You've probably played at making soap bubbles while washing your hands. When we use soap and mix it with water, it forms a kind of bubble mass that is very easy to play with. Well, this happens thanks to a kind of hydrolysis. Do you know what it is? Well, read on to find out more!

• In this article, we will touch on the topic of hydrolysis, what it is and how it takes place.
• You will read about hydrolysis of halogenoalkanes, also called haloalkanes.
• You will learn about the conditions required for the hydrolysis of halogenoalkanes.
• You will see how different halogens affect the rate of hydrolysis of halogenoalkanes.
• You will learn how to check the difference in rates of hydrolysis of different halogenoalkanes in the chemistry laboratory.

What is hydrolysis?

Water molecules break down the bonds of the chemical dissolved in water. Water molecules act as nucleophiles in this type of reaction, and the hydrolysis of halogenoalkanes is a nucleophilic substitution type of reaction.

Halogenoalkanes and water

Let us take the example of a halogenoalkane, chloromethane (CH₃Cl). We know that chlorine is much more electronegative than carbon. The shared electrons between carbon and chlorine get pulled towards the chlorine atom due to this, and the chlorine atom acquires a partial negative charge, while the carbon acquires a partial positive charge.

Fig. 1: Partial positive charge on carbon & partial negative charge on chlorine in chloromethane | Vaia Originals

Now if you look at the definition of nucleophilic substitution again, you can tell that the carbon atom is the electron-poor site in the molecule. Now let's bring a nucleophile in the picture.

A nucleophile is a chemical species that has an extra pair of electrons, and can donate it. The nucleophile involved in a hydrolysis reaction is the hydroxide ion (OH^-). The oxygen of the hydroxide ion has 3 lone pairs of electrons.

Fig. 2: Lewis structure of hydroxide Ion | Vaia Originals

Fig. 3: Hydrolysis of chloromethane produces methanol | Vaia Originals

So, the general reaction of hydrolysis of halogenoalkanes can be written as:

$$ RX + OH^- \rightarrow ROH + X^- $$

Where R is any hydrocarbon chain, and X is any halogen atom.

Hydrolysis of halogenoalkanes: mechanism

It is important to understand how the hydrolysis of halogenoalkanes works at the molecular level, not only because it's frequently asked in exams, but also because it will help you get a better grasp of the concept of hydrolysis.

The nucleophile hydroxide ion attacks the electrophilic site (the electron-poor carbon atom). Now, you know that carbon can only make 4 covalent bonds. Therefore, the bond with the halide ion breaks as the bond with the hydroxide ion is forming. It is important to note that the leaving halogen atom takes with it both the electrons shared in the carbon-halogen bond. This is known as heterolytic fission. The electron needs of the carbon atom (which is now 2 electrons short) are now fulfilled by the hydroxide ion.

Note that in the diagram above, the hydroxide ion is shown to be attacking the electrophilic site from the opposite side of the leaving group. This is not just in the diagram, but that's how the reaction takes place. This is due to the resistance offered by the negative charges on the leaving group as well as the nucleophile.

Hydrolysis of halogenoalkanes: conditions

There are two conditions which are necessary for the hydrolysis of halogenoalkanes to take place.

To carry out the hydrolysis, the reaction is carried out in an aqueous solution of hydroxide ions. This can be obtained by dissolving sodium hydroxide, NaOH, or potassium hydroxide, KOH in water. Now, the problem arises due to the insolubility of halogenoalkanes in water. Being organic compounds, halogenoalkanes can only be dissolved in organic solvents. Therefore, ethanol is added to the solution, which dissolves halogenoalkanes in the solution and enables them to react with the hydroxide ions.

The second condition is that the reaction is heated under reflux to increase the reaction rate, as hydrolysis is very slow at cooler temperatures.

To summarise, the two conditions for the hydrolysis of halogenoalkanes are -

1. Ethanol must be added to dissolve halogenoalkane.
2. Heat must be provided to increase the rate of the reaction.

$$ CH_3Cl_{(l)} + NaOH_{(aq)} \xrightarrow [\text{heat under reflux}]{\text{ethanol solvent}} CH_3OH_{(l)} + NaCl_{(aq)} $$

Rate of halogenoalkane hydrolysis: an experiment with silver nitrate

Different halide groups attached to the same alkane group show different rates of hydrolysis. Let us do an experiment to see how the halide group affects the rate of hydrolysis of halogenoalkanes.

For the experiment, we will take 3 primary halogenoalkanes - 1-chlorobutane, 1-bromobutane, and 1-iodobutane.

Fig. 5: 1-chlorobutane, 1-bromobutane, and 1-iodobutane | Vaia Originals

To compare the rate of hydrolysis for these 3 compounds, we need to detect the release of halide ion, which would indicate that the hydrolysis is complete. We'll be doing this by adding silver nitrate to the solution. Silver nitrate reacts with halide ions to form silver halide, which is a precipitate, and thus can be seen with the naked eye. This is shown in the reaction below, where X is a halogen atom.

$$ AgNO_{3(aq)} + X^-_{(aq)} \rightarrow AgX_{(s)} + NO_3^-\space_{(aq)} $$

Let us do the experiment now. We'll take 3 separate test tubes and add 1 ml of ethanol in each before adding 0.1 ml of each compound in separate test tubes. Remember that ethanol acts as a solvent for halogenoalkanes and allows it to react with water.

We will now keep these test tubes in a water bath at 60^oC. and wait for them to heat up. Remember that it is necessary to heat the reaction because hydrolysis is slower at lower temperatures. It is also necessary to ensure that all 3 test tubes are at the same temperatures so that the comparison of reaction rate is fair.

After the test tubes are heated up, we will add 1 ml of aqueous AgNO₃ to the test tubes. In an aqueous solution, water molecules act as nucleophiles, which will now react with the halogenoalkanes and hydrolysis will occur.

Reactions that will occur in each test tube:

$$ C_4H_9X + OH^- \rightarrow C_4H_9OH + X^- $$

$$ X^-_{(aq)} + Ag^+_{(aq)} \rightarrow AgX_{(s)} $$

Where X is either Cl, Br, or I.

Observations

As the halide ion is released, a precipitate of silver halide will form.

• We observed that the reaction in the test tube with C₄H₉I completed the fastest among the 3 test tubes. That means that the rate of hydrolysis of C₄H₉I was the fastest.
• Rate of hydrolysis of C₄H₉Br was slower than that of C₄H₉I.
• Rate of hydrolysis of C₄H₉Cl was the slowest.

Explanation for observations

Why do you think the hydrolysis for 1-iodobutane was the fastest, and that for 1-chlorobutane was the slowest? It is clear that the rate is increasing as we go down the halogen group of the periodic table.

This is explained by the carbon-halogen bond strength. C-I bond strength is the lowest, and hence it is easier for the leaving group (I^-) during the hydrolysis of 1-iodobutane. Similarly, C-Cl bond strength is the highest, and it takes a lot more energy and time for the leaving group (Cl-) to leave during the hydrolysis of 1-chlorobutane. This will be true for all halogenoalkanes.

So the rate of hydrolysis of halogenoalkanes for different halogen atoms can be written as:

R-Cl < R-Br < R-I

where R is any hydrocarbon chain.