Introduction to Halogenoalkanes (Haloalkanes)

Welcome to the fascinating world of Halogenoalkanes! These are organic molecules where one or more hydrogen atoms in an alkane chain have been replaced by a halogen atom ($F, Cl, Br, I$).


You might know them as solvents, or maybe you've heard of anaesthetics, or perhaps the notorious CFCs (chlorofluorocarbons) that damage the ozone layer. Halogenoalkanes are incredibly important because they are highly reactive intermediate compounds, acting as chemical starting blocks for synthesizing a huge variety of other functional groups (like alcohols, nitriles, and amines).


This chapter is all about understanding why they are reactive and mastering the two main types of reactions they undergo: Substitution and Elimination.


1. Structure and Classification of Halogenoalkanes

1.1 The Crucial C–X Bond

The key to understanding halogenoalkane reactivity lies entirely in the bond between the carbon atom and the halogen atom (\(C-X\)).


  • Halogens (\(X\)) are generally more electronegative than carbon.
  • This difference in electronegativity creates a polar bond, meaning the electrons in the bond are pulled closer to the halogen.
  • This results in a partial negative charge on the halogen, \(\delta-\), and a partial positive charge on the carbon, \(\delta+\).

$$ \mathbf{C}^{\delta+}-\mathbf{X}^{\delta-} $$


The carbon atom now has a partially positive charge, making it an excellent target for species that are seeking a positive centre—these are called nucleophiles.


Key Term: A nucleophile is an electron-pair donor. Nucleophiles are attracted to the positive carbon centre (\(\mathbf{C}^{\delta+}\)) and initiate the primary reactions of halogenoalkanes.


1.2 Classification: Primary, Secondary, and Tertiary

Halogenoalkanes are classified based on the number of alkyl (R) groups attached to the specific carbon atom bonded to the halogen ($X$).


This classification is essential, as it determines which reaction mechanism is preferred (more on $S_N1$ and $S_N2$ later!).


Primary (1°) Halogenoalkanes:
  • The carbon bonded to the halogen is attached to one alkyl (R) group.
  • Example: Bromoethane ($CH_3CH_2Br$).
Secondary (2°) Halogenoalkanes:
  • The carbon bonded to the halogen is attached to two alkyl (R) groups.
  • Example: 2-Chloropropane.
Tertiary (3°) Halogenoalkanes:
  • The carbon bonded to the halogen is attached to three alkyl (R) groups.
  • Example: 2-Bromo-2-methylpropane.

Quick Takeaway: Halogenoalkanes are highly reactive because the $C-X$ bond is polar, creating a positive carbon target ($\mathbf{C}^{\delta+}$) ready for attack by electron-rich nucleophiles.


2. Synthesis of Halogenoalkanes (How to Make Them)

The syllabus requires you to recall three main ways to prepare halogenoalkanes:


2.1 From Alkanes (Free-Radical Substitution)

This reaction occurs when an alkane is reacted with a halogen ($Cl_2$ or $Br_2$) in the presence of ultraviolet (UV) light. It results in a mixture of products, making it less useful for targeted synthesis.


$$ \text{Alkane} + \mathbf{X_2} \xrightarrow{UV\ light} \text{Halogenoalkane} + \mathbf{HX} $$


2.2 From Alkenes (Electrophilic Addition)

Alkenes are unsaturated (contain a $C=C$ double bond). They react quickly with halogens or hydrogen halides at room temperature via electrophilic addition.


  • With a Halogen ($X_2$): The halogen adds across the double bond. (e.g., $C_2H_4 + Br_2 \rightarrow CH_2BrCH_2Br$)
  • With a Hydrogen Halide ($HX$): The hydrogen halide adds across the double bond. (e.g., $C_2H_4 + HBr \rightarrow CH_3CH_2Br$)

2.3 From Alcohols (Substitution)

This is often the most controlled and efficient method for making specific halogenoalkanes.


  • Using Hydrogen Halides ($HX$): Heating an alcohol with $HX(g)$.
  • Using Sulfuric/Phosphoric Acids: Heating an alcohol with $KCl$ and concentrated $H_2SO_4$ or $H_3PO_4$.
  • Using Phosphorus Halides:
    • With $PCl_3$ (and heat): \(3 R-OH + PCl_3 \xrightarrow{heat} 3 R-Cl + H_3PO_3\)
    • With $PCl_5$: \(R-OH + PCl_5 \rightarrow R-Cl + POCl_3 + HCl\)
  • Using Thionyl Chloride ($SOCl_2$):
    • \(R-OH + SOCl_2 \rightarrow R-Cl + SO_2(g) + HCl(g)\)
    • (This is useful because the by-products, $SO_2$ and $HCl$, are gases and easily removed.)

3. Reactions of Halogenoalkanes

Halogenoalkanes primarily undergo two types of reaction dictated by the positive carbon centre: Nucleophilic Substitution and Elimination.


3.1 Nucleophilic Substitution Reactions ($S_N$)

In substitution, the halogen ($X$) is replaced by a nucleophile (Nu:). The $C-X$ bond breaks heterolytically, and the halogen leaves as a halide ion ($X^-$). Since the bond breaking involves the nucleophile attacking the $\mathbf{C}^{\delta+}$, this is often called heterolytic fission.


A. Hydrolysis (Forming Alcohols)
  • Nucleophile: Hydroxide ion (\(OH^-\)) from aqueous sodium hydroxide.
  • Reagents & Conditions: $NaOH(aq)$ and heat (usually heating under reflux).
  • Product: An alcohol.

$$ R-X + OH^- \rightarrow R-OH + X^- $$


B. Formation of Nitriles (Increasing Chain Length)
  • Nucleophile: Cyanide ion (\(CN^-\)).
  • Reagents & Conditions: $KCN$ (or $NaCN$) dissolved in ethanol (ethanolic solution) and heat (reflux).
  • Product: A nitrile ($R-C \equiv N$).

Did you know? This reaction is useful in organic synthesis because it adds an extra carbon atom to the chain, allowing you to build longer molecules.


C. Formation of Primary Amines
  • Nucleophile: Ammonia (\(NH_3\)).
  • Reagents & Conditions: $NH_3$ dissolved in ethanol, heated under pressure in a sealed tube.
  • Product: A primary amine ($R-NH_2$).

Caution: The amine product can also act as a nucleophile and attack another halogenoalkane molecule, leading to a mixture of primary, secondary, and tertiary amines, and even quaternary ammonium salts. Using a large excess of concentrated ammonia minimizes this side reaction, favouring the primary amine.


D. Identification Test (Reaction with Aqueous Silver Nitrate)

This reaction tests for the presence and reactivity of the halogen in the halogenoalkane. The nucleophile is the water molecule, $H_2O$, (or $NO_3^-$ as a catalyst, but water is the dominant nucleophile for hydrolysis).


  • Reagents & Conditions: Aqueous silver nitrate ($AgNO_3$) in ethanol (the ethanol helps mix the organic and aqueous layers). The mixture is heated gently.
  • Reaction: Hydrolysis occurs, producing a halide ion ($X^-$), which then precipitates with the silver ions ($Ag^+$).
  • Observation: The colour and rate of precipitate formation identify the halogen.

    • $$ R-X + H_2O \rightarrow R-OH + H^+ + X^- \\ X^- (aq) + Ag^+ (aq) \rightarrow AgX (s) $$

  • Precipitates:
    • $AgCl$: White precipitate (slowest to form)
    • $AgBr$: Cream precipitate (forms faster)
    • $AgI$: Pale yellow precipitate (forms fastest)

The different rates of precipitation directly relate to the $C-X$ bond strength (see Section 4.3).


3.2 Elimination Reaction (Forming Alkenes)

Halogenoalkanes can react to eliminate a small molecule and form a $C=C$ double bond (an alkene). The conditions for this reaction are critical!


  • Reagents & Conditions: Sodium hydroxide ($NaOH$) dissolved in ethanol (ethanolic conditions) and heat.
  • Role of \(OH^-\): In this case, the hydroxide ion acts as a base, removing a hydrogen atom from the carbon *adjacent* to the halogen-bonded carbon (a $\beta$-elimination).
  • Product: An alkene.

Important Distinction:

$$ \mathbf{NaOH(aq) + heat} \rightarrow \mathbf{Substitution} (\text{Alcohol}) \\ \mathbf{NaOH(ethanol) + heat} \rightarrow \mathbf{Elimination} (\text{Alkene}) $$

Remember: Aqueous means water is present, favoring hydrolysis (substitution). Ethanolic means the $OH^-$ acts as a strong base, favoring elimination.


Quick Takeaway: Halogenoalkanes are versatile. Using aqueous $NaOH$ gives substitution (alcohol); using ethanolic $NaOH$ gives elimination (alkene). $KCN$ and $NH_3$ are substitution nucleophiles.


4. Nucleophilic Substitution Mechanisms: $S_N1$ and $S_N2$

Don't worry if this seems tricky at first! Mechanisms explain how the reaction happens. In the syllabus, you must describe and explain the two main pathways for nucleophilic substitution.


4.1 The $S_N2$ Mechanism (Substitution Nucleophilic Bimolecular)

The number '2' means the rate-determining step involves two species (the halogenoalkane and the nucleophile).


  1. One-Step Process: The nucleophile attacks the $\mathbf{C}^{\delta+}$ from the opposite side (backside attack) to the leaving halogen.
  2. Transition State: A high-energy intermediate state is formed where the nucleophile is partially bonded, and the halogen is partially detached. This state is very crowded.
  3. Product Formation: The $C-X$ bond breaks fully, and the new $C-Nu$ bond forms, resulting in the product and the halide ion ($X^-$).

Steric Hindrance and $S_N2$

Since the nucleophile must attack from the back, crowding near the $\mathbf{C}^{\delta+}$ is a huge problem. This crowding is called steric hindrance.


  • Primary (1°) Halogenoalkanes: Have minimal alkyl groups (small H atoms instead), so there is little steric hindrance. They react readily via the $S_N2$ mechanism.

Memory Aid: $S_N\mathbf{2}$ involves 2 reactants in the slow step, making it a one-step process reliant on backside attack.


4.2 The $S_N1$ Mechanism (Substitution Nucleophilic Unimolecular)

The number '1' means the rate-determining step involves only one species (the halogenoalkane).


  1. Step 1 (Slow/Rate-Determining): The $C-X$ bond breaks first, heterolytically, producing a carbocation ($R^+$) and a halide ion ($X^-$). This is the slow step.
  2. Step 2 (Fast): The nucleophile rapidly attacks the electron-deficient carbocation.

Carbocation Stability and $S_N1$

The ease of Step 1 determines the rate. The stability of the carbocation follows the trend:


$$ \text{Tertiary (3°)} > \text{Secondary (2°)} > \text{Primary (1°)} $$


Why is 3° most stable? Because of the inductive effect of alkyl groups (R groups). Alkyl groups are electron-donating. They push electron density towards the positive carbon, spreading out (delocalizing) the positive charge, which stabilizes the ion.


  • Tertiary (3°) Halogenoalkanes: Form the highly stable tertiary carbocations readily. They react exclusively via the $S_N1$ mechanism.

Memory Aid: $S_N\mathbf{1}$ involves 1 reactant in the slow step, leading to a carbocation intermediate.


4.3 Summary: Structure and Mechanism Preference

The structure dictates the mechanism:

  • Primary (1°): Prefers $S_N2$ (low steric hindrance).
  • Tertiary (3°): Prefers $S_N1$ (stable carbocation).
  • Secondary (2°): Reacts via a mixture of both $S_N1$ and $S_N2$ mechanisms, depending on the specific reactants and conditions.

5. Explaining Reactivity: Bond Strength and Halogen Identity

We know halogenoalkanes are attacked by nucleophiles. The speed of this reaction depends on two main factors:

  1. The ease of attack (steric hindrance, related to P/S/T structure).
  2. The ease with which the $C-X$ bond breaks (bond strength).

5.1 The Role of C-X Bond Strength

Although the $C-F$ bond is the most polar, making the carbon highly \(\mathbf{\delta+}\), the rate of reaction is actually determined by the ease of bond breaking, which is linked to the bond enthalpy (strength).


  • As you go down Group 17 (F $\rightarrow$ I), the halogen atomic radius increases.
  • The $C-X$ bond length increases.
  • The $C-X$ bond enthalpy (strength) decreases.

Because the $C-I$ bond is the weakest, it breaks easiest, leading to the fastest substitution reaction. The $C-F$ bond is very strong and requires a huge amount of energy to break, making fluoroalkanes very unreactive.


Reactivity Trend (fastest to slowest):

$$ \mathbf{R-I} > \mathbf{R-Br} > \mathbf{R-Cl} \gg \mathbf{R-F} $$


This trend is confirmed by the Aqueous Silver Nitrate Test (Section 3.1D):

  • Iodoalkanes ($R-I$) form $AgI$ (pale yellow) immediately.
  • Bromoalkanes ($R-Br$) form $AgBr$ (cream) quickly upon gentle heating.
  • Chloroalkanes ($R-Cl$) form $AgCl$ (white) very slowly, often only upon prolonged heating.

5.2 Halogenoalkanes vs. Halogenoarenes (Extension)

The syllabus requires you to compare the reactivity of a halogenoalkane (like chloroethane) and a halogenoarene (like chlorobenzene).


Halogenoarenes are much less reactive towards nucleophilic substitution than halogenoalkanes.


Reason 1: Partial Double Bond Character

In a halogenoarene (where the halogen is directly attached to a benzene ring), the lone pair of electrons on the halogen atom delocalises into the $\pi$ electron system of the benzene ring. This gives the $C-X$ bond partial double bond character.


A double bond is stronger than a single bond, making the $C-X$ bond in the halogenoarene much harder to break than the single $C-X$ bond in the halogenoalkane.


Reason 2: Leaving Group

The leaving group in a nucleophilic substitution must be able to exist stably on its own. While $Cl^-$ is a good leaving group, in halogenoarenes, the substitution would require breaking the aromatic stability, which is highly energetically unfavourable.


Quick Takeaway: Rate of reaction is controlled by $C-X$ bond strength ($C-I$ weakest $\rightarrow$ fastest reaction). Halogenoarenes are unreactive because the $C-X$ bond has partial double bond character due to resonance with the benzene ring.