2021 H2 Chem P1 Q18 — Halogenoalkane Solution

The answer is B (2 and 3).

Breaking the C-X bond limits the rate in nucleophilic substitution: in \(\text{S}_\text{N}1\) it is the slow step, and in \(\text{S}_\text{N}2\) it breaks at the same time as the C-Nu bond forms, in a single step. The C-Br bond is weaker than the C-Cl bond (Data Booklet: C-Br 280, C-Cl 340 kJ mol\(^{-1}\)), so 2-bromopropane reacts faster. The statements must explain why the C-Br bond is weaker.

Statement 2 explains it. Br has one more full quantum shell of inner electrons than Cl, so the shared pair of electrons in the C-Br bond is more shielded from the halogen nucleus. It is attracted less strongly, so the bond is weaker.
Statement 3 explains it. The Br atom is larger than the Cl atom, so the shared pair in the C-Br bond is further from the halogen nucleus. The bond is longer and weaker.
Statement 1 does not explain it. It is true that bromine has a lower ionisation energy than chlorine, but no electron is removed from a halogen atom in this reaction; the halogen leaves as a halide ion, \(\text{X}^-\).
Statement 4 does not explain it. It is true that Br has more protons, but a larger nuclear charge on its own would attract the bonding pair more strongly and make the bond stronger, not weaker. The extra protons are outweighed by the extra shielding and distance in statements 2 and 3.

• A is wrong: statement 1 is not relevant, as the reaction does not involve ionising a halogen atom.
• C is wrong: statement 4 would predict a stronger C-Br bond, so it does not explain the faster reaction.
• D is wrong: neither statement 1 nor statement 4 explains why the C-Br bond is weaker.