Topics

Substitution reactions of alcohols

Swapping the –OH for a halogen

In a substitution reaction, the –OH group is replaced by a halogen atom, turning the alcohol into a haloalkane.
This is one of the two main things an alcohol can do (the other being oxidation), and it is the route alcohol → haloalkane on the reaction map.

The four reagents

Reagent	Product	Note
$HX$ (HCl, HBr, HI)	the matching haloalkane	the only reagent that lets you choose the halogen
$PCl_3$	a chloroalkane	always gives –Cl
$PCl_5$	a chloroalkane	always gives –Cl
$SOCl_2$	a chloroalkane	always gives –Cl

Three of the four reagents contain chlorine only, so they can only make chloroalkanes.
If a question asks you to make a bromoalkane or an iodoalkane from an alcohol, the reagent must be HBr or HI.

With a hydrogen halide

$CH_3CH_2OH + HBr \rightarrow CH_3CH_2Br + H_2O$

The –OH is replaced by –Br, and the second product is water.
The halogen ends up on the same carbon the –OH was on. The skeleton does not change.

With the chlorinating reagents

$CH_3CH_2OH + PCl_5 \rightarrow CH_3CH_2Cl + POCl_3 + HCl$

The organic product is the chloroalkane — that is what a question is asking for.
The inorganic by-products differ between $PCl_3$ , $PCl_5$ and $SOCl_2$ , and you are not expected to reproduce them.

Working out the product

Find the carbon carrying the –OH.
Replace the whole –OH group with the halogen atom — on that same carbon.
Leave the rest of the molecule alone.
Name the product as a haloalkane, using the same locant the alcohol had.

Worked through:

propan-2-ol + HBr → the –OH on carbon 2 becomes –Br on carbon 2 → 2-bromopropane.
butan-1-ol + $SOCl_2$ → –OH on carbon 1 becomes –Cl on carbon 1 → 1-chlorobutane.

Substitution compared with the other reaction types

	What happens	Products
Substitution	one group swapped for another	organic product + a second product
Addition	a double bond opens, atoms add on	one product only
Elimination	a small molecule is removed, a double bond forms	organic product + the small molecule

Worked ExampleChoosing the right reagent

A chemist has a sample of butan-2-ol. Describe how it could be converted into (a) 2-chlorobutane and (b) 2-bromobutane, giving a reagent and an equation for each.

Step 1 — Identify what has to change

Butan-2-ol is $CH_3CH(OH)CH_2CH_3$ — the –OH is on carbon 2.

Both targets are haloalkanes with the halogen on carbon 2, so in each case the –OH must be substituted by a halogen atom, on the same carbon. The carbon skeleton does not change.

Step 2 — (a) Making the chloroalkane

Any of the chlorinating reagents will do this: $PCl_3$ , $PCl_5$ , $SOCl_2$ — or HCl.

Using $PCl_5$ :

$CH_3CH(OH)CH_2CH_3 + PCl_5 \rightarrow CH_3CHClCH_2CH_3 + POCl_3 + HCl$

The organic product is 2-chlorobutane.

Step 3 — (b) Making the bromoalkane

Here the choice is restricted. $PCl_3$ , $PCl_5$ and $SOCl_2$ all contain chlorine only, so none of them can make a bromoalkane. The reagent must be a hydrogen halide containing bromine: HBr.

$CH_3CH(OH)CH_2CH_3 + HBr \rightarrow CH_3CHBrCH_2CH_3 + H_2O$

The organic product is 2-bromobutane, and the second product is water.

Substitution reactions of alcohols

Covers AS91165

Swapping the –OH for a halogen

In a substitution reaction, the –OH group is replaced by a halogen atom, turning the alcohol into a haloalkane.
This is one of the two main things an alcohol can do (the other being oxidation), and it is the route alcohol → haloalkane on the reaction map.

The four reagents

Reagent	Product	Note
$HX$ (HCl, HBr, HI)	the matching haloalkane	the only reagent that lets you choose the halogen
$PCl_3$	a chloroalkane	always gives –Cl
$PCl_5$	a chloroalkane	always gives –Cl
$SOCl_2$	a chloroalkane	always gives –Cl

Three of the four reagents contain chlorine only, so they can only make chloroalkanes.
If a question asks you to make a bromoalkane or an iodoalkane from an alcohol, the reagent must be HBr or HI.

With a hydrogen halide

$CH_3CH_2OH + HBr \rightarrow CH_3CH_2Br + H_2O$

The –OH is replaced by –Br, and the second product is water.
The halogen ends up on the same carbon the –OH was on. The skeleton does not change.

With the chlorinating reagents

$CH_3CH_2OH + PCl_5 \rightarrow CH_3CH_2Cl + POCl_3 + HCl$

The organic product is the chloroalkane — that is what a question is asking for.
The inorganic by-products differ between $PCl_3$ , $PCl_5$ and $SOCl_2$ , and you are not expected to reproduce them.

Working out the product

Find the carbon carrying the –OH.
Replace the whole –OH group with the halogen atom — on that same carbon.
Leave the rest of the molecule alone.
Name the product as a haloalkane, using the same locant the alcohol had.

Worked through:

propan-2-ol + HBr → the –OH on carbon 2 becomes –Br on carbon 2 → 2-bromopropane.
butan-1-ol + $SOCl_2$ → –OH on carbon 1 becomes –Cl on carbon 1 → 1-chlorobutane.

Substitution compared with the other reaction types

	What happens	Products
Substitution	one group swapped for another	organic product + a second product
Addition	a double bond opens, atoms add on	one product only
Elimination	a small molecule is removed, a double bond forms	organic product + the small molecule

Worked ExampleChoosing the right reagent

A chemist has a sample of butan-2-ol. Describe how it could be converted into (a) 2-chlorobutane and (b) 2-bromobutane, giving a reagent and an equation for each.

Step 1 — Identify what has to change

Butan-2-ol is $CH_3CH(OH)CH_2CH_3$ — the –OH is on carbon 2.

Both targets are haloalkanes with the halogen on carbon 2, so in each case the –OH must be substituted by a halogen atom, on the same carbon. The carbon skeleton does not change.

Step 2 — (a) Making the chloroalkane

Any of the chlorinating reagents will do this: $PCl_3$ , $PCl_5$ , $SOCl_2$ — or HCl.

Using $PCl_5$ :

$CH_3CH(OH)CH_2CH_3 + PCl_5 \rightarrow CH_3CHClCH_2CH_3 + POCl_3 + HCl$

The organic product is 2-chlorobutane.

Step 3 — (b) Making the bromoalkane

$CH_3CH(OH)CH_2CH_3 + HBr \rightarrow CH_3CHBrCH_2CH_3 + H_2O$

The organic product is 2-bromobutane, and the second product is water.