RetroSynthetiX

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!

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I. Introduction

Towards the end of my PhD, I became aware that there were people who had zero idea on how to use pK_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 pK_aH works. Therefore, to really understand this topic, make sure you master what pK_a and pK_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 (pK_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 (C₅H₅N, pK_aH = 5)
• Sodium amide (NaNH₂, pK_aH = 38)

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

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 pK_as 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 K_a (acetylene). From the pK_a table, we know that the pK_a of acetylene is 25. Therefore, K_a (acetylene) is 10^–25 and we can obtain the value of K₂.

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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 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 pK_a (pyridinium) = pK_aH (pyridine) = 5. Hence, K_aH (pyridine) is 10^–5.

Note: I didn’t include the pK_a of pyridinium in the pK_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 K₃ 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 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 pK_as 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 pK_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 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:

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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 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 pK_a (ammonia) = pK_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 K₅ 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 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 pK_aH values. Here they are:

From this we can conclude two things:

• We can gauge the course of the reaction by looking at the pK_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 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 pK_aH of the base is larger than the pK_a of the acid, then the reactions take place.

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

• Sodium methoxide (NaOCH₃, pK_aH = 16)
• tert-Butyllithium ((CH₃)₃C–Li, pK_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 = 10²⁸, reaction takes place
  

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Page last updated: 23i17
RP

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Other Posts in Acid, Base, and pK_a Series

Acid-Base Prologue	Acid-Base I What are Acids and Bases?	Acid-Base II What is pKa and Why is It Important?
Acid-Base III The pKa Table and What Information It Tells You	Acid-Base IV Factors Affecting the Acidity of Organic Compounds	Acid-Base V Using pKa to Predict the Course of a Reaction Part 1
Acid-Base VI Using pKa to Predict the Course of a Reaction Part 2	Acid-Base VII Using pKa to Predict the Course of a Reaction Part 3	Acid-Base VIII Acidity, Basicity, and Solvent Part 1
Acid-Base IX Acidity, Basicity, and Solvent Part 2	Acid-Base X Levelling Effect	Acid-Base XI Hard and Soft Acids and Bases (HSAB) Theory in Organic Chemistry