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Substitution reactions of haloalkanes

Swapping the halogen out

A haloalkane's halogen can be replaced by another group, in a substitution reaction.
At Level 2 there are exactly two reagents to know: aqueous potassium hydroxide and ammonia.

Reagent	Group that replaces the halogen	Product family
KOH(aq) — potassium hydroxide in water	–OH	alcohol
$NH_3$ — ammonia	–NH₂	primary amine

With aqueous potassium hydroxide

$CH_3CH_2Br + KOH \rightarrow CH_3CH_2OH + KBr$

The –Br is replaced by –OH, giving ethanol.
The solvent is critical: the KOH must be dissolved in water.
If the KOH is dissolved in ethanol and heated instead, you get a completely different reaction — elimination to an alkene. That is the next page.

With ammonia

$CH_3CH_2Br + NH_3 \rightarrow CH_3CH_2NH_2 + HBr$

The –Br is replaced by –NH₂, giving the primary amine ethanamine.
This is the only route to an amine in this standard, so a question asking you to make an amine always passes through a haloalkane.

Working out the product

Find the carbon carrying the halogen.
Replace the halogen with –OH (if KOH(aq)) or –NH₂ (if $NH_3$ ) — on the same carbon.
Leave the skeleton unchanged, keeping the same locant.
Name the product.

Worked through:

2-bromopropane + KOH(aq) → –OH on carbon 2 → propan-2-ol.
1-bromopropane + $NH_3$ → –NH₂ on carbon 1 → propan-1-amine.

Where this sits on the reaction map

Haloalkane → alcohol with KOH(aq).
Haloalkane → amine with $NH_3$ .
Both are the reverse direction of routes you already know: an alcohol becomes a haloalkane with HX or $PCl_5$ , and a haloalkane becomes an alcohol with KOH(aq). That two-way link between alcohols and haloalkanes is used constantly in multi-step questions.

Worked ExampleMaking an amine

Starting from 1-bromobutane, describe how propan-1-amine's four-carbon equivalent, butan-1-amine, could be prepared. Give the reagent, the equation, and the reaction type. Then state what would be formed instead if aqueous potassium hydroxide were used.

Step 1 — Identify what must change

The starting material is $CH_3CH_2CH_2CH_2Br$ , with the bromine on carbon 1.

The target, butan-1-amine, is $CH_3CH_2CH_2CH_2NH_2$ — the –NH₂ is on carbon 1, exactly where the bromine was.

So the halogen must be substituted by an –NH₂ group, in place.

Step 2 — Choose the reagent

The reagent that replaces a halogen with –NH₂ is ammonia, $NH_3$ .

Step 3 — Write the equation

$CH_3CH_2CH_2CH_2Br + NH_3 \rightarrow CH_3CH_2CH_2CH_2NH_2 + HBr$

The displaced bromine leaves as HBr.

Step 4 — Name the reaction type

The halogen has been swapped for another group on the same carbon, with a second product formed. This is a substitution reaction.

Step 5 — Answer the second part

With aqueous potassium hydroxide the halogen would be replaced by –OH instead:

$CH_3CH_2CH_2CH_2Br + KOH \rightarrow CH_3CH_2CH_2CH_2OH + KBr$

That gives butan-1-ol, an alcohol.