Alkyl halides (haloalkanes)
An alkyl halide is an alkane with a halogen replacing a hydrogen (R–X, where X = Cl, Br, I). The carbon–halogen bond is polar — the halogen pulls electrons, leaving the carbon slightly positive (δ+) — and that electron-poor carbon is the reactive site, open to attack by nucleophiles.
Their two reactions: substitution and elimination
Nucleophilic substitution: a nucleophile (an electron-pair donor) attacks the δ+ carbon and displaces the halogen. With hydroxide you get an alcohol; with ammonia an amine; with cyanide a nitrile (extending the carbon chain). This is a central route for making other functional groups.
Elimination: with a hot alcoholic solution of a strong base, the halide loses HX to form a double bond — an alkene. Substitution and elimination compete, and conditions (reagent, temperature) decide which wins: aqueous base favours substitution, hot alcoholic base favours elimination.
Same reagent, different conditions
A haloalkane with hydroxide: in warm AQUEOUS solution it substitutes to give an alcohol; in HOT ALCOHOLIC solution it eliminates to give an alkene. The conditions, not just the reagent, choose the product.
Amines
Amines are organic derivatives of ammonia, with one or more of ammonia's hydrogens replaced by carbon chains: primary (one, R–NH₂), secondary (two, R₂NH) or tertiary (three, R₃N).
Like ammonia, the nitrogen carries a lone pair, so amines are bases — they accept a proton (H⁺) to form a positively charged ion, and react with acids to give salts. That lone pair also makes them nucleophiles, which is how the ammonia-plus-haloalkane substitution above builds them.
A worked example
Bromoethane is treated with (a) warm aqueous sodium hydroxide, then (b) hot ethanolic sodium hydroxide. What forms each time?
- (a) Warm aqueous NaOH → nucleophilic substitution. OH⁻ replaces Br.Aqueous conditions favour substitution.
- Product: ethanol (CH₃CH₂OH) + NaBr.Hydroxide substitutes onto the carbon, giving an alcohol.
- (b) Hot ethanolic NaOH → elimination. HBr is removed.Hot alcoholic conditions favour elimination.
- Product: ethene (CH₂=CH₂) + NaBr + water.Loss of HX forms a carbon–carbon double bond.
(a) ethanol (substitution); (b) ethene (elimination) — the conditions decide which reaction dominates.
Nucleophilic substitution in detail
The δ+ carbon of a haloalkane is attacked by whatever nucleophile is supplied, and the halogen leaves — so one reagent opens the door to many products. Hydroxide gives an alcohol; cyanide gives a nitrile, adding a carbon and lengthening the chain; ammonia gives an amine.
Reactivity depends on the carbon–halogen bond strength, not its polarity: the weaker C–I bond breaks most easily, so iodoalkanes react fastest and chloroalkanes slowest (C–I < C–Br < C–Cl in reactivity order, iodo fastest).
Amines: basicity and preparation
Amines are organic relatives of ammonia, and like ammonia the lone pair on nitrogen makes them both bases (accepting H⁺ to form a salt) and nucleophiles. They are classified as primary, secondary or tertiary by how many carbon groups sit on the nitrogen.
They are made by nucleophilic substitution of a haloalkane with excess ammonia, or by reducing a nitrile. Reacting an amine with an acid gives an ammonium-type salt — the basis of many drug formulations, which are made as soluble salts.
Lone pair does double duty
Nitrogen's lone pair is why an amine is both a base (it grabs a proton) and a nucleophile (it attacks δ+ carbon). Spot the lone pair and you can predict the chemistry.
Two mechanisms: SN1 and SN2
Nucleophilic substitution can happen by two different routes. In SN2 the nucleophile attacks the δ+ carbon as the halogen leaves, in a single concerted step — so its rate depends on both the haloalkane and the nucleophile. This is favoured by primary haloalkanes, where the small carbon is easy to approach.
In SN1 the halogen leaves first, forming a carbocation, which the nucleophile then attacks in a second step — so the rate depends only on the haloalkane. This is favoured by tertiary haloalkanes, because the tertiary carbocation is the most stable. Secondary haloalkanes sit in between and can go by either route.
Which route?
Primary haloalkane → SN2 (one step, backside attack). Tertiary → SN1 (via a stable carbocation). The clue is carbocation stability: only tertiary carbocations are stable enough to form readily.