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Cuttoanited(181 YoulubeKINLTICS ANocquilirium Identllying Intcrmcdlates In rcnctlon mechanismConsider the following mechanismn for the formation of carbon tetrachlo...

Question

Cuttoanited(181 YoulubeKINLTICS ANocquilirium Identllying Intcrmcdlates In rcnctlon mechanismConsider the following mechanismn for the formation of carbon tetrachloride:Cl (o)2Cl(g)Cl(g) CHCI,(g)HCI(a) CCI,(9) CCl4(9)CCI;(9) Cl(a)Wite the chemical cquatlon vunnacion0-0Dp 0PNe thete any Intetmnedlale: In this mechanism?DDIf there are intermediates Wntd down (nur chemicz onnuln?commtrocveen each cncinicuntonmulz Lhefe'5 mone thantone

cuttoanited (181 Youlube KINLTICS ANocquilirium Identllying Intcrmcdlates In rcnctlon mechanism Consider the following mechanismn for the formation of carbon tetrachloride: Cl (o) 2Cl(g) Cl(g) CHCI,(g) HCI(a) CCI,(9) CCl4(9) CCI;(9) Cl(a) Wite the chemical cquatlon vunnacion 0-0 Dp 0P Ne thete any Intetmnedlale: In this mechanism? DD If there are intermediates Wntd down (nur chemicz onnuln? commtrocveen each cncinicuntonmulz Lhefe'5 mone thantone



Answers

When 1,3-butadiene $(CH2 =CH-CH= CH_2)$ is treated with HBr, two constitutional isomers are formed,
$CH_3CHBrCH = CH_2$ and $BrCH_2CH = CHCH_3.$ Draw a stepwise mechanism that accounts for the formation of both products.

Hey, guys. So in this question were given to reactions were in part they were given the reaction between benzene and Turk. Beautiful. Ah, tribunal cat eye on And it's a two step electro filic reaction. You ting a substitution mechanism and in part B were given the electro for the condition of hydrochloric acid and Styer ing, which is shown. So I've written out the basic, um, reactions for both of them. And I'll go through the mechanisms and draw in curved arrows where appropriate, to show the movement of electrons to how we're getting each of these products. So in part A, we have our benzene molecule and our Turk beautiful. Ah, cat ion. So the first thing I'm going to do is I like to identify me nuclear files and my electro files. So starting off with our electrify, we know that the Turk beautiful cat ion is an electro vile since it has a positive charge on this carbon right here. And this benzene is our nuclear file since it is rich in electrons because of our double bonds. So now that I have that, I can go ahead and add in my curved arrows to denote the movement of electrons. So whenever we have these types of reactions, we know that the, um, nuclear file is going to be attacking the electric file. So to do that, I'm gonna take one of my bonds, my double bonds, and I'm going to do just that. I'm going to attack the electric file so the double bond is going to be breaking apart, and its electrons are going to be forming a bond with this Turk. Beautiful. So in our next step, we have that here. So we have the formation of a bond between our benzene ring and the Turk Beetle Cat Island. And now that I have this, it forms another car book. Had I on, um, Intermediate because of the breaking of our double bond And just to, um, the ah replaced this double bond and fill in the ah octet of this carbon, I'm going to do a, um, breaking of another bond. So I'm going to take this hydrogen and use its electrons to form another bond. So when I do that, I form my benzene ring again and that we have a Turk Beautiful cat island substitution. So for a next part. We have this reaction again. We have Styer ing and it's forming products from the reaction with hydrogen chloride. So hydrogen in this case is going to be our Electra file. And our nuclear file is this Styer e And I went ahead and broke apart of the hydrogen chloride into the constituent ions that making up because that's how um, it would do it in the actual reaction. So the first thing I'm gonna do is I'm gonna take the electrons from our double bond and they're going to attack this hydrogen here. And according to this rule in organic chemistry called Khar Market Nichols rule, this hydrogen is going to attach to the carbon with the most amount of hydrants. So that's going to be this carbon right here. So that's why this carbon now has three hydrogen ins. And this, um, carbon right now has a positive charge because of the loss of electrons from that double bond. Now that we have that, I can go ahead and have my chlorine I on attach to this positive charge. Teoh, stabilize it. And that yields this product right here where this chlorine is now bonded to this carbon, which was arc horrible cat on before. And our, um ah, hydrogen is bonded to this carbon from the original re agent. So those are our two reactions for this?

Hello Today we're doing problem nine point forty six and this problem asks us to Java organic products formed when pro panel is treated with each of the following regent. So we'LL start from and work our way to eye. So hey, we have propane all reacting with a strong acid. The strong acid in this case is so fury acid. So as you guys know, I lie just to rewrite my strong acid is h A and you know, strong asset fully associates in solution to H plus and a minus. So when we see and alcohol with a strong acid, we know that a dehydration reaction will occur. So as before, as every single problem you'd identify, What is your new cool file? What is your electrified nuclear file being the source of electron? Some has very electron rich. The only thing that's electron rich here is your oxygen alcohol. So you know that's going nuclear. Quickly attack your hydrogen to pick up a proton and get pro nated into oh, age two plus. Now we know that always two plus is not favorable. Option is not like part of charge. It likes to be neutral or negative so this will want to fall off. But in order to fall off, we will use our A minus which is free and solution to Hera. Allow out dehydration reaction to occur to rember dehydration. You go from S P three carbons Espy too. So you lose a degree of saturation So essentially from Al cane, you goto al Keen. So the way you do that is by picking up one of these beta proton. So beta proton, meaning something that is actually adjacent to your leaving group. And as she saw, it's not these ones that debate a proton, the's air, the beta protons here. So if you see here, this alcohol group was attached to this which is called your Alfa Proton and right adjacent to that is this pro this carbon here, which is your beta proton. So your conjure get acid will pick up the protons from that beta carbon. So if I were to just I raised this and show that are conjugated base, I'm confident. I said sorry will pick up the proton making a bond. When you make a bond, you need to break a bond. So Sigma bond between the hydrogen and carbon will break forming our Al Kane, making you bond when you make a Bonnie's Break a bond. So that's Sigma Bond attached to our water will break to give out dehydration, product and free water plus age two. Oh, plus, we reform our strong acid. And because we reform our acid, we know that this is an asset catalyzed reaction because we reform the catalyst. So next for letter B, we see that we have the propane on reacted with sodium hydride, sodium hydrate associates, toe Saudi and plus an H minus story in plus being unexpected ion h minus. Being the hydride, you can act you, Khun. Imagine that H minus is very reactive and would want to pick up another hydrogen to become a TSH to gas. So the only hydrogen that is a label label meaning that is it's susceptible to nuclear Filic attack or to be ripped off is this hydrogen attached to her alcohol? So we make a bond between the two hydrants forming our age to gas and therefore the bond between Owen Rachel Break, giving those two electrons of that signal bon onto options nucleus to make a formal negatively charged. Oh, minus plus H two gas. And now, if you were to continue this reaction, that's all mine. And after the strong nuclear fall to attack any electro file that we add into the mix Sure. So below here we see that we have a primary alcohol with a strong acid. I would associate into h plus and C l minus so as similar to the first one, we know that we're going to protein eight are alcohol because it was nuclear Filic properties. And now because we know we have a primary I'LL call originally, we know that this is going to occur through s into mechanism in which our nuclear file will backside attack our electro hour No electoral Filic carbon making a new bond. And when you make a Bonnie to break upon So now that we broke that bond, we can draw our substitution product plus our age too. Next we have another strong acid. So this exactly going to be the same thing as what we did before HBR Associates in solution to H plus NPR minus oxygen having nuclear Filic properties will attack this proton to put Nate itself Oren always two plus and because the primary I'LL call We know that our BR minus with its nuclear Philip tendencies because of that formal negative charge giving it extra electrons making an electron rich we're gonna back side one eighty Attack that electoral Filic that electron poor carbon and former s into substitution product. When you make a bond, you need to break a bond and reform water as well. Next, it's another way to do a substitution. Been out with Sokal. So I draw out the structure of soccer here we see that we need to first identify a nickel fall. The nuclear fall, as in all these cases, is our oxygen and electro Felix center. Here is the sulfur Adam here. This is the very election poor electron deficit center here because of those electro negative chlorine ions pulling electron density away from the salt sulphur nucleus as well as this carbon. So we'LL attack our electoral filic sulfur making a bond we make about me to break upon the label group Is this carbon old that'LL pop up make giving us formal negative oxygen? An extra loan para lecture on those lone pairs will collapse back down to reformer Carbonell making bond when you make upon you break upon. So this chlorine being a good leaving group because it's a hail ID will pop off plus C l minus. But now we have positively charged oxygen, which once again is not favorable. So we will use period in in the mixture which it with its nuclear filic nitrogen to act as a proton mop. So this mops up any protons to neutralize all your molecules in your reaction vessel and really promotes reactions to go forward to completion. So now you have You're leaving Group, which was originally hi jocks all group that was converted into a very good, very strong leaving good. So now our nuclear file being the seal minus can back side attack that on pop off our leaving group. Next we have PBR three, so same thing occurs substitution reaction. But instead of chlorine, we're talking bro Ming. So we get our brominated product finally to start off with question with letter with letter G, we have our hydroxide with a tosel chlorides or Tosel Cloyd's once again make your leaving group into a poor living group of hydroxide which is very poor living group into something that is a very strong leading group. So if I were to draw the structure but my toss of chloride, we have this just like Sokal Sulfur is the electro file here. It's very electron poor while our nickel follows of election rich oxygen. So we know that we're in a nuclear physically attack that and eventually pop off our very good leading group, which is the chlorine. Now we have our tosel eight, but oxygen is still prone ated. So we use a proton mop with its nuclear filic nitrogen with those long pair electrons to pick up that extra proton to neutralize our product. So now we have something that is a very strong leading group which in future reactions, if you were to react this with a nuclear file, we would get Essen to reaction mechanism. Something to note up until this point, you're sterile. Chemistry is retained. There is no inversion stare chemistry yet. However, if you continue to reacts with your nuclear file, you will get an s into reaction, causing your inversion on stage chemistry. So if we want to question G, you have your sodium hydride that deep throat Nate, your oxygen just like we did before number and letter B. We had our own minus. And like I said, this can act as a very strong nickel foul to react with any electrifies you add in the mixture. And this is our left far. We added here and you see here, bro Ming being the very strong leaving group it is. We attacked the carbon that is bounded to the bro mean making bond When you make upon, you know, break upon So that bro Ming will break off giving us our either and bromide and ion the last question here we see that we have our alcohol reacted with tosel chloride. So just like before, we made hydroxide which was very bad, leaving group into something that is a very strong leaving group and oh tasa wait. Now we reacting it with a nuclear fall We've in sodium that forms a respected ion this assault. So we have any plus and hs minus sulfur being our nuclear file because of its election rich nature. So we can get an s into reaction mechanism from sulphur going to the carbon, making a bond when you make a pawn in its break up on. So now we have our Sile group here, plus our sodium spectator ion plus our tar soit

This is the answer to Chapter 13. Problem number 49 from the Smith Organic Chemistry textbook. In this problem tells us that, um, reaction of t beautiful pencil ether with HBR forms one burma plantain on compound. Be on. Where then? Told the compound B has a molecular ion at 56. Ah, and it gives three particular peaks in its eye. Our spectrum on dso we're asked to propose a structure for B and draw a stepwise mechanism that accounts for its formation. Um, and so it might be easier to actually just start with the stepwise mechanism and then see if our ah spectral data makes sense for what we get. But since this chapter is about Mass spec and I are, we'll start with the mass spec in the I r. And use the mechanism as, ah confirmation. So we have an empty Z of 56. Um, so this suggests perhaps four apartments for 48 on that would leave us with eight hydrogen Sze Ah, and so the hd I assuming that that is the correct formula, are the hd I There would be, um, two times four plus to minus eight eyes gonna be too over too. Eyes going to be one. So we have ah, ring. Or much more likely a double bond. Um, And so when we look at the i r. Data, what we see is CSP to each bonds. So we do probably have a double bond CSP three h bonds, which are ubiquitous. That doesn't tell us anything on. And then we do see a carbon carbon double bond with 16 50 wave number. Um, and so given all of that information and given, um, this piece of the molecule here, I'm gonna go ahead and say that what we end up with is this. Oops. So we end up with this, or I'd prefer I can write it like this. Okay, um and so Ah, that's the correct answer. But we are also asked to draw a mechanism here. And so I'm going to do that on this fresh page. Um, and so let's see. 12345 So here's our starting material. Okay. And so the first step in this reaction, um, treatment with, uh, Hydra Brahma Gassid. Uh, this oxygen is gonna become pro donated. Um, and so it will have a positive charge. Um, and so that is going to Ah, open up, Uh, the way for the bromide ion that we formed in the previous step, which is gonna have four lone pairs and a negative charge. So this is going to attack this carbon, uh, to which the oxygen is bound. Those electrons will revert to that oxygen. So that's going to get us to this one, bro. Mo painting that we were told is a product. Um, and then also to here. So this tert be tac side eye on. Oh, you know what? It's not gonna be a turkey tac side eye on because I forgot that it was pro donated. Their okay, so it's actually just gonna be, uh, t beautiful. So Okay, so this t beautiful can be protein ated once again. So there's still, um, some hydro brahma Gassid floating about. So, uh, this oxygen can pick up another proton and become doubly pro donated. So when that happens, it's going to have just the one lone pair now and a positive charge on it. Um, and so now it's going to be ready to leave as water. And so that is indeed what will happen. So that is going to get us to this cargo cat eye on. Okay, Um and so, uh, from there, what can happen is so I'm actually gonna change the way that I draw one of these. So, um, I'm actually gonna draw the hydrogen is on this one on, and you'll see why Momentarily. So what's gonna happen here is that again we can call on this bromide ion that was generated, um, to snatch, uh, a proton. And so this is going to form a double bond here. And so that is actually going thio get us to our final product. Be okay. And so this is Ah, molecule B. And so that's the stepwise mechanism that we were asked to draw. I'm And then again, if we look back at the first page, this is how we got to that same answer using the spectral data. Since these two answers agree, I would say that's a pretty strong indication that they are the correct answer on Indeed they are. Um, and so again, that's that's how to solve this problem by using special data. So used the the M to Z the molecular eye on that were given to figure out a formula. Calculate in HD eye for that formula. Look at the i R. Data on DSI. What makes sense? It's sort of piece it together from there. Um, And then again, we're asked draw stepwise mechanism. So here. That is on page two. Um, yeah. And so that's the answer to Chapter 13. Problem number 49.

So the question here gives us the photochemical chlorination of chloroform here, where it wants toe basically determine the steady state rate law for the proposed mechanism here. So the elementary where the reaction here is going to be ch three c l, um c h c l three rather, um, plus, C l two gives us, um, CCL four plus hydrochloric acid here. So s so we have a couple of mechanisms here I'm not gonna draw because it's already in the question itself. So what you want to find is going to be the study, um, state rate law. So we're not the steady state rate Law is going to be the derivative of our product here with respect to time. So it's gonna be D c c l four over de t here. So what this is is going to be determined by the elementary reactions here. So what we're gonna get is we're gonna have to find the first mechanism that produces this product. So that is going to be, um, K two. And we're gonna multiply this by a as that's our elementary reaction, and we need to produce chloride to the power of one half in order to balance it out. Um, afterwards, we will then divide this well, actually multiply first by C L two to the one half toe, Balance it out, and then we're gonna divide it by K three year, which is going to be the last mechanism reaction here. And we're going to multiply this by one half again in order to essentially give us the CCL four product here as our final product and nothing else. So we're gonna multiply that by one half to get half of that. We're gonna add, um, I a again in order to fulfill the requirements for the reactions here, which is C l two. So we're gonna get to I A And this is gonna be the study state rate law for the production of carbon Tetrachloride here, which is CCL four


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