Halogenoalkanes

CIE A-Level Chemistry· 10 min read

1. Classification and Nomenclature★★☆☆☆⏱ 10 min

Halogenoalkanes (also called alkyl halides) are alkanes with at least one hydrogen atom replaced by a halogen atom. They are classified based on how many alkyl groups are bonded to the carbon that holds the halogen.

📘 Definition

Classification of Halogenoalkanes

Classification is determined solely by the number of alkyl groups bonded to the halogen-bearing carbon atom, not the total number of carbons in the molecule.

Example:

1-bromopropane (1°), 2-bromopropane (2°), 2-bromo-2-methylpropane (3°)

📐 Worked Example

Classify 1-chloro-2-methylpropane and 2-chloro-2-methylbutane as primary, secondary or tertiary.

1
Step 1: Identify the carbon directly bonded to chlorine in 1-chloro-2-methylpropane. This terminal carbon is only bonded to one other alkyl carbon.
2
Thus, 1-chloro-2-methylpropane is a primary (1°) halogenoalkane.
3
Step 2: Identify the carbon directly bonded to chlorine in 2-chloro-2-methylbutane. This central carbon is bonded to three other alkyl carbons.
4
Thus, 2-chloro-2-methylbutane is a tertiary (3°) halogenoalkane.

Exam tip:

Always classify based on the halogen-bearing carbon, not the most substituted carbon elsewhere in the molecule.

2. Physical Properties★★☆☆☆⏱ 8 min

The C-X bond is polar because halogens are more electronegative than carbon. However, halogenoalkanes cannot form hydrogen bonds with water, so they are immiscible with water. Boiling point depends primarily on molecular size.

Compound	Relative Molecular Mass	Boiling Point (°C)
Chloromethane	50.5	-24
Bromomethane	95	4
Iodomethane	142	43
1-chloropropane	78.5	47

📐 Worked Example

Explain why 1-iodobutane has a higher boiling point than 1-chlorobutane.

1
Step 1: Iodine has a higher atomic number than chlorine, so 1-iodobutane has a higher relative molecular mass than 1-chlorobutane.
2
Step 2: Higher molecular mass leads to stronger instantaneous dipole-induced dipole (London) intermolecular forces between molecules.
3
Step 3: More thermal energy is required to overcome these stronger forces, resulting in a higher boiling point.

3. Nucleophilic Substitution Mechanisms★★★★☆⏱ 15 min

Nucleophilic substitution is the most important reaction of halogenoalkanes. The electronegative halogen pulls electron density away from the carbon, making it electrophilic and open to attack by electron-rich nucleophiles.

📘 Definition

Nucleophilic Substitution

A reaction where a nucleophile replaces the halide leaving group, donating a lone pair of electrons to the electrophilic carbon to form a new covalent bond.

Two mechanisms exist: SN2 (bimolecular, 1 step) for primary halogenoalkanes, and SN1 (unimolecular, 2 steps) for tertiary halogenoalkanes, via a stable carbocation intermediate.

📐 Worked Example

Describe the mechanism for the reaction of bromoethane with aqueous sodium hydroxide.

1
Step 1: Identify the nucleophile: hydroxide ion ( $OH^{-}$ ) from NaOH. The C-Br bond is polar, so the carbon is $δ^{+}$ and attracts the negatively charged nucleophile.
2
Bromoethane is a primary halogenoalkane, so it reacts via an SN2 mechanism (one step):
3
$\ce{OH^- + CH3CH2Br -> HOCH2CH3 + Br^-}$
4
Step 2: To draw the mechanism: draw a curly arrow from a lone pair on the $OH^{-}$ to the $δ^{+}$ carbon, and a second curly arrow from the C-Br bonding pair to the bromine atom, showing the leaving group departing.

✓ Quick check

Test your understanding:

Which of the following undergoes SN1 nucleophilic substitution fastest?
- 1-bromobutane
- 2-bromobutane
- 2-bromo-2-methylpropane
- bromomethane
Reveal answer
2-bromo-2-methylpropane —
Tertiary halogenoalkanes form stable tertiary carbocation intermediates, so they favour SN1 substitution over SN2.

Exam tip:

CIE examiners require curly arrows to start at a lone pair or bonding pair, not at the negative charge on the nucleophile.

4. Elimination Reactions★★★☆☆⏱ 10 min

When halogenoalkanes react with hot ethanolic (not aqueous) potassium hydroxide, elimination (also called dehydrohalogenation) occurs instead of substitution, forming an alkene by removing HX.

📘 Definition

Dehydrohalogenation (Elimination)

A reaction that removes a hydrogen halide (HX) molecule from a halogenoalkane to form a carbon-carbon double bond (alkene).

Example:

2-bromopropane + hot ethanolic KOH → propene + KBr + H₂O

📐 Worked Example

State the major organic product formed when 2-bromobutane reacts with hot ethanolic KOH.

1
Step 1: Elimination of HBr from 2-bromobutane can form two alkene products: but-1-ene and but-2-ene.
2
Step 2: Zaitsev's rule states the more substituted alkene (more alkyl groups attached to the double bond) is the major product.
3
Step 3: But-2-ene has two alkyl groups attached to the double bond, while but-1-ene only has one. Thus, but-2-ene is the major product.

5. Common Pitfalls

Wrong move:

Classifying 1-chloro-2-methylpropane as a secondary halogenoalkane

Why:

Mistakenly counts the branched carbon instead of the carbon directly bonded to the halogen

Correct move:

Only classify based on the carbon directly attached to the halogen: 1-chloro-2-methylpropane is primary

Wrong move:

Explaining higher boiling points of heavier halogenoalkanes by increasing polarity

Why:

Polarity decreases from chlorine to iodine, but boiling point increases, so polarity is not the main factor

Correct move:

Attribute boiling point trends to increasing relative molecular mass and stronger London intermolecular forces

Wrong move:

Drawing a curly arrow from the negative charge of a nucleophile to the electrophilic carbon

Why:

Curly arrows represent the movement of an electron pair, not the negative charge

Correct move:

Always start the curly arrow at a lone pair on the nucleophile atom

Wrong move:

Claiming primary halogenoalkanes undergo SN1 substitution

Why:

Primary carbocations are too unstable to form as intermediates

Correct move:

Primary halogenoalkanes always react via SN2 (1-step) substitution, while tertiary react via SN1

Wrong move:

Using aqueous potassium hydroxide to form an alkene from a halogenoalkane

Why:

Aqueous conditions favour nucleophilic substitution to form an alcohol, not elimination

Correct move:

Use hot, ethanolic potassium hydroxide to carry out elimination and form an alkene

6. Quick Reference Cheatsheet

Property	Primary (1°)	Secondary (2°)	Tertiary (3°)
Halogen-bearing C bonded to	1 other C	2 other C	3 other C
Substitution mechanism	SN2 (1 step)	Mixed SN1/SN2	SN1 (2 step)
Intermediate formed	None	Variable	Carbocation
Product with aq NaOH	Alcohol	Alcohol	Alcohol
Product with hot ethanolic KOH	Alkene	Alkene	Alkene

7. Frequently Asked

How do I remember substitution vs elimination conditions?

Aq = Substitute, Hot Ethanol = Eliminate: Aqueous hydroxide (room/warm temperature) produces alcohols via substitution, while hot ethanolic hydroxide produces alkenes via elimination.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Mechanism of substitution
2023 · 1
Classification and boiling point trends
2021 · 4
Elimination product prediction

What's Next

Halogenoalkanes are fundamental building blocks for organic synthesis, used to introduce key functional groups like alcohols, amines, and alkenes that form the basis of more complex organic molecules. Mastery of their mechanisms is critical for CIE A-Level Chemistry, as mechanism questions frequently appear in both Paper 2 and Paper 4, requiring accurate drawing of curly arrows and identification of intermediates. Understanding the difference between substitution and elimination here will help you predict products of organic reactions across all subsequent topics, from synthesis to aromatic chemistry.