This set of notes covers how to draw arrow-pushing (curved-arrow) mechanisms for SN2 and SN1 reactions and how to sketch their reaction energy coordinate diagrams. SN2 has one step and one transition state. SN1 has two steps, a carbocation intermediate, and two transition states.
Difficulty: Intermediate | Prerequisites: Familiarity with SN1/SN2 reactivity factors (see companion notes), Lewis structures, lone pairs, formal charges.
Knowing that a reaction is SN2 or SN1 is only the first step. The exam will ask you to show how it happens: which electrons move where (arrow pushing) and what the energy landscape looks like along the way (reaction coordinate diagram). These two skills are how you demonstrate that you understand the mechanism, not just the label.
Arrow-pushing is the universal language of organic chemistry. Every mechanism you learn from here forward builds on the same curved-arrow conventions. Energy diagrams, meanwhile, tie the mechanism to thermodynamics and kinetics: where are the transition states, are there intermediates, and is the overall reaction exothermic or endothermic?
Arrow pushing (curved-arrow mechanism)
A notation system in which curved arrows show the movement of electron pairs during a reaction. The arrow tail starts at the electron source (lone pair or bond) and the arrow head points to the electron destination (an atom or a bond).
In simple terms: arrows track where the electrons go, not where the atoms go.
Transition state
A fleeting, maximum-energy arrangement of atoms along the reaction pathway. It cannot be isolated. Drawn in square brackets with a double-dagger symbol (‡).
Think of it as the top of the energy hill the reaction must climb over.
Intermediate
A species that forms during a multi-step reaction and exists in an energy well between two transition states. Unlike a transition state, an intermediate has a finite (though often short) lifetime. In SN1, the carbocation is the intermediate.
In simple terms: a real, if fleeting, molecule that sits in a valley between two energy peaks.
Reaction coordinate diagram (energy diagram)
A plot of free energy (y-axis) against reaction progress (x-axis) that shows starting materials, transition states, intermediates (if any), and products. The height of each peak is the activation energy (Ea) for that step.
Think of it as the topographic profile of the reaction's energy landscape.
Activation energy (Ea)
The energy difference between the starting material (or intermediate) and the transition state of a given step. A higher Ea means a slower step.
In simple terms: the size of the energy barrier the molecules must overcome to react.
Concerted reaction
A reaction in which bond breaking and bond forming happen simultaneously in a single step, with no intermediate. SN2 and E2 are concerted.
In simple terms: everything happens at once.
1-bromopropane reacts with methanethiolate (CH₃S⁻) to give 1-(methylthio)propane, with bromide (Br⁻) as the leaving group.
This is a one-step, concerted mechanism. You need exactly two curved arrows:
Arrow 1: Starts from the lone pair on the sulphur of CH₃S⁻ and points to the electrophilic carbon (the carbon bonded to Br). This arrow shows nucleophilic attack.
Arrow 2: Starts from the C–Br bond and points to Br. This arrow shows the leaving group departing with the bonding electrons.
Both arrows are drawn simultaneously because both events happen in the same step.
The nucleophile attacks from the backside (opposite the leaving group), causing inversion of stereochemistry at the carbon centre (Walden inversion).
Put the nucleophile on one side and the leaving group on the other to make the backside attack visually clear.
The product shows the new C–S bond where the C–Br bond used to be, with the configuration inverted.
SN2 is one step, so the energy diagram has:
One energy peak (one transition state)
No intermediate (no valley between peaks)
Starting materials on the left, products on the right, connected by a single smooth curve that rises to the transition state and then descends
Draw the x-axis (reaction coordinate) and y-axis (free energy).
Place the starting materials (CH₃S⁻ + 1-bromopropane) at a given energy level on the left.
Draw a single hump rising to the transition state. Label it [CH₃S···C···Br]‡ to show partial bonds to both the nucleophile and leaving group.
Descend to the products (1-(methylthio)propane + Br⁻) on the right.
If the reaction is exothermic (as it typically is when a stronger bond replaces a weaker one), draw the product energy level lower than the starting materials.
The height of the peak above the starting materials is the activation energy (Ea). A lower Ea means a faster reaction.
There is no intermediate to stabilise or destabilise. The reaction either goes or it does not.
The overall energy change (ΔG) is the difference between products and starting materials.
A tertiary bromide (bromocyclohexane derivative, specifically a 3° substrate with a fused cyclobutane ring) reacts with methanol (CH₃OH) under heat to give the corresponding methyl ether (OCH₃ product), with loss of HBr.
This is a two-step mechanism. You need arrows for each step separately.
Step 1: Leaving group departure (rate-determining step)
Arrow 1: From the C–Br bond to Br. The bonding electrons leave with the bromine, generating Br⁻ and a carbocation at the tertiary carbon.
Draw the carbocation intermediate explicitly, showing the positive charge on the carbon.
Step 2: Nucleophilic attack by solvent
Arrow 2: From a lone pair on the oxygen of CH₃OH to the carbocation carbon. This forms the new C–O bond.
The oxygen now bears a positive formal charge (it has three bonds plus a lone pair).
Step 3 (often required): Deprotonation
Arrow 3: Another molecule of CH₃OH (or any base present) removes the proton from the positively charged oxygen, giving the neutral ether product and regenerating H⁺ (as CH₃OH₂⁺ or transferred to solvent).
The nucleophile (methanol) is weak and neutral, consistent with SN1 rather than SN2.
Heat favours elimination, but the question specifically shows the substitution product (methyl ether), so you draw the SN1 pathway.
Because the carbocation is planar, the nucleophile can attack from either face. If the starting material were a single enantiomer, you would expect racemisation (mixture of both configurations at the carbon).
SN1 has two steps, so the energy diagram has:
Two energy peaks (two transition states)
One valley between them (the carbocation intermediate)
The first peak is taller than the second because Step 1 (leaving group departure, forming a high-energy carbocation) is the rate-determining step
Draw axes: reaction coordinate (x) and free energy (y).
Place starting materials (the tertiary bromide + CH₃OH) on the left.
Draw the first (taller) peak: transition state 1, where the C–Br bond is partially broken. Label it [R···Br]‡.
Descend to a local minimum: this is the carbocation intermediate (R⁺ + Br⁻). It sits higher in energy than the starting materials because carbocations are inherently unstable.
Draw the second (shorter) peak: transition state 2, where the nucleophile (CH₃OH) is partially bonded to the carbocation. Label it [R···O(H)CH₃]‡.
Descend to the products (the methyl ether + HBr). If the overall reaction is exothermic, products sit lower than starting materials.
The first step has the larger Ea, confirming it is rate-determining. Anything that stabilises the carbocation (more substitution, resonance) lowers this first barrier.
The intermediate sits in a real energy well: it exists, albeit briefly. This is the key difference from the SN2 diagram.
The second step (nucleophilic attack on the carbocation) is typically fast because carbocations are extremely reactive electrophiles.
x-axis: Reaction coordinate (progress of reaction, left to right)
y-axis: Free energy (G) or potential energy
Starting materials go on the left; products on the right
Transition states are peaks (maxima), drawn with ‡ symbol
Intermediates are valleys (local minima) between peaks
Ea = height from reactant (or intermediate) to the next transition state
ΔG = energy of products minus energy of starting materials (negative = exothermic)
One smooth hill: SM → [TS]‡ → Products. No valley.
Two hills with a valley: SM → [TS1]‡ → Intermediate (carbocation) → [TS2]‡ → Products. First hill is taller.
"Show the relative energies of starting materials, products, and intermediate(s)." This means you need to clearly indicate which is highest and lowest in energy. For SN1: TS1 > TS2 > intermediate > SM (often) > products (if exothermic).
Students often draw the carbocation intermediate as a transition state (at the top of a peak). It is not. The intermediate sits in a valley between two peaks. If you put it at the top, you are saying it does not actually exist as a discrete species.
Students sometimes draw arrows from atoms rather than from electron pairs. Curved arrows always start from the electrons (a lone pair or a bond), never from a nucleus.
Students forget the deprotonation step in SN1 when the nucleophile is a neutral molecule (like water or methanol). After the oxygen attacks the carbocation, it carries a positive charge and must lose a proton to give the neutral product.
Students assume the SN2 transition state is an intermediate. It is not. The transition state is the highest-energy point on the diagram and cannot be isolated. Drawing a valley at this point is incorrect.
⚠️ "Draw the arrow-pushing mechanism" is one of the most common exam question formats in organic chemistry. Expect it on every exam.
⚠️ Energy diagrams are frequently paired with mechanism questions. You may be asked to draw the diagram and label all transition states and intermediates.
⚠️ A common follow-up: "Is the overall reaction exothermic or endothermic?" You need to compare product energy to starting material energy on your diagram.
⚠️ The distinction between transition state (peak, fleeting, cannot be isolated) and intermediate (valley, brief lifetime, can in principle be detected) is tested directly in multiple-choice and short-answer formats.
True or false: An SN2 energy diagram has two transition states. Answer: False. SN2 is concerted (one step), so it has exactly one transition state.
Fill in the blank: In an SN1 energy diagram, the carbocation intermediate appears as a ______ between two peaks. Answer: Valley (local energy minimum).
True or false: Curved arrows in a mechanism start from the nuclei of atoms. Answer: False. They start from electron pairs (lone pairs or bonds).
Fill in the blank: The rate-determining step in SN1 is ______, which corresponds to the ______ peak in the energy diagram. Answer: Leaving group departure; taller (first).
True or false: In SN1 with a neutral nucleophile like methanol, you need a deprotonation step to reach the final neutral product. Answer: True.
Q: Draw the arrow-pushing mechanism for the reaction of 2-bromobutane with NaOH in DMSO (a polar aprotic solvent). What mechanism is this?
A: SN2. DMSO is a polar aprotic solvent, which favours SN2. NaOH provides OH⁻, a strong nucleophile. Arrow 1: lone pair on O⁻ attacks the carbon bearing Br from the backside. Arrow 2: C–Br bond electrons move to Br, forming Br⁻. One step, one transition state, inversion of configuration.
Q: For the SN2 reaction above, sketch the energy diagram. Label starting materials, products, and the transition state.
A: One smooth curve from SM (left) rising to a single peak (the transition state, labelled [HO···C···Br]‡) then descending to products (right). Products are lower than SM if the reaction is exothermic (C–O bond formed is typically strong).
Q: Tert-butyl chloride is dissolved in water. The product is tert-butyl alcohol. Draw the full mechanism, including all intermediates.
A: SN1. Step 1: Arrow from C–Cl bond to Cl, forming the tertiary carbocation and Cl⁻. Step 2: Arrow from a lone pair on water's oxygen to the carbocation, forming the C–O bond (oxygen now has a positive charge). Step 3: Arrow from another water molecule's lone pair to one of the O–H bonds, removing a proton to give the neutral alcohol product and H₃O⁺.
Q: How many peaks appear in the energy diagram for the tert-butyl chloride hydrolysis above? Which peak is taller?
A: Two peaks. The first peak (leaving group departure to form the carbocation) is taller because it is the rate-determining step. The second peak (water attacking the carbocation) is shorter because carbocations react readily with nucleophiles.
Arrow-pushing mechanisms are used in every subsequent topic: E1 and E2 eliminations, addition reactions to alkenes and alkynes, carbonyl chemistry, and aromatic substitution. Mastering curved arrows here pays compound interest for the rest of the course.
Energy diagrams connect to kinetics and thermodynamics (Ea, rate constants, equilibrium), to Hammond's postulate (the transition state resembles whichever species it is closer to in energy), and to catalysis (a catalyst lowers Ea without changing ΔG).
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