Difficulty: Intermediate | Prerequisites: Lewis structures, curly arrow notation, carbocation stability, basic energy diagrams.
Mechanisms are the "how" behind every reaction on this exam. This section tests whether you can draw the step-by-step electron movement for electrophilic addition to alkenes and for keto–enol tautomerisation. You also need to read reaction coordinate (energy) diagrams and identify intermediates vs transition states. If you can draw a proper curly arrow pushing electrons from a nucleophilic site to an electrophilic site, and you know what a carbocation rearrangement looks like, you are well prepared.
TL;DR
Electrophilic addition of HCl to an alkene proceeds via protonation (forming a carbocation intermediate, possibly with a hydride or methyl shift) followed by nucleophilic attack of Cl⁻. Acid-catalysed hydration of a terminal alkyne goes through an enol intermediate that tautomerises to the methyl ketone. Both mechanisms require correct curly arrows, intermediates with formal charges, and awareness of which step is rate-determining.
Electrophilic addition
A reaction in which an electrophile (electron-poor species) adds across a π bond. The π electrons act as the nucleophile, attacking the electrophile in the first step.
Think of it as the alkene's electron-rich double bond reaching out to grab a positive or electron-poor species.
Carbocation
A positively charged carbon intermediate with an empty p orbital. Stability: 3° > 2° > 1° > methyl. Carbocations can rearrange via 1,2-hydride shifts or 1,2-methyl shifts to reach a more stable form.
Transition state
The highest-energy point along a reaction step. It represents the geometry where bonds are partially formed and partially broken. Transition states sit at energy maxima on a reaction coordinate diagram and cannot be isolated.
Intermediate
A species that forms during a multi-step reaction, sitting in an energy minimum (valley) between two transition states. Intermediates have finite lifetimes and can sometimes be detected or trapped.
Rate-determining step (RDS)
The slowest step in a multi-step mechanism, with the highest-energy transition state. It determines the overall rate of the reaction.
In simple terms, the RDS is the bottleneck. Everything else is fast by comparison.
Enol
A tautomer of a carbonyl compound bearing a hydroxyl group on a doubly bonded carbon (C=C–OH). Enols are typically less stable than the keto form and interconvert via tautomerisation.
Keto–enol tautomerisation
The equilibrium process by which an enol converts to a ketone (or aldehyde) and vice versa, through proton transfer and π bond migration. Catalysed by acid or base.
1,2-Hydride shift
A carbocation rearrangement in which a hydrogen atom with its bonding electrons migrates from an adjacent carbon to the positively charged carbon, creating a more stable carbocation.
Reading the energy diagram
The exam presents a reaction coordinate diagram with five labelled points (A through E).
Points at energy maxima (peaks) are transition states: A, C and E.
Points at energy minima (valleys) are intermediates: B and D.
The rate-determining step has the highest-energy transition state. In this exam, that is point A, the first transition state (protonation of the alkene to form the carbocation). This makes sense because the first step, breaking the π bond and forming a high-energy carbocation, is the slow step.
The mechanism (step by step):
Step 1: Protonation (the rate-determining step)
The π electrons of the alkene attack the H of HCl.
Curly arrow: from the C=C π bond to the H of H–Cl.
Simultaneously, the H–Cl bond breaks heterolytically: curly arrow from the H–Cl bond to Cl.
This produces a secondary carbocation and Cl⁻.
Step 2: 1,2-Hydride shift (carbocation rearrangement)
If the initial carbocation is secondary but a tertiary carbocation is accessible, a 1,2-hydride shift occurs.
Curly arrow: from the C–H bond on the adjacent carbon to the positively charged carbon.
This converts a 2° carbocation to a more stable 3° carbocation.
The exam mechanism shows this rearrangement explicitly.
Step 3: Nucleophilic attack
Cl⁻ attacks the carbocation.
Curly arrow: from a lone pair on Cl⁻ to the positively charged carbon.
Product: the chloroalkane.
Grading: 1 point per curly arrow, 2 points per correctly drawn intermediate (with formal charges shown). Missing a formal charge on the carbocation or on Cl⁻ loses marks.
Hydration of a terminal alkyne
Reagents: H₂SO₄, H₂O, Hg²⁺ (mercury(II) catalyst).
A terminal alkyne (e.g. HC≡C–CH₃) undergoes Markovnikov addition of water. The OH ends up on the internal carbon (the more substituted end of the triple bond). This initially produces an enol, which is not isolated: it tautomerises to the methyl ketone.
For propyne: HC≡C–CH₃ → enol (CH₃–C(OH)=CH₂) → methyl ketone (CH₃–CO–CH₃, acetone).
Drawing the enol intermediate
The enol has:
A C=C double bond where the triple bond was
An OH group on the carbon that received the water (the internal carbon)
The correct number of hydrogens to satisfy valence
The exam awards 1.5 points if the enol is drawn with the OH and C=C reversed (a common error, but still shows understanding of the concept).
Keto–enol tautomerisation mechanism (acid-catalysed)
This is a two-step process with a protonated intermediate:
Step 1: Protonation of the enol C=C
A lone pair on the π bond of the enol attacks H₃O⁺ (or H⁺ from the acid catalyst).
Curly arrow: from the C=C π bond to H⁺.
This forms a carbocation (protonated ketone, with the positive charge on carbon adjacent to OH).
Step 2: Deprotonation of the OH
Water acts as a base, removing the proton from the OH group.
Curly arrow: from a lone pair on water to the H of the O–H.
This yields the ketone and regenerates H₃O⁺.
Grading: 1 point per curly arrow, 2 points per intermediate. Formal charges must be shown on every intermediate. Missing the positive charge on the protonated species or the lone pairs used in curly arrows costs marks.
Real-world connection: Keto–enol tautomerisation is not just an exam curiosity. It is central to how enzymes catalyse reactions in biochemistry (e.g. glycolysis involves enol intermediates), and it explains why α-hydrogens adjacent to carbonyls are acidic and reactive.
Students draw curly arrows starting from a positive charge or from an atom without available electrons. Curly arrows always go from an electron-rich site (lone pair, π bond, σ bond) to an electron-poor site.
Students confuse intermediates and transition states on energy diagrams. Intermediates are in valleys (minima), transition states are at peaks (maxima). A common exam error is circling a peak when asked for an intermediate.
Students forget to show the 1,2-hydride shift when a rearrangement is needed. If the initial carbocation is secondary and a tertiary carbocation is one shift away, the rearrangement will occur and must be drawn.
Students omit formal charges on intermediates. Every carbocation must show a "+" on the charged carbon, and every anionic species must show "–".
⚠️ The mechanisms section is worth 23 points total. Curly arrows are graded at 1 point each, and each intermediate is worth 2 points. This means a mechanism with 3 arrows and 2 intermediates is worth up to 7 points for one sub-question. Show every arrow and every intermediate.
⚠️ The energy diagram question (identifying intermediates, transition states and the RDS) is 9 points of straightforward marks if you know the definitions. Practise reading energy diagrams until this is automatic.
⚠️ For the tautomerisation mechanism, the exam allows drawing it in either direction (enol to keto or keto to enol). Make sure you are comfortable with both.
⚠️ Formal charges are not optional. They are graded. If a carbocation does not show "+" on the correct carbon, you lose the intermediate credit.
True or false: On an energy diagram, the rate-determining step has the lowest-energy transition state.
False. The RDS has the highest-energy transition state.
Fill in the blank: In electrophilic addition of HCl to an alkene, the first step is attack of the ______ on the H of HCl.
π electrons (or the alkene double bond)
True or false: The enol form of acetone is more stable than the keto form.
False. The keto form is far more stable for simple ketones.
Fill in the blank: A 1,2-hydride shift converts a ____° carbocation into a ____° carbocation.
2° into 3° (or 1° into 2°, depending on the system)
Q: On a reaction coordinate diagram for a two-step reaction, how many transition states and how many intermediates are there?
A: Two transition states (two peaks) and one intermediate (one valley between the peaks).
Q: Draw the curly arrow mechanism for the addition of HBr to propene. Include Markovnikov selectivity.
A: Step 1: π electrons of propene attack H of HBr → secondary carbocation on C2 + Br⁻ (Markovnikov: H adds to the less substituted carbon, positive charge on the more substituted carbon). Step 2: Br⁻ attacks the carbocation → 2-bromopropane.
Q: Draw the enol of butanone (CH₃COCH₂CH₃). Show the O–H and the C=C clearly.
A: CH₃C(OH)=CHCH₃. The OH is on C2 (where the C=O was), and the C=C is between C2 and C3. The geometry around the C=C is trigonal planar.
Q: Why is the first step of HCl addition to an alkene rate-determining?
A: The first step requires breaking the stable π bond of the alkene and forming a high-energy carbocation. This has the highest activation energy barrier. The second step (Cl⁻ attacking the carbocation) is fast because it involves a good nucleophile attacking a very reactive electrophile.
Q: What would happen if you forgot to draw the 1,2-hydride shift in the exam mechanism?
A: You would lose points for the missing intermediate (the rearranged carbocation) and for the missing curly arrow (the hydride shift). You would also draw the wrong final product, costing further marks.
Electrophilic addition is the core mechanism behind nearly every alkene reaction on this exam: HBr addition, HCl addition, hydration, and even the first step of halogenation all begin with the alkene's π electrons attacking an electrophile. Mastering this one mechanism pattern gives you access to a large fraction of the exam.
Keto–enol tautomerisation connects forward to Organic Chemistry II, where enolate chemistry (reactions of the deprotonated enol) becomes a major topic. Understanding tautomerisation now will make carbonyl chemistry much smoother later.
Energy diagrams and the concept of the rate-determining step connect to kinetics and thermodynamics. These ideas appear in every branch of chemistry and biochemistry.
Related Terms / Search Tags
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