# Question #0814e

Feb 14, 2017

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

#### Explanation:

1) With hydroxide ion

Terminal alkynes are slightly acidic ($\text{p"K_"a} = 26$).

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

For water, $\text{p"K_"a} = 16$.

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

$\text{R-C≡C-H" + underbrace("HO"^"-")_color(red)("weaker base") larr underbrace("R-C≡C"^"-")_color(red)("stronger base") + "H"_2"O}$

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, $\text{p"K_"a} = 35$.

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

${\text{CH"_3"CH"_2"-C≡C-H" + underbrace("H"_2"N"^"-")_color(red)("stronger base") rarr underbrace("CH"_3"CH"_2"-C≡C"^"-")_color(red)("weaker base") + "NH}}_{3}$

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