Rank the following compounds from easiest to hardest at removi...

Question

Rank the following compounds from easiest to hardest at removing a proton from its methyl substituent:

Accepted Answer

All right. Hello everyone. So this question is asking us to arrange compounds xy and Z in order of decreasing ease of proton removal from the methyl substituent. So to start off here when describing the ease of proton removal, that's a question of acidity because recall that the more acidic a particular proton happens to be the easier it is to remove. So really this is a question of which methyl substituent is the most acidic and the least acidic. Now, with this in mind, recall also that there's a correlation between acid strength and conjugate based stability. The conjugate base is the species that forms after a particular proton has been removed. So the more stable the conjugate base or CB for short happens to be then the more acidic that compound is in general. So when asking ourselves which compound has the most acidic methyl group, we're really asking ourselves which of these has the most stable conjugate base. Now, all of the three compounds provided are derivatives of pyridine which is an aromatic molecule. So because of this, after each methyl group has been detonated, we're going to see multiple resonance structures involving the pi bonds of the aromatic ring. Now, the catch though is that because of the aromatic ring, there are going to be quite a few resonant structures for each compound. Because of this, I went ahead and I redrew the resonance structures in advance. So I'm going to go ahead and pause this recording to make the resonance structures visible to you. All right. So now up on the screen, we can see not only the conjugate bases for compounds Xy and Z but also their respective resonance structures involving the aromatic R. So let's talk about each of these individually and then go ahead and compare them. Now with respect to the resonance structures themselves recall that resonance describes a movement of electrons from the region of highest electron density to the region of lowest electron density. So because of this, the resonant structures for all three compounds begin at the lone pair of electrons on the carbon atom that has a negative charge, the negative charge indicates that it is the region of highest electron density. So because of that resonance starts off here with the formation of a new pi bond followed by the de localization of that lone pair of electrons across the ring itself. Now what's noteworthy here is that in the structure of compound X nitrogen currently has a positive to so from the very beginning nitrogen has a positive charge that it does not prefer to have due to its high electron negativity. But we can see here that as the lone pair of electrons travels throughout the pyridine ring in one of our resonance structures, that lone pair of electrons falls onto nitrogen, thereby giving it a neutral charge. So essentially nitrogen starts off with a positive charge. But as the electrons begin to delocalized at some point, nitrogen is given a neutral charge after it gains that low pair, that is going to be very favorable. Because once again, nitrogen does not necessarily want that positive charge. So the positively charged nitrogen is very readily going to accept that low impair which once again makes this conjugate base very stable. So now lets compare this to compound why? And once again, resonance starts at the lone pair of electrons on carbon and then travels throughout the ring. Now, unlike the resonance structures described for compound X, notice how the negative charges or rather the one negative charge stays only on carbon atoms at no point in our resonance structures. Does the lone pair of electrons actually fall directly on nitrogen? And carbon notably is less electron than nitrogen. Recall that a negative charge is better stabilized when a highly electron atom such as nitrogen bears that negative charge. So because of this, the conjugate base of compound Y here is less stable, it would be more stable if at some point nitrogen held the negative charge instead of only carbon. So now let's talk about compound Z and its conjugate base with resonance structures. Now, what's noteworthy here is that compounds X and Z both had that methyl group on the opposite side of the nitrogen on the ring itself. This is very important because it's going to affect the de localization of that lone pair across the ring itself. Because similar to the resonance structures for compound X at some point, a lone pair of electrons does fall onto the nitrogen directly, thereby giving it a positive charge. So here nitrogen begins as a neutral atom but then does eventually gain a negative charge. So the conjugate base for compound Z is stabilized because nitrogen, a highly electron atom does in fact gain a negative charge so that stabilizes the charge. So how does this compare to the conjugate base of compound X? Well recall that in compound X nitrogen had a positive charge in the beginning, which is less favorable than neutral nitrogen. So because of this, the conjugate base of X is made more stable than that of Z because a positively charged nitrogen is more readily able to accept a lone pair of electrons because there's an extra incentive there. So to speak, nitrogen does not want to have a positive charge, but it is OK with a neutral charge or no charge at all. So because of this, we've arrived at our final answer. Once again, the stability of the conjugate base determines the acidity of a given proton. The methyl substituent of compound X is going to have the easiest proton to remove, followed by that of Z and then that of Y and there you have it. So with all that being said, if you watch this video all the way through, thank you so very much and I hope you found this helpful.

Rank the following compounds from easiest to hardest at removing a proton from its methyl substituent:

Key Concepts

Acidity and pKa

Inductive Effect

Resonance Stabilization

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