In this screencast we're going to talk about drawing curved arrows to track the flow of electrons in a chemical reaction. So with the red pen, I'm going to draw arrows that show how the electrons move in acid base reactions, which we've seen, and we're also going to see some other types of reactions in this problem. Chapter 3 issue, 35. Our first reaction we're taking formic acid in the presence of sodium, hydroxide we're generating format, anion and water. So as a review let's, go ahead and classify.
The acid and base on each side of the reaction, so we're starting with formic acid. So this is the acid and then hydroxide is the base on the reactant side. Let's, go ahead on the product side and identify the conjugate acid and conjugate base. So the conjugate base on the product side is going to be related to the acid on the reactant side base. So remember that a base is the ability to abstract a hydrogen.
And then this water is going to be the conjugate acid. So we've identified the relationship. Between the two speak between the reactants and the products so let's, look at what's happening here. We are transferring this hydrogen to water in the product.
So how does that occur? A lone pair of electrons from the negatively charged? Oxygen will abstract that proton. Notice how the arrow starts at the lone electron pair and the arrowhead points towards the proton because we're making this oxygen hydrogen bond and hydrogen can only have one bond. We have to break the existing bond to that hydrogen. So the electrons in the oxygen hydrogen bond here will end up on this oxygen.
So we've shown the proton transfer also notice how the charge transfers as well in the reactants it's in the hydroxide. And now in the products it's on the format anion in part b of this problem, we're taking an ester functional group. So we have methyl format in the presence of hydroxide anion. And now we're ending up with this species here, let's, go ahead and identify where the hydroxide is ending up.
So we see hydroxide. Here now it ends up here in the product. So what it is doing is adding to the sp2 hybridized carbon in the reactant. So this is not necessarily an acid base reaction, we're going to introduce new concepts to classify what's happening here. We can also call a base. A nucleophile. A nucleophile is a species that is electron rich nucleophiles.
React with electrifies. An electrified is a specie that is electron deficient. So we know that this oxygen which I'll color in green here ends up being this oxygen. In the product.
So we can see that in the product we're forming an oxygen carbon bond. So the way the arrows will flow is that a lone pair of electrons from oxygen will attack that carbon. We see that in the product that's the bond we have to form. We also see that we have to break this carbon oxygen double bond because the negative charge is transferred from this oxygen to the oxygen in the starting material, which ends up being this oxygen. So it's, conceptually similar to an acid base reaction or. Although we're calling the terms electrified and nucleophile.
So in the next reaction, if we look what's happening, we have the product from this step here is now the starting material in part c. Notice we're only starting with one species that species breaks down to form formic acid and then meth oxide. So this negative charge is transferred from this oxygen onto this oxygen. So we can see that this och3 group ends up as a separate product. So how does that happen? This species will reform the. Carbon oxygen double bond. So anytime we form a bond.
We have to break a bond. So the electrons in that co double bond then end up on this oxygen. Finally, in our last example, in part, d, we have hydroxide anion, plus an alkyl halite. This is called methyl iodide. That is creating methanol plus iodide anion.
If you note what's happening, we're, substituting o h for I concept that we did for part b, where we have a nucleophile and an electrified. So the nucleophile is going to attack this carbon we're. Forming an oxygen carbon bond, therefore, we have to break a carbon iodine bond.
And we can rationalize that happening based on electronegativity. So oxygen is bearing a negative charge. We know from the difference in electronegativity between carbon and iodine that the carbon bears a partial positive, and the iodine bears a partial negative. So in the product, the iodine is bearing the formal negative charge so that's a screencast of acid base and electrified nucleophile reactions showing the curved. Arrow formalism.