## Contents

II. Deprotonation of Acetylene

II.1. With Pyridine (C_{5}H_{5}N) as Base

II.2. With Sodium Amide (NaNH_{2}) as Base

## I. Introduction

Towards the end of my PhD, I became aware that there were people who had zero idea on how to use p*K*_{a} to predict the course of a reaction. How did I know this? Because I was one of them! Yea, well done mate.. Well done..

Having nobody around to teach me, I decided to take the matter into my own hands. And it was actually much much simpler than I thought.

You’ll see in this post how p*K*_{aH} works. Therefore, to really understand this topic, make sure you master what p*K*_{a} and p*K*_{aH} really is. Head over to Acid-Base II and come back here after you’re done. If you’re confident that you know all about them, then feel free to continue.

Without further ado, we shall begin. And for now, let’s not worry about the solvent just yet.

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## II. Deprotonation of Acetylene

The hydrogen attached to the *sp* carbon in an alkyne is quite acidic (p*K*_{a} = 25) and can be deprotonated using the right base. This reaction yields acetylide anion and protonated base:

The question is:

**Does this reaction take place?**

Answer:

** It depends on the equilibrium constant value of the reaction ( K_{eq})**

If the value of *K*_{eq} is large, then yes, the reaction proceeds. Conversely, if the *K*_{eq} value is small, then the reaction does not take place.

Let us consider 2 different bases:

- Pyridine (C
_{5}H_{5}N, p*K*_{aH}= 5) - Sodium amide (NaNH
_{2}, p*K*_{aH}= 38)

Remember that we are using these as bases, therefore we will be looking at these base using their p*K*_{aH}. The values above are rounded to the nearest whole number so you won’t need a calculator to do this.

Can’t remember what conjugate acids/bases are?

Head over to Acid-Base I.

thus:
This method is just a theoretical approximation that gives us a prediction on how the reaction might proceed. It does not take into account other factors and conditions that may be affecting the reaction. That being said, the value of The actual and accurate value of |

Let’s now have a look at each of the bases and whether or not they react with acetylene.

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### II.1. With Pyridine (C_{5}H_{5}N) as Base

The reaction between acetylene and pyridine yields acetylide anion and pyridinium cation. Let’s label it: reaction 1.

The equilibrium constant of this reaction is *K*_{1}:

In the *K*_{1} equation:

- [C
_{2}H^{–}] is the acetylide anion - [C
_{5}H_{5}NH^{+}] is the pyridinium cation - [C
_{2}H_{2}] is the acetylene - [C
_{5}H_{5}N] is the pyridine

So does this reaction take place? To do that, we need to find out the value of *K*_{1} and we can do so by checking the p*K*_{a}s of the species involved.

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#### II.1.1. Acetylene

Acetylene in this reaction acts as an acid, yielding acetylide anion as its conjugate base. This reaction is the acid dissociation reaction of acetylene. Let’s label it: reaction 2.

The equilibrium constant of this reaction is *K*_{2}, which is also *K*_{a} (acetylene). From the p*K*_{a} table, we know that the p*K*_{a} of acetylene is 25. Therefore, *K*_{a} (acetylene) is 10^{–25} and we can obtain the value of *K*_{2}.

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#### II.1.2. Pyridine

Pyridine in this reaction acts as a base, yielding pyridinium cation as its conjugate acid. Let’s label it: reaction 3.

The equilibrium constant of this reaction is *K*_{3}:

Do you recognise this? This is seeing the base through its conjugate acid. Since we know that the conjugate acid is pyridinium cation, we can say that p*K*_{a} (pyridinium) = p*K*_{aH} (pyridine) = 5. Hence, *K*_{aH} (pyridine) is 10^{–5}.

**Note**: I didn’t include the p*K*_{a} of pyridinium in the p*K*_{a} table but you’ll be able to get the value if you google it.

Since reaction 3 is just the reverse reaction of the dissociation of pyridinium cation, the value of *K*_{3} is the reciprocal of *K*_{aH} (pyridine).

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#### II.1.3. Overall

If we now combine reactions 2 and 3, we can generate reaction 1.

Therefore, the value of *K*_{1} can be calculated as follows:

We can also go the long way, by putting in the components of *K*_{2} and *K*_{3} in the equation to show that combination of the two yields the components of *K*_{1}. Notice that they are in agreement with the *K*_{1} expression shown earlier at the beginning of Section II.1..

In both cases, we know that the value of *K*_{1} is 10^{–20}. This means that the forward reaction is unfavourable!

Therefore, based on the value of *K*_{1}, **this reaction does not take place if pyridine is used as the base**.

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### II.2. With Sodium Amide (NaNH_{2}) as Base

The reaction between acetylene and the amide anion (NH_{2}^{–}) yields acetylide anion and ammonia. Let’s label it: reaction 4. We can simply ignore sodium cation here, because it doesn’t really do anything that affects the reaction.

The equilibrium constant of this reaction is *K*_{4}:

In the *K*_{4} equation:

- [C
_{2}H^{–}] is the acetylide anion - [NH
_{3}] is the ammonia - [C
_{2}H_{2}] is the acetylene - [NH
_{2}^{–}] is the amide anion

So does this reaction take place? To do that, we need to find out the value of *K*_{5} and we can do so by checking the p*K*_{a}s of the species involved.

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#### II.2.1. Acetylene

As shown before, acetylene in this reaction acts as an acid, yielding acetylide anion as its conjugate base. This reaction has a p*K*_{a} of 25, or *K*_{a} = 10^{–25}. This reaction has previously been labelled reaction 2, so let’s just stick to it. The value of *K*_{a} is also the value of *K*_{2}. I will just rewrite it again here so that you don’t have to scroll too far and forget where you were before.

So the equilibrium constant of this reaction is still *K*_{2}:

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#### II.2.2. Amide Anion

The amide anion in this reaction acts as a base, yielding ammonia as its conjugate acid. Let’s label it: reaction 5.

The equilibrium constant of this reaction is *K*_{5}:

Again, here we are seeing a base through its conjugate acid. We know that the conjugate acid is ammonia, so we can say that p*K*_{a} (ammonia) = p*K*_{aH} (amide anion) = 38. Hence, *K*_{aH} (amide anion) is 10^{–38}.

Since reaction 5 is just the reverse reaction of the dissociation of ammonia, the value of *K*_{5} is the reciprocal of *K*_{aH} (amide anion).

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#### II.2.3. Overall

If we now combine reactions 2 and 5, we can generate reaction 4.

Therefore, the value of *K*_{5} can be calculated as follows:

We can also go the long way, by putting in the components of *K*_{2} and *K*_{5} in the equation to show that combination of the two yields the components of *K*_{4}. Notice that they are in agreement with the *K*_{4} expression shown earlier at the beginning of Section II.2..

In both cases, we know that the value of *K*_{4} is 10^{13}. This means that this forward reaction is favourable and the reaction goes forward!

Therefore, based on the value of *K*_{4}, **this reaction takes place if sodium amide is used as the base**.

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## III. Conclusion

Let’s have a look at the overall picture here. We’ve tried to deprotonate acetylene using 2 different bases with different p*K*_{aH} values. Here they are:

From this we can conclude two things:

- We can gauge the course of the reaction by looking at the p
*K*_{a}values of the species involved in the reaction. The proper value of*K*_{eq}itself is not very crucial; to predict the course of a reaction, we simply need to see whether it is large or small.

We can also make a formula for the *K*_{eq}, if you wish:

Or, if you combine the last two equations:

**CAUTION**: Be extra careful with the + and – signs in the formulas above. Always remember that p*K* is a negative log function!

Of course, you do not have to memorise the formulas. You can always derive them by following the steps we have outlined. It’s really very simple when you’re used to it.

- If the p
*K*_{aH}of the base is larger than the p*K*_{a}of the acid, then the reactions take place.

The p*K*_{aH} of pyridine is smaller than acetylene and therefore the reaction does not take place. On the other hand, the p*K*_{aH} of amide anion is larger than the acetylene and the reaction takes place.

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## IV. Try It Yourself!

Why don’t you try to do some calculations yourself as a practice?

Try to deprotonate acetylene (p*K*_{a} = 25) using:

- Sodium methoxide (NaOCH
_{3}, p*K*_{aH}= 16) *tert*-Butyllithium ((CH_{3})_{3}C–Li, p*K*_{aH}= 53)

Find what the conjugate acids are, and then find the value for each of the *K*_{eq} and predict if these reactions will work!

Highlight text to reveal the answers:

- If the base is sodium methoxide: conjugate acid is methanol,
*K*_{eq}= 10^{–9}, reaction does not take place

- If the base is
*tert*-butyllithium: conjugate acid is*tert*-butane,*K*_{eq}= 10^{28}, reaction takes place

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Page last updated: 23i17

RP

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