# Why can't the alkoxide ion deprotonate acetylene?

##### 1 Answer

*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 **deprotonation** of acetylene is **favored** by a factor of

That is, it is **disfavored** by a factor of

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** (**deprotonated**, just like the **alcohol** (**deprotonated**.

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

**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

#"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

#"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

#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 ** deprotonation** of acetylene and

**of the alcohol.**

*deprotonation**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.*