Why is imidazole a stronger acid (pK_a~ 14...

Question

Why is imidazole a stronger acid (pK_a~ 14.4) than pyrrole (pK_a ~ 17)?

Accepted Answer

Hello everyone. So in this video, I'm going to explain why Try Asil, which has a PK value of 10.2 is a stronger asset compared to parasol, which has a PK value of 14.2. Also given these two structures of each of our molecules mentioned here in this problem. So the stronger assets have a more stable or less basic conjugate base. So let's consider the residents contributing structure of Try Asil. So we're gonna go ahead and actually bridge all our structure here just so we can go ahead and do some arrow pushing. So we can actually display out the residents here. So again, we're talking about triathlon. Alright. So we have this five member ring with three necks, chins in here. So we'll draw that structure out first. Okay. So we have two double bonds here. And then since we're taking this proton out, because we're talking about the conjugate base here, we'll have a negative charge on our nitrogen without this double bond. Alright. So because we have this negative here is because we have this one extra long pair. So this lone pair here that's causing the nitrogen to have a negative. One formal charge is going to go ahead and come in and make a pie pond that one of the pi bonds in the neighboring nitrogen will go ahead and break to form a long pair on the carbon. So drawing the double ended arrows for our residents, we didn't follow where this arrow is going. Of course, we need to draw this five member ring again. So the nitrogen stay in place is more of the pi bonds and lone pairs are moving around. So now where are nitrogen originally has a long pair will gain a pi bond. And then the carbon with originally a double bond will now have a long pair getting a negative one formal charge. And what's gonna happen next is basically keep going around. So now the carbon with the extra lone pairs that gives it a negative charge will go ahead and form a pi bond, then the pi bond, which we forgot to draw here at the bottom, the pi bond, then we'll go and break and move to the next neighboring carbon to form a long pair. So again, let's redraw this five member ring With the three pathogens attached to it or in the structure rather. All right. So now we have a negative charge here at this bottom right hand corner of our carbon. Alright, again, basically everything is just keep moving and rotating around. So the long pair here will come on to make a pi bond. And then the pi bond between the two nitrogen will go ahead and break and form a long pair on this legend, this times of a carbon because again, we're going all around our ring here again, drunk in my resonance arrows. All right. So we have created a new pi bond and now we have an extra long pair of this nitrogen on top of our five member ring getting a negative one formal charge. Then this extra electron or long pear then again is going to come down and form a pi bond. Then this pi bond here is gonna go ahead and form a extra loan pair. On this bottom left hand corner nitrogen again, double arrows to signify the drawing resonant structures here instead of an actual reaction or continuation of reaction, our last president structure again, we'll just go ahead and draw redraw this five memory ring closing in three nitrogen atoms. We have formed a new pi bond and then we have a extra electron at the bottom left hand nitrogen and that gains again a negative one formal charge. So the conjugate base of try asil is stabilized with five residents contributors. And among them three contributors have a negative charge on the more electro negative nitrogen atom. As for the residents contributing structure of parasol, let's go ahead and draw that in as well. So again, we're talking about the conjugate base. So we're taking out that proton so that nitrogen will now have a negative charge. So again, we're talking about pure Brazil. So we have this five memory ring with just two nations in its ring. This time instead of the three, we still have the two pi bonds. And this right hand nitrogen here will have a lecture electron getting this negative one formal charge. So this actual long pair here is going to go ahead and come down to form a pi bond. And this already existing pi bond is going to go ahead and break and form a long pair on the carbon. So following our air pushing here, let's actually redraw this five memory ring again. Alright. So we have formed a new pi bond between our two nitrogen atoms and the other double bond or the pi bond will remain as is. But we also have put in a new lump air on the carbon which now gives a this carbon a negative one formal charge. Now again, just the same exact sort of style here. This is gonna go and come down and form a pi bond. Many CC double bond gonna go ahead and break and form a lump here on the neighboring carbon over to the right Qianjiang in my five memory during here. Alright. So again, we have formed a new pipe bond and the two pi bonds or the one pi bond between the two nitrogen is going to go ahead and remain as is. And we have an extra electron or two electrons rather a long pair on our carbon. So we have gained a negative one from a charge on this bright hand carbon. So what's gonna come next is again, this lone pairs will come down and form a pi bond. And then the pi bond existing between, between the two nitrogen. Here is gonna go ahead and break and form a Lumpur on our top nitrogen item of our molecule. Again, redrawing our five member here ring here we have formed a new pipe on over to the right and we have a extra electron or two electrons on the nitrogen on top. So this nitrogen will not have a negative one formal charge, same exact pattern, this nitrogen long pair here. So go ahead and come down to form a pi bond and the existing cc double bonds go ahead break and form a long pair on the carbon. Again, we're drawing the five member ID ring, The two Nigerians attached or embedded into this ring. We have yet again created a new pi bond and the last instruction will have a negative one formal charge on the carbon atom. So the conjugate base of parasol is stabilized with five president structures and among them to contributors have a negative charge on the more electro negative nitrogen atom. So due to the more stable conjugate base, which is try as all is going to be more acidic than parasol. So to write our answer in more formally do this on top here. So we can say that triangle is going to be more acidic that appears all because of the lowest peak a of its conjugate acid, which was just simplified to see a and the localized electrons on three nitrogen atoms that's going to be our proper and final answer for this problem again, triangle is more acidic than parasol because of the lowest P K of its conjugate acid and the localized electrons on the nitrogen atoms, which we have three of those in our molecule.

Why is imidazole a stronger acid (pK_a~ 14.4) than pyrrole (pK_a ~ 17)?

Key Concepts

Acidity and pKa

Resonance Stabilization

Electronegativity of Nitrogen

Watch next

Why is imidazole a stronger acid (pKa ~ 14.4) than pyrrole (pKa ~ 17)?

Key Concepts

Acidity and pKa

Resonance Stabilization

Electronegativity of Nitrogen

Watch next

Why is imidazole a stronger acid (pK_a~ 14.4) than pyrrole (pK_a ~ 17)?