• Halogenoalkanes are named by adding fluoro-/chloro-/bromo-/iodo- prefixes to the parent alkane name, with position numbers and alphabetical ordering of substituents.
• They're classified as primary, secondary, or tertiary based on how many carbon (alkyl) groups are attached to the carbon bearing the halogen.
• The C–X bond is polar (Cδ+ — Xδ−) because halogens are more electronegative than carbon — this is *why* nucleophiles attack.
• Nucleophilic substitution reactions swap the halogen (X) for: OH⁻ (→ alcohol), CN⁻ (→ nitrile, extends chain by 1 carbon), or NH₃ (→ primary amine).
• Elimination reactions (with ethanolic NaOH + heat) remove HX to form an alkene instead of substituting.
• The mechanism for substitution always follows the same logic: nucleophile's lone pair attacks the δ+ carbon, the C–X bond breaks heterolytically, X⁻ leaves.
• Reactivity trend: C–I is the weakest bond (breaks fastest) → C–F is the strongest bond (barely reacts at all). This is tested experimentally using silver nitrate precipitation.

Think of a halogenoalkane's name as built in three layers, applied in this order:

1. Start from the parent alkane — count the longest carbon chain and name it as you would an alkane (methane, ethane, propane...).
2. Add the halogen prefix — fluoro-, chloro-, bromo-, or iodo- — with a number showing which carbon it's attached to. Number the chain so the substituent gets the lowest possible number.
3. If there's more than one substituent, list them alphabetically (bromo before chloro, chloro before fluoro, etc.) — not in the order they appear on the chain.

Notice in the last example: bromo is listed first because "b" comes before "c" alphabetically — even though the bromine is on carbon 1 and the chlorine is on carbon 2. Alphabetical order for listing, chain numbering for position — don't mix these two jobs up.

This classification is all about the carbon directly attached to the halogen (call it the "halogen-bearing carbon"). Ask: how many other carbon groups (alkyl groups) is that carbon also attached to?

It's exactly the same logic as classifying alcohols or amines later in the course — so nailing this now pays off repeatedly. The trick students get wrong: they count total carbons in the molecule instead of counting only the groups attached to that one specific carbon.

Before diving into each reaction, here's the one concept that unlocks the whole topic:

The four "species" that can act as the nucleophile in this chapter, and what each one turns the halogenoalkane into:

Conditions: warm aqueous NaOH or KOH, with a little ethanol added purely as a co-solvent to help the halogenoalkane (which doesn't mix well with water) dissolve.

Why must water be the main solvent? If you used ethanol as the main solvent instead of water (i.e., "ethanolic" conditions), you'd get elimination instead — forming an alkene, not an alcohol. Same reagent (NaOH), completely different product, just because of the solvent. This is a favourite exam trap.

Why OH⁻ beats H₂O as a nucleophile: OH⁻ carries a full formal negative charge, while the oxygen in H₂O only carries a partial negative charge (δ−). A full negative charge is a much stronger electron-pair donor, so the reaction with OH⁻ is dramatically faster than with plain water.

If water alone reacts (very slowly), the mechanism still produces an alcohol, but also releases H⁺ and X⁻:

You can then detect the X⁻ ions formed by adding silver nitrate (in ethanol) — the Ag⁺ reacts instantly with any halide ion present to give a precipitate: Ag⁺(aq) + X⁻(aq) → AgX(s). This idea becomes hugely important later in Section 4 (measuring rates of hydrolysis).

Conditions: ethanolic potassium cyanide (KCN dissolved in ethanol), heated under reflux. Ethanol is needed here as the solvent because both the halogenoalkane and KCN dissolve well in it — unlike the OH⁻ reaction, there's no elimination side-reaction to worry about because CN⁻ is a much better nucleophile than base in this context.

Conditions: excess ethanolic ammonia, heated under pressure (a sealed tube, since ammonia is a gas at room temperature and pressure and would otherwise just escape).

Why excess ammonia? Here's the subtlety that trips people up: the product (the primary amine) is actually a better nucleophile than ammonia itself (the nitrogen's lone pair is more available once alkyl groups are attached, due to their electron-donating effect). So if there's still unreacted halogenoalkane floating around, the primary amine you just made will react with it again, giving a secondary amine — and this can even cascade further to tertiary amines and quaternary ammonium salts. Using a large excess of NH₃ makes it statistically far more likely that any given halogenoalkane molecule collides with an NH₃ molecule (which is in huge excess) rather than with another amine molecule, keeping the primary amine as the major product.

Conditions: ethanolic NaOH (the opposite solvent choice to the alcohol-forming reaction!), heated. Here, OH⁻ isn't acting as a nucleophile at all — it acts as a base, plucking off a hydrogen atom from the carbon adjacent to the C–X carbon. This causes the C–X bond to break heterolytically, and a C=C double bond forms between the two carbons.

Net result: HBr is effectively "eliminated" from the molecule (hence the name "elimination reaction") — one H from the neighbouring carbon and the Br together.

A nucleophilic substitution reaction is one where a nucleophile attacks a carbon atom carrying a partial positive charge, and an atom/group with a partial negative charge (here, the halogen) is kicked out and replaced.

Because carbon is less electronegative than any halogen, the shared electrons in the C–X bond sit closer to X. This gives carbon a δ+ charge — making it an easy target for anything carrying a lone pair of electrons (a nucleophile).

Here's how to draw it step by step, using bromoethane as the example:

Reading the curly arrows:

1. A curly arrow starts at the lone pair on OH⁻ and points toward the δ+ carbon — this shows the nucleophile donating its electron pair to form a new C–O bond.
2. A second curly arrow starts on the C–Br bond and points toward the Br atom — this shows the bond breaking heterolytically (both electrons go to Br, since Br is more electronegative).
3. The result: a new C–OH bond has formed, and Br has left as a free Br⁻ ion (with a full octet, hence the three lone pairs shown as :Br⁻).

This one is slightly more involved because it happens in two stages. Using chloromethane as the example:

What's actually happening:

1. The lone pair on NH₃'s nitrogen attacks the δ+ carbon, and the C–Cl bond breaks heterolytically (Cl leaves with both electrons as Cl⁻). This forms an intermediate ammonium ion [CH₃NH₃]⁺ — notice nitrogen now has 4 bonds and a positive charge, because it "spent" its lone pair forming the new C–N bond.
2. A second molecule of NH₃ then acts as a base — its lone pair grabs one of the H atoms off the ammonium intermediate's N–H bond, restoring a neutral amine (CH₃NH₂) and creating NH₄⁺ as a byproduct.

This is why excess ammonia matters twice over: once to favour primary amine formation over further substitution (see Section 2), and once because a second NH₃ molecule is literally required in Stage 2 to complete the mechanism.

This section is about a specific practical technique: using acidified aqueous silver nitrate to measure how fast different halogenoalkanes hydrolyse (react with water). It's a classic exam-practical question.

The Method

1. Set up three test tubes containing a mixture of ethanol (to help everything dissolve) and acidified silver nitrate solution, all sitting in a 50°C water bath (gentle heat speeds up the otherwise-slow reaction with water).
2. Add a few drops of a chloroalkane, a bromoalkane, and an iodoalkane (same alkyl chain, only the halogen differs) into three separate tubes and start a stopwatch simultaneously.
3. Time how long each one takes to form a visible precipitate.

Why does a precipitate form at all? As each halogenoalkane slowly reacts with water (hydrolysis), it releases halide ions (Cl⁻, Br⁻, or I⁻) into solution. These immediately react with the Ag⁺ ions already present to form an insoluble silver halide precipitate. The faster the precipitate appears, the faster the C–X bond is being broken.

Result: the iodoalkane forms its precipitate fastest, and the chloroalkane forms its precipitate slowest. This directly reflects the strength of the C–X bond — see Section 5 for why.

The same silver nitrate method can be adapted to compare rates of hydrolysis for primary, secondary, and tertiary halogenoalkanes — but this time you must keep the halogen and the overall molecular formula the same across all three, changing only where the halogen sits on the chain (its "position" determines primary/secondary/tertiary).

Result:

The full explanation for why (it comes down to whether the mechanism is S_N1 or S_N2) isn't required at AS level — you just need to know and state the trend. It becomes examinable at full A-level.

This is where the whole chapter ties together with one key number: bond enthalpy (the energy needed to break a bond). Since every substitution reaction requires breaking the C–X bond, the strength of that bond directly controls how reactive the halogenoalkane is.

Notice the trend goes the opposite way to what you might guess from electronegativity alone. Electronegativity actually decreases down Group 7 (F is the most electronegative, I the least), so you might expect the C–F bond to be the most polar and therefore fastest to react. But it's the exact opposite in practice — and here's the key exam point:

• I can name a halogenoalkane correctly, including alphabetical ordering of multiple substituents.
• I can classify a halogenoalkane as primary, secondary, or tertiary by examining the halogen-bearing carbon.
• I understand why the C–X bond is polar and can explain it using electronegativity.
• I know the reagents and conditions for converting a halogenoalkane into an alcohol (aqueous NaOH/KOH, warm).
• I can explain why OH⁻ is a better nucleophile than H₂O.
• I know the reagents/conditions for making a nitrile (ethanolic KCN, reflux) and why it extends the carbon chain.
• I know the reagents/conditions for making a primary amine (excess ethanolic NH₃, heat + pressure) and why excess is needed.
• I can distinguish aqueous NaOH (substitution → alcohol) from ethanolic NaOH (elimination → alkene).
• I can draw the full curly-arrow mechanism for substitution with OH⁻, including correct arrow origins.
• I can describe the two-step mechanism for the reaction with ammonia, including the ammonium intermediate.
• I can describe the silver nitrate practical method for measuring rate of hydrolysis, including precipitate colours.
• I know why AgF doesn't form a visible precipitate (solubility, not lack of reaction).
• I can state and explain the reactivity trend (I > Br > Cl > F) using bond enthalpy data.
• I understand that bond enthalpy — not polarity — determines this reactivity trend.
• I know the rate trend for primary vs secondary vs tertiary halogenoalkanes (3° fastest, 1° slowest).