The Chemistry of the Haloalkanes (OCR A-Level Chemistry A): Revision Notes

The Chemistry of the Haloalkanes

Introduction to haloalkanes

Haloalkanes (also called halogenoalkanes) are organic compounds that contain carbon, hydrogen, and at least one halogen atom. The halogen elements that appear in these compounds are fluorine (F), chlorine (Cl), bromine (Br), and iodine (I). These compounds form an important class of organic molecules with distinctive chemical properties.

When we name haloalkanes, we add a prefix to identify which halogen is present. Each halogen has its own specific prefix that appears before the main chain name:

The haloalkanes we study most commonly are aliphatic, meaning the halogen atom is attached to either a straight or branched carbon chain. Similar to alcohols, we can classify these compounds as primary, secondary, or tertiary, depending on how many carbon atoms are bonded to the carbon that carries the halogen.

Polarity of the carbon-halogen bond

The carbon-halogen bond in haloalkanes possesses a distinctive property - it is polar. This polarity arises from the difference in electronegativity between carbon and halogen atoms. Halogen atoms are more electronegative than carbon atoms, which means they have a stronger attraction for electrons in the bond.

As a result of this electronegativity difference, the electron pair in the carbon-halogen bond sits closer to the halogen atom than to the carbon atom. We represent this unequal sharing using partial charges: the carbon atom becomes slightly positive (written as $\delta^+$) whilst the halogen atom becomes slightly negative (written as $\delta^-$).

Nucleophiles and nucleophilic substitution

A nucleophile is defined as an atom or group of atoms that possesses a lone pair of electrons and is attracted to an electron-deficient (partially positive) carbon atom. When a nucleophile encounters such a carbon, it donates its pair of electrons to form a new covalent bond. The term "nucleophile" literally means "nucleus-loving", reflecting the attraction to positive centres.

When a haloalkane reacts with a nucleophile, a substitution reaction occurs. In this process, the nucleophile replaces the halogen atom, and a new compound containing a different functional group is produced. The reaction mechanism for this process is called nucleophilic substitution.

The mechanism proceeds through several key steps. The nucleophile approaches the carbon atom that is bonded to the halogen, attacking from the opposite side to the halogen atom. This direction of attack is crucial - it minimises repulsion between the nucleophile (which is either negatively charged or has a lone pair) and the halogen atom (which is partially negative). As the nucleophile donates its lone pair to form a new bond with the carbon, the carbon-halogen bond breaks through heterolytic fission, with both electrons going to the halogen atom. This produces a halide ion as one of the products.

Hydrolysis of haloalkanes

Hydrolysis is a specific type of chemical reaction involving water or an aqueous solution of a hydroxide. The term literally means "breaking with water". In hydrolysis reactions, water (or hydroxide ions from an aqueous solution) causes the breaking of a bond in a molecule, resulting in the molecule being split into two products.

When we carry out the hydrolysis of a haloalkane, the halogen atom is replaced by an $-\text{OH}$ group. This is an example of nucleophilic substitution where the nucleophile is the hydroxide ion ($\text{OH}^-$).

The mechanism of hydrolysis

Practical hydrolysis with sodium hydroxide

To convert haloalkanes to alcohols in the laboratory, we use aqueous sodium hydroxide as the source of hydroxide ions. At room temperature, this reaction proceeds very slowly. To obtain a reasonable yield of product in a sensible timeframe, we heat the mixture under reflux. Reflux ensures that any volatile components don't escape whilst allowing the reaction mixture to be heated.

Bond strength and rate of hydrolysis

The rate at which different haloalkanes undergo hydrolysis depends fundamentally on the strength of the carbon-halogen bond. During hydrolysis, this bond must be broken, so the strength of the bond directly affects how quickly the reaction can occur.

The bond enthalpy (bond energy) tells us how strong a bond is - the larger the bond enthalpy value, the more energy is required to break the bond, and therefore the stronger the bond. We can compare the bond enthalpies of different carbon-halogen bonds:

From this data, we can see a clear pattern:

• The $\text{C}-\text{F}$ bond is the strongest carbon-halogen bond (approximately $460 \text{ kJ mol}^{-1}$)
• The $\text{C}-\text{I}$ bond is the weakest carbon-halogen bond (approximately $240 \text{ kJ mol}^{-1}$)
• Less energy is required to break the $\text{C}-\text{I}$ bond than any other carbon-halogen bond

This trend can be explained by considering the size of the halogen atoms. As we move down Group 7 from fluorine to iodine, the halogen atoms become larger. The larger the halogen atom, the further its outer electrons are from the carbon atom, and the weaker the orbital overlap in the bond. This results in a weaker bond that is easier to break.

Experimental investigation of hydrolysis rates

We can test the predictions about relative rates of hydrolysis by carrying out a practical experiment comparing three primary haloalkanes: 1-chlorobutane, 1-bromobutane, and 1-iodobutane.

General equation for hydrolysis with water

The general equation for the hydrolysis of these haloalkanes with water is:

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{X} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{H}^+ + \text{X}^-$

where X represents any of the halogens (Cl, Br, or I).

Following the reaction with silver nitrate

The rate of each hydrolysis reaction can be monitored by carrying out the reaction in the presence of aqueous silver nitrate. As the hydrolysis takes place, halide ions ($\text{X}^-$) are produced. These halide ions react with silver ions ($\text{Ag}^+$) from the silver nitrate to form a precipitate of the silver halide:

$\text{Ag}^+(\text{aq}) + \text{X}^-(\text{aq}) \rightarrow \text{AgX}(\text{s})$

The nucleophile in this reaction is water, which is present in the aqueous silver nitrate solution. Haloalkanes are insoluble in water, so we carry out the reaction in the presence of ethanol as a solvent. The ethanol allows the water and the haloalkane to mix together and produce a single homogeneous solution rather than two separate layers.

Experimental procedure

Observations and results

The observations clearly show a pattern in the rate of precipitate formation:

• 1-chlorobutane: white precipitate forms very slowly
• 1-bromobutane: cream precipitate forms at a moderate speed (slower than iodobutane but faster than chlorobutane)
• 1-iodobutane: yellow precipitate forms rapidly

These observations are explained perfectly by considering the bond enthalpies we examined earlier. The compound with the slowest rate of reaction is the one with the strongest carbon-halogen bond, whilst the fastest reaction occurs with the weakest carbon-halogen bond:

• 1-Chlorobutane reacts slowest because the $\text{C}-\text{Cl}$ bond is the strongest of the three
• 1-Iodobutane reacts fastest because the $\text{C}-\text{I}$ bond is the weakest

Hydrolysis of primary, secondary, and tertiary haloalkanes

Whilst bond strength is clearly important in determining the rate of hydrolysis, it is not the only factor that influences how quickly haloalkanes react. This becomes evident when we examine structural isomers of bromoalkanes with the molecular formula $\text{C}_4\text{H}_9\text{Br}$.

Comparing rates for structural isomers

When we compare the hydrolysis rates of three structural isomers of bromoalkanes, we observe something surprising:

All three compounds have the same $\text{C}-\text{Br}$ bond, so bond strength cannot explain these different rates. The key difference lies in whether the haloalkane is primary, secondary, or tertiary.

When we examine the structures carefully, we see that:

• One haloalkane is primary - the carbon bearing the bromine is bonded to only one other carbon
• One is secondary - the carbon bearing the bromine is bonded to two other carbons
• One is tertiary - the carbon bearing the bromine is bonded to three other carbons

Different reaction mechanisms

The key to understanding the different rates is recognising that tertiary carbocations are more stable than primary carbocations. The increased stability of the tertiary carbocation intermediate makes it easier to form, which means the two-step mechanism for tertiary haloalkanes proceeds faster than the one-step mechanism for primary haloalkanes.

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