The following compounds can all react as acids.

Question

Structural formulas of five compounds: acetic acid, acetamide, methyl sulfonic acid, methyl sulfinic acid, and fluorosulfonic acid.

b. Rank the conjugate bases in the order you would predict, from most stable to least stable.

Accepted Answer

Alright. Hello everyone. So this question is asking us to rank the conjugate base of the following molecules from most stable to least stable. And here we have five different compounds labeled a through E which I've gone ahead and redrawn on the bottom of the screen here. Now for this question, this question is based around the Brony Larry acid based definition in which an acid if you recall is a species that donates a proton. On the other hand, the bronzed dari base is the species that accepts the proton. Now to be a little bit more specific, the conjugate base is the species formed when the bronzed acid loses its proton. So when drawing the conjugate base of a given compound, first, the most acidic proton is removed and then the atom that was initially connecting to that proton is going to gain a lone pair of electrons which decreases its charge by one. Now, what does it mean when I refer to a most acidic proton while recall it protons attached to electron elements such as oxygen or nitrogen are going to be the most acidic because of how electron deficient they are. So before we go ahead and draw out the conjugate basis for these compounds. Let me just go ahead and add lone pairs of electrons on elements like oxygen and nitrogen. All right. So now let's start off here with compound A. Now compound A has an oh group, which means that the proton attached to the electron oxygen atom is going to be the most acidic and therefore, is the one that gets removed. So for the conjugate base of compound A, I'm going to remove that proton attached to oxygen, which is going to make oxygen gain an extra loan pair of electrons. Because of that extra loan pair, the charge is going to decrease from zero to negative one. Now, from here compound CD and E all have hydroxy groups or oh groups that are going to be derotated. So for these compounds, I would go ahead and draw the conjugate base the same way, which means that all oxygen atoms in their conjugate bases are going to have three lone pairs of electrons and therefore a negative charge. Now, in this case, the atom that's going to be detonated happens to be a nitrogen. So even if the result is going to look different, the principle is still the same. So first one proton gets removed from nitrogen, bringing the total down from zero to just one. And the nitrogen is now going to have two loan pairs of electrons that gives it a negative charge. But now we have all five conjugate bases. So now let's talk about factors that affect the stability of said conjugate bases. Recall that there are a few different factors. We can first talk about the electron negativity of an element or en for short. And we can also talk about factors like the size of an element bearing a negative charge. We can talk about any inductive effects by substituents behaving as electron donating or electron withdrawing groups. We can talk about hybridization and we can also talk about how many resonant structures a conjugate base has. If any. Now, in this particular case, hybridization is not going to affect our ranking because all of the atoms that bear a negative charge are going to be sp three hybridized. So that's not going to help us compare. However, I do want to point out here that some of these compounds do have electron withdrawing substituents take for example, compound E compound is conjugate base contains a chlorine substituent which behaves as an electron withdrawing group due to chlorine's electronegativity. Now, recalled electron withdrawing groups stabilize a negative charge by actually pulling electron density away from the atom bearing the negative charge, which makes the conjugate base more stable since that charge is relatively delocalized. This means that the conjugate base of compound E is going to be the most stable. And another thing that I want to point out here when understanding the stability of the conjugate base E is that oxygen the oxygen that has a negative charge can participate in resonance with the two double bonds between sulfur and oxygen. That increases the number of resonance structures, which also increases the stability of the conjugate base overall. So similar to compound e the conjugate base of compound D also has three resonant structures. Because the negative charge on the oxygen can be delocalized across the two sulfur oxygen double bonds. So as the number of resonance structures increases, so does stability because again, that negative charge is delocalized, this means that the second most stable conjugate base would be that of compound D. So now we must compare the conjugate bases of compounds A B and C because all of them only have a total of two resonance structures. Each. Now another difference that I want to point out here is that we can distinguish between equivalent and non equivalent resonance forms. Take for example, compounds A and C, both of these have an oxygen atom that bears a negative charge after re de localization with the carbonel, excuse me, not the carbonel, but the carbonel and the sulfur oxygen double bond. In the next resonance structure, oxygen is once again going to have that negative charge. On the other hand, in the case of the conjugate base for compound E, the negative charge is actually delocalized between nitrogen and oxygen. So this means that the conjugate base of compound B is going to have non equivalent resonance structures because the negative charge is delocalized across different elements. This means that the conjugate base of compound B is going to be the least stable because compounds A and C have equivalent resonance structures, they are more stable than compound B. So now let's go ahead and break the tie between A and C recall that electron negativity also plays a role in the stability of a conjugate base. The only difference between the two structures here is that in compound a carbon is the central atom and in compound C sulfur, its now it just so happens that carbon and sulfur have the same electron negativity. So we can't actually compare them using electron negativity. However, we can compare the size, we can compare the size because sulfur is larger than carbon and electrons surrounding a larger atom are going to be made more stable because they have more space to separate from each other on that atom. So because of this, the conjugate base of compound C is more stable than that of compound A. And so that breaks the tie that makes the conjugate base of C third place overall and the conjugate base of 1/4 place overall and there you have it. So now let's go ahead and write our ranking more neatly from most stable to least stable. We have the conjugate base of E followed by D, followed by sea, followed by A and last but not least we have B and there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

The following compounds can all react as acids.

b. Rank the conjugate bases in the order you would predict, from most stable to least stable.

Key Concepts

Acid-Base Chemistry

Conjugate Base Stability

Inductive and Resonance Effects

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