 Let's summarize the reactions of ethanoic acid. So, ethanoic acid is the one which has carboxylic acid group. It's ethanoic acid because there are two carbon atoms. In the first reaction, we'll make it react with an alcohol, methanol, in the presence of an acid. What's going to happen? Well, the acid will be useful in removing the H from here, OH from here, to give you H2O. And this will be the theme for all the reactions, as you will see. And then what's left now would be CH3COO, because the H is removed. But to it, the CH3 group gets attached. And this compound, or sorry, this molecule is now called an ester. An ester is any molecule in which you have a COO group, and that is on both either sides you have some carbon chains. Another, you know, on both sides, you can have long carbon chains if you want. An ester is a sweet-smelling compound, and therefore it is useful in your perfumes, for example. And this reaction is therefore called esterification reaction, because it helps you in getting esters. Okay, what do you want to stop there? Now, for this particular ester, we're going to make it react with NaOH. What's going to happen? Well, NaA being very reactive, it's going to kick this CH3 out. And what we now get is CH3COO Na. And then what else we'll get? This CH3 will combine with OH to give you back your initial alcohol, so you get your alcohol back. And this now, this particular molecule, it's the sodium salt of your carboxylic acid. It has lost its hedge, and it has now become a salt, and this is useful in creating soaps, and therefore this reaction is called the saponification reaction. All right, in the next one, we're going to take ethanoic acid and directly react with NaOH. What do you think is going to happen? Can you pause and think of what's going to happen? Very similar to what we just saw. All right, again, Na being very reactive, it's going to go and kick this hydrogen out. So what do you get? You get CH3COO Na. Again, you get the sodium salt of carboxylic acid. And if this hedge is gone from here, it can now combine with OH to give you water. Very similar results, isn't it? All right, now let's see what happens when you have ethanoic acid reacting with sodium carbonate. Again, why don't you pause and think about what's going to happen? Very, very similar result. All right, so again, the sodium is going to go kick this hydrogen out. So we'll get the same thing again, CH3COO Na. But this time, you have a carbonate over here. So you have a CO3 over here, and you have this H over here. So what compounds can be formed? Well, clearly hydrogen from the hydrogen and the oxygen over here, you can get an H2O. But there's a carbon. What is a common molecule that you would get from carbon? You'll get carbon dioxide. And carbon dioxide will be released. Now, of course, this equation is not very balanced. The main reason it's not balanced is because there are two of Na over here, and there's only one Na over here. So I can just, like, multiply this by two. Now if I multiply this by two, I need to multiply this by two. And I think everything is now balanced. You can check that. But we can now go to the last reaction. What happens when you have ethanolic acid reacting with sodium bicarbonate? Again, very similar story. Pause and try. All right, so Na is going to kick this hydrogen out. So you get CH3COO Na. And what do we get? Well, again, we have a H that comes out. We have a H here. You can definitely get a H2O. And again, since there's a C over there, I can guess there's going to be carbon dioxide released as well. This time, the equation is actually balanced. The H H and the O from here gives you a H2O, and the CO2 remains, and everything looks nice and balanced. And so the common theme over here is that this hydrogen, the poor hydrogen from the CSTCOH, from the ethanolic acid keeps getting kicked out, and that helps us give some very interesting results.