NaH or NaOH for a Williamson synthesis?

Resposta e explicação

1. Resposta correta Hydroxide is a weaker base than cyclohexoxide, so it converts only a small fraction of the alcohol (K ≈ 0.005) and the remaining hydroxide reacts with CH₃I; hydride deprotonates completely and H₂ escapes.

   K = 10^(15.7 − 18) ≈ 0.005 for hydroxide, but about 10^18 for hydride, which is also irreversible because H₂ leaves as a gas.

2. Esta não NaOH would protonate the alcohol instead of deprotonating it.

   Hydroxide is a base, far too weak an acid to protonate an alcohol. The issue is that it is too weak a base to deprotonate cyclohexanol fully.

3. Esta não Hydride is the nucleophile that attacks CH₃I and forms the ether.

   Hydride only removes the O–H proton; the cyclohexoxide oxygen attacks CH₃I.

4. Esta não NaOH works for phenol and therefore must also work for cyclohexanol, so NaH is chosen only because it is cheaper.

   Phenol (pKa 10) is about 10⁸ times more acidic than cyclohexanol, so hydroxide deprotonates phenol completely but not cyclohexanol. NaH is chosen for that reason, not cost.