Haloalkanes — OCR A-Level Chemistry

Haloalkanes contain a polar carbon–halogen bond, so the carbon is electron-poor and open to attack by a nucleophile. In a substitution reaction the nucleophile replaces the halogen, which leaves as a halide ion. With aqueous hydroxide ions and heat, the nucleophile is OH⁻ and the product is an alcohol, for example CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻. With ammonia in excess, the nucleophile is NH₃ and an amine forms. With cyanide ions, a nitrile forms and the carbon chain lengthens by one carbon. The reaction is nucleophilic substitution, and the rate depends on the strength of the carbon–halogen bond, so iodoalkanes react fastest and fluoroalkanes slowest.

Hydrolysis of a haloalkane is nucleophilic substitution: the halogen is replaced by a nucleophile and the C–X bond breaks. The nucleophile depends on the reagent. With aqueous sodium or potassium hydroxide, warm, the nucleophile is the hydroxide ion, OH⁻, and the product is an alcohol plus the halide ion. With water alone, the nucleophile is a water molecule, H₂O, which attacks the carbon and later loses a proton, so water is the nucleophile and not a source of OH⁻. The rate depends on the C–X bond strength, which falls down Group 7 as bonds get weaker, so iodoalkanes hydrolyse fastest and fluoroalkanes slowest. Primary haloalkanes react mainly by SN2, tertiary mainly by SN1.

A haloalkane is hydrolysed by aqueous alkali in a nucleophilic substitution. Warm the haloalkane with aqueous sodium hydroxide or potassium hydroxide; the hydroxide ion, OH⁻, is the nucleophile and attacks the electron-poor carbon that carries the halogen. The carbon–halogen bond breaks heterolytically and the halide ion leaves. With a primary haloalkane the main organic product is a primary alcohol, for example bromoethane with aqueous NaOH gives ethanol and NaBr. The reaction is slow at room temperature, so heating under reflux is used. Aqueous conditions matter: if the alkali is dissolved in ethanol the dominant pathway becomes elimination, giving an alkene. Recognise that the halogen is replaced, not removed to leave a bare carbon, and that the hydroxide supplies the new O–H group of the alcohol.

This test hydrolyses a haloalkane with water, using ethanol as a solvent to mix the organic halide with the aqueous silver nitrate, and silver nitrate to detect the halide ion released. Water supplies the nucleophile and the carbon–halogen bond breaks, releasing a halide ion. The Ag⁺ ions then precipitate that halide as a silver halide, whose appearance and colour identify which halogen was present. The time taken for the precipitate to appear measures the rate of hydrolysis. A faster precipitate means a weaker, more easily broken carbon–halogen bond. Rates increase down the halogen group: iodoalkanes react fastest, then bromoalkanes, then chloroalkanes, because the C–I bond is weakest and most polarisable. Compare haloalkanes fairly by keeping concentration, temperature and volume the same.

A nucleophile is a species that donates a lone pair of electrons to an electron-deficient centre, forming a new covalent bond. In haloalkane chemistry the carbon bonded to the halogen carries a partial positive charge (δ⁺) because the halogen is more electronegative, so nucleophiles such as OH⁻, CN⁻, NH₃ and H₂O attack that carbon. The term is used to explain and predict reaction mechanisms: the nucleophile supplies the electron pair, and the curly arrow starts at its lone pair and points to the δ⁺ carbon. Recognise that a nucleophile must have a lone pair available for donation, so not every negative ion is automatically a good nucleophile, and that the definition concerns electron-pair donation rather than simply being negatively charged.

Hydrolysis of a primary haloalkane with aqueous alkali, for example aqueous sodium hydroxide, proceeds by nucleophilic substitution. The hydroxide ion acts as the nucleophile: its lone pair attacks the δ⁺ carbon bonded to the halogen, and a curly arrow runs from the lone pair to that carbon. A second curly arrow shows the C–X bond breaking, with both electrons moving onto the halogen to form a halide ion. The product is an alcohol, and the halide leaves as X⁻. This is a substitution because OH⁻ replaces the halogen on the carbon skeleton. Heating under reflux with aqueous alkali drives the reaction. Recognise that the mechanism shows one step in which bond formation and bond breaking occur together, and that the halogen departs with the bonding electron pair.

Primary haloalkanes hydrolyse when warm aqueous alkali, typically aqueous sodium hydroxide, replaces the halogen with an –OH group. The rate follows the strength of the carbon–halogen bond being broken: the weaker that bond, the faster the hydrolysis. Bond enthalpies fall down the group from C–F (strongest, roughly 467 kJ mol⁻¹) through C–Cl and C–Br to C–I (weakest, roughly 238 kJ mol⁻¹). A stronger bond needs more energy to break, so fluoroalkanes react slowest and iodoalkanes fastest. The trend is therefore C–F < C–Cl < C–Br < C–I in rate. A good answer links the measured rate to the specific bond enthalpy of the C–X bond broken, not to the C–H bonds or to halogen size alone.

Organohalogen compounds are useful as solvents, refrigerants, pesticides and in polymers, but their persistence and mobility create environmental problems. Chlorofluorocarbons (CFCs) are unreactive near the ground, so they drift into the stratosphere, where ultraviolet radiation breaks the C–Cl bond to form chlorine radicals. A chlorine radical catalyses breakdown of ozone, and one radical can destroy many ozone molecules before it is removed, thinning the ozone layer that screens ultraviolet radiation. Organohalogens also accumulate in fatty tissue and can persist in soil and water. International agreements such as the Montreal Protocol phased out CFCs, replaced by HCFCs and then HFCs, though HFCs are potent greenhouse gases. Answers should name a specific compound class, the mechanism of harm and a management response.

In the upper atmosphere, ultraviolet radiation supplies enough energy to break the carbon–chlorine bond in a chlorofluorocarbon such as CF₂Cl₂. Homolytic fission produces two radicals, for example CF₂Cl• and Cl•. The chlorine radical attacks ozone: Cl• + O₃ → ClO• + O₂. The chlorine monoxide radical then reacts with an oxygen atom: ClO• + O → Cl• + O₂. Because Cl• is regenerated in the second step, one chlorine radical destroys many ozone molecules in a catalytic cycle, so the ozone layer that absorbs harmful UV is depleted. A correct answer identifies UV as the radical source, writes radical equations with single-headed curly arrows or dot notation, and explains catalysis by showing the radical is reformed.

Halogen radicals form when a halogen-containing molecule absorbs enough energy to break a carbon–halogen bond. In the upper atmosphere, ultraviolet radiation provides this energy, and the bond breaks by homolytic fission so that each atom takes one electron from the shared pair. For a CFC such as CF₂Cl₂, the equation is CF₂Cl₂ → CF₂Cl• + Cl•. The key features are the energy source (UV), the type of fission (homolytic) and the products (two radicals, each shown with a dot). A correct answer recognises that radicals are species with an unpaired electron and that the C–Cl bond is the one broken in CFCs.

Chlorine radicals, Cl•, formed when UV light breaks C–Cl bonds in chloroalkanes such as CFCs, catalyse ozone breakdown in the stratosphere. In the first step a chlorine radical reacts with ozone to form chlorine monoxide and oxygen: Cl• + O₃ → ClO• + O₂. In the second step the chlorine monoxide radical reacts with an oxygen atom to regenerate the chlorine radical: ClO• + O → Cl• + O₂. Adding these two steps gives the overall equation O₃ + O → 2O₂, and the chlorine radical is regenerated, so it acts as a catalyst. Nitrogen oxide radicals, •NO, work in a similar catalytic cycle, for example •NO + O₃ → •NO₂ + O₂ followed by •NO₂ + O → •NO + O₂.