3.3.3 Halogenoalkanes

The halogen is more electronegative than carbon, so a halogenoalkane has a polar Cδ⁺–Xδ⁻ bond. The δ⁺ carbon is therefore susceptible to attack by an electron-pair donor.

Polarity explains where a nucleophile attacks. It does not by itself explain the reactivity order down Cl, Br and I; for that comparison, C–X bond enthalpy is the decisive course argument.

For the three nucleophilic substitutions here, tie each nucleophile to its product: OH⁻ gives an alcohol, CN⁻ gives a nitrile and NH₃ gives an amine. In each case the nucleophile attacks the carbon bonded to the halogen and X⁻ leaves.

CN⁻ adds one carbon to the skeleton because the carbon of –C≡N becomes part of the product. With NH₃, excess ammonia helps favour the primary amine over further substitution.

With OH⁻ or CN⁻ substitution, start the curly arrow at the nucleophile lone pair and point it to the δ⁺ carbon. The second arrow goes from the C–X bond to X so the leaving group departs as X⁻.

With NH₃, the first substitution product is protonated, so include the subsequent proton-transfer step needed to give the neutral primary amine.

For comparable primary halogenoalkanes, hydrolysis becomes faster from chloroalkane to bromoalkane to iodoalkane because C–X bond enthalpy decreases down the group. The weaker C–I bond is easier to break.

Do not use bond polarity to predict this order: C–Cl is the most polar of the three but chloroalkanes are not the fastest to substitute.

With 2-bromopropane and OH⁻, I’d deliberately ask what role OH⁻ is playing. If it attacks carbon it acts as a nucleophile and gives propan-2-ol; if it removes H⁺ from an adjacent carbon it acts as a base and gives propene.

Aqueous hydroxide favours substitution; hot ethanolic hydroxide favours elimination. The same reagent can therefore play two different roles depending on what its electron pair attacks.

The same OH⁻ ion can do two different jobs, and I’d keep those electron-pair moves separate. Donate the lone pair to carbon and it is acting as a nucleophile; use it to remove H⁺ from a neighbouring carbon and it is acting as a base.

With OH⁻, make the competing bond changes visibly different. In substitution, OH⁻ attacks carbon as C–X breaks; in elimination, the base removes a β-hydrogen while C=C forms and X⁻ leaves.

These mechanisms are not interchangeable diagrams. The substitution pathway forms C–O; the elimination pathway forms C=C, so the curly arrows must account for different bond changes.

Start with the radiation when you explain why stratospheric ozone matters: ozone absorbs higher-energy UV before it reaches the lower atmosphere. Less ozone therefore means more biologically damaging UV reaches the surface.

UV light can generate chlorine radicals from CFCs in the stratosphere. The chlorine radical is regenerated in a catalytic cycle that converts ozone to oxygen, so one radical can destroy many ozone molecules.

CFCs can survive long enough to reach the stratosphere because they are relatively unreactive lower in the atmosphere. There, higher-energy UV can split a C–Cl bond homolytically and generate a chlorine radical, Cl•.

That first Cl• is important because it enters a catalytic radical cycle: the radical is regenerated, so one Cl• can participate in the destruction of many ozone molecules.

When you show chlorine-catalysed ozone destruction, make sure Cl• comes back. It reacts with O₃ to form ClO•, then a later step regenerates Cl•, so the chlorine species cancel from the overall cycle.

A regenerated radical can repeat the cycle many times, which is why a relatively small amount of chlorine-containing material can have a much larger effect on ozone.

The CFC–ozone example shows how a chemical explanation becomes persuasive when several kinds of evidence agree: laboratory photochemistry, atmospheric measurements and observations of ozone change can all test parts of the same mechanism.

The course point is how evidence supports a causal model strongly enough to guide decisions. You do not need to pretend one single experiment established the entire atmospheric mechanism.

To judge a CFC replacement, removing chlorine solves the specific ozone-radical problem, but I would not stop the evaluation there. You still need to consider things such as greenhouse impact, toxicity, flammability and persistence.

“Does not contain chlorine” answers the ozone mechanism, but it is not a complete environmental assessment of a replacement refrigerant or propellant.

Add the two radical steps and cancel species that appear on both sides. For Cl• + O₃ → ClO• + O₂ and ClO• + O₃ → 2O₂ + Cl•, both chlorine radicals cancel.

The net equation is therefore 2O₃ → 3O₂. Cancellation also shows why Cl• acts as a catalyst: it participates in the steps but is regenerated overall.