Halogenoalkanes

Q: Why do boiling points of halogenoalkanes increase in the order \(\text{R}-\text{F}?

Boiling points increase because iodine has the greatest number of electrons, making it the most polarizable and resulting in significantly stronger van der Waals (London dispersion) forces between molecules.

Q: Why are halogenoalkanes insoluble in water despite possessing a polar \(\text{C}-\text{X}\) bond?

Halogenoalkanes cannot form hydrogen bonds with water, and the energy released by forming new dipole-dipole attractions is insufficient to break the strong hydrogen bonds between water molecules.

Q: Why is bond enthalpy rather than bond polarity the main factor determining the reactivity of halogenoalkanes?

Although the \(\text{C}-\text{F}\) bond is the most polar, the \(\text{C}-\text{I}\) bond is much weaker (lowest bond enthalpy), making it significantly easier to break during reactions.

Q: What is a nucleophile and what role does it play in nucleophilic substitution of halogenoalkanes?

A nucleophile is an electron-pair donor that uses a lone pair to attack the electron-deficient \(\text{C}^{\delta+}\) carbon atom, displacing the halogen as a halide leaving group.

Q: What product and conditions are involved when a halogenoalkane reacts with aqueous alkali?

It forms an alcohol when warmed under reflux with aqueous sodium or potassium hydroxide (\(\text{NaOH}\) or \(\text{KOH}\)).

Q: Why is the nucleophilic substitution reaction of halogenoalkanes with cyanide ions synthetically important?

It produces a nitrile and extends the carbon chain length by one carbon atom.

Q: Why must excess concentrated ethanolic ammonia be used to form a primary amine from a halogenoalkane?

Using excess ammonia prevents further substitution reactions by ensuring ammonia, rather than the newly formed primary amine nucleophile, attacks the halogenoalkane.

Q: How can aqueous silver nitrate be used to distinguish the rate of hydrolysis between chloro-, bromo-, and iodoalkanes?

The rate of formation of a silver halide precipitate (\(\text{AgI}\) yellow fastest, \(\text{AgBr}\) cream moderate, \(\text{AgCl}\) white slowest) indicates the relative rate of hydrolysis.

Q: What reagents and conditions are required for a halogenoalkane to undergo elimination instead of substitution?

Potassium hydroxide (\(\text{KOH}\)) or \(\text{NaOH}\) dissolved in pure ethanol, heated under reflux at high temperature.

Chemistry 1110 · CCEA AS Level · 9 min read · Updated: 31 Aug 2026

Introduction to Halogenoalkanes

Welcome to one of the most versatile and exciting chapters in AS Chemistry! Halogenoalkanes (also frequently called haloalkanes) are saturated organic compounds where one or more hydrogen atoms in an alkane have been replaced by halogen atoms (\(\text{F}\), \(\text{Cl}\), \(\text{Br}\), or \(\text{I}\)).

Think of halogenoalkanes as the "stepping stones" of organic chemistry. Unreactive alkanes are difficult to turn into other chemicals, but halogenoalkanes react readily to form alcohols, nitriles, amines, and alkenes. Mastering this chapter gives you the master key to synthetic organic chemistry!

1. Classification and Nomenclature

Just like alcohols, halogenoalkanes are classified based on the environment of the carbon atom bonded directly to the halogen.

Types of Halogenoalkanes

• Primary (\(1^\circ\)) Halogenoalkane: The carbon atom carrying the halogen is attached to only one other alkyl group (or none, as in \(\text{CH}_3\text{Cl}\)).
Example: 1-chloropropane, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}\)

• Secondary (\(2^\circ\)) Halogenoalkane: The carbon atom carrying the halogen is attached to two other alkyl groups.
Example: 2-bromopropane, \(\text{CH}_3\text{CH}(\text{Br})\text{CH}_3\)

• Tertiary (\(3^\circ\)) Halogenoalkane: The carbon atom carrying the halogen is attached to three other alkyl groups.
Example: 2-chloro-2-methylpropane, \((\text{CH}_3)_3\text{CCl}\)

Memory Trick: Count how many carbon "neighbors" the halogen-bearing carbon has. 1 neighbor = Primary (\(1^\circ\)), 2 neighbors = Secondary (\(2^\circ\)), 3 neighbors = Tertiary (\(3^\circ\)).

Key Takeaway:

Halogenoalkanes are categorized as primary, secondary, or tertiary depending on whether the carbon bonded to the halogen is attached to one, two, or three other carbon atoms.

2. Physical Properties & The Nature of the Carbon–Halogen Bond

Boiling Points

The boiling points of halogenoalkanes depend on molecular size, halogen type, and chain branching:
• Chain length: As the carbon chain gets longer, boiling points increase due to stronger van der Waals forces (London dispersion forces) between larger molecules with more electrons.
• Halogen identity: For a given alkyl chain, boiling points increase in the order: \(\text{R}-\text{F} < \text{R}-\text{Cl} < \text{R}-\text{Br} < \text{R}-\text{I}\). Iodine has the largest number of electrons, making it the most polarizable, which leads to significantly stronger London forces.
• Branching: Branched isomers have lower boiling points than straight-chain isomers because branching reduces surface contact area between molecules.

Solubility

Even though the \(\text{C}-\text{X}\) bond is polar, halogenoalkanes are insoluble in water. They cannot form hydrogen bonds with water molecules because the energy released when new dipole-dipole attractions form between halogenoalkanes and water is not enough to break the strong hydrogen bonds between water molecules.

Bond Polarity vs. Bond Enthalpy: What Controls Reactivity?

This is one of the most common exam questions at AS Level, so let's break it down carefully!

Halogens are more electronegative than carbon, which creates a polar bond with a partial positive charge on the carbon atom and a partial negative charge on the halogen atom: \(\text{C}^{\delta+}-\text{X}^{\delta-}\).

• Argument 1 (Bond Polarity): Fluorine is the most electronegative halogen, so \(\text{C}-\text{F}\) is the most polar bond. If bond polarity controlled reactivity, \(\text{C}-\text{F}\) would have the most electron-deficient carbon and react fastest.
• Argument 2 (Bond Enthalpy / Strength): The \(\text{C}-\text{F}\) bond is very short and exceptionally strong (\(467\text{ kJ mol}^{-1}\)), while the \(\text{C}-\text{I}\) bond is much longer and weaker (\(238\text{ kJ mol}^{-1}\)).

The Verdict: Bond enthalpy is the dominant factor! The \(\text{C}-\text{I}\) bond is the easiest to break, so iodoalkanes react fastest. Fluoroalkanes are virtually inert under standard lab conditions.

Reactivity Order:

\(\text{R}-\text{I} > \text{R}-\text{Br} > \text{R}-\text{Cl} > \text{R}-\text{F}\) (Fastest \(\rightarrow\) Slowest)

Key Takeaway:

Reactivity increases down Group 7 from chloro- to iodoalkanes because the \(\text{C}-\text{X}\) bond enthalpy decreases, making weaker bonds much easier to break.

3. Nucleophilic Substitution Reactions

What is a Nucleophile?

A nucleophile is an electron-pair donor. It is an electron-rich species (often negatively charged or possessing a lone pair of electrons) that is attracted to an electron-deficient region, such as a \(\text{C}^{\delta+}\) carbon atom.

The three main nucleophiles at AS Level are:
1. Hydroxide ion: \(:\!\text{OH}^-\)
2. Cyanide ion: \(:\!\text{CN}^-\)
3. Ammonia: \(:\!\text{NH}_3\)

Mechanism Overview: How Nucleophilic Substitution Works

Don't worry if mechanisms seem intimidating at first! Follow these simple steps:
1. The nucleophile uses a lone pair of electrons to attack the partially positive carbon atom (\(\text{C}^{\delta+}\)).
2. Simultaneously, the \(\text{C}-\text{X}\) bond breaks heterolytically, with both electrons in the bond leaving with the halogen atom to form a halide ion (\(:\!\text{X}^-\)), known as the leaving group.
3. A new covalent bond forms between the nucleophile and the carbon atom.

Reaction 1: Hydrolysis with Aqueous Alkali (Formation of Alcohols)

• Reagent: Aqueous sodium hydroxide (\(\text{NaOH}\)) or aqueous potassium hydroxide (\(\text{KOH}\))
• Conditions: Warm under reflux
• General Equation:
\(\text{R}-\text{X} + \text{OH}^- \rightarrow \text{R}-\text{OH} + \text{X}^-\)
Example: Bromoethane reacts with aqueous \(\text{NaOH}\) to form ethanol:
\(\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{NaBr}\)

Reaction 2: Reaction with Cyanide Ions (Formation of Nitriles)

• Reagent: Potassium cyanide (\(\text{KCN}\)) dissolved in ethanol/water mixture
• Conditions: Heat under reflux
• General Equation:
\(\text{R}-\text{X} + \text{CN}^- \rightarrow \text{R}-\text{C}\equiv\text{N} + \text{X}^-\)
Example: 1-chloropropane produces butanenitrile:
\(\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{KCN} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + \text{KCl}\)

Important Synthetic Tip: This reaction is extremely useful because it increases the carbon chain length by one carbon atom! When naming the product nitrile, make sure to count the carbon in the \(-\text{CN}\) group as part of the longest carbon chain.

Reaction 3: Reaction with Ammonia (Formation of Amines)

• Reagent: Excess concentrated ethanolic ammonia (\(\text{NH}_3\))
• Conditions: Heat in a sealed tube (to prevent volatile ammonia gas from escaping under pressure)
• Reaction Steps:
Step 1: The ammonia molecule attacks to form an intermediate alkylammonium salt:
\(\text{R}-\text{X} + \text{NH}_3 \rightarrow [\text{R}-\text{NH}_3]^+\text{X}^-\)
Step 2: A second ammonia molecule acts as a base to remove a proton (\(\text{H}^+\)), liberating the free primary amine:
\([\text{R}-\text{NH}_3]^+\text{X}^- + \text{NH}_3 \rightarrow \text{R}-\text{NH}_2 + \text{NH}_4^+\text{X}^-\)
• Overall Equation:
\(\text{R}-\text{X} + 2\text{NH}_3 \rightarrow \text{R}-\text{NH}_2 + \text{NH}_4\text{X}\)
Example: Bromoethane forms ethylamine (aminoethane):
\(\text{CH}_3\text{CH}_2\text{Br} + 2\text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + \text{NH}_4\text{Br}\)

Why use excess ammonia? The primary amine formed is itself a nucleophile and can react further with remaining halogenoalkane to produce secondary and tertiary amines. Using an excess of ammonia ensures that the primary amine is the main organic product.

Key Takeaway:

Nucleophiles (\(:\!\text{OH}^-\), \(:\!\text{CN}^-\), \(:\!\text{NH}_3\)) attack the \(\text{C}^{\delta+}\) carbon, displacing the halide ion to produce alcohols, nitriles, and primary amines respectively.

4. Experimental Comparison: Rates of Hydrolysis

We can experimentally verify that iodoalkanes react faster than bromoalkanes and chloroalkanes using silver nitrate solution.

The Experiment Step-by-Step

1. Place equal amounts of 1-chlorobutane, 1-bromobutane, and 1-iodobutane in three separate test tubes.
2. Add ethanol to each tube (ethanol acts as a mutual solvent to allow water and halogenoalkanes to mix).
3. Add aqueous silver nitrate (\(\text{AgNO}_3\)) to each tube and place them in a warm water bath at \(50\ ^\circ\text{C}\).
4. Time how quickly a precipitate of silver halide (\(\text{AgX}\)) appears.

Observations and Results

• 1-iodobutane: Rapidly forms a yellow precipitate of silver iodide (\(\text{AgI}\)).
• 1-bromobutane: Forms a cream precipitate of silver bromide (\(\text{AgBr}\)) at a moderate rate.
• 1-chlorobutane: Slowly forms a white precipitate of silver chloride (\(\text{AgCl}\)).

Precipitate Identification & Confirming with Ammonia

To confirm the identity of the precipitates:
• \(\text{AgCl}\) (White ppt): Dissolves readily in dilute aqueous \(\text{NH}_3\).
• \(\text{AgBr}\) (Cream ppt): Insoluble in dilute \(\text{NH}_3\), but dissolves in concentrated \(\text{NH}_3\).
• \(\text{AgI}\) (Yellow ppt): Insoluble in both dilute and concentrated \(\text{NH}_3\).

Key Takeaway:

The rate of precipitate formation follows: \(\text{R}-\text{I} > \text{R}-\text{Br} > \text{R}-\text{Cl}\). This confirms that bond enthalpy, not bond polarity, determines the rate of reaction.

5. Elimination Reactions

Halogenoalkanes can undergo an entirely different type of reaction called elimination to form alkenes.

Reagents and Conditions

• Reagent: Potassium hydroxide (\(\text{KOH}\)) or sodium hydroxide (\(\text{NaOH}\)) dissolved in ethanol (no water present!)
• Conditions: High temperature / heat under reflux
• Role of \(\text{OH}^-\): In this reaction, \(\text{OH}^-\) acts as a base (proton acceptor) rather than a nucleophile.

General Equation

\(\text{R}-\text{CH}_2-\text{CH}_2-\text{X} + \text{OH}^- \xrightarrow{\text{ethanol, heat}} \text{R}-\text{CH}=\text{CH}_2 + \text{H}_2\text{O} + \text{X}^-\)

Example: 2-bromopropane produces propene:
\(\text{CH}_3\text{CH}(\text{Br})\text{CH}_3 + \text{KOH} \xrightarrow{\text{ethanol, heat}} \text{CH}_3\text{CH}=\text{CH}_2 + \text{KBr} + \text{H}_2\text{O}\)

Substitution vs. Elimination: The Solvent Rule

Students often mix these two up! Remember this golden rule:
• \(\text{OH}^-\) in Aqueous solution (\(\text{H}_2\text{O}\)) \(\rightarrow\) Nucleophilic Substitution \(\rightarrow\) Alcohol formed
• \(\text{OH}^-\) in Ethanolic solution (\(\text{Ethanol}\)) + Heat \(\rightarrow\) Elimination \(\rightarrow\) Alkene formed

Memory Trick: Aqueous makes Alcohol; Ethanolic makes Ethene (Alkene).

Key Takeaway:

Treating a halogenoalkane with ethanolic \(\text{KOH}\) under heat causes elimination of \(\text{HX}\), producing an alkene and water.

6. Environmental Chemistry: CFCs and Ozone Depletion

What are CFCs?

Chlorofluorocarbons (CFCs) are compounds containing only chlorine, fluorine, and carbon (e.g., \(\text{CCl}_3\text{F}\) or \(\text{CCl}_2\text{F}_2\)). Because of their non-toxicity, non-flammability, and low reactivity, they were widely used as refrigerants, air conditioning propellants, and aerosol sprays.

The Ozone Layer Problem

The ozone layer (\(\text{O}_3\)) in the stratosphere absorbs harmful ultraviolet radiation (UV-B and UV-C) from the Sun, protecting living organisms on Earth from skin cancer and crop damage.

Because CFCs are unreactive in the lower atmosphere, they slowly drift up into the stratosphere, where intense high-energy UV light breaks the relatively weak \(\text{C}-\text{Cl}\) bond (via homolytic fission) to produce reactive chlorine free radicals (\(\text{Cl}^\bullet\)):

Initiation:
\(\text{CCl}_3\text{F} \xrightarrow{\text{UV light}} ^\bullet\text{CCl}_2\text{F} + \text{Cl}^\bullet\)

Catalytic Ozone Depletion Cycle

The chlorine radical acts as a homogeneous catalyst, destroying thousands of ozone molecules in a repeating chain reaction:

Propagation Step 1:
\(\text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2\)

Propagation Step 2:
\(\text{ClO}^\bullet + \text{O} \rightarrow \text{Cl}^\bullet + \text{O}_2\)

Overall Equation:
\(\text{O}_3 + \text{O} \rightarrow 2\text{O}_2\)

Notice how the \(\text{Cl}^\bullet\) radical is regenerated at the end of Step 2! A single chlorine radical can destroy up to \(100,000\) ozone molecules before chain termination occurs.

Safer Alternatives

To protect the ozone layer, international agreements (like the Montreal Protocol) phased out CFCs in favor of Hydrofluorocarbons (HFCs) (e.g., \(\text{CH}_2\text{FCF}_3\)).
• HFCs contain no chlorine atoms (\(\text{C}-\text{Cl}\) bonds), so they do not produce \(\text{Cl}^\bullet\) radicals and cannot destroy ozone.
• The \(\text{C}-\text{F}\) bond is too strong to be broken by stratospheric UV light.

Key Takeaway:

CFCs produce \(\text{Cl}^\bullet\) radicals under stratospheric UV light, which catalytically break down \(\text{O}_3\) into \(\text{O}_2\). HFCs are used as ozone-friendly substitutes because they lack chlorine.

7. Quick Review Summary

• Classification: Primary (\(1^\circ\)), secondary (\(2^\circ\)), and tertiary (\(3^\circ\)) based on the number of carbons bonded to the \(\text{C}-\text{X}\) carbon.
• Reactivity: \(\text{R}-\text{I} > \text{R}-\text{Br} > \text{R}-\text{Cl} > \text{R}-\text{F}\) (governed by \(\text{C}-\text{X}\) bond enthalpy, not bond polarity).
• Substitution with \(\text{OH}^-\) (aq): Produces alcohols.
• Substitution with \(\text{CN}^-\) (ethanolic): Produces nitriles (adds one carbon to the chain).
• Substitution with \(\text{NH}_3\) (excess, ethanolic): Produces primary amines.
• Elimination with \(\text{OH}^-\) (ethanolic, heat): Produces alkenes.
• Testing with \(\text{AgNO}_3\): \(\text{AgCl}\) (white), \(\text{AgBr}\) (cream), \(\text{AgI}\) (yellow).
• Ozone Depletion: Stratospheric UV generates \(\text{Cl}^\bullet\) radicals from CFCs, which act as catalysts in breaking down \(\text{O}_3\).

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

How are halogenoalkanes classified as primary (\(1^\circ\)), secondary (\(2^\circ\)), or tertiary (\(3^\circ\))?

They are classified based on the number of alkyl groups (carbon atoms) directly attached to the carbon bearing the halogen atom: one (or none) for primary, two for secondary, and three for tertiary.

Why do boiling points of halogenoalkanes increase in the order \(\text{R}-\text{F} < \text{R}-\text{Cl} < \text{R}-\text{Br} < \text{R}-\text{I}\) for a given alkyl chain?

Boiling points increase because iodine has the greatest number of electrons, making it the most polarizable and resulting in significantly stronger van der Waals (London dispersion) forces between molecules.

Why are halogenoalkanes insoluble in water despite possessing a polar \(\text{C}-\text{X}\) bond?

Halogenoalkanes cannot form hydrogen bonds with water, and the energy released by forming new dipole-dipole attractions is insufficient to break the strong hydrogen bonds between water molecules.

Why is bond enthalpy rather than bond polarity the main factor determining the reactivity of halogenoalkanes?

Although the \(\text{C}-\text{F}\) bond is the most polar, the \(\text{C}-\text{I}\) bond is much weaker (lowest bond enthalpy), making it significantly easier to break during reactions.

What is a nucleophile and what role does it play in nucleophilic substitution of halogenoalkanes?

A nucleophile is an electron-pair donor that uses a lone pair to attack the electron-deficient \(\text{C}^{\delta+}\) carbon atom, displacing the halogen as a halide leaving group.

What product and conditions are involved when a halogenoalkane reacts with aqueous alkali?

It forms an alcohol when warmed under reflux with aqueous sodium or potassium hydroxide (\(\text{NaOH}\) or \(\text{KOH}\)).

Why is the nucleophilic substitution reaction of halogenoalkanes with cyanide ions synthetically important?

It produces a nitrile and extends the carbon chain length by one carbon atom.

Why must excess concentrated ethanolic ammonia be used to form a primary amine from a halogenoalkane?

Using excess ammonia prevents further substitution reactions by ensuring ammonia, rather than the newly formed primary amine nucleophile, attacks the halogenoalkane.

How can aqueous silver nitrate be used to distinguish the rate of hydrolysis between chloro-, bromo-, and iodoalkanes?

The rate of formation of a silver halide precipitate (\(\text{AgI}\) yellow fastest, \(\text{AgBr}\) cream moderate, \(\text{AgCl}\) white slowest) indicates the relative rate of hydrolysis.

What reagents and conditions are required for a halogenoalkane to undergo elimination instead of substitution?

Potassium hydroxide (\(\text{KOH}\)) or \(\text{NaOH}\) dissolved in pure ethanol, heated under reflux at high temperature.

How do chlorofluorocarbons (CFCs) cause the catalytic destruction of the stratospheric ozone layer?

UV light breaks the weak \(\text{C}-\text{Cl}\) bond generating \(\text{Cl}^\bullet\) radicals, which catalytically convert \(\text{O}_3\) into \(\text{O}_2\) and are regenerated at the end of the cycle.

Why are hydrofluorocarbons (HFCs) safe alternatives to CFCs for the ozone layer?

HFCs contain no chlorine atoms to form \(\text{Cl}^\bullet\) radicals, and the \(\text{C}-\text{F}\) bond is too strong to be broken by stratospheric UV light.

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