Question #0814e

1 Answer
Feb 14, 2017

1) No reaction; 2) no reaction; 3) formation of acetylide ion.


1) With hydroxide ion

Terminal alkynes are slightly acidic (#"p"K_"a" = 26#).

Thus, they can be deprotonated by a strong enough base.

For water, #"p"K_"a" = 16#.

Thus, water is a stronger acid than the alkyne, and hydroxide ion is a weaker base than acetylide ion.

#"R-C≡C-H" + underbrace("HO"^"-")_color(red)("weaker base") larr underbrace("R-C≡C"^"-")_color(red)("stronger base") + "H"_2"O"#

The position of equilibrium is far to the left, so there is effectively no reaction.

b) With ammonia

Ammonia is a weaker base than hydroxide ion, so there will be no reaction with but-1-yne.

c) With amide ion

For ammonia, #"p"K_"a" = 35#.

Ammonia is a weaker acid than a terminal alkyne, so its conjugate base is stronger than an acetylide ion.

#"CH"_3"CH"_2"-C≡C-H" + underbrace("H"_2"N"^"-")_color(red)("stronger base") rarr underbrace("CH"_3"CH"_2"-C≡C"^"-")_color(red)("weaker base") + "NH"_3#

The position of equilibrium will be to the right, forming an acetylide ion.