What would you expect from the reaction of 1-butanol with NaOCH3?

Question

BlackTom AI · Accepted Answer

Consider the reaction context: a primary alcohol (1-butanol) reacting with sodium methoxide (NaOCH3) in a typical alcohol/alkoxide system.
Option A: 'no reaction' — In principle, alkoxides are bases and can deprotonate alcohols. Since pKa values for methanol (~15.5) and 1-butanol (~16) are close, the equilibrium can reasonably favor deprotonation of the alcohol by the alkoxide, giving the corresponding alkoxide (BuO−) and methanol. Therefore, saying there is no reaction ignores this feasible acid–base process.
Option B: 'an acid base reaction' — This is the most accurate description. NaOCH3 acts as a base and abstracts a proton from 1-butanol, forming the butoxide ion (BuO−) and methanol (MeOH). This is a typical acid–base equilibrium between a strong enough base and a relatively acidic alcohol, and it explains observed species in such systems.
Option C: 'an elimination' — Elimination from an alcohol under basic conditions would require a suitable leaving group and a substrate set that supports β-elimination to form an alkene. A primary alcoholLike 1-butanol does not readily undergo E2/Elimination in this context with methoxide; there’s no good leaving group being expelled to form an alkene here.
Option D: 'a substitution' — A classic substitution (SN) on an alcohol would require conversion of the hydroxyl into a better leaving group and a suitable electrophile. In this setup, NaOCH3 is not acting as a nucleophile displacing the OH group on 1-butanol; the reaction is not a typical SN process on the alcohol carbon.
In summary, the most plausible and supported interpretation is that the system undergoes an acid–base reaction, with deprotonation of 1-butanol by sodium methoxide to form the corresponding alkoxide and methanol.

What would you expect from the reaction of 1-butanol with NaOCH3?单项选择题

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