Topics

Elimination reactions

Covers AS91165

What elimination is

In an elimination reaction a small molecule is removed from a compound, and a C=C double bond forms where it left.
The product is always an alkene.
Two starting materials do this at Level 2:
- alcohols lose water (a dehydration),
- haloalkanes lose a hydrogen halide.
Elimination is the reverse of addition — addition opens a double bond, elimination creates one.

Elimination from an alcohol: dehydration

Conditions: concentrated sulfuric acid, heated (hot aluminium oxide also works).
The –OH leaves from one carbon and an H leaves from an adjacent carbon. Together they make water.

$CH_3CH_2OH \xrightarrow[\text{heat}]{\text{conc. } H_2SO_4} CH_2{=}CH_2 + H_2O$

Ethanol → ethene.
This is exactly the reverse of hydrating an alkene to make an alcohol.

Elimination from a haloalkane

Conditions: KOH dissolved in ethanol, heated.
The halogen leaves from one carbon and an H leaves from an adjacent carbon, together forming a hydrogen halide.

$CH_3CH_2Br + KOH \xrightarrow{\text{ethanol, heat}} CH_2{=}CH_2 + KBr + H_2O$

The conditions decide everything

This is the most important contrast in the whole standard, and it is a favourite exam question:

Conditions	Reaction	Product
KOH in water	substitution	an alcohol
KOH in ethanol, heated	elimination	an alkene

Same haloalkane, same reagent — only the solvent differs, and the product changes family.

Major and minor products in elimination

If the hydrogen can be taken from either side of the functional-group carbon, two different alkenes are possible.
The major product is the more substituted alkene — the one with more carbon groups attached to the C=C carbons.

Worked through for butan-2-ol, $CH_3CH(OH)CH_2CH_3$ :

The –OH is on carbon 2. A hydrogen can be removed from carbon 1 or from carbon 3.
Removing it from carbon 3 puts the double bond between carbons 2 and 3 → but-2-ene. Each C=C carbon carries a carbon group — more substituted. MAJOR.
Removing it from carbon 1 puts the double bond between carbons 1 and 2 → but-1-ene. Carbon 1 carries only hydrogens — less substituted. MINOR.

The method

Find the carbon carrying the –OH or the halogen.
List the adjacent carbons that have at least one hydrogen.
For each, remove that H and the functional group, and put a double bond between the two carbons.
Compare the alkenes: the one with more carbon groups on the C=C is the major product.
Name both, and label which is which.

Tips and Tricks

Give the conditions every time, because they are what distinguishes elimination from substitution. For alcohols write "concentrated $H_2SO_4$ , heated"; for haloalkanes write "KOH in ethanol, heated". An answer saying only "react with KOH" cannot be given the elimination marks.

Do not confuse the two major-product rules. For addition to an alkene it is Markovnikov — the H joins the carbon that already has more hydrogens. For elimination it is the more substituted alkene that wins. They are different rules for different reactions, and swapping them is a reliable way to get an Excellence question wrong.

Finally, check that the hydrogen you remove is on a carbon adjacent to the functional group. Taking one from further along the chain would put the double bond in the wrong place.

Worked ExampleElimination with two possible products

2-bromobutane is heated with potassium hydroxide dissolved in ethanol. Identify the major and minor products, name both, and justify which is major.

Step 1 — Identify the reaction from the conditions

The KOH is dissolved in ethanol and the mixture is heated. These are elimination conditions, so a hydrogen halide (HBr) will be removed and an alkene will form.

(Had the KOH been in water, this would instead have been a substitution giving butan-2-ol.)

Step 2 — Find where the hydrogen can come from

2-bromobutane is $CH_3CHBrCH_2CH_3$ . The bromine is on carbon 2.

The hydrogen must come from a carbon next to carbon 2 — so from carbon 1 or carbon 3. Both have hydrogens, so two products are possible.

Step 3 — Work out both alkenes

Taking the H from carbon 3: the double bond forms between carbons 2 and 3.

$CH_3CH{=}CHCH_3 \quad \text{but-2-ene}$

Taking the H from carbon 1: the double bond forms between carbons 1 and 2.

$CH_2{=}CHCH_2CH_3 \quad \text{but-1-ene}$

Step 4 — Decide which is major

Count the carbon groups attached to the two C=C carbons:

But-2-ene — carbon 2 carries a $CH_3$ and carbon 3 carries a $CH_3$ . Two carbon groups — more substituted.
But-1-ene — carbon 1 carries only two hydrogens; carbon 2 carries one $CH_2CH_3$ . One carbon group — less substituted.

The more substituted alkene is the major product.

Elimination reactions

Covers AS91165

What elimination is

In an elimination reaction a small molecule is removed from a compound, and a C=C double bond forms where it left.
The product is always an alkene.
Two starting materials do this at Level 2:
- alcohols lose water (a dehydration),
- haloalkanes lose a hydrogen halide.
Elimination is the reverse of addition — addition opens a double bond, elimination creates one.

Elimination from an alcohol: dehydration

Conditions: concentrated sulfuric acid, heated (hot aluminium oxide also works).
The –OH leaves from one carbon and an H leaves from an adjacent carbon. Together they make water.

$CH_3CH_2OH \xrightarrow[\text{heat}]{\text{conc. } H_2SO_4} CH_2{=}CH_2 + H_2O$

Ethanol → ethene.
This is exactly the reverse of hydrating an alkene to make an alcohol.

Elimination from a haloalkane

Conditions: KOH dissolved in ethanol, heated.
The halogen leaves from one carbon and an H leaves from an adjacent carbon, together forming a hydrogen halide.

$CH_3CH_2Br + KOH \xrightarrow{\text{ethanol, heat}} CH_2{=}CH_2 + KBr + H_2O$

The conditions decide everything

This is the most important contrast in the whole standard, and it is a favourite exam question:

Conditions	Reaction	Product
KOH in water	substitution	an alcohol
KOH in ethanol, heated	elimination	an alkene

Same haloalkane, same reagent — only the solvent differs, and the product changes family.

Major and minor products in elimination

If the hydrogen can be taken from either side of the functional-group carbon, two different alkenes are possible.
The major product is the more substituted alkene — the one with more carbon groups attached to the C=C carbons.

Worked through for butan-2-ol, $CH_3CH(OH)CH_2CH_3$ :

The –OH is on carbon 2. A hydrogen can be removed from carbon 1 or from carbon 3.
Removing it from carbon 3 puts the double bond between carbons 2 and 3 → but-2-ene. Each C=C carbon carries a carbon group — more substituted. MAJOR.
Removing it from carbon 1 puts the double bond between carbons 1 and 2 → but-1-ene. Carbon 1 carries only hydrogens — less substituted. MINOR.

The method

Find the carbon carrying the –OH or the halogen.
List the adjacent carbons that have at least one hydrogen.
For each, remove that H and the functional group, and put a double bond between the two carbons.
Compare the alkenes: the one with more carbon groups on the C=C is the major product.
Name both, and label which is which.

Tips and Tricks

Finally, check that the hydrogen you remove is on a carbon adjacent to the functional group. Taking one from further along the chain would put the double bond in the wrong place.

Worked ExampleElimination with two possible products

2-bromobutane is heated with potassium hydroxide dissolved in ethanol. Identify the major and minor products, name both, and justify which is major.

Step 1 — Identify the reaction from the conditions

The KOH is dissolved in ethanol and the mixture is heated. These are elimination conditions, so a hydrogen halide (HBr) will be removed and an alkene will form.

(Had the KOH been in water, this would instead have been a substitution giving butan-2-ol.)

Step 2 — Find where the hydrogen can come from

2-bromobutane is $CH_3CHBrCH_2CH_3$ . The bromine is on carbon 2.

The hydrogen must come from a carbon next to carbon 2 — so from carbon 1 or carbon 3. Both have hydrogens, so two products are possible.

Step 3 — Work out both alkenes

Taking the H from carbon 3: the double bond forms between carbons 2 and 3.

$CH_3CH{=}CHCH_3 \quad \text{but-2-ene}$

Taking the H from carbon 1: the double bond forms between carbons 1 and 2.

$CH_2{=}CHCH_2CH_3 \quad \text{but-1-ene}$

Step 4 — Decide which is major

Count the carbon groups attached to the two C=C carbons:

But-2-ene — carbon 2 carries a $CH_3$ and carbon 3 carries a $CH_3$ . Two carbon groups — more substituted.
But-1-ene — carbon 1 carries only two hydrogens; carbon 2 carries one $CH_2CH_3$ . One carbon group — less substituted.

The more substituted alkene is the major product.