Haloalkanes

In a nutshell

Haloalkanes are alkanes with one or more hydrogen atoms replaced by a halogen. The polar C-X bond makes the carbon a target for nucleophiles, so haloalkanes react by nucleophilic substitution, for example hydrolysis to an alcohol.

The second half of the subtopic is environmental: UV radiation breaks C-Cl bonds in CFCs in the upper atmosphere, and the chlorine radicals released catalyse the breakdown of ozone.

Assumed knowledge: Basic concepts of organic chemistry (curly arrows, bond fission, radicals), Alkanes, Bonding and structure (electronegativity and polar bonds).

Core content

Substitution reactions: hydrolysis by aqueous alkali

A haloalkane contains the C-X functional group, where X is F, Cl, Br or I.

The halogen is more electronegative than carbon, so the C-X bond is polar: C^δ+-X^δ-. The δ+ carbon is the part that gets attacked.

Hydrolysis is a reaction with water (or aqueous hydroxide ions) that breaks a bond. Here the C-X bond breaks and the halogen is replaced by an OH group, so the haloalkane becomes an alcohol.

Because one atom (X) is swapped for a group (OH), this is a substitution reaction.

For 1-bromobutane:

CH₃CH₂CH₂CH₂Br + NaOH → CH₃CH₂CH₂CH₂OH + NaBr

As an ionic equation (Na⁺ is a spectator ion):

CH₃CH₂CH₂CH₂Br + OH^- → CH₃CH₂CH₂CH₂OH + Br^-

The product is butan-1-ol: the OH goes on the same carbon the halogen left.

Substitution reactions: nucleophiles

A nucleophile is an electron pair donor.

Every nucleophile has a lone pair of electrons it can donate to an electron-poor (δ+) atom to form a new covalent bond.

The two nucleophiles in this subtopic both use a lone pair on oxygen:

• hydroxide ion, OH^-: negatively charged, so it is a strong nucleophile (aqueous alkali)
• water, H₂O: neutral, so it is a much weaker nucleophile and reacts more slowly

The nucleophile is attracted to the δ+ carbon of the C-X bond.

Substitution reactions: the nucleophilic substitution mechanism

You need the mechanism for the hydrolysis of a primary haloalkane with aqueous alkali. A primary haloalkane has the halogen on a carbon joined to only one other carbon (for example 1-bromobutane).

What a full mechanism must show:

1. The dipole on the C-X bond: δ+ on carbon, δ- on the halogen.
2. A curly arrow from the lone pair (or negative charge) on the O of OH^- to the δ+ carbon.
3. A curly arrow from the C-X bond to the halogen atom.
4. The products: the alcohol and the halide ion, X^-, with its negative charge.

The C-X bond breaks by heterolytic fission: the halogen takes both electrons from the bonding pair, so it leaves as a halide ion.

Still don't get it? · reading the curly arrows

Think of a game of musical chairs for one pair of electrons. A newcomer (the hydroxide ion) arrives holding a spare pair of electrons and wants to sit next to the carbon. The carbon only has room if someone leaves, so the halogen walks off, taking the pair of electrons it was sharing with carbon.

Now step by step:

• The halogen pulls the shared electrons in the C-X bond towards itself. That leaves the carbon slightly positive (δ+) and the halogen slightly negative (δ-).
• OH^- has spare lone pairs on its oxygen. A negative ion is attracted to the δ+ carbon.
• Arrow 1 shows the oxygen's lone pair moving towards the carbon to make a new C-O bond. It starts at the lone pair, because that is where the electrons are now.
• Carbon cannot hold five bonds, so the C-X bond must break as the new bond forms.
• Arrow 2 shows the pair of electrons in the C-X bond moving onto the halogen. The halogen now owns both electrons, so it leaves as X^-.
• What is left is the alcohol, with OH on the carbon where X used to be.

The exam version: a curly arrow shows the movement of an electron pair. Draw the dipole, an arrow from the lone pair on OH^- to the δ+ carbon, an arrow from the C-X bond to X, and the products alcohol + X^-. The C-X bond breaks by heterolytic fission.

Substitution reactions: why the rate depends on C-X bond enthalpy

The rate of hydrolysis increases from fluoroalkanes to iodoalkanes, because the C-X bond enthalpy decreases down the group.

Values differ slightly between data sources. An exam question will give you any values it wants you to use; what you must know is the order.

The explanation, as a chain:

1. Going from C-F to C-I, the bond enthalpy decreases (the bond gets weaker).
2. So less energy is needed to break the C-X bond.
3. So the C-X bond breaks more easily, and the rate of hydrolysis increases.

The trend is not explained by polarity. C-F is the most polar bond, so polarity alone would predict fluoroalkanes react fastest. They react slowest, so bond enthalpy is the deciding factor.

Still don't get it? · bond enthalpy beats polarity

Imagine trying to steal a ball from four people. One holds it loosely but is waving it about so you notice it easily; another grips it tightly. It does not matter how much the ball catches your eye if you cannot prise it out of their hands.

Step by step:

• The nucleophile is attracted to the δ+ carbon. A bigger dipole (C-F is the most polar) makes that carbon more attractive to attack. That is the "noticing the ball" part.
• But for substitution to happen, the C-X bond has to break. That is the "prising it out" part.
• C-F is by far the strongest bond (highest bond enthalpy), so it hardly ever breaks. C-I is the weakest, so it breaks most easily.
• Experiments show iodoalkanes react fastest and fluoroalkanes slowest. So breaking the bond matters more than the size of the dipole.

The exam version: the rate of hydrolysis depends on the C-X bond enthalpy. The C-I bond has the lowest bond enthalpy, so it breaks most easily and iodoalkanes hydrolyse fastest. Do not answer with polarity or electronegativity.

PAG7: comparing rates of hydrolysis with silver nitrate

This practical (part of PAG7, qualitative analysis of organic functional groups) compares how fast the C-Cl, C-Br and C-I bonds are hydrolysed by water, using aqueous silver nitrate and ethanol.

The chemistry

Water is the nucleophile. When the C-X bond breaks, a halide ion is released:

CH₃CH₂CH₂CH₂I + H₂O → CH₃CH₂CH₂CH₂OH + H⁺ + I^-

The halide ion meets the silver ions and forms a silver halide precipitate:

Ag⁺(aq) + X^-(aq) → AgX(s)

A precipitate can only form after the C-X bond has broken, so the faster the precipitate appears, the faster the hydrolysis.

You do not need the mechanism for hydrolysis by water.

Method

1. Put the same volume of ethanol into three test tubes and add the same number of drops of 1-chlorobutane, 1-bromobutane and 1-iodobutane (one per tube).
2. Stand these tubes, and a tube of aqueous silver nitrate, in a water bath at about 50 °C so they reach the same temperature.
3. Add the same volume of warm silver nitrate solution to each tube, start a stopwatch, and shake.
4. Record the time taken for a precipitate to appear and its colour.

Why each condition matters:

• Ethanol is a solvent that mixes with both the haloalkane and the aqueous silver nitrate. Haloalkanes do not dissolve in water, so without ethanol the reactants stay in separate layers.
• Same chain length and all primary (all 1-halobutanes): the only variable that changes is the halogen.
• Same temperature, volumes and concentration: a fair test.
• Water bath, not a flame: ethanol and the haloalkanes are flammable.

Results

The order matches the bond enthalpies: C-I is weakest, so 1-iodobutane hydrolyses fastest.

Since rate is inversely proportional to time, you can compare rates using 1/time.

Environmental concerns: why ozone matters and why CFCs were used

Ozone, O₃, in the upper atmosphere (the stratosphere) absorbs much of the Sun's harmful UV radiation. Too much UV reaching the ground increases the risk of skin cancer.

CFCs (chlorofluorocarbons) are haloalkanes containing carbon, chlorine and fluorine only, for example CF₂Cl₂.

They were widely used because they are:

• unreactive (stable)
• non-toxic
• non-flammable
• volatile (low boiling points)

That made them ideal as refrigerants, aerosol propellants and blowing agents for foam plastics.

Their low reactivity is also the problem: CFCs survive long enough in the lower atmosphere to drift up into the stratosphere.

Environmental concerns: production of halogen radicals

In the upper atmosphere, UV radiation breaks the C-Cl bond in a CFC by homolytic fission, forming two radicals:

CF₂Cl₂ → CF₂Cl• + •Cl

It is the C-Cl bond that breaks, not C-F, because C-Cl has a lower bond enthalpy (the same idea as the hydrolysis trend).

A radical is a species with an unpaired electron, shown by a dot, •. OCR requires the dot on every species with a single unpaired electron.

Environmental concerns: catalysed breakdown of ozone by Cl• and •NO

The chlorine radical destroys ozone in two steps:

•Cl + O₃ → •ClO + O₂

•ClO + O → •Cl + O₂

Add the steps and cancel what appears on both sides (•Cl and •ClO):

Overall: O₃ + O → 2O₂

The oxygen atoms (O) are present in the stratosphere because UV radiation also splits O₂ and O₃ molecules.

Cl• is a catalyst:

• it is used up in step 1 but regenerated in step 2
• so it does not appear in the overall equation
• so one chlorine radical can break down thousands of ozone molecules before it is removed

Other radicals do the same job. Nitrogen monoxide, •NO, forms from nitrogen and oxygen at high temperatures (lightning, aircraft engines):

•NO + O₃ → •NO₂ + O₂

•NO₂ + O → •NO + O₂

Overall: O₃ + O → 2O₂

The pattern for any radical R• (you may be asked to write these for a radical you have not met; for •NO, •RO is •NO₂):

Environmental concerns: evidence, legislation and alternatives

How the science changed what we do:

• 1974: chemists Molina and Rowland proposed that chlorine from CFCs could destroy stratospheric ozone.
• 1985: scientists of the British Antarctic Survey reported large seasonal losses of ozone over Antarctica (the "ozone hole").
• 1987: governments agreed the Montreal Protocol, which phased out the production and use of CFCs.

The evidence was accepted and acted on, so this is often used as an example of science leading to government legislation.

CFCs have been replaced, for example by HFCs (hydrofluorocarbons):

• HFCs contain no C-Cl bonds, so they do not release chlorine radicals and do not break down ozone.
• Many HFCs are powerful greenhouse gases, so they are now being phased down too.

CFCs are very long-lived, so CFCs released years ago are still breaking down ozone today.

Worked examples

Example 1: predicting a hydrolysis product and its mechanism

1-iodopropane, CH₃CH₂CH₂I, is heated under reflux with aqueous potassium hydroxide. Give the product, the equation and what the mechanism must show.

1. The OH group replaces the I on the same carbon, so the organic product is propan-1-ol, CH₃CH₂CH₂OH.
2. Equation: CH₃CH₂CH₂I + KOH → CH₃CH₂CH₂OH + KI
3. Ionic equation: CH₃CH₂CH₂I + OH^- → CH₃CH₂CH₂OH + I^-
4. Mechanism (same shape as Figure 1, with I in place of Br):
   • δ+ on the carbon bonded to I, δ- on I
   • curly arrow from a lone pair on the O of OH^- to the δ+ carbon
   • curly arrow from the C-I bond to I
   • products: CH₃CH₂CH₂OH and I^-

Where students slip: drawing the product with the carbon bonded to the H of OH, and leaving the charge off the iodide ion.

Example 2: model answer on the rate trend

1-chlorobutane and 1-iodobutane are each warmed with ethanol and aqueous silver nitrate. Predict and explain the observations.

A full answer makes these linked points, in order:

1. 1-iodobutane gives a yellow precipitate (AgI); 1-chlorobutane gives a white precipitate (AgCl).
2. The yellow precipitate appears first: 1-iodobutane is hydrolysed faster.
3. The C-I bond enthalpy is lower than the C-Cl bond enthalpy.
4. So the C-I bond breaks more easily, releasing I^- faster to react with Ag⁺.

Where students slip: explaining with polarity or electronegativity (C-Cl is more polar, yet reacts slower), or saying "iodine is more reactive" (as an element, iodine is the least reactive of the three halogens).

Example 3: radical equations for an unfamiliar radical

Bromotrifluoromethane, CF₃Br, was used in fire extinguishers. Write equations to show how it forms a radical in the upper atmosphere and how that radical catalyses the breakdown of ozone.

1. Which bond breaks? C-Br has a lower bond enthalpy than C-F, so UV breaks C-Br by homolytic fission:

   CF₃Br → CF₃• + •Br

2. Follow the pattern for Cl•, with Br in place of Cl:

   •Br + O₃ → •BrO + O₂

   •BrO + O → •Br + O₂

3. Overall (•Br and •BrO cancel): O₃ + O → 2O₂

Check each equation balances: in step 1, one Br and three O atoms appear on each side; in step 2, one Br and two O atoms.

Where students slip: leaving dots off the radicals, writing Br^- (an ion) instead of •Br (a radical), and including •Br in the overall equation.

Common exam mistakes

• Explaining the hydrolysis trend with polarity or electronegativity. The mark needs bond enthalpy: C-I has the lowest bond enthalpy, so it breaks most easily. Polarity predicts the wrong order (C-F is the most polar but the slowest).
• Saying iodoalkanes are fastest "because iodine is the most reactive halogen". It is the reverse for the elements; the reason is the weak C-I bond.
• Defining a nucleophile as an "electron donor" or "something that attacks a positive carbon". The definition is an electron pair donor; "pair" is essential.
• Starting the first curly arrow from the H of OH^-, or from the O atom with no lone pair or charge shown. It must start from a lone pair (or the negative charge) on the oxygen and end on the δ+ carbon.
• Drawing the second curly arrow from the carbon to the halogen, or from the halogen outwards. It starts from the middle of the C-X bond and ends on X.
• Missing dipoles or the halide ion's charge. OCR expects relevant dipoles (δ+ C, δ- X) and the product as X^-, not X or X₂.
• Joining OH to the carbon through the H in the product. The C-O bond must be drawn to the oxygen.
• Leaving out "aqueous" in the conditions. The reagent is aqueous NaOH (or KOH), heated under reflux.
• Forgetting "precipitate" or mixing up the colours in the silver nitrate experiment: AgCl white, AgBr cream, AgI yellow.
• Not explaining why ethanol is added: the haloalkane does not mix with water, and ethanol is a solvent for both.
• Missing radical dots. OCR requires a dot on every species with one unpaired electron: Cl•, •ClO, •NO, CF₂Cl•.
• Saying UV breaks the C-F bond in a CFC. It breaks the C-Cl bond, which has the lower bond enthalpy.
• Not explaining why Cl• is a catalyst. Say it is regenerated in the second step (so one radical breaks down many ozone molecules), and do not include Cl• in the overall equation.
• Confusing ozone depletion with global warming. Ozone depletion lets more UV reach the Earth; it is a different problem from the greenhouse effect.

Key definitions

• Nucleophile: an electron pair donor.
• Substitution: a reaction in which an atom or group of atoms is replaced by a different atom or group of atoms.
• Hydrolysis: a reaction in which water (or aqueous hydroxide ions) breaks a chemical bond; here, the C-X bond is broken and the halogen is replaced by OH.
• Curly arrow: shows the movement of an electron pair, either in heterolytic fission or in the formation of a covalent bond.
• Heterolytic fission: bond breaking in which one bonding atom receives both electrons from the bonded pair.
• Homolytic fission: bond breaking in which each bonding atom receives one electron from the bonded pair, forming two radicals.
• Radical: a species with an unpaired electron.

Specification

• I can describe the hydrolysis of a haloalkane by aqueous alkali as a substitution reaction, with reagent, conditions and equation. (4.2.2(a)(i))
• I can describe how water, with aqueous silver nitrate and ethanol, is used to compare the rates of hydrolysis of C-Cl, C-Br and C-I bonds, and give the precipitate colours. (4.2.2(a)(ii), PAG7)
• I can define a nucleophile as an electron pair donor and identify OH^- and H₂O as nucleophiles. (4.2.2(b))
• I can draw the mechanism of nucleophilic substitution for the hydrolysis of a primary haloalkane with aqueous alkali, with dipoles, lone pair, curly arrows and products. (4.2.2(c))
• I can explain the trend in the rates of hydrolysis of primary haloalkanes using the bond enthalpies of C-F, C-Cl, C-Br and C-I. (4.2.2(d))
• I can write an equation for the production of halogen radicals from a CFC by UV radiation in the upper atmosphere. (4.2.2(e)(i))
• I can write equations for the catalysed breakdown of ozone by Cl• and •NO, and construct similar equations for another stated radical. (4.2.2(e)(ii))
• I can outline the benefits of CFCs and how the evidence on ozone depletion led to legislation against their use. (4.2.2(e), HSW)

Related notes

• Basic concepts of organic chemistry: curly arrows, bond fission and radicals
• Alkanes: radical substitution, the other place UV and radicals appear
• Alcohols: the products of hydrolysis, and making haloalkanes from alcohols
• Organic synthesis: heating under reflux and preparing a haloalkane (PAG5)
• The halogens: the silver nitrate test for halide ions
• Reaction rates: catalysts, including Cl• as a homogeneous catalyst
• Bonding and structure: electronegativity and bond polarity