[PoetryInMotion]

Problem. What products would you expect when CH₃CH₂CH₂Br reacts with NaC≡CH?

My thoughts. HC≡C:^- anion is both a strong nucleophile and base. Thus, both an E2 and SN2 reaction could be expected and the products would be hex-1-yne and but-1-ene. According to the book, however, only hex-1-yne is formed. Does anyone know why?

[spirochete]

Because the book is over simplifying things. With most strong, non bulky bases mixed with primary halides the conditions can be tweaked to give mostly substitution. IE polar aprotic solvent, relatively low temperature etc. Conditions can also be tweaked to give a significant amount of elimination, but it would be more ideal to use a strong and bulky base for this purpose.

[orgopete]

Problem. What products would you expect when CH₃CH₂CH₂Br reacts with NaC≡CH?

My thoughts. HC≡C:^- anion is both a strong nucleophile and base. Thus, both an E2 and SN2 reaction could be expected and the products would be hex-1-yne and but-1-ene. According to the book, however, only hex-1-yne is formed. Does anyone know why?

Why indeed. There is a seeming paradox(s) in substitution reaction. Anions are better nucleophiles than uncharged ones.* However, weaker bases are better than strong bases.

Let me advance this question to a secondary bromide than primary. Primary halides generally give a substitution product as the textbook predicts. If the reaction were with a secondary halide, then an acetylide would give elimination and essentially no substitution product. If a thiolate were the nucleophile, then substitution would predominate.

Rather than answering the question why, I stick to what data teaches, primary gives mainly substitution, tertiary-elimination, and secondary-both. I look at secondary examples to see which give elimination and which substitution.

*If bonds are formed from electron pairs, then the charge of an electron pair is exactly the same despite the net number of protons, i.e., the electrons of ammonia have the same charge as those of fluoride even though fluoride may have lost a proton connected to one of its four valence pairs. There is more to this than charge as ammonia is also a stronger base than any of the halides despite their charge.