Difficulty: Intermediate to Advanced | Prerequisites: Acid-base chemistry, resonance, orbital theory
Organic reactions follow a handful of core mechanisms, and the exam tests whether you can identify which one applies to a given reaction, what factor makes one reaction faster than another, and which of a pair of reactions proceeds more quickly. The four things to examine are the substrate structure, the nucleophile or base, the leaving group, and the solvent.
SN1 (Nucleophilic Substitution, Unimolecular)
A two-step mechanism: the leaving group departs first (forming a carbocation), then the nucleophile attacks. Rate depends only on the substrate concentration. Favoured by tertiary substrates, weak nucleophiles, and polar protic solvents.
SN2 (Nucleophilic Substitution, Bimolecular)
A one-step mechanism: the nucleophile attacks at the same time as the leaving group departs (backside attack). Rate depends on both substrate and nucleophile concentrations. Favoured by primary substrates (methyl best), strong nucleophiles, and polar aprotic solvents.
E1 (Elimination, Unimolecular)
A two-step mechanism: the leaving group departs first (forming a carbocation), then a base removes a beta-hydrogen. Competes with SN1 and is favoured by higher temperatures and weaker bases.
E2 (Elimination, Bimolecular)
A one-step mechanism: the base removes a beta-hydrogen at the same time as the leaving group departs. The H and the leaving group must be anti-periplanar. Favoured by strong, bulky bases.
Electrophilic addition
A mechanism where an electrophile adds to a pi bond (typically an alkene). The pi electrons attack the electrophile first, forming a carbocation intermediate, then a nucleophile attacks the carbocation. Classic example: HCl or HBr adding across a double bond.
Electrophilic aromatic substitution (EAS)
A mechanism where an electrophile replaces a hydrogen on an aromatic ring. The ring's pi electrons attack the electrophile, forming an arenium ion intermediate, then a base removes a proton to restore aromaticity.
Nucleophilic acyl substitution
A mechanism at a carbonyl bearing a leaving group (esters, acid chlorides, anhydrides, amides). A nucleophile attacks the carbonyl carbon, the carbonyl re-forms, and the leaving group is expelled. Think of it as addition then elimination at the carbonyl.
Leaving group
The atom or group that departs with the bonding electrons during a substitution or elimination. Better leaving groups are weaker bases (more stable once they leave). Tosylate (OTs) and halides are common good leaving groups.
Nucleophile
An electron-rich species that donates a pair of electrons to an electron-poor centre. Stronger nucleophiles have a more available lone pair. In simple terms, the species that does the attacking.
Electrophile
An electron-poor species that accepts a pair of electrons. In simple terms, the species with a partial or full positive charge that gets attacked.
Ask yourself these questions in order:
What type of substrate is reacting? Is it an alkene (electrophilic addition), an aromatic ring (EAS), a carbonyl with a leaving group (nucleophilic acyl substitution), or a carbon with a leaving group (substitution or elimination)?
Is there a strong nucleophile or base present? Strong nucleophile + primary substrate = SN2. Strong, bulky base = E2.
Is the substrate tertiary? Tertiary substrates cannot do SN2 (too much steric hindrance). They go through SN1 or E1 (weak nucleophile/base, polar protic solvent) or E2 (strong base).
What is the solvent? Polar aprotic solvents (DMF, DMSO, acetone) favour SN2. Polar protic solvents (water, alcohols) favour SN1/E1.
When an exam gives you two similar reactions and asks which is faster, one of four factors is being tested:
A. Structure of the organic substrate. More substituted carbocations are more stable. A tertiary carbocation forms faster than a secondary one, so any reaction that proceeds through a carbocation intermediate (SN1, E1, electrophilic addition) is faster with a more substituted substrate.
B. Structure of the nucleophile, electrophile, or base. A stronger nucleophile makes SN2 faster. A better electrophile makes electrophilic addition faster. A stronger base makes E2 faster. Nucleophilicity increases with charge (RS- > RSH) and going down the periodic table in protic solvents (I- > Br- > Cl-).
C. Structure of the leaving group. A better leaving group (more stable anion) accelerates both substitution and elimination. Tosylate (OTs) > I- > Br- > Cl-. Phenoxide is a better leaving group than ethoxide because the phenoxide ion is resonance-stabilised.
D. Nature of the solvent. Polar protic solvents stabilise carbocations and favour SN1/E1. Polar aprotic solvents do not solvate nucleophiles as much, keeping them more reactive for SN2.
Pair 1: HCl + propene vs. HCl + 2-methylpropene
Mechanism: Electrophilic addition (H+ adds to the double bond, forming a carbocation, then Cl- attacks).
Products: Markovnikov addition gives 2-chloropropane and 2-chloro-2-methylpropane respectively.
Key factor: A (substrate structure). 2-Methylpropene forms a more stable tertiary carbocation.
Faster reaction: 2 (the tertiary carbocation forms more readily).
Pair 2: CH3COCH2CH3 + CH3O-/CH3OH vs. CH3CO-Ph + CH3O-/CH3OH
Mechanism: Nucleophilic acyl substitution (the nucleophile CH3O- attacks the carbonyl carbon, the leaving group departs).
Key factor: C (leaving group). Phenoxide (PhO-) is a better leaving group than ethoxide (CH3CH2O-) because phenoxide is resonance-stabilised.
Faster reaction: 2 (the better leaving group is expelled more easily).
Pair 3: Cyclohexyl-OTs + CH3O-Na/DMF vs. Cyclohexyl-OTs + CH3S-Na/DMF
Mechanism: SN2 (strong nucleophile, polar aprotic solvent, secondary substrate with a good leaving group).
Key factor: B (nucleophile). Thiolate (CH3S-) is a better nucleophile than methoxide (CH3O-) because sulfur is larger and more polarisable.
Faster reaction: 2.
Pair 4: Tertiary alcohol + H2SO4 (two substrates compared)
Mechanism: E1 (acid-catalysed dehydration of a tertiary alcohol). Protonation of OH converts it to a good leaving group (water), then the carbocation forms, then a base removes a beta-hydrogen.
Key factor: A (substrate). The substrate that can form a carbocation stabilised by an adjacent phenyl group (benzylic stabilisation) reacts faster.
Faster reaction: 2 (benzylic carbocation stabilisation).
This mechanism is tested in depth on the exam (Question 8). The specific reaction is transesterification: a methyl ester reacting with sodium ethoxide (catalytic) in ethanol to give the ethyl ester.
Mechanism step by step:
The ethoxide ion (CH3CH2O-) attacks the carbonyl carbon of the methyl ester, forming a tetrahedral intermediate. The carbonyl C=O becomes C-O- .
The tetrahedral intermediate collapses: the C=O reforms and the methoxide leaving group (CH3O-) departs.
Product: the ethyl ester + methoxide ion.
Why the ethoxide is catalytic:
The methoxide ion expelled in step 2 is itself a base/nucleophile of comparable strength to ethoxide. In ethanol solvent, methoxide deprotonates ethanol to regenerate ethoxide: CH3O- + CH3CH2OH -> CH3OH + CH3CH2O-. The ethoxide is thereby regenerated, so only a catalytic amount is needed.
Students often label any reaction at a carbonyl as "nucleophilic addition." If the carbonyl has a leaving group (ester, acid chloride, anhydride, amide), the mechanism is nucleophilic acyl substitution, not simple addition. Nucleophilic addition applies to aldehydes and ketones, which have no leaving group.
Confusing nucleophilicity with basicity is extremely common. A good nucleophile is not always a strong base. For example, thiolate (RS-) is an excellent nucleophile but a weak base, while tert-butoxide is a strong base but a poor nucleophile (too bulky for SN2).
Students sometimes assume SN2 can occur on tertiary substrates. It cannot. The three bulky groups block the backside attack. Tertiary substrates undergo SN1, E1 or E2.
Forgetting to check the solvent is a common slip. Two otherwise identical reactions can follow different mechanisms if one uses a polar protic solvent and the other uses a polar aprotic solvent.
⚠️ The "compare two reactions" format (identify the mechanism, the key factor, and which is faster) is a high-value question type worth 12% of this exam. Practise it with varied pairs.
⚠️ Nucleophilic acyl substitution and its mechanism (tetrahedral intermediate) is worth 10% on this exam. Be able to draw every intermediate and explain why catalytic base works.
⚠️ Know the difference between SN1/SN2 and E1/E2 cold. The exam will not always tell you which it is; you have to read the substrate, the reagent and the solvent.
⚠️ EAS (electrophilic aromatic substitution) often appears alongside Friedel-Crafts reactions (CH3Cl / AlCl3). Know the directing effects of substituents already on the ring.
True or false: SN2 reactions are faster with tertiary substrates. (False: SN2 is fastest with methyl and primary substrates.)
Fill in the blank: In SN1, the rate-determining step is the formation of the ______. (Carbocation.)
True or false: Polar aprotic solvents favour SN1 reactions. (False: they favour SN2.)
Fill in the blank: In nucleophilic acyl substitution, the intermediate is called the ______ intermediate. (Tetrahedral.)
True or false: Thiolate (CH3S-) is a better nucleophile than methoxide (CH3O-) in a polar aprotic solvent. (True: sulfur is larger and more polarisable.)
Q: Two reactions both involve a secondary substrate with a tosylate leaving group in DMF. Reaction 1 uses CH3O-Na+ and reaction 2 uses CH3S-Na+. Identify the mechanism, the key factor, and which is faster.
A: Both are SN2 (strong nucleophile, polar aprotic solvent, secondary substrate). The key factor is B (nucleophile structure). CH3S- is a better nucleophile than CH3O- because sulfur is larger and more polarisable. Reaction 2 is faster.
Q: A methyl benzoate reacts with catalytic sodium ethoxide in ethanol. What is the product, what is the mechanism, and why is only a catalytic amount of base needed?
A: The product is ethyl benzoate + methanol. The mechanism is nucleophilic acyl substitution: ethoxide attacks the ester carbonyl, a tetrahedral intermediate forms, then methoxide departs. Ethoxide is regenerated because the expelled methoxide deprotonates ethanol solvent to reform ethoxide.
Q: HCl adds to propene and to 2-methylpropene. Which reaction is faster and why?
A: The addition to 2-methylpropene is faster. Both proceed by electrophilic addition. H+ adds to the less substituted carbon (Markovnikov), forming a carbocation. 2-Methylpropene gives a tertiary carbocation, which is more stable and forms faster than the secondary carbocation from propene.
Q: An ester with an ethyl leaving group reacts with methoxide, and a separate ester with a phenyl leaving group reacts with methoxide under the same conditions. Which is faster?
A: The ester with the phenyl leaving group reacts faster. Phenoxide is a better leaving group than ethoxide because phenoxide is resonance-stabilised by the aromatic ring.
Mechanism identification underpins all of the synthesis and product-prediction questions on the exam. If you cannot identify the mechanism, you cannot predict the product or its stereochemistry. This connects directly to stereochemistry (SN2 gives inversion, SN1 gives racemisation, E2 requires anti-periplanar geometry) and to the reagent toolkit covered in the synthesis notes.
SN1, SN2, E1, E2, nucleophilic substitution, elimination reaction, electrophilic addition, electrophilic aromatic substitution, nucleophilic acyl substitution, Markovnikov, anti-Markovnikov, carbocation stability, leaving group ability, nucleophilicity, polar protic solvent, polar aprotic solvent, DMF, DMSO, transesterification, tetrahedral intermediate, Friedel-Crafts, reaction rate factors, CHEM 2700