Class 12 Haloalkanes and Haloarenes — Substitution Reactions Made Clear

The haloalkanes chapter is where organic chemistry stops being a list of names and starts being a set of rules you can actually apply. Almost every question in this chapter reduces to one skill: given a halide and a reagent, name the product. Once you can do that reliably, the conversions, the reasoning questions and the mechanism questions all become easier. This guide gives you every substitution product with a balanced equation, explains why haloarenes behave so differently, and works through calculations you may be asked to attach to these reactions.

The general reaction

A halogen atom bonded to carbon is a leaving group. The carbon carrying it is slightly positive, so an electron-rich species — a nucleophile — attacks that carbon and pushes the halide out.

What each symbol means:

• R — an alkyl group (or, in haloarenes, an aryl ring).
• X — F, Cl, Br or I. It leaves carrying the bonding pair, so it becomes X⁻.
• Nu⁻ — the nucleophile: OH⁻, CN⁻, NH₃, R'O⁻, NO₂⁻ and so on. It must have a lone pair or a negative charge.

Two facts control everything in this chapter. First, the C–X bond enthalpy falls in the order C–F > C–Cl > C–Br > C–I, so the reactivity of the halides towards substitution rises in the order R–F < R–Cl < R–Br < R–I. Second, the carbon must be sp³ and free to be attacked. In haloarenes it is sp², and that single change shuts most of this chemistry down.

Every product you are expected to know

Learn this as a table, not as scattered equations. The reagent decides the product, and in three cases the counter-ion decides it too.

Balanced examples, atom by atom:

Check the last one: the left side has 4 C, 10 H, 2 Br and 2 Na; the right side has butane (4 C, 10 H) plus 2 NaBr (2 Na, 2 Br). It balances. Get into the habit of counting like this — examiners deduct for unbalanced equations even when the organic product is right.

Ambident nucleophiles — the KCN/AgCN trap

An ambident nucleophile has two different atoms that can attack. The cyanide ion can attack through carbon or through nitrogen.

• KCN is largely ionic. The free CN⁻ attacks through carbon, because a C–C bond is stronger than a C–N bond. Product: a nitrile.
• AgCN is largely covalent. The carbon is tied up with silver, so only the nitrogen lone pair is available. Product: an isocyanide.

The nitrite ion behaves the same way: KNO₂ gives the alkyl nitrite (attack through O), AgNO₂ gives the nitroalkane (attack through N). Remember the pattern as "silver gives the nitrogen product" for both pairs.

Which mechanism, S_N1 or S_N2?

For Class 12 you mainly need to predict which pathway a substrate prefers, and what that means for the stereochemistry of the product.

The two orders are exact opposites, and that is not a coincidence: bulky groups block a backside attack but stabilise a carbocation. Allylic and benzylic halides are the exception that proves the rule — they are unusually reactive by S_N1 because their carbocations are resonance-stabilised. For the full mechanism with energy profiles, see our separate article on S_N1 vs S_N2.

Why haloarenes refuse to react

Chlorobenzene and chloroethane look similar on paper. In the laboratory they are worlds apart: chlorobenzene ignores aqueous KOH completely. Four reasons, all worth writing in an answer:

1. Resonance. A lone pair on the halogen delocalises into the ring, giving the C–X bond partial double-bond character. A double bond is much harder to break.
2. Hybridisation. The ring carbon is sp² (33% s character) instead of sp³ (25%). The sp² orbital is smaller and holds the halogen closer and more tightly.
3. Unstable intermediate. An S_N1 route would need a phenyl cation, which cannot be stabilised by the ring's π system. It simply does not form under ordinary conditions.
4. Repulsion. The π electron cloud of the ring repels the approaching electron-rich nucleophile.

Force the reaction and it does happen. In the Dow process, chlorobenzene is heated with aqueous NaOH at about 623 K and 300 atm:

In practice excess NaOH neutralises the HCl formed. The reaction becomes far easier when strongly electron-withdrawing groups sit ortho or para to the halogen: they pull electron density away and stabilise the intermediate. So 4-nitrochlorobenzene needs milder conditions than chlorobenzene, and 2,4-dinitrochlorobenzene reacts with warm aqueous NaOH. A nitro group at the meta position gives no such help — the negative charge in the intermediate never reaches that carbon.

One more contrast students mix up: towards electrophilic substitution, halogens are deactivating but ortho/para directing. The −I effect slows the whole ring down, while the +R effect still puts extra electron density specifically at the ortho and para positions. Both statements are true at the same time.

Worked example 1 — Finkelstein reaction and percentage yield

Worked example 2 — one halide, four reagents

Worked example 3 — percentage of halogen by mass

Common mistakes that cost marks

Why this chapter matters

These reactions are not a dead end — they are the standard gateway into the next three chapters. Almost every "conversion" question in the board paper passes through a halide.

Chapter weightings and question patterns change from year to year, so plan your revision against the syllabus document your board has published for your own session rather than against last year's paper.