Dehydration and Substitution of Alcohols

Beyond oxidation, alcohols have two more important synthetic transformations at A-Level: dehydration (losing water to give an alkene) and substitution (replacing the –OH with a halogen to give a haloalkane). Both are workhorses of organic synthesis — dehydration gives alkenes for addition reactions and polymers, while substitution opens the door to the haloalkane chemistry of the next few lessons.

This lesson covers the OCR A-Level Chemistry A (H432) specification points 4.2.1 (d)–(e): dehydration of alcohols to alkenes using concentrated phosphoric acid or concentrated sulfuric acid; conversion of alcohols to haloalkanes by substitution with hydrogen halides.

1. Dehydration of Alcohols

Dehydration is the elimination of a water molecule from an alcohol to form an alkene. It is formally the reverse of the acid-catalysed hydration of an alkene (covered in the alkenes lesson of the previous course).

$R\text{-}CH_2\text{-}CH_2\text{-}OH \xrightarrow{\text{acid, heat}} R\text{-}CH\text{=}CH_2 + H_2O$

1.1 Reagents and Conditions

Two acids are used as dehydration catalysts:

Acid	Typical conditions
Concentrated phosphoric acid, H₃PO₄	Heat under reflux, ~170 °C
Concentrated sulfuric acid, H₂SO₄	Heat under reflux, ~170 °C

Phosphoric acid is the OCR-preferred reagent for a "clean" dehydration because sulfuric acid sometimes over-oxidises the alcohol, giving unwanted side-products (SO₂, CO₂ and blackening from charring).

The acid catalyst protonates the –OH to give a good leaving group (water), and then water leaves and an H is lost from the adjacent carbon — the definition of an elimination reaction.

1.2 Reaction Type and Mechanism Summary

Dehydration of an alcohol is a β-elimination: the leaving group (H₂O) leaves from one carbon, and an H⁺ is lost from the adjacent (β) carbon, forming the new C=C double bond.

graph LR
  A[Alcohol<br/>R-CH2-CH2-OH] -->|Step 1<br/>protonation| B[R-CH2-CH2-OH2+]
  B -->|Step 2<br/>water leaves| C[R-CH2-CH2+<br/>carbocation]
  C -->|Step 3<br/>beta-H lost| D[R-CH=CH2 alkene]
  D -->|+ H2O| E[Alkene product]

You are not required to know the full mechanism at OCR A-Level, but understanding it helps you predict products when there is more than one possible alkene.

1.3 Dehydrating Ethanol

The simplest example:

$C_2H_5OH \xrightarrow{\text{conc H}_3\text{PO}_4,\, 170^\circ C} C_2H_4 + H_2O$

Ethanol loses water to form ethene. Only one product is possible because ethanol is symmetric.

1.4 Unsymmetric Alcohols — Multiple Products

When the alcohol is not symmetric, dehydration can produce more than one alkene, depending on which β-hydrogen is lost. Consider butan-2-ol:

Losing an H from C1 gives but-1-ene (CH₂=CH–CH₂–CH₃).
Losing an H from C3 gives but-2-ene (CH₃–CH=CH–CH₃). This can exist as E or Z isomers.

The major product follows Zaitsev's rule: the alkene with the more substituted (higher-degree) double bond is preferred because it is more thermodynamically stable. Here, but-2-ene is the major product and but-1-ene the minor.

graph TD
  A[Butan-2-ol<br/>CH3-CH OH -CH2-CH3] --> B[Lose H from C1]
  A --> C[Lose H from C3]
  B --> D[But-1-ene<br/>minor]
  C --> E[E and Z But-2-ene<br/>major]

Tip: Unless asked specifically, you should draw all possible alkenes, including stereoisomers (E/Z) where relevant. Exam questions love to give you three marks here — one for each alkene.

1.5 Worked Example — Dehydration of 2-Methylbutan-2-ol

Draw all possible alkenes produced by the dehydration of 2-methylbutan-2-ol.

2-Methylbutan-2-ol is (CH₃)₂C(OH)–CH₂–CH₃. The C–OH carbon is bonded to three other carbons: two CH₃ groups and a CH₂CH₃ group. Each of those three groups carries β-hydrogens.

Losing an H from a CH₃: the other CH₃ stays, the CH₂CH₃ stays → (CH₃)(CH₂CH₃)C=CH₂, i.e. 2-methylbut-1-ene.
Losing an H from the CH₂ of the ethyl group: → (CH₃)₂C=CH–CH₃, i.e. 2-methylbut-2-ene.

(The two CH₃ groups are equivalent, so there is only one product from losing a methyl H.) Therefore there are two products — 2-methylbut-1-ene and 2-methylbut-2-ene. Zaitsev predicts that 2-methylbut-2-ene (trisubstituted double bond) is the major product.

2. Substitution of Alcohols: Forming Haloalkanes

Alcohols can be converted into haloalkanes by substituting the –OH with a halogen atom. The general reaction is:

$R\text{-}OH + HX \rightarrow R\text{-}X + H_2O \quad (X = Cl, Br, I)$

The reagent is usually generated in situ from a sodium halide and a mineral acid, because hydrogen halide gases are hard to handle:

NaBr + H₂SO₄ → NaHSO₄ + HBr (then HBr attacks the alcohol)
NaCl + H₂SO₄ → NaHSO₄ + HCl
NaI cannot be used with concentrated H₂SO₄ because H₂SO₄ oxidises HI to I₂; use NaI + H₃PO₄ instead.

2.1 Example: Ethanol to Bromoethane

$C_2H_5OH + HBr \rightarrow C_2H_5Br + H_2O$

Lesson 3: Dehydration and Substitution of Alcohols