Difficulty: Intermediate | Prerequisites: Chapter 5 (substitution reactions from alcohols), acid-base chemistry, Lewis structures, electronegativity trends.
Nucleophilic substitution is one of the most fundamental reaction types in organic chemistry. In Chapter 5, you learnt how alcohols undergo substitution; now Chapter 6 shifts the starting material to alkyl halides (R-X), which are better electrophiles because halogens are inherently good leaving groups without needing acid activation. The two competing mechanisms, S_N1 and S_N2, each have a distinct set of conditions that favour them, and your job on an exam is to read those conditions (substrate class, nucleophile strength, solvent type) and pick the right pathway. If you can do that reliably, you can predict products, stereochemistry, and even whether a reaction will occur at all.
Alkyl halides react with nucleophiles through either S_N1 (two-step, carbocation intermediate) or S_N2 (one-step, backside attack). Which pathway wins depends mainly on the substrate's carbon class (primary, secondary, tertiary), the nucleophile's strength, and whether the solvent is protic or aprotic. Learn to read those three factors and you can predict the product, the stereochemistry, and whether the reaction happens at all.
Nucleophile
A Lewis base that donates an electron pair to an electrophilic carbon. In simple terms, it is the "attacker" in a substitution reaction, the species with electrons to share.
Electrophile (substrate)
A Lewis acid that accepts an electron pair. In substitution reactions, this is the alkyl halide or alcohol carbon bearing the leaving group. Think of it as the carbon that gets attacked.
Leaving group
The atom or group that departs with the bonding electrons when the nucleophile arrives. Good leaving groups are weak bases (e.g. Cl⁻, Br⁻, I⁻, tosylate). In simple terms, the better a group is at leaving, the more easily the reaction proceeds.
S_N2 (bimolecular nucleophilic substitution)
A one-step mechanism where the nucleophile attacks the electrophilic carbon at the same time the leaving group departs. The rate depends on both substrate and nucleophile concentrations. Think of it as a direct swap: the nucleophile pushes the leaving group out in a single motion, always from the back side.
S_N1 (unimolecular nucleophilic substitution)
A two-step mechanism. Step 1: the leaving group departs on its own to form a carbocation. Step 2: the nucleophile attacks the carbocation. The rate depends only on the substrate concentration. Think of it as "leave first, attack second."
Carbocation
A positively charged carbon intermediate with only three bonds and an empty p orbital. Stability order: tertiary > secondary > primary > methyl. In simple terms, more alkyl groups around the positive carbon means more stable, because alkyl groups donate electron density through hyperconjugation.
Protic solvent
A solvent that has O-H or N-H bonds and can donate hydrogen bonds (e.g. water, ethanol, methanol). Protic solvents stabilise ions through H-bonding, which favours S_N1 by stabilising the carbocation intermediate.
Aprotic solvent
A solvent that cannot donate hydrogen bonds (e.g. acetone, DMSO, DMF, acetonitrile, DCM). Aprotic solvents do not cage the nucleophile in a solvation shell, so strong nucleophiles stay "naked" and reactive, which favours S_N2.
Tosylate (OTs)
A sulfonate ester leaving group formed by reacting an alcohol with tosyl chloride (TsCl) and a base such as pyridine. It converts the poor leaving group -OH into an excellent one without changing the carbon skeleton. In simple terms, tosylation is the workaround when you need to do a substitution on an alcohol but -OH will not leave on its own.
Stereochemistry (inversion vs racemisation)
S_N2 gives inversion of configuration (Walden inversion) because the nucleophile attacks from the back side. S_N1 gives a racemic mixture (or close to it) because the planar carbocation can be attacked from either face.
Nucleophile strength matters because strong nucleophiles push a reaction toward S_N2, while weak nucleophiles leave the door open for S_N1. Here are the rules for ranking, assuming an aqueous (protic) solvent unless stated otherwise.
Rule 1: Charge beats neutral
A negatively charged species is a stronger nucleophile than its conjugate acid. The extra electron density makes it more eager to donate.
⁻OH is stronger than H₂O
⁻NH₂ is stronger than NH₃
RS⁻ is stronger than RSH
Rule 2: Down a column in the periodic table, nucleophilicity increases in protic solvents
Larger atoms are less tightly solvated by protic solvents, so their electron pairs are more available. This is why I⁻ is a stronger nucleophile than Br⁻ in water, even though Br⁻ is a stronger base.
I⁻ > Br⁻ > Cl⁻ > F⁻ (in protic solvent)
Rule 3: Polarisability matters in protic solvents
Sulfur is larger and more polarisable than oxygen. In protic solvents, R-SH is a stronger nucleophile than R-OH, and RS⁻ is stronger than RO⁻.
R-SH > R-OH
Worked examples from the problem set (aqueous solvent)
Br⁻ vs I⁻: I⁻ wins (larger, less solvated, more polarisable)
H₂O vs ⁻OH: ⁻OH wins (charged beats neutral)
⁻NH₂ vs NH₃: ⁻NH₂ wins (charged beats neutral)
R-SH vs R-OH: R-SH wins (sulfur is more polarisable than oxygen)
Solvents have a lower priority than substrate class or nucleophile identity when deciding S_N1 vs S_N2, but they still influence reaction rates and can tip borderline cases.
How to tell them apart
Protic solvents have an O-H or N-H bond and can donate hydrogen bonds. Aprotic solvents cannot donate hydrogen bonds, even if they are polar.
Protic solvents (favour S_N1)
Water (H₂O)
Methanol (CH₃OH)
Ethanol (CH₃CH₂OH)
Acetic acid (CH₃COOH)
Protic solvents stabilise ions through hydrogen bonding. This helps stabilise the carbocation intermediate in S_N1, speeding up the rate-determining step. They also solvate (cage) strong nucleophiles, slowing S_N2.
Aprotic solvents (favour S_N2)
Acetone, CH₃COCH₃
DMSO, (CH₃)₂SO
DMF, HC(O)N(CH₃)₂
Acetonitrile (ACN), CH₃CN
Dichloromethane (DCM), CH₂Cl₂
Aprotic solvents do not hydrogen-bond to nucleophiles, leaving them "naked" and highly reactive. This benefits S_N2, where a strong, unsolvated nucleophile can attack effectively.
The quick test
Look at the structure. If the solvent has an H directly bonded to O or N, it is protic. If not, it is aprotic, regardless of how polar it might be. DMSO and DMF are very polar but still aprotic.
When you face a substitution problem, work through these factors in order of priority.
1. Substrate class (highest priority)
Methyl and primary substrates: S_N2 (almost always). These carbons have minimal steric hindrance for backside attack, and the corresponding carbocations would be far too unstable to form.
Tertiary substrates: S_N1 (almost always). The carbon is too hindered for backside attack, and the tertiary carbocation is stable enough to form.
Secondary substrates: could go either way. Look at the nucleophile and solvent to decide.
2. Nucleophile identity (second priority)
Strong nucleophile (charged, good electron donor): favours S_N2.
Weak nucleophile (neutral, poor electron donor like H₂O or ROH): favours S_N1.
3. Solvent (third priority, tiebreaker)
Polar aprotic solvent: favours S_N2 (nucleophile stays reactive).
Polar protic solvent: favours S_N1 (stabilises carbocation, solvates nucleophile).
4. Leaving group quality
A good leaving group is needed for either mechanism. If the leaving group is poor (e.g. -OH in neutral or basic conditions), no substitution occurs at all. This is why alcohols need activation (protonation, or conversion to a tosylate) before they can undergo substitution.
"No reaction" scenarios to watch for
An alcohol treated with a nucleophile in a non-acidic, aprotic environment: -OH is not a leaving group under these conditions. No amount of strong nucleophile will force the substitution.
A neopentyl or very bulky primary substrate with a weak nucleophile: S_N2 is blocked by steric hindrance, and the primary carbocation is too unstable for S_N1.
Students often assume that a strong nucleophile automatically means S_N2. It does not. If the substrate is tertiary, S_N1 wins regardless of how strong the nucleophile is, because steric hindrance blocks backside attack.
Students frequently confuse "polar" with "protic." DMSO and DMF are highly polar but still aprotic. The test is whether the solvent has an H bonded to O or N, not whether it dissolves salts.
Some students think -OH is a leaving group. It is not, unless it is first protonated to water (in acidic conditions) or converted to a tosylate. An alcohol plus a nucleophile in basic or neutral aprotic conditions gives no reaction.
Students sometimes forget that S_N1 can produce rearranged products. If a more stable carbocation can form via a hydride or methyl shift, it often will. Always check whether the carbon skeleton has changed between starting material and product.
⚠️ Predicting S_N1 vs S_N2 is one of the most commonly tested skills in Organic Chemistry I. Expect to be given a reaction and asked to identify the mechanism, predict the product, and explain your reasoning.
⚠️ "No reaction" is a valid answer and professors test for it. If the leaving group is poor (e.g. -OH without acid or tosylation) or conditions are mismatched, write "no reaction" and state why.
⚠️ Stereochemistry matters. S_N2 always gives inversion. S_N1 gives racemisation (or near it). If a problem says "include stereochemistry," you must show this.
⚠️ Solvent classification (protic vs aprotic) appears as its own question type. Memorise the common examples: water, ethanol, methanol = protic; acetone, DMSO, DMF, ACN, DCM = aprotic.
⚠️ Tosylation of alcohols is a frequently tested technique. Know that TsCl + pyridine converts -OH into -OTs (a good leaving group) with retention of configuration at the carbon, because the C-O bond is not broken in that step.
True or False: A tertiary alkyl halide reacts primarily via S_N2 when treated with a strong nucleophile. (False, tertiary substrates undergo S_N1 regardless of nucleophile strength.)
Fill in the blank: In protic solvents, nucleophilicity increases as you go ______ a column in the periodic table. (Down.)
True or False: DMSO is a protic solvent because it is very polar. (False, DMSO is polar but aprotic; it has no O-H or N-H bond.)
Fill in the blank: S_N2 gives ______ of configuration at the electrophilic carbon. (Inversion.)
True or False: An alcohol (R-OH) treated with NaCl in DMSO will undergo substitution to give R-Cl. (False, -OH is not a leaving group under these conditions.)
Q: Predict the stronger nucleophile in aqueous solution: Br⁻ or I⁻. Explain your reasoning.
A: I⁻ is the stronger nucleophile. In protic (aqueous) solvent, larger atoms are less tightly solvated and more polarisable, so nucleophilicity increases going down a group. I⁻ is larger than Br⁻.
Q: A secondary alkyl chloride is treated with NaSH in DMSO. Will the reaction proceed via S_N1 or S_N2? Explain.
A: S_N2. The substrate is secondary (could go either way), but NaSH provides a strong, charged nucleophile (HS⁻), and DMSO is a polar aprotic solvent that keeps the nucleophile reactive. Both factors push toward S_N2.
Q: An alcohol is treated with NaCl in DMSO. Predict whether a reaction occurs and explain.
A: No reaction. The hydroxyl group (-OH) is a poor leaving group, and there is no acid present to protonate it. DMSO is aprotic, so there is no way to convert -OH into a better leaving group under these conditions.
Q: Why does treatment of an alcohol with TsCl and pyridine allow a subsequent substitution to occur?
A: TsCl converts the -OH into a tosylate (-OTs), which is an excellent leaving group (the conjugate base of a strong acid). Pyridine acts as a base to absorb the HCl produced. The C-O bond is not broken during tosylation, so the configuration at carbon is retained, setting up a clean S_N1 or S_N2 in the next step.
Q: A reaction converts a secondary alkyl chloride (with a cyclopentane ring) into two methyl ether products with an expanded ring. What mechanism is operating, and what accounts for the ring expansion?
A: S_N1. The secondary carbocation intermediate undergoes a 1,2-alkyl shift (ring expansion) to form a more stable tertiary carbocation. Methanol, a weak nucleophile, then attacks the rearranged carbocation from either face, giving the methyl ether products.
This material connects directly to elimination reactions (E1 and E2), which compete with substitution under many of the same conditions. When you reach elimination, you will need the same substrate/nucleophile/solvent framework, with the added question of whether the reagent acts as a base (elimination) or a nucleophile (substitution).
The acid-base chemistry from earlier in the course underpins leaving-group ability: good leaving groups are weak bases, and nucleophile strength tracks with basicity (with important exceptions in protic solvents).
Stereochemistry concepts from Chapter 4 (R/S assignments, chirality centres) are essential for drawing S_N2 inversion products correctly.
Related Terms / Search Tags
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