Nucleophilic substitution or elimination

Note: Before reading this page, make sure you are familiar with nucleophilic substitution reactions involving halogenalkanes. If you need a quick recap, click the links above.

Alcohols are formed when halogenalkanes are warmed with a warm aqueous solution of sodium hydroxide or potassium hydroxide. Here the hydroxide ion (OH^-) is acting as a nucleophile and attacking the δ⁺ carbon in the polar C-X bond present in the halogenalkane molecule 2-bromopropane to form the secondary alcohol propan-2-ol. This is outlined in the image below:

Nucleophilic substitution: hydroxide ions attack a halogenalkane to form an alcohol.

Hydroxide ions: base and nucleophile

You are probably most familiar with sodium hydroxide and potassium hydroxide from the work you will have done on acids and alkalis, where they act as Brønsted–Lowry (H⁺ acceptors). However the hydroxide ion (OH^-) is not only base but it can also behave as a nucleophile. There are lots of similarities between nucleophiles and bases, for example they both have lone pairs of electrons for a start. Strong bases are often good nucleophiles too, but whether the hydroxide ion will act as a base or a nucleophile depends a lot on the reaction conditions. With hydroxide ions you can push the reaction towards acting mainly as a base or mainly as a nucleophile by changing the solvent and the temperature at which the reaction takes place.

Sodium hydroxide and a bottle of distilled water, aqueous solutions of bases favours nucleophilic substitution reactions.

Hot alcoholic solutions of strong bases favour elimination reactions.

Nucleophilic substitution reactions- base in aqueous conditions

If a halogenalkane is gently warmed with sodium hydroxide or potassium hydroxide dissolved in water (so an aqueous solution), the hydroxide ion behaves as a nucleophile. This leads to a nucleophilic substitution reaction, producing an alcohol.

Elimination reaction

However by changing the reaction conditions we can alter the reaction that takes place, for example if potassium hydroxide or sodium hydroxide is dissolved in a hot ethanol (an alcoholic solution), then a different type of reaction will take place, this time rather than a nucleophilic substitution reaction occurring an elimination reaction will take place. Under these hot alcoholic conditions the hydroxide ion (OH^-) acts mainly as a base (H⁺ acceptor) rather than as a nucleophile. At A-level this is usually described as a one-step elimination reaction. Key features of a typical elimination reaction you should know are:

Elimination reactions form alkenes (unsaturated molecules with a carbon-carbon double bond, C=C).

The base (OH^-) removes a proton (H⁺) from a carbon atom adjacent to the carbon bonded to the halogen atom.

As the base (OH^-) removes the proton (H⁺) a new C=C bond starts to form between the carbon atoms.

As the C=C bond forms, the C-X bond breaks and the halogen leaves with both electrons from the C-X bond to form a halide ion.

This is shown below:

Elimination mechanism: a halogenalkane forms an alkene when heated with hot alcoholic sodium or potassium hydroxide solution.

Elimination from an unsymmetrical halogenalkane

🧠 Zaitsev’s Rule (Elimination)

In base-induced elimination reactions of haloalkanes (for example with ethanolic KOH), more than one alkene can sometimes form.

Zaitsev’s rule helps you predict which one is formed most 👇

🔑 The major product is the more substituted alkene
This alkene has fewer hydrogens on the C=C bond
⚖️ More substituted alkenes are more stable

📝 Exam tip: Zaitsev’s rule works best for secondary and tertiary haloalkanes.

In the example above using 2-bromopropane (a symmetrical halogenalkane), it would not make any difference which adjacent carbon you remove a hydrogen from, because you still end up with alkene, propene in this case.

However if you use an unsymmetrical halogenalkane then more than one alkene can be formed. As an example consider the products formed from heating 2-bromobutane (an unsymmetrical secondary haloalkane) with hot alcoholic potassium hydroxide.

Predicting the exact products of elimination reactions can initially be a little trickier than those formed during nucleophilic substitution reactions. However to find all possible products that can be formed simply locate the halogen atom in the haloalkane molecule and remove one hydrogen atom from each of the carbon atoms adjacent to the one bonded to the halogen. Now elimination reactions often results in a mixture of alkenes.

For 2-bromobutane, the base (OH^-) can remove hydrogen atoms on either adjacent carbon bonded to the halogen; as outlined in the mechanism below; so two different alkene molecules can be formed, now one of the alkene molecules formed can exist as a pair of geometric isomers- this "little trick" is a very common exam question; as many students fail to recognise the possibility of geometric isomerism occurring. This is shown in the mechanism below:

Elimination from an unsymmetrical halogenalkane gives a mixture of alkene products.

Predicting the main product formed: Zaitsev's rule

Even though a mixture of products is formed they are not produced in equal amounts during these base-induced elimination reactions. Now we can use Zaitsev's rule (see the brain box above) to help in predicting which product will be the major one produced and which product is produced in smaller amounts. According to Zaitsev's rule the more substituted alkene; that is the one with the least number of hydrogen atoms bonded to the C=C bond is usually the major product. In the example above but-2-ene is more substituted than but-1-ene so it will be the main product of this reaction.

However nucleophilic substitution reactions can compete with elimination reactions; especially with secondary halogenalkanes and so you may get some alcohol products as well. In general elimination reactions are favoured most strongly with tertiary halogenalkanes; while primary halogenalkanes are more likely to give substitution products; that is alcohols. Note: Primary halogenalkanes can still undergo elimination reactions under hot ethanolic conditions, but nucleophilic substitution is often a big competitor here.

Reaction conditions also help decide whether you get mainly nucleophilic substitution or mainly elimination reactions:

Elimination reactions are more likely as the basicity of the reagent increases. That is the stronger the base the more likely the reaction will be an elimination reaction.
Elimination is favoured by hot ethanolic conditions. (In ethanol there can also be some ethoxide, CH₃CH₂O^- which is a stronger base than hydroxide.)
Nucleophilic substitution is more likely in warm aqueous conditions.

Summary table

Why not use the summary table below to create some flashcards to help you recall the reaction conditions and the products formed!

🧾 Summary: Substitution vs Elimination

What you’re doing 🧪	Conditions 🔥	What OH^- acts as 👇	Main product ✅	Quick clue 💡
Nucleophilic substitution	Warm aqueous NaOH / KOH (OH^- in water)	Nucleophile attacks the δ⁺ carbon	Alcohol 🧴	Water solvent → substitution is favoured
Elimination	Hot ethanolic NaOH / KOH (OH^- in ethanol)	Base removes H⁺ from adjacent carbon	Alkene (C=C) ⚡	Hot + ethanol → elimination is favoured
Elimination (unsymmetrical haloalkane)	Same: hot ethanolic base	Base	Mixture of alkenes 🧩 often one is major	Use Zaitsev’s rule: more substituted alkene is usually the major product ⭐

Self-check: Try the quick quiz below to test your understanding of elimination and nucleophilic substitution reactions

Click the button below to open the quiz.

Key Points

⚠️ Exam Trap: Substitution or Elimination?

You see: KOH / NaOH + a halogenoalkane and the question asks for the product…

Do you instantly write an alcohol or an alkene? 😬

💧 Aqueous hydroxide (in water) → mostly substitution → alcohol
🔥 Ethanolic hydroxide (in ethanol) + heat → mostly elimination → alkene
🧩 Secondary haloalkanes can give a mixture of products — conditions decide what dominates
⭐ If it’s unsymmetrical, check for a mixture of alkenes and apply Zaitsev’s rule

📝 Quick check: if the solvent isn’t stated, don’t assume — the examiner may be testing this exact trap.

Halogenalkanes can form either alcohols or alkenes with hydroxide ions, depending on whether they undergo nucleophilic substitution or an elimination reaction.
Primary halogenalkanes usually favour nucleophilic substitution, while tertiary halogenalkanes usually favour elimination. Secondary halogenalkanes often give a mixture of substitution and elimination products.
Substitution is more likely with a warm aqueous solution of hydroxide, whereas elimination is more likely with hot ethanolic hydroxide.
Elimination reactions can produce a mixture of alkenes from unsymmetrical halogenalkanes where the more substituted alkene is usually the major product.