Module 4: Core Organic Chemistry · Year 12
4.2.2 Haloalkanes
Use the polar C–X bond to explain nucleophilic substitution and hydrolysis rates, then connect radical chemistry to ozone depletion.
What you need to know
Open a line for a quick recap. If it feels obvious, move straight to the linked practice.
4.2.2 a(i) Describe hydrolysis of haloalkanes by aqueous alkali. Quick revision
If you warm a haloalkane with aqueous NaOH or KOH, the halogen is replaced by –OH and you form an alcohol. The hydroxide ion is the reacting nucleophile.
At equation level, keep charge balanced as well as atoms: R–X + OH⁻ → R–OH + X⁻. This is hydrolysis because the C–X bond is broken in a reaction involving aqueous conditions.
4.2.2 a(ii) Describe how AgNO3 and ethanol are used to compare rates of hydrolysis for different carbon-halogen bonds. Quick revision
To compare haloalkane hydrolysis rates experimentally, use ethanol so the organic haloalkane mixes with the aqueous reagents, then add silver nitrate and keep the conditions the same between samples. As hydrolysis releases halide ions, a silver halide precipitate forms.
Time how long it takes for cloudiness or precipitate to appear. A shorter time means faster hydrolysis. You are comparing the C–X bonds, so temperature, concentrations and amounts need to be controlled.
4.2.2 b Define a nucleophile as an electron-pair donor and apply the idea in mechanisms. Quick revision
A nucleophile is an electron-pair donor. When you use that idea in a mechanism, look for the species with a lone pair or negative charge that can form a new covalent bond to an electron-deficient centre.
For OH⁻ attacking a haloalkane, start the curly arrow from a lone pair or the negative charge on oxygen and finish it at the carbon attached to the halogen. That carbon is δ+ because the C–X bond is polar.
4.2.2 c Draw the nucleophilic substitution mechanism for hydrolysis of primary haloalkanes with aqueous alkali. Quick revision
When you draw the OCR Year 12 hydrolysis mechanism, show the C–X bond dipole with Cδ+ and Xδ−. OH⁻ donates an electron pair to the δ+ carbon, so draw the attacking curly arrow from oxygen to that carbon.
At the same time, the C–X bonding pair moves to X, giving X⁻. Check the final carbon valency: the new C–O bond forms as the C–X bond breaks.
4.2.2 d Explain the hydrolysis-rate order for primary fluoro-, chloro-, bromo- and iodoalkanes by comparing the relevant carbon–halogen bond enthalpies. Quick revision
For comparable primary haloalkanes, hydrolysis gets faster as the C–X bond gets easier to break. C–F has the highest bond enthalpy, while C–I is much weaker, so iodoalkanes hydrolyse fastest and fluoroalkanes slowest.
If you are explaining the trend, make the causal link explicit: lower C–X bond enthalpy means less energy is needed to break that bond during substitution.
- C–F strongest → slowest
- C–Cl
- C–Br
- C–I weakest → fastest
4.2.2 e(i) Write equations showing production of chlorine radicals from CFCs by UV radiation. Quick revision
High-energy UV radiation can break a C–Cl bond in a CFC homolytically, producing chlorine radicals. The important idea is that the bonding pair splits equally, so one electron stays with each fragment.
You do not need to turn this into an ionic mechanism. Once Cl• has formed in the stratosphere, it can enter a radical chain that destroys ozone.
4.2.2 e(ii) Explain and write equations for radical-catalysed ozone breakdown by chlorine and other radicals. Quick revision
The key feature is that the radical is regenerated, so one radical can destroy many ozone molecules. For chlorine, Cl• reacts with O₃ to make ClO•; the ClO• then reacts and regenerates Cl•.
When you add the propagation steps, cancel the radical intermediates to see the overall ozone-destruction reaction. That regeneration is why we describe the radical as acting catalytically.