Topics

Reactions of haloalkanes

Two reactions, decided by the solvent

A haloalkane does one of two things depending on what the base is dissolved in. This is the most heavily examined pair of conditions in the standard.

$\text{NaOH}(aq)$ or $\text{KOH}(aq)$ , warmed → substitution → an alcohol
KOH dissolved in alcohol, heated → elimination → an alkene
Same base, different solvent, completely different product. Quote the solvent every time.

Substitution with hydroxide

The halogen is replaced by an –OH group.

$\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{NaBr}$

Conditions: aqueous sodium or potassium hydroxide, warmed.
The product is an alcohol, and the halogen leaves as a halide ion in the salt.

Elimination with KOH in alcohol

A hydrogen and the halogen are removed from adjacent carbons, and a double bond forms between them.

$\text{CH}_3\text{CH}_2\text{Br} + \text{KOH} \rightarrow \text{CH}_2{=}\text{CH}_2 + \text{KBr} + \text{H}_2\text{O}$

Conditions: KOH dissolved in ethanol, heated.
The product is an alkene.
If the two neighbouring carbons are different, two alkenes form — see the major and minor products page.

Substitution with ammonia

Concentrated ammonia replaces the halogen with an –NH₂ group, giving a primary amine.

$\text{CH}_3\text{CH}_2\text{Br} + \text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + \text{HBr}$

Conditions: concentrated ammonia, heated in a sealed tube, with the ammonia in excess.
The sealed tube is needed because ammonia is a gas and would otherwise escape on heating.
Excess ammonia limits further substitution — the amine product can itself react with more haloalkane.

Substitution with a primary amine

A primary amine can act in place of ammonia, giving a secondary amine.

$\text{CH}_3\text{Br} + \text{CH}_3\text{NH}_2 \rightarrow (\text{CH}_3)_2\text{NH} + \text{HBr}$

You will not be asked to name the secondary amine systematically, but you may be asked to draw it or to identify the reaction type.

Reactivity of the halogens

The order of reactivity is iodo > bromo > chloro > fluoro.
The C–I bond is the weakest and the C–F bond the strongest, so the iodoalkane reacts fastest and the fluoroalkane is essentially unreactive.

Tips and Tricks

Name the solvent. If you write "react with KOH" you have described two different reactions at once and cannot be given the mark. Write "KOH(aq)" for substitution and "KOH in alcohol" for elimination.

Note which reagent goes with which starting material. KOH in alcohol eliminates from a haloalkane; concentrated $\text{H}_2\text{SO}_4$ eliminates from an alcohol. Swapping them is a common error.

For the ammonia reaction, say concentrated, excess and sealed tube. Each is worth stating, and the sealed tube has a reason you may be asked for.

When you write an equation, balance the halogen out properly — as $\text{NaBr}$ with sodium hydroxide, or as $\text{HBr}$ with ammonia.

Elimination needs a hydrogen on a carbon next to the one carrying the halogen. A haloalkane with no such hydrogen cannot eliminate at all.

Worked ExamplePredicting products from the same starting material

Give the structural formula and name of the organic product formed when 2-bromopropane, $\text{CH}_3\text{CHBrCH}_3$ , is treated with each of the following.

(a) $\text{NaOH}(aq)$ , warmed (b) KOH dissolved in ethanol, heated (c) concentrated $\text{NH}_3$ , heated in a sealed tube

(a) $\text{NaOH}(aq)$ , warmed

Step 1 — Identify the reaction type. The hydroxide is aqueous, so this is a substitution.

Step 2 — Make the substitution. The bromine is replaced by an –OH, in the same position on the chain.

$\text{CH}_3\text{CHBrCH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{CH}(\text{OH})\text{CH}_3 + \text{NaBr}$

(b) KOH in ethanol, heated

Step 1 — Identify the reaction type. The KOH is dissolved in alcohol, so this is an elimination.

Step 2 — Find the carbons that lose the atoms. The bromine is on carbon 2. A hydrogen must be removed from a neighbouring carbon — either carbon 1 or carbon 3.

Step 3 — Check whether the two choices give different products. Carbon 1 and carbon 3 are both $\text{CH}_3$ groups — they are identical. So it makes no difference which one loses a hydrogen: the same alkene results either way.

$\text{CH}_3\text{CHBrCH}_3 + \text{KOH} \rightarrow \text{CH}_3\text{CH}{=}\text{CH}_2 + \text{KBr} + \text{H}_2\text{O}$

(c) Concentrated $\text{NH}_3$ , heated in a sealed tube

Step 1 — Identify the reaction type. Ammonia is a substituting reagent, replacing the halogen.

Step 2 — Make the substitution. The bromine is replaced by an –NH₂ group, again on carbon 2.

$\text{CH}_3\text{CHBrCH}_3 + \text{NH}_3 \rightarrow \text{CH}_3\text{CH}(\text{NH}_2)\text{CH}_3 + \text{HBr}$

Reactions of haloalkanes

Covers AS91391

Two reactions, decided by the solvent

A haloalkane does one of two things depending on what the base is dissolved in. This is the most heavily examined pair of conditions in the standard.

$\text{NaOH}(aq)$ or $\text{KOH}(aq)$ , warmed → substitution → an alcohol
KOH dissolved in alcohol, heated → elimination → an alkene
Same base, different solvent, completely different product. Quote the solvent every time.

Substitution with hydroxide

The halogen is replaced by an –OH group.

$\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{NaBr}$

Conditions: aqueous sodium or potassium hydroxide, warmed.
The product is an alcohol, and the halogen leaves as a halide ion in the salt.

Elimination with KOH in alcohol

A hydrogen and the halogen are removed from adjacent carbons, and a double bond forms between them.

$\text{CH}_3\text{CH}_2\text{Br} + \text{KOH} \rightarrow \text{CH}_2{=}\text{CH}_2 + \text{KBr} + \text{H}_2\text{O}$

Conditions: KOH dissolved in ethanol, heated.
The product is an alkene.
If the two neighbouring carbons are different, two alkenes form — see the major and minor products page.

Substitution with ammonia

Concentrated ammonia replaces the halogen with an –NH₂ group, giving a primary amine.

$\text{CH}_3\text{CH}_2\text{Br} + \text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + \text{HBr}$

Conditions: concentrated ammonia, heated in a sealed tube, with the ammonia in excess.
The sealed tube is needed because ammonia is a gas and would otherwise escape on heating.
Excess ammonia limits further substitution — the amine product can itself react with more haloalkane.

Substitution with a primary amine

A primary amine can act in place of ammonia, giving a secondary amine.

$\text{CH}_3\text{Br} + \text{CH}_3\text{NH}_2 \rightarrow (\text{CH}_3)_2\text{NH} + \text{HBr}$

You will not be asked to name the secondary amine systematically, but you may be asked to draw it or to identify the reaction type.

Reactivity of the halogens

The order of reactivity is iodo > bromo > chloro > fluoro.
The C–I bond is the weakest and the C–F bond the strongest, so the iodoalkane reacts fastest and the fluoroalkane is essentially unreactive.

Tips and Tricks

For the ammonia reaction, say concentrated, excess and sealed tube. Each is worth stating, and the sealed tube has a reason you may be asked for.

When you write an equation, balance the halogen out properly — as $\text{NaBr}$ with sodium hydroxide, or as $\text{HBr}$ with ammonia.

Elimination needs a hydrogen on a carbon next to the one carrying the halogen. A haloalkane with no such hydrogen cannot eliminate at all.

Worked ExamplePredicting products from the same starting material

Give the structural formula and name of the organic product formed when 2-bromopropane, $\text{CH}_3\text{CHBrCH}_3$ , is treated with each of the following.

(a) $\text{NaOH}(aq)$ , warmed (b) KOH dissolved in ethanol, heated (c) concentrated $\text{NH}_3$ , heated in a sealed tube

(a) $\text{NaOH}(aq)$ , warmed

Step 1 — Identify the reaction type. The hydroxide is aqueous, so this is a substitution.

Step 2 — Make the substitution. The bromine is replaced by an –OH, in the same position on the chain.

$\text{CH}_3\text{CHBrCH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{CH}(\text{OH})\text{CH}_3 + \text{NaBr}$

(b) KOH in ethanol, heated

Step 1 — Identify the reaction type. The KOH is dissolved in alcohol, so this is an elimination.

Step 2 — Find the carbons that lose the atoms. The bromine is on carbon 2. A hydrogen must be removed from a neighbouring carbon — either carbon 1 or carbon 3.

$\text{CH}_3\text{CHBrCH}_3 + \text{KOH} \rightarrow \text{CH}_3\text{CH}{=}\text{CH}_2 + \text{KBr} + \text{H}_2\text{O}$

(c) Concentrated $\text{NH}_3$ , heated in a sealed tube

Step 1 — Identify the reaction type. Ammonia is a substituting reagent, replacing the halogen.

Step 2 — Make the substitution. The bromine is replaced by an –NH₂ group, again on carbon 2.

$\text{CH}_3\text{CHBrCH}_3 + \text{NH}_3 \rightarrow \text{CH}_3\text{CH}(\text{NH}_2)\text{CH}_3 + \text{HBr}$

Reactions of haloalkanes

Two reactions, decided by the solvent

Substitution with hydroxide

Elimination with KOH in alcohol

Substitution with ammonia

Substitution with a primary amine

Reactivity of the halogens

(a) NaOH(aq)\text{NaOH}(aq)NaOH(aq), warmed

(b) KOH in ethanol, heated

(c) Concentrated NH3\text{NH}_3NH3​, heated in a sealed tube

Reactions of haloalkanes

Two reactions, decided by the solvent

Substitution with hydroxide

Elimination with KOH in alcohol

Substitution with ammonia

Substitution with a primary amine

Reactivity of the halogens

(a) NaOH(aq)\text{NaOH}(aq)NaOH(aq), warmed

(b) KOH in ethanol, heated

(c) Concentrated NH3\text{NH}_3NH3​, heated in a sealed tube

(a) $\text{NaOH}(aq)$ , warmed

(c) Concentrated $\text{NH}_3$ , heated in a sealed tube

(a) $\text{NaOH}(aq)$ , warmed

(c) Concentrated $\text{NH}_3$ , heated in a sealed tube