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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?

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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.

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  • $\begingroup$ 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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