2. Indicate which compounds would be more than 99% deprotonate...

Question

2. Indicate which compounds would be more than 99% deprotonated by a solution of sodium ethoxide in ethanol.

Accepted Answer

All right. Hi everyone. So for this question you must determine which of the following compounds would be fully depreciated by a solution of sodium with oxide in ethanol. So this question is essentially asking us about acidity for each of the following compounds that we have here and recall that acidity is directly correlated to the stability of the resulting conjugate base. And there are a number of factors that affect the stability of the conjugate base. 1st 1 is going to be residents. Right? Because if you're an ion is stabilized by resonance, then that's going to make your acid more acidic. The next is going to be electro negativity because um protons that are next to electro negative atoms like oxygen tend to be more acidic. Next up we have hybridization we have hybridization, the size of the atom and inductive effects. So all of these factors we have to take into account when trying to decide how stable the conjugate base is going to be here. So, to go more into detail about resonance really quickly recall that um certain alpha carbons are going to be a lot more acidic than others. And what I mean by alpha carbon is if I draw a generic ketone on the side here, the carbonnel is going to be our reference point. Right so the alpha carbons are the carbons that are directly adjacent to it. So in this case we have to now recall that alpha protons hydrogen next to or rather hydrogen is connected to alpha carbons. They tend to be a lot more acidic because they're stabilized via resonance. Right? So if I go ahead and I. d. protein eight um one of these alpha hydrogen is that I have here I'm going to end up with an an eye on over here and that's going to stabilize via and in a late ion. So if I draw a resonance structure like this then are in a late ion is going to have an o minus and an all keen in between the reference and the alpha carbon. So this resonance this back and forth between um between the bullet and the car bottle that's going to stabilize your conjugate base. So this is especially true. Um if we have an alpha carbon in between two car bottles. So if we go ahead and we look at molecule one here molecule one has an alpha hydrogen in between our two car bottles. So um the stabilization that in a late formation that I mentioned earlier, that's going to happen um twice full. Right? Or to fold. Not sure how to say that. But the point is is that there's even more resonant structures possible because we have um two car bottles instead of one. So compound one is actually going to um d protein it fully because its conjugate base is stabilized via resonance. Alright so with that being said, let's go ahead and focus on um molecule to really quickly which only has one alpha hydrogen. Right? Because we have an ester. So the thing about Esther's is that um recall that the alpha hydrogen of an ester is not going to be acidic because of the resonant structure. So the oxygen of your Esther, that's not the carbonnel. Right. That's going to contribute to the residents as well by donating a pair of electrons. Right? So if that donates a pair of electrons, then that's going to push the carbon pi electrons towards the oxygen of the carbon and that's going to leave. If we were to go ahead and d protein eight this alpha hydrogen, then we would actually end up with two negative charges and one positive charge. So that's not really ideal. And that's why the alpha hydrogen of esters tend to not be that acidic. But molecule three is different. Right? Because we do have an ester but we have um we have another car bottle. Right? So because we have this alpha hydrogen in between two car bottles, we're gonna see the same phenomenon that we talked about in molecule a right, So um that resulting an ion is going to be stabilized via resonance as well. And as for molecule four. Well, molecule four, we have a carb oxalic acid group. Right? And carve oxalic acids are specific. So this is also going to be fully deep rotated by sodium oxide. Right? Because recall that the hydrogen of our crab oxalic acid, it's bonded directly to an electro negative atom making it more acidic. And molecule five well molecule five is a fennel. Right? And finals are going to be relatively acidic because the resulting um dark oxide. An ion is going to be stabilized via a resident with the benzene ring. Right? So that oh minus that results from the deep coordination of a final. That's going to contribute to residents with the benzene ring. So that's very stable and that's going to be acidic Now. The same isn't really true for molecule um From molecules six and 7 here, Right? Because molecule six is an ordinary alcohol and if we go ahead and dip rotate it that's going to be an ordinary Alcock side. So it can't really be stabilized by residents. So but it is bonded directly to an electro negative atom. Right? So it might be protein eight. But not fully like the question is asking. So that's not an answer. And the same is true for molecule seven. Because um The same is true for molecules seven because we only have one car bottle, we only have one car bottle. So that resonance with the late ion is not going to be as strong And therefore it would D protein eight. Just not necessarily fully like the question is asking. So with that being said, the molecules that are going to fully d protein eight in sodium with oxide are going to be um compounds 134 and five. And with that being said this is going to be it for this question. So thank you very much for watching and I hope you found this helpful.

2. Indicate which compounds would be more than 99% deprotonated by a solution of sodium ethoxide in ethanol.

Key Concepts

Deprotonation

Acidity and pKa

Resonance Stabilization

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