Dehydration and Substitution of Alcohols

Spec Mapping — OCR H432 Module 4.2.1(d)-(e) — Dehydration and substitution of alcohols, covering acid-catalysed dehydration of alcohols to alkenes using concentrated phosphoric(V) acid or concentrated sulfuric acid at ~170 °C; substitution of the alcohol -OH by a halide ion in the presence of a strong acid (HX generated in situ from NaBr/H₂SO₄, NaCl/H₂SO₄ or NaI/H₃PO₄) to give a haloalkane; the difference between intramolecular dehydration (one molecule → alkene + water) and intermolecular dehydration (two alcohols → ether + water — a brief aside); Zaitsev product selectivity for unsymmetric alcohols (refer to the official OCR H432 specification document for exact wording).

Beyond oxidation, alcohols have two more synthetically critical transformations at A-Level: dehydration (intramolecular loss of water to give an alkene) and substitution (replacement of -OH by halide to give a haloalkane). Both reactions exploit the same trick — the alcohol's -OH is converted into a good leaving group by acid, then either a β-elimination (dehydration) or a halide nucleophile (substitution) takes over. Dehydration is the laboratory synthesis of small alkenes for addition reactions and polymers and underpins the industrial cracking-then-polymerising route to polythene. Substitution is the entry point to all haloalkane chemistry (Lessons 4-6) — the conversion of butan-1-ol to 1-bromobutane is the textbook PAG 5 protocol. The intramolecular-vs-intermolecular dichotomy in dehydration (alkene at high T vs ether at lower T) is briefly addressed, but the OCR examined focus is alkene formation. This lesson develops both reactions with their reagents, conditions, mechanisms, product selectivity, and worked yield/atom-economy calculations.

Key Mechanism: dehydration proceeds via a three-step acid-catalysed E1 route — (1) protonation of the -OH to give -OH₂⁺, (2) loss of water to give a carbocation, (3) loss of a β-H to give the alkene. Substitution proceeds via the same first two steps but with a halide nucleophile attacking the carbocation in step 3 instead of a base removing a β-H. Whether substitution or elimination dominates depends on the substrate, the nucleophile, and the temperature (developed in Lessons 4-6).

Dehydration of alcohols

Dehydration is the acid-catalysed elimination of water from an alcohol to form an alkene. It is formally the reverse of the acid-catalysed hydration of an alkene that you met in the alkenes module of the basic-organic course.

$R-CH_2-CH_2-OH \xrightarrow[\text{170 °C}]{\text{conc. H}_3\text{PO}_4} R-CH=CH_2 + H_2O$

Reagents and conditions

Two acids are used as dehydration catalysts:

Acid	Concentration	Temperature	Notes
Concentrated phosphoric(V) acid, H₃PO₄	~85 %	~170 °C	OCR-preferred — gives a clean alkene without over-oxidation
Concentrated sulfuric acid, H₂SO₄	~98 %	~170 °C	Acceptable for ethanol, but can over-oxidise the alcohol to SO₂, CO₂ and blackened tar — especially with secondary alcohols

The acid catalyst protonates the -OH to convert it from a bad leaving group (-OH⁻ has a high pKa, ~16) to a good one (water, leaving group conjugate-acid pKa ~ -1.7). After water has left, the molecule is a carbocation, and loss of a β-H to a base (water, HSO₄⁻ or H₂PO₄⁻) gives the alkene.

The E1 mechanism

flowchart LR
  A[R-CH2-CH2-OH] -->|Step 1: protonation H+| B[R-CH2-CH2-OH2+]
  B -->|Step 2: water leaves| C[R-CH2-CH2+ carbocation]
  C -->|Step 3: lose beta-H to base| D[R-CH=CH2 + H2O + H+]

You are not required to draw curly arrows for the full mechanism at OCR A-Level (the spec asks for the overall change), but understanding the carbocation intermediate is essential for predicting (i) the Zaitsev major product and (ii) why tertiary alcohols dehydrate faster than primary (tertiary carbocations are more stable).

Dehydrating ethanol

$\text{C}_2\text{H}_5\text{OH} \xrightarrow[\text{170 °C}]{\text{conc. H}_3\text{PO}_4} \text{C}_2\text{H}_4 + \text{H}_2\text{O}$

Ethanol loses water to give ethene. Only one product is possible because the molecule is symmetric — either CH₃ hydrogen can be lost and the result is the same.

Unsymmetric alcohols — multiple products (Zaitsev's rule)

When the alcohol is unsymmetric, dehydration can produce more than one alkene depending on which β-hydrogen is lost. Consider butan-2-ol, CH₃-CH(OH)-CH₂-CH₃:

Losing an H from C1 (a CH₃) gives but-1-ene, CH₂=CH-CH₂-CH₃.
Losing an H from C3 (the CH₂) gives but-2-ene, CH₃-CH=CH-CH₃, which can be either E (trans) or Z (cis).

The major product follows Zaitsev's rule: the more substituted (higher-degree) double bond is favoured because it is more thermodynamically stable. Disubstituted but-2-ene (one alkyl on each end) wins over monosubstituted but-1-ene (no alkyl on one end). Among the two but-2-ene stereoisomers, the E isomer is generally favoured (lower steric strain), although in laboratory practice you get a mixture of E and Z in roughly equal amounts.

flowchart TD
  A[Butan-2-ol] --> B[Protonate OH; lose H2O; gives secondary carbocation]
  B --> C[Lose H from C1 - CH3]
  B --> D[Lose H from C3 - CH2]
  C --> E[But-1-ene - minor monosubstituted]
  D --> F[E-but-2-ene + Z-but-2-ene - major disubstituted]

Worked example 1 — Dehydration of 2-methylbutan-2-ol

Draw all possible alkenes from acid-catalysed dehydration of 2-methylbutan-2-ol, $(\text{CH}_3)_2\text{C(OH)CH}_2\text{CH}_3$ . Predict the major product.

The C-OH carbon (C2) is bonded to two equivalent CH₃ groups and one CH₂CH₃ group. β-hydrogens are available on (i) the two methyls and (ii) the CH₂ of the ethyl branch.

Loss of an H from a methyl → $(\text{CH}_3)(\text{CH}_2\text{CH}_3)\text{C=CH}_2$ = 2-methylbut-1-ene (disubstituted).
Loss of an H from the ethyl CH₂ → $(\text{CH}_3)_2\text{C=CHCH}_3$ = 2-methylbut-2-ene (trisubstituted).

The two methyls are equivalent, so there is no distinct product from one versus the other. Zaitsev: 2-methylbut-2-ene is the more substituted (trisubstituted) alkene and is therefore the major product. 2-methylbut-1-ene is the minor product.

Worked example 2 — Dehydration of cyclohexanol

Cyclohexanol is dehydrated in concentrated H₃PO₄ at 170 °C. Write an equation and name the product. Why is this a common student preparation?

$\text{C}_6\text{H}_{11}\text{OH} \rightarrow \text{C}_6\text{H}_{10} + \text{H}_2\text{O}$

The product is cyclohexene. Only one alkene is possible because cyclohexanol is symmetric (all six β-hydrogens are equivalent by the C₆ rotational symmetry). The preparation is a popular school practical because: (i) the starting material is a cheap, low-toxicity solid, (ii) the product distils at 83 °C (clean separation), and (iii) the cyclohexene gives an immediate decolourisation of bromine water as a confirmatory test for the alkene. PAG 5 anchor.

Substitution of alcohols — forming haloalkanes

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

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

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

Target haloalkane	Reagents in flask	In situ HX
Chloroalkane	NaCl + concentrated H₂SO₄	HCl
Bromoalkane	NaBr + concentrated H₂SO₄	HBr
Iodoalkane	NaI + concentrated H₃PO₄	HI (H₂SO₄ oxidises HI to I₂, so must use phosphoric)

The acid serves two purposes: (i) generating HX in situ, and (ii) protonating the alcohol's -OH to make water a good leaving group. The halide ion then attacks the carbon.

Worked example 3 — Bromoethane from ethanol

$\text{C}_2\text{H}_5\text{OH} + \text{HBr} \rightarrow \text{C}_2\text{H}_5\text{Br} + \text{H}_2\text{O}$

PAG 5 protocol:

Add NaBr (~25 g) to ethanol (~25 cm³) in a round-bottom flask with a few anti-bumping granules.
Cool the flask in an ice bath and add concentrated H₂SO₄ (~15 cm³) dropwise (exothermic — risk of charring if too fast).
Set up reflux apparatus and heat for ~30 minutes.
Reconfigure for distillation; collect the fraction boiling at ~38 °C (bromoethane).
Wash the distillate with dilute Na₂CO₃ to remove acidic impurities, then dry with anhydrous calcium chloride.
Redistil to purify; collect bromoethane.

Reactivity order of HX

The order of reactivity towards alcohols is HI > HBr > HCl (≫ HF, which does not work for substitution). This is the opposite order of electronegativity (F > Cl > Br > I) but matches the H-X bond strength:

Hydrogen halide	Bond enthalpy of H-X (kJ mol⁻¹)
HF	568
HCl	432
HBr	366
HI	298

A weaker H-X bond is more easily heterolysed to give X⁻ (the actual nucleophile) and H⁺. The same trend matters in the next lesson when we consider hydrolysis of haloalkanes — C-I bonds hydrolyse fastest because they are weakest.

Mechanism overview

OCR does not require curly arrows for alcohol substitution at A-Level, but the mechanism is two steps:

Protonation of the -OH by the acid: $R-OH + H^+ \rightarrow R-OH_2^+$ . Water is now a good leaving group.
Halide attack on the carbon, water leaves: $R-OH_2^+ + X^- \rightarrow R-X + H_2O$ .

Primary alcohols proceed by an SN2-like concerted attack (water leaves as the halide attacks); tertiary alcohols proceed by SN1 (water leaves first to give a carbocation, which is then captured by X⁻); secondary alcohols can do either. You will meet the SN1/SN2 nomenclature explicitly in Lesson 4.

Worked example 4 — Iodoethane preparation

Write an equation for the formation of iodoethane from ethanol. Explain why concentrated phosphoric(V) acid must be used in place of sulfuric acid.

$\text{C}_2\text{H}_5\text{OH} + \text{HI} \rightarrow \text{C}_2\text{H}_5\text{I} + \text{H}_2\text{O}$

HI is generated in situ from NaI + H₃PO₄. Concentrated sulfuric acid cannot be used because H₂SO₄ is a strong oxidising agent that oxidises HI to I₂ (purple-brown vapours), preventing the substitution. Phosphoric acid is non-oxidising under these conditions, so the HI survives long enough to react with ethanol.

The same logic applies to HBr but more weakly — small amounts of Br₂ form when NaBr/H₂SO₄ is used, giving a faint orange tinge to the reaction mixture. For HI the oxidation is essentially quantitative, hence the need for the alternative acid.

Worked example 5 — Alternative halogenating reagents

The OCR specification mentions HX-from-NaX-and-acid as the standard route, but several cleaner alternatives are used in research synthesis:

Reagent	What it does	Byproducts
PCl₃	Converts -OH to -Cl	H₃PO₃
PCl₅	Converts -OH to -Cl	POCl₃ + HCl
SOCl₂ (thionyl chloride)	Converts -OH to -Cl	SO₂ + HCl (both gases)
PBr₃	Converts -OH to -Br	H₃PO₃
Red P + I₂	Converts -OH to -I	H₃PO₃

Thionyl chloride is favoured in real laboratories because the byproducts are gases that simply escape, leaving no inorganic waste in the flask. This is one of the cleaner routes to chloroalkanes from alcohols, and a frequent context for "atom economy" or "green chemistry" exam-question themes.

Lesson 3: Dehydration and Substitution of Alcohols