a. Predict the relative pKa values of cyclopentadiene and cyclohe...

Question

a. Predict the relative pKa values of cyclopentadiene and cycloheptatriene.

Structures of cyclopentadiene (A) and cycloheptatriene (B) for pKa comparison.

Accepted Answer

Hello everyone. Let's do this problem. It says suggest the relative P K A values of the following compounds. Compound A is called in and compound B is called dial. All right. So how can we determine the relative P K A values of this or these compounds? What does P K A even mean? We know P K A is the acid dissociation constant. So the P K A value tells us how good of an acid A compound is the lower the P K A value, the higher the acidity or the better the acid and it's a better acid because it'll be more likely to donate a proton, which is the whole job of an acid, right? And therefore, if a compound is more likely to donate a proton, that means it has a more stable conjugate base. So we can look at these compounds draw the structure of their conjugate base by removing a proton and see how stable it is and compare the stability of those conjugate bases, which in turn, we can determine their relative P K A values. So since these compounds are both cyclic, they have several pie bonds in there, I think we can compare the stability in terms of aromatic. So stability by aromatic, the ranking goes as follows. Aromatic compounds are the most stable followed by non aromatic compounds. And then anti aromatic compounds are the least stable. So how do we know whether a compound is aromatic, anti aromatic or non aromatic? Let's look at the rules or the criteria. So for an aromatic compound, we have huckle rule and the three criteria to be aromatic or first, the compound needs to be cyclic and planer or flat, then we must have a fully conjugated pie system. And finally, the number of pi electrons must satisfy the expression four N plus two. All right. What about anti aromatic compounds? Their set of rules are called Brelo's rule. And the criteria are similar in that the structure needs to be cyclic and planer. It also needs to be fully conjugated, but they differ by the number of pi electrons. An anti aromatic compound satisfies the expression for N for the number of pi electrons. And lastly, a non aromatic compound is going to fail any of the criteria mentioned. So if the compound is not cyclic, not cleaner or not fully conjugated, then it will be non aromatic. All right. So let's look at our structures, remove a proton and see the aromatic of the conjugate base. So structure A in dean is a benzene ring fused with a five member ring that has one double bond. So what is the aromatic of this compound. Well, it's cyclic and planer. Is it fully conjugated? No, we have some carbons that do not have pi electrons. So this is going to be non aromatic. OK. Now, if we remove a proton or rather the compound donates a proton, since it is acting as an acid, we will get the following conjugate base where we have a loan pair and a negative charge. Now, on carbon, let's see, 123456789 on carbon nine. So is this compound cyclic and planar? Yes. Now, is it fully conjugated? Yes, because these lone pair electrons can act as pi electrons and notice how they alternate between these two double bonds. So this is aromatic. Those lone pair electrons can participate in resonance. That's a good way to ask yourself if the compound is fully conjugated. All right. Oh I forgot to count the number of pi electrons 2468 10. So since it satisfies four N plus two, that's how we know it's aromatic. If we had say eight pi electrons, that would be anti aromatic. So fully conjugated four N plus two electrons, this is an aromatic conjugate base. All right, let's look at compound B dial in, we have a benzene ring fused with another six membered ring with a double bond on carbon 1234567. So is this compound aromatic non aromatic or anti aromatic? It is not fully conjugated. So this is non aromatic and that is because of carbons eight and nine, right? There are two carbons between those pie bonds. Sorry, I meant carbons nine and 10, not eight and nine. So it is not fully conjugated. So now our compound donates a proton to give us this conjugate base where now there is a loan pair and a negative charge on carbon. Number 123456789 10. So is this compound cyclic and planer? Yes, is it fully conjugated? No, because we have this extra carbon between the lone pair and the double bond. So this is going to be non aromatic. OK. So which conjugate base is more stable? The aromatic one or the non aromatic one, the aromatic conjugate base for structure A. So which compound will have a higher P K A, the more stable one or the less stable one, the less stable one higher P K a more stable will be a lower P K A, right, better acid. So to summarize the relationship of the P K A values for these set of compounds, compound B has a higher P K A value then compound A. That's it for this problem. I hope this video was helpful and I'll see you next time.

a. Predict the relative pKa values of cyclopentadiene and cycloheptatriene.

Key Concepts

pKa and Acidity

Conjugate Base Stability

Ring Strain and Hybridization

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