What happens when a halogenoalkane reacts with aqueous hydroxide?

What happens when a halogenoalkane reacts with aqueous hydroxide?

• A. R-X + OH− → R−H + X−; hydrogen is formed.
• B. R-X + OH− → R−OH + X−; alcohol is formed.
• C. R-X + OH− → R−O− + HX.
• D. R-X + OH− → R−X + OH−.

Answer: (B)

When a halogenoalkane meets aqueous hydroxide, a nucleophilic substitution occurs. The hydroxide ion attacks the carbon bonded to the halogen, the halogen leaves as a halide ion, and the oxygen from the hydroxide ends up attached to the carbon. After a quick proton transfer from water, this gives the neutral alcohol. So the overall result is an alcohol plus a halide ion in solution. The other ideas don’t fit because hydrogen would have to replace the halogen, which isn’t the role of hydroxide here; the reaction instead replaces X with OH. Forming an alkoxide and HX would require a different proton transfer pattern not supported by this base in water, and saying there’s no reaction ignores the common SN2 substitution that occurs with good leaving groups and a strong nucleophile like OH−.

When a halogenoalkane meets aqueous hydroxide, a nucleophilic substitution occurs. The hydroxide ion attacks the carbon bonded to the halogen, the halogen leaves as a halide ion, and the oxygen from the hydroxide ends up attached to the carbon. After a quick proton transfer from water, this gives the neutral alcohol. So the overall result is an alcohol plus a halide ion in solution.

The other ideas don’t fit because hydrogen would have to replace the halogen, which isn’t the role of hydroxide here; the reaction instead replaces X with OH. Forming an alkoxide and HX would require a different proton transfer pattern not supported by this base in water, and saying there’s no reaction ignores the common SN2 substitution that occurs with good leaving groups and a strong nucleophile like OH−.