OCR Specification focus:
‘Convert alcohols to haloalkanes via substitution with halide ions in the presence of acid, e.g. NaBr/H₂SO₄; mechanism not required.’
Alcohols undergo acid-catalysed substitution to form haloalkanes, an essential transformation in synthetic chemistry. This process replaces the hydroxyl group with a halide ion, enabling further reactions.
Conversion of Alcohols to Haloalkanes: Core Concepts
Alcohols react with halide ions in the presence of an acid to yield haloalkanes, compounds containing at least one carbon–halogen bond.
Generic reaction showing an alcohol reacting with a hydrogen halide to form an alkyl halide and water, illustrating the overall substitution process required by the OCR specification. Source
When discussing haloalkanes, it is crucial to understand the role of a halide ion when acting as a nucleophile.
Nucleophile: A species that donates an electron pair to an electron-deficient atom to form a new covalent bond.
A single sentence ensures separation between definition and later structures.
Reagents for Substitution to Haloalkanes
Generating the Hydrogen Halide In Situ
The specification emphasises the use of mixtures such as sodium bromide (NaBr) with sulfuric acid (H₂SO₄) to form hydrogen bromide (HBr).
Reaction scheme showing butan‑1‑ol reacting with sodium bromide and sulfuric acid to form 1‑bromobutane, illustrating the NaBr/H₂SO₄ substitution conditions specified by OCR. Source
Students must understand why these reagents are chosen.
Key points:
Halide salts supply the halide ion (Br⁻, Cl⁻, I⁻).
Strong acids protonate the alcohol and generate the corresponding hydrogen halide.
The hydrogen halide then substitutes the –OH group, forming the haloalkane.
Why Alcohols Need an Acid
Alcohols contain a poor leaving group: the hydroxide ion, OH⁻. Direct displacement is unlikely because hydroxide is strongly basic. Protonation converts –OH into water, a far better leaving group. This increases the rate and likelihood of substitution.
Formation of the Hydrogen Halide
The reaction between a halide salt and acid produces the hydrogen halide used for substitution.
Hydrogen Bromide Formation (HBr) = NaBr + H₂SO₄ → HBr + NaHSO₄
NaBr = Source of bromide ions
H₂SO₄ = Acid used to liberate hydrogen bromide
HBr = Hydrogen bromide, the reactive species for substitution
This sentence separates the equation from subsequent explanations.
Reaction Conditions and Practical Considerations
General Conditions
The conversion of alcohols to haloalkanes requires the following conditions:
Presence of halide ions (e.g., Br⁻, Cl⁻, I⁻).
A strong acid catalyst, most commonly concentrated H₂SO₄ or sometimes H₃PO₄.
Heating, often under reflux, to increase reaction rate.
Reaction mixtures are typically prepared in aqueous solution, though the haloalkane product often forms a distinct organic layer.
Why Mechanism Is Not Required
Although substitution mechanisms such as SN1 and SN2 underlie these transformations, OCR does not assess the mechanistic pathway here. Students only need to understand the practical process and the essential role of the reagents.
Factors Affecting the Reaction
Structure of the Alcohol
Primary, secondary and tertiary alcohols do not react at identical rates.
Tertiary alcohols react fastest because protonation easily forms a stable carbocation.
Secondary alcohols react moderately.
Primary alcohols react more slowly because they resist carbocation formation.
Although the mechanism is not examined, awareness of differing reactivity is helpful when predicting outcomes in synthetic pathways.
Nature of the Halide Ion
The halide ion’s ability to substitute depends partly on its size and polarisation.
General trend in ease of substitution:
I⁻ (most effective)
Br⁻
Cl⁻
This trend relates to bond strengths and nucleophilicity.
Process Outline: Substitution of Alcohols to Haloalkanes
Step-by-Step Flow
The following bullet points provide a clear representation of the transformation process:
Combine the alcohol with a halide salt (e.g., NaBr).
Add a concentrated acid such as H₂SO₄ to generate the hydrogen halide.
Protonation converts the –OH group into water, improving leaving group ability.
The halide ion attacks the carbon previously bonded to –OH.
Haloalkane forms, often separating as an organic layer due to lower polarity.
Typical Observations
Students may observe:
Release of vapours from hydrogen halide formation.
Formation of a second layer when the haloalkane separates.
Changes in reaction mixture appearance as substitution proceeds.
Applications in Organic Synthesis
Haloalkanes produced from alcohols act as versatile intermediates:
They readily undergo nucleophilic substitution to form amines, nitriles or further alcohols.
Their reactivity allows synthetic routes involving multiple steps.
Substitution expands the functional group repertoire available from a relatively simple alcohol.
Safety, Handling and Laboratory Practice
Working with Strong Acids
Strong mineral acids such as H₂SO₄ are corrosive and must be handled using:
Protective gloves
Eye protection
Adequate ventilation
Managing Haloalkane Products
Haloalkanes may be volatile and toxic.
Use fume cupboards when possible.
Keep containers sealed to reduce inhalation risk.
Dispose of waste following laboratory guidelines.
FAQ
Sulfuric acid is more commonly used because it strongly protonates the alcohol and effectively liberates hydrogen halides from halide salts.
Phosphoric acid is milder and avoids side reactions such as oxidation or dehydration that sulfuric acid may promote, making it preferred when forming iodoalkanes that might otherwise decompose.
Tertiary alcohols form a highly stabilised intermediate once protonated, allowing the halide ion to substitute quickly.
Although OCR does not require mechanism knowledge, understanding steric effects and carbocation stability helps explain observed rate differences during laboratory synthesis.
Purity is affected by:
Competing reactions such as elimination or oxidation
Presence of unreacted alcohol or hydrogen halide
Efficiency of separation of the organic layer
Drying agents and careful distillation improve purity without altering the underlying substitution chemistry.
Haloalkanes formed from secondary or tertiary alcohols may favour elimination because protonation creates a structure that can lose water and form an alkene.
High temperatures or strongly acidic conditions make elimination more competitive with substitution.
Polar solvents help stabilise the ions formed during protonation and halide attack, increasing reaction efficiency.
A biphasic mixture may form, where the haloalkane separates as an organic layer, aiding both reaction progress and product isolation.
Practice Questions
(2 marks)
A student converts an alcohol to a bromoalkane using sodium bromide and concentrated sulfuric acid.
(a) State the role of sulfuric acid in this reaction.
(b) Explain why the hydroxyl group of the alcohol must first be protonated before substitution occurs.
(a)
Sulfuric acid generates hydrogen bromide (HBr) from sodium bromide. (1)
Acts as an acid to protonate the alcohol / provide acidic conditions. (1)
(b)
Protonation converts the poor leaving group OH⁻ into water. (1)
Water is a much better leaving group, allowing substitution to occur. (1)
(Any two correct points; maximum 2 marks total.)
Butan-2-ol is reacted with sodium iodide and phosphoric acid under heated conditions to form 2-iodobutane.
(a) Explain why iodide ions are effective nucleophiles in this substitution process.
(b) Outline the steps that occur in the conversion of butan-2-ol to 2-iodobutane under these conditions.
(c) Suggest one observation the student might make during this reaction.
(5 marks)
(a)
Iodide ions are large and highly polarisable. (1)
They form weaker C–I bonds, making substitution easier. (1)
(b)
Award one mark for each distinct step (max 3):
Acid protonates the OH group of the alcohol, forming a better leaving group. (1)
Water leaves, forming a reactive intermediate (appropriate description accepted, no mechanism required). (1)
Iodide ion attacks the carbon atom to form 2-iodobutane. (1)
(c)
Formation of a second organic layer / immiscible layer appears. (1)
Release of fumes from hydrogen halide formation. (1)
(Maximum 1 mark for part (c).)
Total: 5 marks
