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For example, can $\ce{NaOH}$ ionize ethanol into ethoxide?

My textbook (Organic Chemistry, 6th edition, by M. Loudon and J. Parise, Roberts & Company 2016) suggests yes — it uses it in its mechanism for the intramolecular substitution of 4-bromo-1-butanol in the presence of $\ce{NaOH}$.

But I was taught that $\ce{^-OH}$ could only ionize thiols, not alcohols. After all, if simple $\ce{NaOH}$ can create alkoxides, why would chemists bother to use $\ce{NaH}$?

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  • $\begingroup$ If you're going to use NaOH to ionize ethanol then you'd need anhydrous ethanol. To get and keep a solvent anhydrous requires a bit of care. $\endgroup$
    – MaxW
    Commented Dec 11, 2017 at 6:17
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    $\begingroup$ In a polar non-protic solvent such as 1,2-dimethoxyethane, it may be possible to get complete deprotonation of the alcohol by a hydroxide base. The hydroxide ion is much more basic than it may appear at first, because its reactivity is strongly levelled in water and other protic solvents. See this past answer of mine for background $\endgroup$
    – Nicolau Saker Neto
    Commented Dec 11, 2017 at 13:10

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Yes, and no.

Acid and base chemistry is a neat balance between equilibria. As you well know, water dissociates readily by itself. A tendency that can be observed in any compound with $\ce{-OH}$ groups, yet not to a similar extend.
In other words, a hydroxyl group tends to be slightly acidic. Given a base, i.e. sodium hydroxide in this particular case, the following equilibrium will develop: $$\ce{EtOH + NaOH <<=> EtO- + Na+ + H2O}$$

While the equilibrium will mostly be on the left side, depending on what other substrates are available, a tiny amount of alkoxide might already be enough to drive the reaction forward. In this case it is important to look at what actually drives the reaction.

In the example of 4-bromo-1-butanol that your book uses I am assuming there is a cyclisation happening. $$\ce{BrH2C(CH2)2CH2OH + NaOH <=>> THF + H2O + NaBr}$$ Unfortunately I cannot quickly find a reference, but sodium bromide is not very soluble in diethylether, so I would assume it's not very soluble in THF either. The driving force here would be having a well enough leaving group and the continuous removal of product, basically making it a one-way reaction.

In some cases you need a higher concentration of alkoxide, and that's where hydrides come into play. $$\ce{EtOH + NaH <=>> EtO^-\bond{~}Na+ + H2 ^}$$

In this case dihydrogen, as it is a gas, will be removed from the reaction mixture, basically ensuring full conversion to the alkoxide.

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You are asking for a clear yes/no answer in a situation where there is none. See, it is all about equilibrium. True, $\ce{OH-}$ is a weaker base than $\ce{RO-}$ (in other words, alcohols are weaker acids than water), but not by much, hence the reaction is still possible, if only a little. That is, an equilibrium is much closer to the starting compounds, but still it contains some products.

So the full answer to your question goes along the following lines: Yes, if ionizing one molecule in a hundred is good enough for you; otherwise no.

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