Halogenoalkanes

In a nutshell

Halogenoalkanes are alkanes with one or more hydrogen atoms swapped for a halogen. That one swap makes them far more reactive than alkanes, because the carbon-halogen bond is polar and can be broken.

This subtopic covers the two ways they react (nucleophilic substitution and elimination), why the rate depends on carbon-halogen bond enthalpy, and how chlorine atoms from CFCs destroy ozone. Mechanism marks here turn up on almost every organic paper.

Assumed knowledge: Introduction to organic chemistry (naming, curly arrows, E-Z isomers), Alkanes (free radicals), Energetics (bond enthalpy).

Core content

Nucleophilic substitution: the polar carbon-halogen bond

Every reaction in this subtopic starts from one fact: halogenoalkanes contain polar bonds.

• The halogen is more electronegative than carbon.
• So the bonding pair is pulled towards the halogen: the carbon becomes δ+ (electron deficient) and the halogen δ−.
• The δ+ carbon is attacked by a nucleophile, which donates a lone pair to it.

A nucleophile is an electron pair donor. Every nucleophile you need has a lone pair:

In substitution, the halogen atom is replaced by the nucleophile: the C-X bond breaks, a new bond forms between the carbon and the nucleophile, and the halogen leaves as a halide ion.

Name the mechanism with both words: nucleophilic substitution.

Nucleophilic substitution: with hydroxide ions

Reagent and conditions: sodium hydroxide or potassium hydroxide, aqueous (water as the solvent), warm or heat under reflux.

Product: an alcohol.

CH₃CH₂CH₂Br + NaOH → CH₃CH₂CH₂OH + NaBr

1-bromopropane → propan-1-ol. As an ionic equation: CH₃CH₂CH₂Br + OH^- → CH₃CH₂CH₂OH + Br^-

Two arrows, and each one earns a mark:

1. From the lone pair on the O of the negatively charged OH^- to the δ+ carbon.
2. From the middle of the C-Br bond to the Br atom.

When water does the same job (without OH^-), the reaction is called hydrolysis: splitting a molecule (breaking a C-X bond) using water. Water is a weaker nucleophile, so the reaction is slower.

Nucleophilic substitution: with cyanide ions

Reagent and conditions: potassium cyanide (KCN), dissolved in aqueous ethanol (ethanol and water both present), heat under reflux.

Product: a nitrile. The carbon chain is one carbon longer, because the CN group adds a carbon.

CH₃CH₂CH₂Br + KCN → CH₃CH₂CH₂CN + KBr

1-bromopropane → butanenitrile (four carbons, counting the C in C≡N). The nitrile carbon is always carbon 1 when you number the chain.

Nucleophilic substitution: with ammonia

Reagent and conditions: excess ammonia, dissolved in ethanol, heated in a sealed container (under pressure).

Product: a primary amine.

CH₃CH₂CH₂Br + 2NH₃ → CH₃CH₂CH₂NH₂ + NH₄Br

1-bromopropane → propylamine. Notice the two NH₃: one is the nucleophile, the second removes H⁺ and ends up as ammonium bromide.

This mechanism has two stages:

1. NH₃ acts as a nucleophile. The lone pair on N attacks the δ+ carbon and the C-Br bond breaks. This forms an alkylammonium ion, CH₃CH₂CH₂NH₃⁺, with the positive charge on the N.
2. A second NH₃ acts as a base. Its lone pair takes an H⁺ from the N of the alkylammonium ion. The curly arrow goes from the N-H bond to the N atom.

Why excess ammonia? The amine product also has a lone pair on N, so it is itself a nucleophile and can react with more halogenoalkane. A large excess of NH₃ makes it far more likely that the halogenoalkane meets ammonia, so further substitution is minimised.

Summary of the three substitutions

Nucleophilic substitution: bond enthalpy and rate

Down Group 7, the carbon-halogen bond gets weaker (lower bond enthalpy).

Values differ slightly between data sources; only the order is examinable.

The C-X bond must break for substitution to happen. So:

• Lower C-X bond enthalpy → less energy needed to break the bond → faster reaction.
• Rate of substitution (and hydrolysis): iodoalkane > bromoalkane > chloroalkane > fluoroalkane.
• Fluoroalkanes are very unreactive because the C-F bond is so strong.

The answer must compare the two named bonds: "the C-Br bond is weaker (has a lower bond enthalpy) than the C-Cl bond, so it breaks more easily and the bromoalkane reacts faster."

Polarity does not explain this trend. C-F is the most polar bond, yet fluoroalkanes react the slowest. Bond enthalpy wins.

Still don't get it? · why bond enthalpy, not polarity, decides the rate

Think of two locked doors. One has a big "PUSH" sign that attracts lots of people (that is polarity: it attracts the attacker). The other has a flimsy lock. Which one do people actually get through fastest? The one with the flimsy lock. Being attractive does not help if the lock will not break.

Step by step:

1. Every halogenoalkane has a polar C-X bond, so every one has a δ+ carbon that attracts the nucleophile. That part works for all of them.
2. But attracting the nucleophile is not enough. To finish the reaction, the C-X bond has to break so the halide ion can leave.
3. Breaking a bond needs energy. The bond enthalpy tells you how much.
4. C-I needs the least energy to break, C-F needs the most.
5. So the iodoalkane finishes the reaction fastest and the fluoroalkane barely reacts, even though C-F is the most polar.

Exam wording: "The C-I bond has a lower bond enthalpy than the C-Br bond, so it is weaker / breaks more easily, so the iodoalkane reacts faster." Always name both bonds, and always use a comparative for both the bond and the rate.

Practical skill: comparing rates of hydrolysis

You compare how fast different halogenoalkanes hydrolyse by watching for a silver halide precipitate.

Method

1. Put the same volume of each halogenoalkane (for example 1-chlorobutane, 1-bromobutane, 1-iodobutane) into separate test tubes, each with the same volume of ethanol. Ethanol is a solvent that both the halogenoalkane and the aqueous solution mix with.
2. Stand the tubes, and a tube of aqueous silver nitrate, in a water bath at the same fixed temperature (about 50 °C).
3. Add the same volume of silver nitrate solution (of the same concentration) to each tube and start a timer.
4. Record the time for a precipitate to first appear.
5. Rate ∝ 1/time: the shorter the time, the faster the hydrolysis.

What is happening

Water hydrolyses the halogenoalkane, releasing halide ions:

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

The halide ions then meet silver ions:

Ag⁺(aq) + X^-(aq) → AgX(s)

Making it a fair test: the only thing that changes is the halogenoalkane. Keep temperature, volumes and concentration of silver nitrate the same.

Practical skill: preparing and purifying a chloroalkane

A liquid chloroalkane can be made in the lab and purified using a separating funnel and distillation. A common example is making 2-chloro-2-methylpropane by shaking 2-methylpropan-2-ol with concentrated hydrochloric acid:

(CH₃)₃COH + HCl → (CH₃)₃CCl + H₂O

Purifying the crude product:

1. Separating funnel: the chloroalkane does not mix with water and is less dense, so it forms the upper layer. Run off and discard the lower aqueous layer.
2. Wash with sodium hydrogencarbonate solution to remove leftover acid. CO₂ is produced, so stopper, invert and open the tap regularly to release the pressure. Discard the aqueous layer again.
3. Dry the organic layer with an anhydrous drying agent (for example anhydrous sodium sulfate) until the liquid is clear, then decant or filter it.
4. Distil, collecting the fraction that boils at the chloroalkane's boiling point.

Elimination: the hydroxide ion as a base

The same reagent, KOH, can give a completely different product if you change the conditions.

Reagent and conditions: potassium hydroxide, dissolved in ethanol (ethanolic conditions), hot (heat under reflux).

Product: an alkene. H and Br are removed from neighbouring carbon atoms and a C=C forms.

CH₃CHBrCH₃ + KOH → CH₂=CHCH₃ + KBr + H₂O

2-bromopropane → propene.

Here the hydroxide ion acts as a base: a proton acceptor. It removes an H⁺ from a carbon next to the C-Br carbon.

The three arrows, in order:

1. From the lone pair on the O of OH^- to an H on a carbon adjacent to the C-Br carbon.
2. From that C-H bond to the C-C bond, forming the C=C.
3. From the C-Br bond to the Br atom.

Name it elimination (AQA also accepts "base elimination"). Never write "nucleophilic elimination".

Unsymmetrical halogenoalkanes give more than one alkene. In 2-bromopropane both neighbouring carbons are CH₃, so propene is the only alkene. In 2-bromobutane the H can be removed from carbon 1 or carbon 3, giving but-1-ene or but-2-ene (and but-2-ene exists as E and Z isomers). See Worked example 4.

Elimination: substitution and elimination at the same time

When a halogenoalkane reacts with KOH, substitution and elimination happen together (concurrently). The reagent can play two roles:

You never get 100% of one product, but changing the conditions shifts which one is major.

Still don't get it? · one reagent, two roles

Picture a dog that has been trained to fetch two things: a ball and a stick. Throw both at once and sometimes it grabs the ball, sometimes the stick. Change the setting (say, long grass hides the ball) and it brings back the stick far more often. The dog has not changed; the setting has.

Step by step:

1. The hydroxide ion has a lone pair. A lone pair can be donated to anything that wants electrons.
2. The δ+ carbon wants electrons. If the lone pair goes there, a new C-O bond forms and Br leaves. That is substitution, and OH- is acting as a nucleophile.
3. An H atom on the next carbon along can also be pulled off as H+. If the lone pair grabs that H+, the C-H electrons swing across to make a C=C, and Br leaves. That is elimination, and OH- is acting as a base.
4. Both happen in the same flask. Water as solvent and gentle warming tip the balance towards substitution; ethanol as solvent and strong heating tip it towards elimination.

Exam wording: in substitution the hydroxide ion acts as a nucleophile / electron pair donor; in elimination it acts as a base / proton acceptor. Give the whole reagent (KOH or NaOH, not "OH^-") and the solvent (aqueous or ethanolic).

Ozone depletion: why ozone matters and where Cl• comes from

Ozone (O₃) forms naturally in the upper atmosphere. It is beneficial because it absorbs ultraviolet (UV) radiation, reducing the harmful UV reaching the Earth's surface (UV can cause skin cancer).

Chlorofluorocarbons (CFCs) were widely used as refrigerants, aerosol propellants and solvents. They are very unreactive, so they last long enough to drift into the upper atmosphere. There:

• UV radiation breaks the C-Cl bond (it is weaker than C-F) to form chlorine atoms, Cl•.
• For example: CF₂Cl₂ → CF₂Cl• + Cl•

The dot in Cl• represents an unpaired electron: a chlorine atom is a free radical.

Ozone depletion: how chlorine atoms catalyse ozone breakdown

Chlorine atoms catalyse the decomposition of ozone in two steps:

Cl• + O₃ → ClO• + O₂

ClO• + O₃ → 2O₂ + Cl•

Add the two steps and cancel the species that appear on both sides (Cl• and ClO•):

Overall: 2O₃ → 3O₂

Why this counts as catalysis:

• Cl• is used up in step 1 but regenerated in step 2, so it does not appear in the overall equation.
• Cl• provides an alternative route with a lower activation energy.
• Because it is regenerated, one chlorine atom can destroy many thousands of ozone molecules (a chain reaction). This contributes to the hole in the ozone layer.

The same two-step pattern works for other halogen atoms, for example Br• + O₃ → BrO• + O₂ then BrO• + O₃ → 2O₂ + Br•. Write the equations for whichever atom the question gives you.

Ozone depletion: legislation and alternatives

• Research by different groups of scientists gave independent evidence that CFCs were depleting ozone.
• That shared evidence led to international legislation (the Montreal Protocol) that banned the use of CFCs as solvents and refrigerants.
• Chemists have developed chlorine-free alternatives, for example hydrofluorocarbons (HFCs) such as CH₂FCF₃, and hydrocarbons such as pentane.
• These do not deplete ozone because they contain no C-Cl bonds, so no Cl• forms. The C-F bond is too strong to be broken by UV in the upper atmosphere.

Worked examples

Worked example 1: percentage yield from a liquid volume

1-bromobutane is hydrolysed with aqueous sodium hydroxide:

CH₃CH₂CH₂CH₂Br + NaOH → CH₃CH₂CH₂CH₂OH + NaBr

A student uses 3.00 cm³ of 1-bromobutane (density 1.27 g cm^-3, M_r = 136.9) with excess NaOH and obtains 1.52 g of butan-1-ol (M_r = 74.0). Calculate the percentage yield.

Step 1: turn the volume of liquid into a mass, using density.

$m = \text{density} \times V = 1.27\ \text{g cm}^{-3} \times 3.00\ \text{cm}^{3} = 3.81\ \text{g}$

Step 2: moles of 1-bromobutane (the limiting reagent, because NaOH is in excess).

$n = \frac{m}{M_r} = \frac{3.81\ \text{g}}{136.9\ \text{g mol}^{-1}} = 0.02783\ \text{mol}$

Step 3: the ratio is 1 : 1, so the theoretical moles of butan-1-ol are the same. Convert to the theoretical mass.

$m_{\text{theoretical}} = 0.02783\ \text{mol} \times 74.0\ \text{g mol}^{-1} = 2.059\ \text{g}$

Step 4: percentage yield.

$\%\ \text{yield} = \frac{1.52\ \text{g}}{2.059\ \text{g}} \times 100 = 73.8\%$

Answer: 73.8% (3 significant figures, matching the data).

Where marks are lost:

• Dividing the volume (3.00) by M_r instead of first converting to mass with the density. Examiners report this "volume and density" step beats many students.
• Using the moles of NaOH, which is in excess, instead of the halogenoalkane.
• Rounding the moles to 1 significant figure mid-calculation, which throws the final answer outside the accepted range. Keep at least 3-4 figures until the end.

Worked example 2: mass of product from a given yield

Bromoethane reacts with excess ammonia to make ethylamine:

CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br

Calculate the mass of ethylamine (M_r = 45.0) made from 21.8 g of bromoethane (M_r = 108.9) if the yield is 60.0%.

Step 1: moles of bromoethane.

$n = \frac{21.8\ \text{g}}{108.9\ \text{g mol}^{-1}} = 0.2002\ \text{mol}$

Step 2: 1 mol bromoethane gives 1 mol ethylamine (the 2NH₃ does not change this). Apply the yield.

$n_{\text{actual}} = 0.2002\ \text{mol} \times \frac{60.0}{100} = 0.1201\ \text{mol}$

Step 3: mass of ethylamine.

$m = 0.1201\ \text{mol} \times 45.0\ \text{g mol}^{-1} = 5.40\ \text{g}$

Answer: 5.40 g (3 significant figures).

Where marks are lost:

• Using the 1 : 2 ratio with ammonia to change the moles of amine. The amine is 1 : 1 with the halogenoalkane.
• Multiplying by 100/60.0 (making the answer bigger than the theoretical mass). A yield below 100% must give less product than theoretical.
• Giving "5.4 g". The data are to 3 significant figures, so the answer is 5.40 g.

Worked example 3: number of molecules in a CFC replacement

A car air-conditioning unit contains 0.850 kg of the HFC 1,1,1,2-tetrafluoroethane, CH₂FCF₃ (M_r = 102.0). Calculate the number of molecules. (L = 6.022 × 10²³ mol^-1)

Step 1: convert kg to g.

$0.850\ \text{kg} \times 1000 = 850\ \text{g}$

Step 2: moles.

$n = \frac{850\ \text{g}}{102.0\ \text{g mol}^{-1}} = 8.333\ \text{mol}$

Step 3: number of molecules.

$N = n \times L = 8.333\ \text{mol} \times 6.022 \times 10^{23}\ \text{mol}^{-1} = 5.02 \times 10^{24}$

Answer: 5.02 × 10²⁴ molecules (3 significant figures, no unit).

Where marks are lost: examiners report students not converting kg to g and giving too many significant figures (5.018333 × 10²⁴).

Worked example 4: predicting the products of 2-bromobutane with KOH

Task: work out every organic product when 2-bromobutane, CH₃CHBrCH₂CH₃, reacts with KOH.

Step 1: substitution (OH^- as a nucleophile). OH replaces Br on carbon 2: butan-2-ol, CH₃CH(OH)CH₂CH₃.

Step 2: elimination (OH^- as a base). Find every carbon adjacent to the C-Br carbon that has an H. Carbon 2 has two neighbours, carbon 1 and carbon 3.

• H removed from carbon 1 → C=C between C1 and C2 → but-1-ene, CH₂=CHCH₂CH₃.
• H removed from carbon 3 → C=C between C2 and C3 → but-2-ene, CH₃CH=CHCH₃.

Step 3: check for E-Z isomers. But-2-ene has two different groups (CH₃ and H) on each C=C carbon, so it exists as E-but-2-ene and Z-but-2-ene. But-1-ene has two H on carbon 1, so it has no E-Z isomers.

Answer: butan-2-ol (substitution), plus but-1-ene, E-but-2-ene and Z-but-2-ene (elimination). Ethanolic KOH and heat make the alkenes the major products; aqueous KOH makes butan-2-ol the major product.

Where marks are lost: explaining the two structural isomers using carbocations. That is the wrong mechanism here; the mark comes from saying the H is removed from different carbon atoms (C1 or C3).

Worked example 5: model answer on relative rates

Why is 1-bromobutane hydrolysed faster than 1-chlorobutane? The full-mark answer has two linked points:

1. The C-Br bond has a lower bond enthalpy than the C-Cl bond (C-Br is weaker).
2. So less energy is needed to break the C-Br bond, and the bromoalkane reacts faster.

That is the full answer. Do not mention the reactivity of the elements bromine and chlorine, electronegativity, or atom size: these are ignored at best and can contradict the answer.

Common exam mistakes

• Naming the mechanism with one word. Writing "nucleophilic" or "substitution" alone loses the mark; it is nucleophilic substitution, both words.
• Putting a prefix on elimination. "Nucleophilic elimination" and "electrophilic elimination" are rejected. Write elimination (or "base elimination").
• Getting the role of KOH the wrong way round. In elimination it is a base / proton acceptor, not a nucleophile. In substitution it is a nucleophile / electron pair donor, not a base or an "oxidising agent". Writing "nucleophilic substitution" as the type does not earn a separate mark for the role.
• Defining a nucleophile as an "electron donor". The word pair is essential: an electron pair donor.
• Giving an incomplete reagent. "OH^-", "hydroxide ion" or "CN^-" is not a reagent. A reagent is something you take out of a bottle: NaOH or KOH, KCN. Without a correct reagent, the condition mark is often lost too.
• Missing or wrong conditions. Substitution with hydroxide needs aqueous; elimination needs ethanol as solvent. Writing "ethanoic" for ethanolic is a different chemical and is not treated as a spelling slip. "Reflux" on its own does not earn the elimination condition.
• Forgetting "excess" ammonia. The condition that makes the primary amine the major product is excess NH₃.
• Drawing NH₃ with a negative charge. Ammonia is a neutral molecule. A minus sign on NH₃ loses the first mechanism mark, and examiners report it even in strong scripts.
• Misplacing the + on the alkylammonium ion. The positive charge must be on the N atom, not on or next to an H. The final arrow goes from the N-H bond to the N, not towards the H.
• Thinking NH₄⁺ is a nucleophile. It has no lone pair. Over half of students got this wrong in one multiple-choice question.
• Sloppy curly arrows on hydroxide. Missing the lone pair or the negative charge on OH^-, drawing a covalent K-OH or Na-OH, or adding an arrow from Br to K⁺ all lose marks.
• A full + on the carbon of C-Br. That carbon is only δ+. A formal charge, or wrong partial charges, loses the C-Br bond-breaking mark.
• Drawing a bond as a line plus two dots. It is penalised. A bond is a line; dots are only for lone pairs.
• Elimination arrow to the wrong place. OH^- must attack an H on a carbon adjacent to the C-Br carbon. Attacking the carbon, or an H on the C-Br carbon itself, loses marks.
• Explaining rate with the elements or with polarity. "Bromine is more reactive than chlorine" is penalised. "Bromomethane has weaker bonds" is too vague. In one question only 28.6% of students scored this mark: you must compare the C-Br and C-Cl bonds specifically and use a comparative for both bond strength and rate.
• Thinking the fluoroalkane reacts fastest. It reacts slowest: C-F has the highest bond enthalpy, even though it is the most polar.
• "Chloromethane is polar" instead of "the C-Cl bond is polar". The mark is for the polar bond and the δ+ carbon, not a polar molecule.
• Defining hydrolysis as "reaction with water" or "adding water". Neither is credited. It is splitting a molecule (breaking a bond) using water.
• Calling AgI cream. Silver iodide is yellow only; AgBr is cream; AgCl is white.
• Saying fluoroalkanes are left out of the AgNO₃ experiment "because they don't react". The mark is for AgF being soluble, so no precipitate would form.
• Weak fair-test answers. "Same person" or "same stopwatch" earns nothing. Control the temperature, the volumes, and the concentration of silver nitrate.
• Ozone: making Cl• from Cl₂. The chlorine atoms come from UV breaking the C-Cl bond in a CFC. Generating Cl• from Cl₂ is penalised, and F• from a CFC is wrong.
• Ozone: leaving off the radical dot, or writing •O₃. Ozone is not a radical; Cl• and ClO• are.
• Ozone: "Cl• is a catalyst, not used up" with no mention of regeneration. State that Cl• is regenerated (reformed in step 2).
• Nitrile names and formulae. Count the C of C≡N in the chain: 1-bromopropane gives butanenitrile, not propanenitrile. Ethanenitrile is CH₃CN.
• Amine equation with one NH₃. The balanced equation needs 2NH₃ and gives NH₄Br.
• Numbering the double bond in methylpropene. Elimination from 1-bromo-2-methylpropane gives methylpropene (2-methylpropene is also accepted). A name with a number for the double-bond position, such as "2-methylprop-1-ene", was rejected.

Key definitions

• Nucleophile: an electron pair donor (a species that uses a lone pair to form a new covalent bond).
• Nucleophilic substitution: a reaction in which a nucleophile replaces the halogen atom; the C-X bond breaks and the halogen leaves as a halide ion.
• Hydrolysis: the splitting of a molecule (breaking a bond) using water.
• Elimination: a reaction in which a small molecule (here H and Br, as HBr) is removed from an organic compound, forming a C=C double bond.
• Base (in elimination): a proton acceptor.
• Polar bond: a covalent bond with an unequal share of the electron pair, because one atom (the halogen) is more electronegative; the carbon is δ+.
• Bond enthalpy: the enthalpy needed to break a covalent bond (here the carbon-halogen bond).
• Free radical: a species with an unpaired electron, shown by a dot, for example Cl•.
• Catalyst (Cl• in ozone depletion): speeds up the reaction by providing an alternative route with a lower activation energy, and is regenerated (not used up).

Specification

• I can explain why halogenoalkanes contain a polar C-X bond and why the carbon is attacked by nucleophiles.
• I can define a nucleophile and identify OH^-, CN^- and NH₃ as nucleophiles.
• I can give the reagent, conditions, product and balanced equation for the substitution of a halogenoalkane by OH^-, CN^- and NH₃.
• I can outline the nucleophilic substitution mechanisms with OH^-, CN^- and NH₃, with correct curly arrows, lone pairs, partial charges and the alkylammonium ion.
• I can explain why the carbon-halogen bond enthalpy influences the rate of reaction, comparing named C-X bonds.
• I can describe how to compare rates of hydrolysis of halogenoalkanes using aqueous silver nitrate, including observations and fair-test variables.
• I can describe how a liquid chloroalkane is purified using a separating funnel and distillation.
• I can describe the concurrent substitution and elimination reactions of a halogenoalkane such as 2-bromopropane with potassium hydroxide, and the conditions that favour each.
• I can explain the role of the hydroxide ion as both a nucleophile and a base.
• I can outline the elimination mechanism and deduce all the alkenes formed from an unsymmetrical halogenoalkane.
• I can state that ozone in the upper atmosphere is beneficial because it absorbs ultraviolet radiation.
• I can explain how UV radiation breaks C-Cl bonds in CFCs to form chlorine atoms.
• I can use the equations Cl• + O₃ → ClO• + O₂ and ClO• + O₃ → 2O₂ + Cl• to explain how chlorine atoms catalyse the decomposition of ozone.
• I can explain how research by different scientific groups led to legislation banning CFCs, and why chlorine-free alternatives do not deplete ozone.

Related notes

• Introduction to organic chemistry
• Alkanes
• Alkenes
• Alcohols
• Energetics
• Kinetics
• Group 7(17), the halogens
• Amount of substance