When we mix 1-bromopropane and $\ce{NaNH2}$, what happens? Which reactions wins out to make the major product?

My textbook says that primary alkyl halides give mainly SN2 products unless the base is hindered ($\ce{t-BuO-}$) (where E2 would win out).

But it contradicts itself when you have for example 1,2-dichlorobutane and you add three molar equivalents of $\ce{NaNH2}$ and then $\ce{NH4Cl}$ to have an alkyne. Why is there no SN2 going on in that case?


IIRC, Sodamide is generally insoluble, except in liquid ammonia. And why do people like to prepare amines with ammonia as a nucleophile? The basicity complicates things, as does the fact that the primary amine initially formed will react further to give secondary and tertiary amines, even quaternary ammonium compounds. It's not a reaction useful in preparative chemistry.

That said, the balance of nucleophilicity and basicity is also influenced by the solvent. It's well known that KOt-Bu acts mostly as a base in THF and mostly as a nucleophile in DMSO, there's plenty of examples in the literature. Hydrogen-bonding solvents tip the balance towards elimination.

  • $\begingroup$ So, that means that the major product will be the alkyne because the immensely great basicity of NH2- will win out even though there could be some NH2- that could do Sn2? $\endgroup$ – yolo123 Oct 31 '14 at 17:28
  • $\begingroup$ My gut feeling is that from 1-bromopropane and NaNH2 in liquid ammonia you'd get mostly propene. $\endgroup$ – Abel Friedman Oct 31 '14 at 18:06

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