Alcohols contain a polar carbon-oxygen bond, making them highly versatile starting materials in synthetic organic chemistry. In this lesson, we study three key transformations: dehydration (an elimination pathway to form alkenes), halide substitution (to yield halogenoalkanes), and esterification (to synthesise sweet-smelling esters).
🔑 Key Principle
The dehydration of an unsymmetrical alcohol can produce a mixture of isomeric alkenes. This occurs because the hydrogen atom eliminated can be lost from either of the carbon atoms adjacent to the C-OH carbon.
Dehydration (Elimination) of Alcohols
Under acidic conditions and heat, alcohols undergo a dehydration reaction, losing a water molecule to form a carbon-carbon double bond (alkene):
An elimination reaction in which a molecule of water is removed from a reactant, transforming an alcohol into an alkene.
A type of organic reaction in which two substituents are removed from a molecule in a mechanism, forming a double or triple bond.
This reaction requires specific conditions:
- Catalyst: Concentrated phosphoric acid (\( \text{H}_3\text{PO}_4 \)) or concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)). Alternatively, vapourised alcohol can be passed over a hot aluminium oxide (\( \text{Al}_2\text{O}_3 \)) catalyst.
- Temperature: Heated under reflux (typically around \( 170^\circ\text{C} \)).
Dehydration is an elimination reaction. In exam questions, if you are asked to identify the type of reaction, always write 'elimination'. Concentrated phosphoric acid (\( \text{H}_3\text{PO}_4 \)) is preferred in the laboratory over concentrated sulfuric acid because it is a safer acid that does not produce toxic sulfur dioxide gas by-products.
Dehydration Mechanism
The elimination mechanism proceeds via three main steps:
- Protonation: The oxygen atom of the hydroxyl group uses one of its lone pairs to accept a proton (\( \text{H}^+ \)) from the acid catalyst, forming a protonated intermediate.
- Loss of Water: The C-O bond breaks heterolytically, releasing a water molecule and leaving behind a carbocation intermediate.
- Deprotonation: A hydrogen atom on a carbon atom adjacent to the positive carbon is lost as \( \text{H}^+ \), and the bonding electron pair collapses to form the C=C double bond, regenerating the catalyst.
Solution:
The structural formula of butan-2-ol is \( \text{CH}_3\text{-CH}_2\text{-CH(OH)-CH}_3 \). In the elimination mechanism, the hydroxyl group is lost along with a hydrogen atom from an adjacent carbon atom (either carbon-1 or carbon-3):
- If the hydrogen is lost from carbon-1 (\( \text{-CH}_3 \)): The double bond forms between carbon-1 and carbon-2, producing but-1-ene:
\( \text{CH}_3\text{CH}_2\text{CH=CH}_2 \) - If the hydrogen is lost from carbon-3 (\( \text{-CH}_2- \)): The double bond forms between carbon-2 and carbon-3, producing but-2-ene:
\( \text{CH}_3\text{CH=CH-CH}_3 \)
Because but-2-ene has stereocentres and exhibits stereoisomerism, it exists as a pair of geometric isomers: cis-but-2-ene (Z-but-2-ene) and trans-but-2-ene (E-but-2-ene). This brings the total number of alkene products in the mixture to three.
Halide Substitution Reactions
Alcohols can undergo acid-catalysed nucleophilic substitution reactions when treated with halide ions (such as bromide, chloride, or iodide) in the presence of an acid. This converts the hydroxyl group into a halogen atom, forming a halogenoalkane.
For example, to synthesise bromoethane from ethanol, the alcohol is heated under reflux with potassium bromide (\( \text{KBr} \)) and concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)). The acid reacts with potassium bromide in situ to produce hydrogen bromide (\( \text{HBr} \)), which then reacts with the alcohol:
Esterification
Alcohols react with carboxylic acids when heated in the presence of a strong acid catalyst to form esters and water. This is a reversible condensation reaction:
A reversible reaction in which an alcohol and a carboxylic acid combine to form an ester and water, typically catalysed by concentrated sulfuric acid.
Reaction conditions: Heated under reflux with a concentrated sulfuric acid catalyst (\( \text{H}_2\text{SO}_4 \)).
When naming an ester, always identify the alcohol part and the carboxylic acid part first. The name is composed of two words:
1. The alkyl prefix derived from the alcohol (e.g. ethanol becomes ethyl).
2. The carboxylate suffix derived from the carboxylic acid (e.g. propanoic acid becomes propanoate).
Thus, the reaction of ethanol and propanoic acid yields ethyl propanoate. The concentrated sulfuric acid catalyst acts both as a proton source to speed up the reaction and as a dehydrating agent to shift the equilibrium position towards the products.
Solution:
Carboxylic acid: Ethanoic acid (\( \text{CH}_3\text{COOH} \))
Alcohol: Methanol (\( \text{CH}_3\text{OH} \))
During the reaction, the hydroxyl group is lost from the acid, and the hydrogen atom is lost from the alcohol's hydroxyl group to form a water molecule:
\[ \text{CH}_3\text{COOH} + \text{CH}_3\text{OH} \rightleftharpoons \text{CH}_3\text{COOCH}_3 + \text{H}_2\text{O} \]The ester contains a methyl group from the alcohol and an ethanoate group from the acid. The IUPAC name is methyl ethanoate.
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