Would the following reactions go to the right or the left? Dra...

Question

Would the following reactions go to the right or the left? Draw the products and label ALL species. Provide the full mechanism.

Chemical reaction showing reactants and arrow indicating direction for acidity ranking.

Accepted Answer

Let's go ahead and solve this first one. So what's the first rule that I want to follow when I'm predicting equilibrium? Well, what I want to do is I want to identify the base and the acid. And remember that there are some times that it's difficult to determine which one is the acid or the base, but in this case, it's a really easy question because I have a formal charge. And remember that formal charges are always really good indicators of acids and bases. Remember that we said if it's a positive, then that's an acid and if it's a negative, then that's a base. Okay? So in this case, I actually see a positive and a negative, but only one of them is important. Why? Because the positive is on a k. Remember that a k is a spectator ion. So I'm just going to cross that out. I don't care about it. What I do care about is the fact that I have a negative on an o, which means that this species or this molecule is my base. And if that's my base, that means that the other side must be or the other molecule must be my acid. Okay? So it was that easy. Sometimes it's going to be more difficult to predict the acid and the base, but in this case since we had a full formal charge, it was easy. Now the next thing that we want to do is predict the conjugates. So sometimes the conjugates are going to be drawn for you. Sometimes you have draw them yourself. If you have to draw them yourself, like in this case, it's not that hard. Just think about what you need to do. For example, let's start with the acid. Remember that the acid is a proton donor. So what it should do is it should make a base, a conjugate base, that just is missing a proton pair, but it's going to has that lone pair, but it's going to have an additional lone pair from the one that it's going to receive after giving away the hydrogen. So that's going to give me a negative charge. Okay? Then what's the other conjugate? Well, the other conjugate should be this o negative that's going to gain an h, so I should have plus a molecule that looks like this, o h, where one of the lone pairs is donated, so this should be a neutral compound. So now I should have an alcohol on the other side. So this should be my conjugate base and this should be my conjugate acid. Cool so far? Now when we were doing these questions before and they were already drawn out, I always start off with the mechanism. Okay? But it turns out for these questions, I actually don't want to start off with the mechanism. The reason is because I don't know what direction the equilibrium is going in yet. And remember that when I'm drawing a mechanism, I want to make sure that the mechanism and the products are on different sides. So if the products, let's say, if the major products are on this side, let's say this is the product side here, that means that this is the favored side, then I would want to make sure that my mechanism is here. Okay? Since I don't know the equilibrium yet, I shouldn't draw the mechanism until I know that. Okay? So we're going to hold off on drawing the mechanism and we're going to do our last step, which is determining the equilibrium. So how do I do that? Do you guys remember? I would compare the pKa's of the acids. Okay. And that's if I know the pKa's. So in this case, do I know these pKa's? Yes, I do. This acid on the left hand side would be 38. This would be an amine. And then my conjugate acid has a pKa of 16. Okay? So what I always like to say is that you want to make sure your number is going from lower to bigger. Okay. Because you want to make sure you're going from a stronger acid to a weaker acid. So am I going from lower to bigger? No, I'm not. I'm going to scratch out this arrow and I'm actually going to draw equilibrium arrows. The way we're going to draw them is like this. It's going to be a big equilibrium arrow going towards the left and then a small one going towards the right. You know what that means? That means that even though these two sets of products are in equilibrium with each other, there's much more on the left hand side. Much more of it is favored to look like what's on the left side because that's the weaker base and the weaker acid. Okay? So just by drawing those equilibrium arrows like that, I've already answered the question of to the right or to the left. So now all we have to do is provide the full mechanism. So now that I know that my left is the one that is favored. Alright, so this would be my products area. This would be the favored area because this is the one that's actually making the thing that is going to exist most of the time. Okay. That means that like I was saying before, I want my mechanism to be on the other side, so my mechanism should actually be over here because what I want to show is that I'm making a reaction that goes in this direction. Okay? So if I had to draw my mechanism on the right hand side like I have right on top of me, where would I start that mechanism? And the answer is I would start always from the same place, from the negative charge to the acidic h. So I would do this. Okay. So I have from negative to H. Now I have grabbed an H. Is that it? No, that's not it because of the fact that h only wants to have one bond, not 2. So even though there isn't a bond drawn there, we know that there actually is a bond between the o and the h. Right? This is just a condensed structure. So I'm going to take the 2 electrons from there and jump them onto the o. That's what's going to give me my o negative on the other side and my NH on the other side. Does that make sense? So if you guys said that the answer was to the left, you got this one right. Alright? So let's go ahead and move on to the next one.

Organic Chemistry 때문에 힘들어하고 계신가요?

Would the following reactions go to the right or the left? Draw the products and label ALL species. Provide the full mechanism.

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