What products are formed when each of the following reacts with H...

Question

What products are formed when each of the following reacts with HO− ?

e. +CH3

f. FeBr3

Accepted Answer

Welcome back, everyone determine the products that are formed in a reaction between each of the species given below with metox side and ion number one methyl carrion, number two aluminum bro bromide. So for the first reaction, we first will have our methyl ion, we can draw our central carbon atom bonded to three hydrogen atoms. And we understand that there is a positive charge on carbon because it is electron deficient. It doesn't have an o and essentially the methyl ion undergoes a reaction with the metox side and ion. So we have a negative charge on oxygen. We can immediately state that according to the Lewis acid based theory, the metox M I N will be our base due to the fact that oxygen is electron rich, there's a negative charge which indicates that oxygen has three lone pairs present. And the methyl will be our Lewis acid because it's electron deficient and it can accept electrons. So that carbon gets an Octa, meaning we can simply show how we are forming our abduct. So the arrow shows the abduct formation, simply speaking, one of the lone pairs on oxygen will be used in the formation of our carbon oxygen bond and combining the two ions into one, we get the following. First of all, we have our C H three fragments. We are going to write down H three C because we are primarily interested in showing the bond formed and we are forming it between carbon and oxygen. So now oxy is bonded to carbon and it also has a methyl group bonded set, there is no charge on any of the atoms because carbon now has four bonds, oxygen has two bonds and two lone pairs. So there are no formal charges that we want to include. And we simply end up with a structure of metox methane. Even if we don't show that bond, we can just write down that we have H three C O C H three. But for the purposes of this problem, it would be better to show the bond that we are forming. So we can clearly see what the answer is and how we obtained it. For the second problem. Our starting material is aluminum bromide. So we have aluminum bonded to three bromine atoms. And because aluminum only has six von selections around it, we know that it can also accept a pair of electrons into its free orbital. So once again, we're treating aluminum bromite, in this case with metox side. And because aluminum has a pre orbital, it can act as a Lewis acid, right? Because Lewis acids are lone pair acceptor and aluminum has a pre orbital in which it can accept that long pair. And our loose base similarly to the previous problem will be metox because it's an electron pair donor. So we can easily show how we are forming another a duct. So we have a Lewis acid based reaction once again. But in this case, if we draw our resultant aluminum oxygen bond, we will notice that oxygen will have two lone pairs and two bonds. So there will be no formal charge on oxygen. But aluminum which has three bromine atoms bonded to it will now have a formal negative charge because it has four bonds. So we can simply check that the formal charge of aluminum will be equal to its number of ions electrons, which is three minus four bonds minus zero alone pair electrons. So that indeed gives us a negative one formal charge on aluminum. And of course, the charge is conserved. So on the left hand side, the net charge is negative one. Meaning on the right side, we must also end up with a net negative charge of one that will be it. We got our abducts and we are ready to conclude the video. Thank you for watching and I hope to see you in the next video.

What products are formed when each of the following reacts with HO− ? e. +CH3 f. FeBr3

Watch next