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QX-a: Ustmg thc Tigure; evaluate tke folloning pradlucts in D)an group (draw all Ibe steps} (2 poinbv)0.6C,0'c; ~G67,0Cz...

Question

QX-a: Ustmg thc Tigure; evaluate tke folloning pradlucts in D)an group (draw all Ibe steps} (2 poinbv)0.6C,0'c; ~G67,0Cz

QX-a: Ustmg thc Tigure; evaluate tke folloning pradlucts in D)an group (draw all Ibe steps} (2 poinbv) 0.6 C,0' c; ~G 67,0 Cz



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\text { Whirc phosphorous on reacrion wirh Ca(OF l) gives } (a) $\mathrm{Ca}_{3}\left(\mathrm{PO}_{\mathrm{i}}\right)_{2}$ (b) $\mathrm{PH}_{\mathrm{z}}$ (c) $\mathrm{Ca}\left(\mathrm{ll}_{2} \mathrm{PO}_{2}\right)_{2}$ (d) $\mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{1}\right)_{2}$

Kill everyone. Today we're doing Chapter eleven Problem ten. And this phone asked us to draw the products form from this out kind when we treat it. Sorry, when we treat this out kind with two currents of di, atomic bomb e and or one equivalent a diatonic chlorine. So let's go through the mechanism to understand why we get the products we get. So as you guys remember, you're al kind is a left turn rich. Each of these pie bonds general is formed by two electrons. So this very electron rich Morley here and this can actually act as your nuclear fall in chemical reactions. So if we have diatonic Brahmins and there's going to be a nickel father, we know that one of these bombings is going to be our electro files. So I'll drop my arrow from the source of electrons to form these Bowman Adams making a bond when you make a bond need to break a bond. So I'm going to break one of these the signal bond here and now we form our intermediate. But if you remember, you specify which Adam you're going through. So I'm gonna go through this carbon atom first. So this signal bond is going to be generated on this carbon atom. And we're losing one of these pie bonds because we used it in this formation of the signal bombs and are left with Bella Bond. San means adjusting to it. We have our Carvel Catalan Intermediate. What else do we have? What we have are wrong in on ion which now Khun actors are nuclear file. So this being very left, Sean, rituals extra long pre elections can attack our car. Okada on intermediates being election poor carbon accidents are electoral fall. And now we have Miss Germinal General meaning to Rome means on adjacent carbons Intermediate. But now we have another addition. Another equivalence sorry of this Titanic roaming So the same thing is gonna happen. So you obviously would one former clinic clubs rule. But in this case, there's no Prothom. So my problem of cardinal rule doesn't really apply, so it doesn't really matter which carbon we go through there both equivalent carbons in terms of stability. So we'll just ever truly go through one of them picking up this pro mean making a bonding and make a bond break a bond. Now I have one carbon with two row means the other adjacent carbon as one roaming on the carpet Cata Intermediate. Then we have our BR minus connect which connect us. I'm new profile. Why? Because they picked up those two lecture electrons from the Sigma Bond. So it's going after the nickel foul and attack our electoral Filic carpal cattle and intermediate to generate our Tetris substituted product. Now we have our Tetris substituted. Hey lied Uncle haloed on This happened this form when we have two equivalence of this br too. But now what happens if you only have one equivalent of our diatonic hairline? Well, once again we had identified my cabinet colleagues will, which side has more protons? But this case each side has zero proton So these air equivalents carbon So we arbitrarily say that the nuclear Filic pie bond electrons will go through one of these carbon atoms attacking our electoral Filic Chlorine making important you make a bond you to break a bond on to the signal bomb. Teens to chlorine atoms will break generating our intermediate. Now we have I'm clearing out of here Double bond crap! A cat, an intermediate here and we also have our chloride ion has generated. So now using this chloride ion, we need two nuclear physically attack the carb Okada and intermediate, which is our electro file. But now we have two options. We can hit from the same side or from opposite sides. Well, to minimize hysteric bulk this they're repulsion between these two highly electoral negative Hey light. It's best toe adamant on the furthest distance away So furthest distance away is if they're going to be trans from each other Nazis or transmitting opposite sides It's just meaning same size. So if they're trans, they're gonna have the furthest distance apart from each other compared to if they persist, they will be right next to each other. So this cleaners, you know, Trans I'm a carbon Canyon intermediate form now trans products here And to make it very obvious is trans You going actually redraw these bonds just to make it very obvious that this is Trans John like this. So this generates our trans products. We maximize difference the distance between these two metre negative chloride Adams. So we minimize the repulsion, so increases stability of this product. So this will be your favor full product. So we have to equivalent of a diatonic. Hey, Light added, Added to an al kind. We get Tetra edition when we have one equivalent of a diatonic. Hail, I'd adding to an al kind we get Trans edition.

So for this question here we had a product from a process, and we basically have to show the residents. So here, by ch two to have C six h five here and I ring and then here on this one, we have our seats too. We had to put a positive church there to get it to that ch two. Um, okay, So for this problem, we basically are gonna move this positive church here around the circle and make on the structure stuff. Um, so we always we don't start with the positive church we are. We're always gonna use or start at the pie pond at the positive here at the double bind. So we take this, we move it like that Positive charge. So basically, that's going to create this structure here is that these will be the same. This will be the same. This is where it changes. So this has then moved up to here because basically, we're rotating it around this point here. And basically what that does is it creates positive charge. We picked up some dirt from here, put it in a hole over here, but now we have a whole over here. That's what my old professor uses. So then we could keep moving this around so we could do the same again. So taking this double bond here and moving over to here. So this is what we call an Olympic. Can I on? Because a little it means that it ah, positive charge account I am that is attached to a carbon that is double bonded. So here we have another look, can I? And that's one of the things that you can see if you can do residents. So the next one that we would see here is something like So this is where it would change. This would still be the same in this pond here, and this would still be the same because nothing changed with these. This is the only part that change. So now we have our positive charge for here because we picked up the dirt from here if you want like that and we moved it over to here. But now that this is filled up, we have a That's where a positive charges. Then we take our pi bon here, do the same thing again so that we have Sorry, for my messy Well, heck scenes. Um Okay, so this is the same. This is the same. This is where we changed. So instead of this being here, we moved in here. That does is we picked up the ached extra or the electrons here. We moved in here, but we still have a lack of electrons. If there's any time in your structure where you create a an extra charge. So maybe if we think that we did something and we created a negative charge here, this wouldn't work out because we would have now a neutral Adam When? At the very beginning. And throughout all of these structures, we've always had just one positive charge. Technically, yes. I guess you could have something that has, like, a positive and the negative and another positive that would create an overall positive. These how in the end. But, um, usually that's a bad time that you created an unstable structure. So what I do is I just try and keep it normal. Keep it as normal as possible. Um, the most amount of charges that I think that I've ever had was positive and negative going from a neutral Adam to deposit minutes. Still overall neutral. But I don't think I've ever used a structure where I have three or more charges lingering throughout my resident structures. Um, so then Ah, yeah. I think that that's our last structure. Because if we were to and move it back, we would have this back. This so positive, hard you'd be out here now is it? Moved the electrons from here to here. So this would be where trying to be allowed to be here. I like to be here. Technically, yes, it is different from this one because they've all shifted over one place here. So let's see how it goes here and here and here. And how appears in a different spot? Um, that's usually that's a given. Um, some people even draw their things like this with the circle inside of them, just to show that they'll always be rotating. So some professors won't make you through all of this one some. Well, so, um, it's all up to your professors preference. Really? And how you want to do it, how they want you to do it. And if they're gonna greet you on certain things,

So now we'LL work on Problems sixty eight from Chapter twenty one. So here we're given names of benzene compounds, benzene containing compounds and were asked to draw the structure. So that's going to start with a. We were given Ethel Benzie as the name of the structure. And so we know that benzene is, of course, an aromatic six member dring and an F A group. We know it's ch to ch three, so that's simply all we have to do for that. You get the answer there so we could move to be where it says one. I ah, no. Two methyl benzene. Yeah, so we can draw this over here. We can go and draw in our benzene ring here and now we can choose any position to be one. Usually we can take the top here for convenience sake. So that's one I auto, too, to represent the iodine here. And then we go down to the second Carbon and put a method group, and that is C H three. And that's our structure there. So if we go to part C were given a stretcher, Pera di Ethyl Benzine one. So let's write this name in here so we can dry an hour. Benzene bring here. And then we have to ethel groups which are parent to each other. Which means that their three three carbon separated so they could be at the one and four positions. Aaron means that they're across from each other in a benzene ring.

Let's try naming more benzene compounds. So the first one, we have unethical benzene. You can put the Ethel Group anywhere, but for the sake of space, I'm going to put it on that carbon. And that's our Ethel bending one iota. Two metal bending ever bending. You can pick any carbon you want as carbon one, but make sure that you have your method group wow on carbon to relative to carbon one. I could have put the metal group right here, and it would have still been perfectly fine. It would just be numbered the opposite direction as it is right now. And that is totally fine, because it's a cyclical molecule para di. A full benzene para means 14 which means no matter what, I pick us my one I'm going to pick this one for space. And this one because one and four, it needs to be on the opposite side. So di ethyl ch two ch. Three. Same thing here. Yeah,


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