Alcohol Elimination Reactions: Mechanisms and Alkene Formation Revision Notes | A Level AQA Chemistry

1. Why Elimination Matters: Alcohols to Alkenes

Alcohols are one of the most versatile functional groups in organic chemistry, and one of their most important reactions is the ability to lose water and form an alkene. This process, called dehydration or acid-catalysed elimination, is the reverse of the hydration of alkenes you may have met earlier in the course.

There is a compelling reason to care about this reaction beyond the mechanism itself. Alkenes produced from alcohols can be polymerised into addition polymers — and if the alcohol comes from a renewable source such as fermented sugars, the entire polymer chain is produced without any dependence on crude oil. That makes this reaction a genuine green chemistry story, and AQA examiners test it directly.

2. Conditions for Elimination

To drive an alcohol towards elimination rather than substitution, you need two things: an acid catalyst and heat. The standard AQA conditions are concentrated sulphuric acid (or concentrated phosphoric acid) and heat. Phosphoric acid is often preferred in industrial contexts because it is less corrosive and produces a cleaner product.

The overall equation for the dehydration of ethanol is a clean starting point: $\text{CH}_3\text{CH}_2\text{OH}_{(l)} \xrightarrow{\text{conc. H}_2\text{SO}_4,\, \Delta} \text{CH}_2{=}\text{CH}_2{}_{(g)} + \text{H}_2\text{O}_{(l)}$. One molecule of water is lost for every molecule of alcohol, and the acid catalyst is not consumed overall.

3. The Elimination Mechanism: Step by Step

The mechanism has three distinct stages. Understanding each stage — not just memorising the arrows — is what separates a grade-A answer from a grade-C one. Picture the alcohol molecule sitting in concentrated acid: the first thing that happens is protonation of the oxygen.

Worked example: Tracing the Elimination Mechanism for Propan-2-ol

Outline the mechanism for the acid-catalysed elimination of propan-2-ol to form propene. Identify the type of intermediate formed.

1. OH$^-$ is a poor leaving group. The acid protonates the oxygen: $\text{CH}_3\text{CH(OH)CH}_3 + \text{H}^+ \rightarrow \text{CH}_3\text{CH(OH}_2^+\text{)CH}_3$ — Draw a curly arrow from a lone pair on oxygen to the incoming H+. This converts OH into OH2+, which is water — an excellent leaving group.
2. $\text{CH}_3\text{CH(OH}_2^+\text{)CH}_3 \rightarrow \text{CH}_3\overset{+}{\text{CH}}\text{CH}_3 + \text{H}_2\text{O}$ — Draw a curly arrow from the C-O bond to the oxygen. Both electrons leave with the water molecule, leaving a secondary carbocation on C-2. This is the rate-determining step in an E1 mechanism.
3. $\text{CH}_3\overset{+}{\text{CH}}\text{CH}_3 \xrightarrow{\text{base}} \text{CH}_3\text{CH}{=}\text{CH}_2 + \text{H}^+$ — Draw a curly arrow from a C-H bond on an adjacent carbon to the C-C bond, forming the pi bond. The proton is removed by a base (water or HSO4-). The acid catalyst is regenerated.
4. Intermediate: secondary carbocation $\text{CH}_3\overset{+}{\text{CH}}\text{CH}_3$ — AQA mark schemes specifically ask you to name the intermediate. A secondary carbocation is more stable than a primary one because two alkyl groups donate electron density towards the positive charge.

Answer: The mechanism proceeds via a secondary carbocation intermediate. The three stages are: protonation of OH, loss of water to form the carbocation, and deprotonation of an adjacent carbon to form the C=C double bond (propene).

Notice that the curly arrows in step 2 start from the C-O bond — not from the carbon atom itself. This is a critical curly arrow rule: arrows must originate from a bond or a lone pair, never from an atom label. Examiners penalise arrows drawn from the wrong source.

4. Predicting the Major Product: Zaitsev's Rule

For simple primary alcohols such as ethanol or propan-1-ol, there is only one possible alkene product. Some branched primary alcohols (e.g. 2-methylpropan-1-ol) may give more than one alkene product. With secondary and tertiary alcohols, however, the carbocation intermediate has adjacent carbons on more than one side, and a proton could be removed from either. This is where Zaitsev's rule becomes essential.

Consider butan-2-ol: the carbocation forms at C-2. Removing H from C-1 gives but-1-ene (one alkyl substituent on the double bond). Removing H from C-3 gives but-2-ene (two alkyl substituents). By Zaitsev's rule, but-2-ene is the major product. You can also note that but-2-ene exists as cis and trans isomers — both are formed, with the trans isomer typically predominating due to reduced steric strain.

5. Carbocation Stability and Why It Matters

The stability of the carbocation intermediate directly influences both the rate of elimination and the product distribution. Tertiary carbocations are the most stable, secondary are intermediate, and primary carbocations are the least stable. This is why tertiary alcohols undergo elimination most readily. Primary alcohols do not readily form primary carbocations (which are too unstable for a straightforward E1 pathway); instead, they may undergo elimination via an E2 (concerted) mechanism or require more forcing conditions.

Alkyl groups attached to the positively charged carbon donate electron density through inductive effects and hyperconjugation, delocalising the positive charge and lowering the energy of the intermediate. A more stable intermediate means a lower activation energy for its formation, which is why tertiary alcohols react faster.

6. Alkenes from Alcohols: The Green Chemistry Connection

Here is where the chemistry connects to a much bigger picture. Traditionally, alkene monomers for addition polymerisation — ethene for poly(ethene), propene for poly(propene) — are obtained by cracking crude oil fractions. This is a finite, non-renewable resource. The elimination route offers an alternative: ferment glucose (from plant material) to produce ethanol, then dehydrate the ethanol to ethene, then polymerise the ethene.

The resulting poly(ethene) is chemically identical to the petroleum-derived version. The difference is entirely in the feedstock: one comes from a renewable biological source, the other from fossil fuels. AQA specifically requires you to be able to discuss this as a reason why alkene formation from alcohols is significant — it is a route to addition polymers that does not depend on crude oil.

• Glucose (from crops or waste biomass) is fermented by yeast to produce ethanol: $C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2$
• Ethanol is dehydrated over concentrated acid and heat to produce ethene: $C_2H_5OH \rightarrow CH_2{=}CH_2 + H_2O$
• Ethene undergoes addition polymerisation to produce poly(ethene): $n\,CH_2{=}CH_2 \rightarrow (-CH_2-CH_2-)_n$
• The entire chain uses renewable plant material rather than crude oil as the carbon source

8. Exam Technique and Common Mistakes

Mechanism questions on elimination are among the most mark-rich and most frequently dropped sections in AQA organic chemistry papers. The marks are very specific about arrow placement, intermediate labelling, and product naming. Here is what the mark scheme is actually looking for.