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Major and minor products of elimination

Covers AS91391

When two alkenes are possible

Elimination removes a hydrogen and a leaving group from adjacent carbons, forming a double bond between them.
If the carbon carrying the –OH or halogen has two different neighbouring carbons, the hydrogen can come from either, so two different alkenes form.
The standard calls these the major and minor products, and it names the source as asymmetric alcohols and haloalkanes.
If both neighbours are identical, only one alkene is possible and there is no major/minor distinction.

The rule

The major product is formed by removing the hydrogen from the carbon with the FEWEST hydrogen atoms attached.

The minor product comes from removing a hydrogen from the carbon with more hydrogens.
Both products do form — the question is which one forms in the greater amount.

Working one out

Find the carbon carrying the –OH or the halogen.
List its neighbouring carbons and count the hydrogens on each.
Major product: form the double bond towards the neighbour with fewer hydrogens.
Minor product: form the double bond towards the neighbour with more hydrogens.
Name both alkenes.

The standard example

For 2-bromobutane, $\text{CH}_3\text{CH}_2\text{CHBrCH}_3$ :

Neighbour	Hydrogens on it	Product	Which
C3 ( $\text{CH}_2$ )	2	$\text{CH}_3\text{CH}{=}\text{CHCH}_3$ , but-2-ene	MAJOR
C1 ( $\text{CH}_3$ )	3	$\text{CH}_2{=}\text{CHCH}_2\text{CH}_3$ , but-1-ene	minor

Note that the numbering is from the end that gives the bromine the lower locant, so the $\text{CH}_3$ next to the CHBr is C1 and the $\text{CH}_2$ is C3.

The same rule for alcohols

Alcohols dehydrated with concentrated $\text{H}_2\text{SO}_4$ follow exactly the same rule.
For pentan-2-ol, $\text{CH}_3\text{CH}(\text{OH})\text{CH}_2\text{CH}_2\text{CH}_3$ $CH_{3} CH (OH) CH_{2} CH_{2} CH_{3}$ :
- C3 is a $\text{CH}_2$ with 2 hydrogens → pent-2-ene, the major product
- C1 is a $\text{CH}_3$ with 3 hydrogens → pent-1-ene, the minor product

Cis and trans among the products

If the major alkene has two different groups on each carbon of its double bond, it will itself exist as cis and trans forms.
But-2-ene, for instance, forms as a mixture of cis-but-2-ene and trans-but-2-ene.
You will not normally be asked which of those predominates, but you may be asked to note that both exist.

Tips and Tricks

Count the hydrogens and write the numbers down. The rule is easy to state and easy to apply backwards under pressure. Writing "C1 has 3 H, C3 has 2 H" next to the structure takes five seconds and fixes the answer.

The major product comes from the carbon with FEWER hydrogens. If you remember it as "more hydrogens", every answer in the question will be wrong.

Give both products when a question asks for the products of an elimination from an asymmetric substrate, and label which is which. Naming only the major one loses half the marks.

Check first whether the two neighbours are actually different. 2-bromopropane and 2-bromo-2-methylpropane both have identical neighbours, so each gives only one alkene and the rule does not apply.

A carbon with no hydrogens on it cannot be the source, so exclude it before you start counting.

Worked ExamplePredicting major and minor products

2-methylbutan-2-ol, $(\text{CH}_3)_2\text{C}(\text{OH})\text{CH}_2\text{CH}_3$ , is heated with concentrated sulfuric acid. Draw and name the major and minor organic products, and justify your assignment.

Step 1 — Draw the skeleton and number it

The parent chain is butan (four carbons), the –OH is on carbon 2, and there is a methyl branch also on carbon 2.

$\overset{1}{\text{C}}\text{H}_3 - \overset{2}{\text{C}}(\text{CH}_3)(\text{OH}) - \overset{3}{\text{C}}\text{H}_2 - \overset{4}{\text{C}}\text{H}_3$

Step 2 — Identify the carbon bearing the –OH and list its neighbours

The –OH is on carbon 2. The carbons adjacent to it are:

carbon 1, a $\text{CH}_3$ group
carbon 3, a $\text{CH}_2$ group
the methyl branch on carbon 2, also a $\text{CH}_3$ group

Step 3 — Count the hydrogens on each

Neighbour	Type	Hydrogens
carbon 1	$\text{CH}_3$	3
the methyl branch	$\text{CH}_3$	3
carbon 3	$\text{CH}_2$	2

Note that carbon 1 and the methyl branch are equivalent — both are methyl groups attached to carbon 2 — so removing a hydrogen from either gives the same alkene. There are therefore only two distinct products, not three.

Step 4 — Apply the rule

The major product comes from the neighbour with the fewest hydrogens, which is carbon 3 with two.

Forming the double bond between C2 and C3:

$\text{CH}_3-\text{C}(\text{CH}_3){=}\text{CH}-\text{CH}_3$

The minor product comes from a methyl neighbour, which has three hydrogens.

Forming the double bond between C2 and C1:

$\text{CH}_2{=}\text{C}(\text{CH}_3)-\text{CH}_2-\text{CH}_3$

Step 5 — State the justification

The major product is 2-methylbut-2-ene because it is formed by removing a hydrogen from carbon 3, which carries only two hydrogens — fewer than the three on either methyl group.