Nucleophilic acyl substitution

Voiceover: Let's look at the general mechanism for a nucleophilic acyl substitution reaction. Here we have our carboxylic acid derivative and we know that this carbon right here is our electrophilic portion of the molecule. It's partially positive. The oxygen's withdrawing some electron density. We talked about the relative reactivities of carboxylic acid derivatives in the last video. That carbon is where our nucleophile is going to attack. Our nucleophile attacks here, these electrons kick off onto the oxygen. Let's go ahead and draw the result of that. We would have an R group over here, we would have a carbon, we would have on this left side here an oxygen with three lone pairs of electrons, giving it a negative one formal charge. Let's say that these electrons in here in magenta move off onto our oxygen, a negative one formal charge. Over here on the right we have our Y substituent and now our nucleophile is bonded to our carbon. Let's show those electrons in blue here. These electrons in blue attacked our carbonyl carbon and formed this bond. To go from our tetrahedral intermediate here to our final product, we must reform our carbonyl. These electrons would have to move into here and then that would push these electrons off onto our Y substituent and we would form, if we started with a - This would be a Y negative now. We have a negative one charge on our Y when those two electrons move off on it. Let's show those electrons here in green. These electrons over here move off onto the Y to give us a negative one formal charge. This is our leaving group right here. We can see the end result. The end result is to substitute our nucleophile for our Y substituent and this portion is called an acyl group. We have nucleophilic acyl substitution, where our nucleophile substitutes for the Y group. There are several aspects to this mechanism that we need to talk about. We've already talked about the reactivity of carboxylic acid derivatives in the previous video. What does the substituent do to the reactivity? How does it affect the partial positive charge on the carbonyl carbon? We saw that acyl chlorides are the most reactive. We compared the inductive effects versus resonance effects for our carboxylic acid derivatives. Other things that could affect this mechanism, one thing could be stearic entrance. Thinking about this R group on our carboxylic acid derivative. If the R group was a methyl versus a tert-butyl, the tert-butyl group would have increase stearic entrance, it's much bigger. That could prevent the nucleophile from attacking. Stearic effects are something to think about. The reactivity of your carboxylic acid derivative is something to think about. The strength of the nucleophile is another thing to think about. You want a strong nucleophile to attack your carbonyl carbon. In this video, we're going to focus mostly on the leaving group, the stability of the leaving group. You want something that's stable with a negative one formal charge on it. It's much more likely to leave if it's stable. Let's look at an example where we try to identify the leaving group, so which one is the most stable and why? Let's look at this reaction here. We have an acyl chloride. We have acetyl chloride here and then we're going to react that with sodium formate. Acetyl chloride, this is our reactive portion of the molecule, so oxygen's more electronegative, chlorine is more electronegative, so that means that this carbon right here is partially positive. That's our electrophile. Our nucleophile is sodium formate. Negative one formal charge on the oxygen, so this is going to be our nucleophile and attack our electrophile. These electrons are going to come off onto the oxygen, so let's go ahead and draw our intermediate. We would have our oxygen here, now has three lone pairs of electrons, so negative one formal charge. Let's follow those electrons in here. In magenta, these electrons move off onto our oxygen and what else do we have bonded to this carbon? We have a CH3 over here on the left. On the right we have a chlorine. Let me go ahead and put in lone pairs of electrons on the chlorine, and then now this oxygen is bonded to this carbon. Let's go ahead and finish this off here and put in some lone pairs of electrons on our oxygen. Let's be consistent. Let's show these electrons in here in blue, forming this bond between this carbon and this oxygen. We have our tetrahedral intermediate. We know the next step in the mechanism for nucleophilic acyl substitution is to reform our carbonyl and when we reform our carbonyl we can't have five bonds to carbon, so we have to lose something as a leaving group. We have several possibilities here. One possibility would be to have these electrons come off onto the chlorine to form the chloride anion as a leaving group. Let me go ahead and draw that over here as one of our possibilities. The chloride anion could leave, negative one formal charge on our chloride anion. That's one of our possibilities. Another possibility is when we reform our carbonyl, these electrons in blue come back off onto the oxygen, giving us our formate anion back, so that's another possibility. Let me go ahead and sketch that in here. Another possibility would be the formate anion as a leaving group. Let me go ahead and draw that in with our H here. Then, another possibility would be for these electrons to come off onto CH3, so we form a carb anion. Let me go ahead and draw in CH3, lone pair of electrons, negative one formal charge. Those are our three possible leaving groups. To think about which one of those is the best leaving group, a good way of doing it is thinking about the conjugate acid. The conjugate acid to the chloride anion would of course be H-cl. Let me go ahead and draw in H-cl here. I can put in lone pairs of electrons. The conjugate acid to the formate anion would be formic acid, so let me go ahead and draw in formic acid here. The conjugate acid to our carb anion would of course be methane, so we have CH4. When we think about the pKa values for these acids, the pKa of hydrochloric acid is approximately negative seven, the pKa of formic acid is approximately five, and the pKa of methane or alkanes in general is somewhere around 50. So we have these different values for our pKas. Remember what the pKa value tells you. The lower the pKa, the stronger the acid. As you go this way, you have increasing acidity. Hydrochloric acid is the most acidic out of those three acids that we just talked about and why is it the most acidic? It's the most acidic because it's most willing to donate a proton. It's most willing to donate a proton because it's conjugate base, the chloride ion is extremely stable on its own. The stability of the chloride anion means that hydrochloric acid is most likely to donate a proton. So we have our answer. We know that the chloride anion is the most stable out of these three possible leaving groups. We got that by thinking about the pKa of the conjugate acid. Going back to our mechanism, the chloride anion is going to leave. These electrons in here in green are going to come off onto the chlorine and we can go ahead and draw our final product now, so when we reform our carbonyl over here on the left we had a CH3 and then over here on the right we had an oxygen and that was - and then we had over here and then this is a hydrogen. This is our final product and then let's go ahead and draw in our chloride anion as our leaving group. Negative one formal charge on our chloride anion. Electrons in green came off onto here as our leaving group and we formed our final product, which is an acid anhydride. This is, if we were to name this acid anhydride, we would have acetic formic anhydride. We formed an acid anhydride from an acyl chloride and we thought about the stability of the leaving groups, so the chloride anion is the most stable leaving group. In general, we can look at the pKa values for the conjugate acid. We could call these pKa and then, if you wanted to, H values and that just means the pKa of the conjugate acid. Thinking about your leaving groups, if you think about the pKa of the conjugate acid, the lower the pKaH, the better the leaving group. As you go this way, you increase in terms of better leaving groups. So the chloride anion is a better leaving group than the formate anion, which is a better leaving group than our carb anion here, as well. This is just a nice way of thinking about leaving groups in a mechanisms. Think about the conjugate acid and think about the pKa.