Organic Chemistry: Halogenoalkanes, Alcohols and Spectra
Edexcel International A-Level Chemistry· 10.1–10.23 (10A–10D)· 50 min read
1. Foundations of Organic Reaction Mechanisms★★☆☆☆⏱ 10 min
Organic reaction mechanisms show the movement of electron pairs during reactions, using double-barbed curly arrows. Heterolysis (polar bond breaking) produces charged ions, which form the basis of substitution and elimination reactions, and creates electron-deficient centres that attract nucleophiles.
Nucleophile
An electron pair donor that is attracted to an electron-deficient (δ+) centre. Common examples include hydroxide ions (OH⁻), ammonia (NH₃) and cyanide ions (CN⁻).
Classify the reaction of bromoethane with warm aqueous KOH and identify the nucleophile.
- 1
First identify reactants and products: bromoethane + KOH(aq) → ethanol + KBr. The Br group is replaced by an OH group, so this is a nucleophilic substitution reaction.
- 2
The nucleophile is the OH⁻ ion, which donates a lone pair of electrons to the partially positive carbon atom bonded to Br.
Exam tip:
You will be penalised for drawing curly arrows starting from bonds or atoms without lone pairs; always start arrows from electron pairs, either on nucleophiles or covalent bonds.
2. Halogenoalkane Structure and Reactions★★★☆☆⏱ 15 min
Halogenoalkanes are classified as primary (1°), secondary (2°) or tertiary (3°) based on the number of alkyl groups attached to the carbon bonded to the halogen atom. They undergo a range of reactions depending on reagent conditions.
Warm aqueous KOH: nucleophilic substitution → alcohol
Hot ethanolic KOH (reflux): elimination → alkene
AgNO₃ in ethanol: hydrolysis forms silver halide precipitate (rate test)
Alcoholic NH₃ (high pressure): substitution → amine
Alcoholic KCN: substitution → nitrile (lengthens carbon chain by 1)
Draw the nucleophilic substitution mechanism for the reaction of 1-chloropropane with warm aqueous KOH.
- 1
Label the polar C-Cl bond: the C atom is δ+, the Cl atom is δ−.
- 2
Draw a curly arrow from the lone pair on the OH⁻ nucleophile to the δ+ carbon atom.
- 3
Draw a second curly arrow from the C-Cl covalent bond to the Cl atom, forming a Cl⁻ leaving group.
- 4
The final products are propan-1-ol and Cl⁻ ion.
Hydrolysis rates increase from 1° → 2° → 3° halogenoalkanes, and from chloro → bromo → iodo halogenoalkanes. This is due to two factors: the stability of intermediate species for higher-degree halogenoalkanes, and the decreasing C-X bond enthalpy down group 7, with C-I bonds breaking most easily.
Which halogenoalkane will have the fastest hydrolysis rate?
1-chlorobutane
2-bromobutane
2-iodo-2-methylpropane
Reveal answer
2-iodo-2-methylpropane —Tertiary iodoalkanes have both a 3° structure and the weakest C-I bond, so they hydrolyse fastest.
Exam tip:
When explaining hydrolysis rate differences, always reference both the 1°/2°/3° classification and C-X bond enthalpy if asked, to earn all available marks.
3. Alcohol Structure, Reactions and Oxidation★★★☆☆⏱ 15 min
Alcohols are classified as 1°/2°/3° based on the number of alkyl groups attached to the carbon bonded to the hydroxyl (-OH) group. Key reactions include combustion, substitution to form halogenoalkanes, elimination to form alkenes, and oxidation with acidified potassium dichromate(VI).
Reflux
Heating technique using a vertical open condenser to return volatile reactants to the reaction vessel, allowing extended heating without loss of product. Used for full oxidation of primary alcohols to carboxylic acids.
PCl₅: produces steamy fumes of HCl (test for -OH group)
50% H₂SO₄ + KBr: substitution → bromoalkane
Concentrated H₃PO₄: elimination → alkene
Acidified K₂Cr₂O₇: oxidation, product depends on conditions and alcohol class
Predict the products of oxidation of propan-1-ol under (a) distillation, (b) reflux with excess acidified K₂Cr₂O₇, and state observations for each reaction.
- 1
(a) Distillation removes the volatile product as it forms, so partial oxidation of the 1° alcohol produces propanal (aldehyde). Observation: Orange K₂Cr₂O₇ solution turns green.
- 2
Propanal will produce a red precipitate when heated with Fehling's or Benedict's solution, confirming it is an aldehyde.
- 3
(b) Reflux with excess oxidising agent allows full oxidation to propanoic acid (carboxylic acid). Observation: Orange solution turns green, and product reacts with NaHCO₃ to produce CO₂ gas bubbles.
Secondary alcohols oxidise to ketones under any conditions, and tertiary alcohols do not react with acidified K₂Cr₂O₇, as there is no hydrogen atom attached to the carbon with the -OH group to remove.
Exam tip:
Always specify reaction conditions (distil vs reflux) when predicting oxidation products of primary alcohols, as this is a frequently tested mark point.
4. Interpreting Mass Spectra and Infrared Spectra★★☆☆☆⏱ 10 min
Mass spectrometry (MS) is used to find the relative molecular mass of a compound via the molecular ion peak (M+), the highest m/z value on the spectrum. Fragmentation peaks can be used to identify alkyl groups present in the molecule.
Infrared (IR) Spectroscopy
Analytical technique that measures absorption of IR radiation by bonds in functional groups. Each bond type has a characteristic wavenumber absorption range, allowing identification of functional groups in unknown compounds.
An organic compound has a molecular ion peak at m/z = 74, an IR absorption at 1700 cm⁻¹, and a broad absorption between 2500–3300 cm⁻¹. Identify the functional group and suggest a possible structure.
- 1
The 1700 cm⁻¹ peak indicates a C=O bond, and the broad 2500–3300 cm⁻¹ peak indicates an O-H bond in a carboxylic acid group.
- 2
The carboxylic acid group (-COOH) has a mass of 45, so the remaining mass of 74 - 45 = 29 corresponds to an ethyl group (C₂H₅).
- 3
The compound is propanoic acid, with structure CH₃CH₂COOH.
Exam tip:
IR wavenumber data is provided in the exam data booklet, so you do not need to memorise values, but you must be able to match peaks to the correct functional groups accurately.
5. Common Pitfalls
Wrong move:
Using aqueous KOH for elimination reactions of halogenoalkanes
Why:
Aqueous conditions favour nucleophilic substitution, not elimination
Correct move:
Use hot ethanolic KOH heated under reflux for elimination to form alkenes
Wrong move:
Predicting carboxylic acid product from primary alcohol oxidation under distillation
Why:
Distillation removes volatile aldehyde product before it can be further oxidised
Correct move:
Predict aldehyde for distillation conditions, carboxylic acid only for reflux with excess oxidising agent
Wrong move:
Stating chloroalkanes hydrolyse faster than iodoalkanes
Why:
C-Cl bonds have higher bond enthalpy than C-I bonds, so break more slowly
Correct move:
State hydrolysis rate increases from chloro → bromo → iodo due to decreasing C-X bond enthalpy
Wrong move:
Labelling substitution mechanisms as SN1 or SN2 for Unit 2 questions
Why:
SN1/SN2 distinction is out of scope for Unit 2, only covered in Unit 4
Correct move:
Label the mechanism as nucleophilic substitution, show correct curly arrow notation without SN1/SN2 labels
Wrong move:
Assuming all C=O IR peaks indicate carboxylic acids
Why:
Aldehydes and ketones also have C=O peaks around 1700 cm⁻¹, only carboxylic acids have the broad 2500–3300 cm⁻¹ O-H peak
Correct move:
Use additional test results (Fehling's/Benedict's) or IR peaks to distinguish between carbonyl group types
6. Quick Reference Cheatsheet
Compound Type | Reaction Type | Conditions | Product |
|---|---|---|---|
Primary halogenoalkane | Nucleophilic substitution | Warm aqueous KOH | Primary alcohol |
Any halogenoalkane | Elimination | Hot ethanolic KOH, reflux | Alkene |
Primary alcohol | Partial oxidation | Acidified K₂Cr₂O₇, distil | Aldehyde |
Primary alcohol | Full oxidation | Excess acidified K₂Cr₂O₇, reflux | Carboxylic acid |
Secondary alcohol | Oxidation | Acidified K₂Cr₂O₇, any conditions | Ketone |
IR peak 1700 cm⁻¹ | Functional group indicator | N/A | C=O (carbonyl group) |
7. Frequently Asked
What is the difference between aqueous and ethanolic KOH reactions with halogenoalkanes?
Warm aqueous KOH triggers nucleophilic substitution to form alcohols, while hot ethanolic KOH heated under reflux triggers elimination to form alkenes.
How do I distinguish between aldehyde and ketone oxidation products?
Aldehydes produce a red precipitate when heated with Fehling's or Benedict's solution, while ketones show no reaction. Aldehydes can also be further oxidised to carboxylic acids under reflux.
Going deeper
- Official GuideEdexcel IAL Chemistry Unit 2 SpecificationRefer to section 10 for full topic content
- ResourceEdexcel IAL Chemistry Data BookletContains IR wavenumber and periodic table data for exams
What's Next
Now that you have mastered halogenoalkanes, alcohols and analytical spectra for Edexcel IAL Chemistry Unit 2, you can progress to more advanced organic chemistry topics in Unit 4, including detailed SN1/SN2 mechanism distinctions, aromatic chemistry, and nuclear magnetic resonance (NMR) spectroscopy. This topic accounts for 15–20% of Unit 2 marks, so practise past paper questions regularly to build confidence with mechanism drawing and spectrum interpretation. Make sure you also revise the core practicals for this topic, as practical-based questions appear in every Unit 2 exam paper. Familiarising yourself with the data booklet IR and mass spec reference tables will also help you save time during the exam.