In the following acid–base reactions,
1. draw Lewis structures of...

Question

In the following acid–base reactions, 1. draw Lewis structures of the reactants and the products. 2. determine which species are acting as electrophiles (acids) and which are acting as nucleophiles (bases). 3. use the curved-arrow formalism to show the movement of electron pairs in these reactions, as well as the imaginary movement in the resonance hybrids of the products. 4. indicate which reactions are best termed Brønsted–Lowry acid–base reactions. b. acetaldehyde
+ [CH3O]- —> [CH3C(O)H(OCH3)]-

+ [CH3O]- —> [CH3C(O)H(OCH3)]-

Accepted Answer

Hey friends. Let's solve this problem. And says consider the following acid base reaction. We have NH three plus B F three yields an H three, B f three for all the reactant and products draw lewis structures, identify the nuclear file based in electro file acid in the reaction, draw curved arrows to show the flow of electrons and determine if the reaction can be termed a bronze did Lowry acid base reaction. So it looks like a lot of steps. But let's just take it one step at a time. So first we are drawing the Lewis structures for these reactant and products. And to do that first we will total the number of valence electrons for each atom in the compound. And we'll draw just a sigma bond framework for the compound. Then we will use any remaining valence electrons to complete the octet. And we'll do that by adding those electrons as lone pairs. And lastly we'll calculate formal charge which is the valence electrons for that adam minus the touching electrons and the touching electrons are just half of the bonding electrons for that atom and the non bonding electrons for that atom total together. Alright, so let's do this. So we have NH three and that has eight valence electrons. Right? Five for nitrogen, three for the three hydrogen will put nitrogen in the center because it has the highest bonding preference or three hydrogen. And that takes 246 electrons. So we'll add the remaining two as a lone pair. And that nitrogen has 12345 electrons touching it nitrogen likes to have five electrons. So we are good. There are no formal charge BF three. We have three electrons from Boron And seven from each flooring. So that is going to give us 24 valence electrons. So it's pretty much the same structure except foreign atoms have all these lone pairs. So 246, 8, 10, 12, 14, 16, 18, 22, 24. So that is all of our electrons. So we can't complete bar on's octet but it is what it is. So let's keep going. So we have NH three plus and B f three minus. So when we're totaling these electrons we're going to subtract one for the plus and add one for the minus. Right, add one electron subtract one electron. So we have seven here. Right, Because five for nitrogen three for the hydrogen. Right? So it's basically this value minus one. This value plus one, so eight minus one is 7, 24 plus one is 25. That gives us 32 valence electrons. So we're basically going to draw these two but connected together. So NH three B F three and we'll add the lone pairs to the flooring and that will use all of our electrons, oops, and we see that we have the plus and the minus and the pluses written next to the n H three. The BF three has the minus. So I'm pretty sure those are going to be formal charges on our central atoms. That's usually the case. So let's calculate that. So nitrogen likes to have five valence electrons. It has +1234 touching it, yep. So that gives us our plus one formal charge in boron likes to have three but it has 1234 touching it. So that is negative one. So our negative charge is in fact on our boron. Okay, so those are all the lewis structures. So now going on to step two, identify the nuclear file and electric file. So what are nuclear files And electro files? Well recall that nuclear files are either negative or they have a lone pair. They could donate. I'm just going to write a negative sign. But it doesn't have to have a negative charge. And it wants a positive charge or an electron deficient atom. And electro files are the opposite. They are electron deficient or positive charge. And they want a more electron rich substance. Right. So looking at our reactant is here. Which one do you think is a nuclear file? And which one is the electro file? Well, NH three has that lone pair. Right, so that's going to be our nuclear file because it can donate that lone pair to a bond. And B. F three is going to be our electro file because it has an incomplete octet. So that's going to make it more electron deficient. It wants some more electrons there. All right. And also the boron is attached to these very electron electro negative flooring atoms. So, the electron density is lingering outside around those flooring atoms making boron again more electron deficient. So let's draw our reaction so we can draw the curved arrows. So, I'm quickly going to redraw our lewis structures. So we have NH three with the lone pair and boron with the flooring anthems. and that gives us an H. three bonded to B. F. Three. And again, we have our plus formal charge on negative formal charge. Our loan pairs. Alright, so drawing are curved arrows. Where do we start from? I know I'm forming a bond between the nitrogen and the boron. But which direction is my arrow going? Well, we always want the arrow to go from the negative towards the positive. So starting at the nuclear file going towards the electro file. So our nuclear file is NH three. So, we're going to start at that lone pair and draw the curved arrow towards the boron. Okay, last step, is this a bronze ted Lowry acid based reaction? Well, what does that mean in a bronze did larry acid based reaction? We are looking at a proton transfer happening. Are we transferring a proton in this reaction? No. So let me move this down. So this is not a Bronston. Lowry acid based reaction. Brown celery. Which makes choice B The correct answer for this problem. That's it for this one. I hope this video is helpful. And thank you for watching.

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