Why can't the alkoxide ion deprotonate acetylene?

Because the pKa of an alcohol is generally lower than the pKa of acetylene. That means the alcohol is generally a stronger acid than acetylene and is disfavored from becoming protonated.

This is an equilibrium as follows:

#"RO"^(-) + stackrel("pKa" = 25)overbrace("H"-"C"-="C"-"H") rightleftharpoons stackrel("pKa" ~~ 16)overbrace("ROH") + "H"-"C"-="C":^((-))#

We will show why it is that the equilibrium lies heavily on the side of the not-deprotonated acetylene, i.e. why an alkoxide "cannot" deprotonate acetylene.

In reality, it does, but it's that it does not deprotonate a large enough percentage of acetylene molecules in a given sample that it is considered significant.

A shortcut to determine this factor is:

#\mathbf(K_(a,"acid")/(K_(a,"conj. acid")) = 10^("pKa"_"conj. acid" - "pKa"_"acid")#

#= 10^(16 - 25)#

#= color(blue)(10^(-9))#

This tells you that since the pKa of the alcohol (the conjugate acid) is #9# pKa units lower than the pKa of acetylene (the acid), the deprotonation of acetylene is favored by a factor of #10^(-9)#.

That is, it is disfavored by a factor of #10^9#. From this we can draw the basic rule that says:

The equilibrium lies on the side of the weaker acid.

Let's figure out how I knew that, using the Henderson-Hasselbalch equation.

Suppose that we compare two equilibria in solution, where acetylene (#"H"-"C"-="C"-"H"#) tries to get deprotonated, just like the alcohol (#"ROH"#) tries to get deprotonated.

Then we will have two competing acid-base equilibria going on at the same time and get a single solution with a single #"pH"#.

ACETYLENE EQUILIBRIUM

Here is acetylene acting like an acid in solution.

#\mathbf("H"-"C"-="C"-"H" + "H"_2"O"(l) rightleftharpoons "H"-"C"-="C":^((-)) + "H"_3"O"^(+)(aq))#

#color(green)(K_(a1) = (["acetylide"]["H"_3"O"^(+)])/(["acetylene"])# (1a)

where #K_(a1)# represents the deprotonation of acetylene.

#"pH" = "pKa"_"acid" + log\frac(["A"^(-)])(["HA"])#

#= 25 + log\frac(["acetylide"])(["acetylene"])# (1b)

ALCOHOL EQUILIBRIUM

And the other one is the alcohol acting like an acid in the same solution.

#\mathbf("ROH" + "H"_2"O"(l) rightleftharpoons "RO"^(-) + "H"_3"O"^(+)(aq))#

#color(green)(K_(a2) = (["RO"^(-)]["H"_3"O"^(+)])/(["ROH"])# (2a)

where #K_(a2)# represents the deprotonation of the alcohol.

#"pH" = "pKa"_"conj. acid" + log\frac(["A"^(-)])(["HA"])#

#= 16 + log\frac(["RO"^(-)])(["ROH"])# (2b)

RELATING THE TWO EQUILIBRIA: WHICH IS MORE FAVORABLE?

Now, we can equate (1b) and (2b) since they have the same #"pH"# (by being in the same solution):

#16 + log\frac(["RO"^(-)])(["ROH"]) = 25 + log\frac(["acetylide"])(["acetylene"])#

#-9 = log\frac(["acetylide"])(["acetylene"]) - log\frac(["RO"^(-)])(["ROH"])#

#log\frac(["ROH"]["acetylide"])(["RO"^(-)]["acetylene"]) = -9#

#\frac(["ROH"]["acetylide"])(["RO"^(-)]["acetylene"]) = 10^(-9)#

But remember that the equilibrium expressions were acquired as (1a) and (2a) already:

#K_(a1) = (["acetylide"]["H"_3"O"^(+)])/(["acetylene"])# (1a)

#K_(a2) = (["RO"^(-)]["H"_3"O"^(+)])/(["ROH"])# (2a)

When we divide (1a) by (2a), we get:

#K_(a1)/K_(a2) = (["acetylide"]cancel(["H"_3"O"^(+)]))/(["acetylene"])*(["ROH"])/(["RO"^(-)]cancel(["H"_3"O"^(+)]))#

#= \frac(["ROH"]["acetylide"])(["RO"^(-)]["acetylene"]) = 10^(-9)#

So, that means:

#K_(a1)/K_(a2) = 10^(-9) = 10^(16 - 25) = 10^("pKa"_"conj. acid" - "pKa"_"acid")#

#color(blue)(10^(9)K_(a1) = K_(a2))#

Recall that #K_(a1)# represents the deprotonation of acetylene and #K_(a2)# represents the deprotonation of the alcohol.

Therefore, this is saying that the deprotonation of acetylene is one billion times as hard as the deprotonation of the alcohol. In other words, acetylene would rather not be deprotonated by the alkoxide.