Draw the products of the following reactions: a.

Question

b.

c.

Accepted Answer

Hey, everyone. And welcome back to another video, draw the Lewis acid bases adduct produced by each of the reactions listed below. A metox ethane reacts with boron trihydrate or Bora and B Ethan amine reacts with aluminum tri fluoride. Let's consider reaction a, whenever we have a Lewis acid bases, adduct formation, we have to consider our acid and our base because the arrow is pointing from the ether, we're going to say that the ether or metox ethane would be our Lewis base because Louis basis are lone pair donors. And that's exactly what we're seeing. We're seeing a lone pair going towards the free orbital of bo. So that is our le asset where in this case, boring as the whole structure. So if we want to perform the formation of an A duct, we're first of all going to redraw the structure of the ether, we're going to add the two lone pairs on oxygen. Now we're going to draw boring, let's try a free orbital on and let's add the arrow showing that we're using the lone pair and oxygen in the formation of the oxygen boron bond. And as a result, our final product will correspond to boring bonded to oxygen from the ether. So on one side, oxygen is bonded to the methyl group. And then on the other side, it has an ethyl group. Now, it's really important for us to add appropriate formal charges, oxygen only has one pair, one lump pair remaining. And that means because it has three bonds, it will have a formal positive charge. Now has a total of four bonds as a result, it will have a formal negative charge. Recalled that the formal charge formula states that we are subtracting the valence electrons minus the number of bonds minus the number of lone pair electrons. So for oxygen, that would be six minus three minus two, that gives us positive charge one, that would be five minus, I'm sorry, that will be three, right? Because we have three electrons for boron minus four bonds minus zero long pair electrons that gives us a formal negative charge moving on to B we have Ethan amine. Once again, we can state that the is our base because they see a lone pair and eye at the chin and aluminum tri fluoride is our Lewis acid because it's accepting the electron pair donated by the base. And once again, we're forming an adduct, meaning we are combining the two structures into one. Let's draw how that happens. So the pair on Niro is used in the formation of the nitrogen aluminum bond. Specifically, we have a free orbital with an aluminum. And now if we draw an a duct to using the same principle as before, let's draw aluminum tri fluoride. And let's indicate the new aluminum nitrogen bond nitrogen will still have its two hydrogen atoms as well as and said, since nitrogen has a total of four bonds, five minus four, that gives us a formal positive charge on nitrogen. For aluminum, we have a formal negative charge because three minus four bonds gives us a formal negative charge. Well done. So we have drawn two adducts for this problem. Adducts are basically a Lewis acid Louis base reaction products. Whenever we combine the two into one structure. Thank you for watching and I hope to see you in the next video.

Draw the products of the following reactions: a. b. c.

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Draw the products of the following reactions: a.
b.
c.