## Contents

I. Introduction

II. Deprotonation of Acetylene

II.1. With Pyridine (C5H5N) as Base

II.1.1. Acetylene

II.1.2. Pyridine

II.1.3. Overall

II.2. With Sodium Amide (NaNH2) as Base

II.2.1. Acetylene

II.2.2. Amide Anion

II.2.3. Overall

III. Conclusion

IV. Try It Yourself!

## I. Introduction

Towards the end of my PhD, I became aware that there were people who had zero idea on how to use pKa 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 pKaH works. Therefore, to really understand this topic, make sure you master what pKa and pKaH 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 (pKa = 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?

It depends on the equilibrium constant value of the reaction (Keq)

If the value of Keq is large, then yes, the reaction proceeds. Conversely, if the Keq value is small, then the reaction does not take place.

Let us consider 2 different bases:

• Pyridine (C5H5N, pKaH = 5)
• Sodium amide (NaNH2, pKaH = 38)

Remember that we are using these as bases, therefore we will be looking at these base using their pKaH. 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.

 The Important Box This method of calculating Keq is an approximation method. thus: The value of Keq that we obtain through this method is only approximate. 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 Keq that we obtain through this method, although not accurate, is enough for us to predict the course of a reation. The actual and accurate value of Keq can only be obtained experimentally.

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 (C5H5N) 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 K1: In the K1 equation:

• [C2H] is the acetylide anion
• [C5H5NH+] is the pyridinium cation
• [C2H2] is the acetylene
• [C5H5N] is the pyridine

So does this reaction take place? To do that, we need to find out the value of K1 and we can do so by checking the pKas 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 K2, which is also Ka (acetylene). From the pKa table, we know that the pKa of acetylene is 25. Therefore, Ka (acetylene) is 10–25 and we can obtain the value of K2. *** Back to Contents ***

#### 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 K3: 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 pKa (pyridinium) = pKaH (pyridine) = 5. Hence, KaH (pyridine) is 10–5.

Note: I didn’t include the pKa of pyridinium in the pKa 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 K3 is the reciprocal of KaH (pyridine). *** Back to Contents ***

#### II.1.3. Overall

If we now combine reactions 2 and 3, we can generate reaction 1. Therefore, the value of K1 can be calculated as follows: We can also go the long way, by putting in the components of K2 and K3 in the equation to show that combination of the two yields the components of K1. Notice that they are in agreement with the K1 expression shown earlier at the beginning of Section II.1.. In both cases, we know that the value of K1 is 10–20. This means that the forward reaction is unfavourable!

Therefore, based on the value of K1, this reaction does not take place if pyridine is used as the base.

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### II.2. With Sodium Amide (NaNH2) as Base

The reaction between acetylene and the amide anion (NH2) 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 K4: In the K4 equation:

• [C2H] is the acetylide anion
• [NH3] is the ammonia
• [C2H2] is the acetylene
• [NH2] is the amide anion

So does this reaction take place? To do that, we need to find out the value of K5 and we can do so by checking the pKas 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 pKa of 25, or Ka = 10–25. This reaction has previously been labelled reaction 2, so let’s just stick to it. The value of Ka is also the value of K2. 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 K2: *** Back to Contents ***

#### 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 K5: Again, here we are seeing a base through its conjugate acid. We know that the conjugate acid is ammonia, so we can say that pKa (ammonia) = pKaH (amide anion) = 38. Hence, KaH (amide anion) is 10–38.

Since reaction 5 is just the reverse reaction of the dissociation of ammonia, the value of K5 is the reciprocal of KaH (amide anion). *** Back to Contents ***

#### II.2.3. Overall

If we now combine reactions 2 and 5, we can generate reaction 4. Therefore, the value of K5 can be calculated as follows: We can also go the long way, by putting in the components of K2 and K5 in the equation to show that combination of the two yields the components of K4. Notice that they are in agreement with the K4 expression shown earlier at the beginning of Section II.2.. In both cases, we know that the value of K4 is 1013. This means that this forward reaction is favourable and the reaction goes forward!

Therefore, based on the value of K4, 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 pKaH values. Here they are: From this we can conclude two things:

• We can gauge the course of the reaction by looking at the pKa values of the species involved in the reaction. The proper value of Keq 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 Keq, 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 pK 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 pKaH of the base is larger than the pKa of the acid, then the reactions take place.

The pKaH of pyridine is smaller than acetylene and therefore the reaction does not take place. On the other hand, the pKaH 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 (pKa = 25) using:

• Sodium methoxide (NaOCH3, pKaH = 16)
• tert-Butyllithium ((CH3)3C–Li, pKaH = 53)

Find what the conjugate acids are, and then find the value for each of the Keq and predict if these reactions will work!

Highlight text to reveal the answers:

• If the base is sodium methoxide: conjugate acid is methanol, Keq = 10–9, reaction does not take place
• If the base is tert-butyllithium: conjugate acid is tert-butane, Keq = 1028, reaction takes place

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Page last updated: 23i17
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