By the end of this guide,
You’ll know how to rank molecules from the least to the most basic by understanding their stability. We’ll guide you through the essential factors that affect the stability of bases, building a strong foundation for solving related chemistry problems with confidence. Whether you’re studying for an exam or just keen to deepen your understanding of this key chemistry concept, you’ve come to the right place.
🖐️ Before we start, make sure you’re familiar with:
- Factors that affect the stability of conjugate bases
Challenge 1: Ranking negatively charged bases by strength
Question:
Rank the following bases in order of increasing basicity
Our strategy is to determine which base is the least basic and which is the most basic based on the stability of its negative charge.
Here’s how we’ll do it
Step 1: Evaluate and rank the stability of the bases
We’ll start by ranking bases based on their stability. Remember this simple rule: the less stable a base, the stronger it is. By identifying the less stable bases, we can determine which ones are stronger.
To evaluate stability, consider all five factors in this order (Charge, Atom, Resonance, Induction, Orbitals), taking into account exceptions to the order of priority.
Charge
What is the charge on each base?
Each base carries a negative one (-1) charge
Conclusion: Factor 1 does not indicate which base is more stable.
Atom
Which atom is holding the negative charge?
In bases A, B, and D, the negative charge is located on oxygen atoms, while in base C, it is on a nitrogen atom.
Both nitrogen and oxygen are in the same row of the periodic table, so electronegativity is a determining factor here. Oxygen is more electronegative than nitrogen, therefore it can stabilize a negative charge more effectively.
As a result, the negative charges on oxygen in bases A, B, and D are more stable compared to the negative charge on nitrogen in base C.
Therefore, base C, with the least stable negative charge among the four, is the strongest base.
Conclusion: ? < ? < ? < C
Resonance
Is the negative charge delocalized over multiple atoms?
Base A
The negative charge is delocalized over one oxygen atom and three carbon atoms.
Base B
The negative charge is ‘stuck’ on the oxygen atom (localized).
Base D
The negative charge is delocalized over one oxygen atom and three carbon atoms.
In bases A and D, the negative charge is delocalized over one oxygen atom and three carbon atoms. This delocalization allows for the distribution of the negative charge across several atoms, enhancing the stability of both bases.
The negative charge in base B is not resonance stabilized.
Therefore, we expect base B to be a stronger base than both A and D due to its lower stability.
Conclusion: ? < ? < B < C
Since bases A and D are similarly stabilized by resonance, we need to examine the next factor — induction — to further distinguish their stability and basicity. This will help us identify which of these two bases is more stable and consequently the weaker base.
Induction
Are there electronegative atoms nearby that stabilize the charge?
Base D
The negative charge in base D is significantly stabilized by the inductive effects of the nearby fluorine atom.
Base A
Base A lacks the presence of electronegative atoms in proximity to the negatively charged oxygen. As a result, there is less stabilization from inductive effects compared to base D.
We can conclude that base A is less stable than base D, making it a stronger base.
Conclusion: D < A < B < C
Orbitals
How close is the orbital that houses the charge to the nucleus?
This factor is not relevant in this case as we have successfully ranked each base using the previous factors.
Step 2: Summarize findings
After evaluating the factors, our final ranking from least basic to most basic is D, A, B, C.
