and conditions.
- Treating inductive, mesomeric and steric effects as interchangeable.
- Drawing arrows in an H1 8873 response when the question asks only for reagent, condition and product.
Exam Tips
- Label the nucleophile and electrophile when asked, but let the arrows prove their roles.
- Show lone pairs and charges wherever they are the source or destination of an arrow.
- If kinetic data are supplied, use the rate equation to test the proposed rate-determining step.
- When more than one product is possible, connect the major product to intermediate stability or steric accessibility.
- For catalytic mechanisms, show how the catalyst is regenerated.
- Finish by counting atoms and total charge on both sides of every step.
Mind Stretchers
Mind stretcher 1: Auditing an impossible arrowExtension
A proposed carbonyl mechanism draws an arrow from the δ + carbonyl carbon to CN⁻ and leaves the C = O bond unchanged. Diagnose both errors and give the correct electron-flow sequence.
Show Hint
Identify the electron-rich species first, then ask where the carbonyl pi pair must move when carbon gains a new bond.
Show Answer
The electron pair cannot start at the electron-poor carbon; it starts at the carbon lone pair/negative charge of CN⁻ and points to the carbonyl carbon. As the new carbon–carbon bond forms, a second full-headed arrow moves the C = O pi pair to oxygen. The resulting alkoxide is then protonated.
Mind stretcher 2: Using rate evidence to reject a mechanismExtension
Hydrolysis of a halogenoalkane has rate equation rate = k[RX] and is unchanged when hydroxide concentration doubles. Explain what this evidence says about the slow step and why a concerted bimolecular proposal is inconsistent.
Show Hint
A reactant present in the rate-determining elementary step normally appears in the rate equation.
Show Answer
Only the halogenoalkane concentration affects rate, so the slow step involves RX but not hydroxide. This is consistent with slow heterolytic C-X cleavage to a carbocation followed by fast nucleophilic attack. A concerted bimolecular step would involve both RX and OH⁻ and would normally predict hydroxide dependence.
Syllabus and review details
- GCE A-Level H2 Chemistry 9476-2027 · 9476-2027
9476 (2027), complete syllabus
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