Contents
II. Amidation of an Ester and Esterification of an Amide
III. Aldol Condensation of Acetophenone
I. Introduction
In Acid-Base VI, we saw an example of an organic reaction that wasn’t an obvious acid-base reaction. In this post, we will see and talk about two more reactions like that but with increasing complexity.
*** Back to Contents ***
II. Amidation of an Ester and Esterification of an Amide
Amidation of an ester is the process of making an amide from an ester, whilst esterification of an amide is the reverse process, i.e. the process of making an ester from an amide. Only one of the two reactions can happen: the amidation reaction. Why?
Let’s use the amidation of ethyl benzoate using ammonia and the esterification of benzamide using ethanol as example reactions.
Amidation Reaction
Esterification Reaction
You should notice that both reactions go through the exact same tetrahedral intermediate. Therefore, we should focus ourselves on this intermediate.
When the tetrahedral intermediate collapses to form carbonyl, the carbon has a choice. It can either kick out the amide anion (NH2–), ethoxide anion (EtO–), or phenyl anion (C6H5–). Using the same method as before, we ought to give each of the three species equal chance to be a leaving group. We then analyse the pKaH values of each leaving group and find the leaving group with the smallest pKaH value. Only then are we able to predict which reaction can take place.
Ok, let’s start with the amidation of ester. To achieve this, the tetrahedral intermediate has to kick out the ethoxide anion (EtO–), which has a pKaH of 16.
Next is the esterification of amide. To achieve this, the tetrahedral intermediate has to kick out the amide anion (NH2–), which has a pKaH of 38.
We can now compare the two, but hang on! We still have one more candidate for a leaving group: the phenyl group. Let’s see what happens when the tetrahedral intermediate kicks out the phenyl anion (C6H5–), which has a pKaH of 43.
Let’s now compile all three reactions with their respective leaving group’s pKaH.
| No | LG | pKaH LG |
Reaction |
| 1 | EtO– | 16 | Amidation of ester |
| 2 | NH2– | 38 | Esterification of amide |
| 3 | C6H5– | 43 | – |
Note: LG stands for leaving group.
Remember that the best leaving group has the smallest pKaH, and amongst the three potential leaving groups, ethoxide anion has the smallest pKaH.
This is why when the tetrahedral intermediate collapses, ethoxide anion is kicked out. This explains why the amidation of ester takes place but not the esterification of amide.
*** Back to Contents ***
III. Aldol Condensation of Acetophenone
We have seen and proved – twice – that a good leaving group has to have a small pKaH. So let’s have a look at the final example that is the second (and final) step in the aldol condensation of acetophenone.
The first step is the self-condensation reaction of 2 molecules of acetophenone in basic condition which yields 3-hydroxy-1,3-diphenylbutan-1-one (let’s call it compound 1 for short).
Compound 1 then condenses to form the final product: 1,3-diphenylbut-3-en-1-one (let’s call it compound 2) in the second step by losing OH–, driven by the deprotonation of the α-carbon by the base. This step is an elimination reaction.
Note: We’re not concerned about the stereoselectivity of aldol reactions here. Which is why I did not put the stereochemical assignments in compounds 1 and 2. We’re only concerned about looking at these reactions through the eyes of acid-base chemistry.
So why does compound 1 kick out OH– in the second step? There are two reasons that make this question a valid question:
- I believe when you learnt nucleophilic substitution, you were always told that OH is a poor leaving group. That’s why when we do nucleophilic substitution to a OH-containing compound, we have to make the OH a good leaving group first by using acids or sulfonic acid compounds. But here, we don’t do such things to the OH and yet it leaves. Weird huh?
- When the deprotonation of the α-carbon takes place in compound 1, the electron forms a double bond between α and β carbons. In this step, the β carbon has a choice: it can kick out either OH–, CH3–, or C6H5– to make room for the double bond. So why does the β carbon choose to kick out OH– in particular?
So let’s address this question then: why OH–?
To explain this, let’s pretend again that we don’t know that OH– is the leaving group. So, we must give all three of these groups equal chance to be a leaving group. We then analyse the pKaH values of each leaving group and find the leaving group with the smallest pKaH value. Only then are we able to predict which reaction can take place.
First, let’s consider OH– as a leaving group. Its pKaH value is 15.7.
Now, let’s consider CH3– as a leaving group. Its pKaH value is 50.
Finally, let’s consider C6H5– as a leaving group. Its pKaH value is 43.
Let’s now compile all three reactions with their respective leaving group’s pKaH.
| No | LG | pKaH LG |
| 1 | OH– | 15.7 |
| 2 | CH3– | 50 |
| 3 | C6H5– | 43 |
Note: LG stands for leaving group.
Again, remember that the best leaving group has the smallest pKaH, and amongst the three potential leaving groups, hydroxide anion has the smallest pKaH.
This is why when the deprotonation takes place, hydroxide anion is kicked out. This is also why, although usually OH– is not a good leaving group, it is the least evil amongst the three potential leaving groups in this particular example.
*** Back to Contents ***
IV. Conclusion
I would like to remind you again that the best leaving group has the smallest pKaH. We have now proved this with two more examples.
We have seen how we can determine the course of a reaction by using the pKa values of the species involved. I hope you can understand now how valuable pKas are in predicting the course of a reaction.
*** Back to Contents ***
Page last updated: 03ii17
RP
Other Posts in Acid, Base, and pKa Series
RetroSynthetiX 