Shown below is the ¹H NMR spectrum of the alkyl bro...

Question

Shown below is the ¹H NMR spectrum of the alkyl bromide used to make the phosphonium ylide that reacts with a ketone in a Wittig reaction to form a compound with molecular formula C₁₁H₁₄. What product is obtained from the Wittig reaction?
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Accepted Answer

Welcome back everyone to another video, an alkyl bromide with the following proton and A R spectrum undergoes a whig reaction with acetone to form a compound with the molecular formula C 11 H 14, determine the structure of the alkyl bromide. And the product obtained from the wiig reaction. We're given our spectrum. And what we want to do is just work backwards. We're given the molecular formula of the product of the with reaction. And what we want to do is just calculate the hydrogen deficiency index which is one half multiplied by two, multiplied by the number of carbon atoms plus two minus the number of hydrogen atoms. And if we substitute our givens, that'll be two multiplied by 11 carbon atoms plus two minus 14 hydrogen atoms. And we get 10 divided by two which is five. We immediately take a look at our spectrum and we notice that at around 7.1 parts per million. At this chemical shift, we have four hydrogens as a single, which is a classical peak for benzene, right? Whenever we have something between seven and eight, sometimes 6.7 and eight, we can say that mm the reason why our hygiene deficiency index is that high is because four degrees of saturation are responsible for benzene, one ring and three double bonds. And there will be an additional degree of saturation which will be our double bonds, not a ring, right? Because the whit reaction will form an additional double bond. So we understand that our starting material or our alky Halide should have a benzene ring. What we can do is just start analyzing our spectrum. And what we can do is just draw a benzene ring to begin with and think about a possible substitution pattern. We noticed that there are 7.1 PM, there are four hydrogens. And there's also a clo the reason why Syle is very simple in order to have a Syle, we must have a vertical line of symmetry and a horizontal line of symmetry. If we have these two lines of symmetry, then that means all of our hygen will be equivalent to each other. And this is a classical pattern in which we have a parady substituted benzene. If we have a one C four, that is substituted benzene, all the four hydrogens will be equivalent to each other due to the presence of a horizontal and a vertical line of symmetry. And that's how we get a singlet. So now we understand that based on this peak, we have a one comma four di substituted pattern. Now to identify our subscriptions, we're going to consider our other signals two H and three H. So we can start with three H at around 2.4 P pm. We have an integration of three. That's also ayla Aylett essentially means that there will be no neighboring hydrogens. Now, the only way to get this is to add a metal group. And in that case, we will not have any neighboring hydrogen atoms. In addition to that, we noticed that the shift is 2.4 P pm, which is roughly 1.5 P pm greater than the standard position of a metal group bonded to an L kal chain. And this is due to the fact that it must be bonded directly to the benzene rank, which is the shielding. So that is sensible and now we have two at around 4.6 P pm. The reason why the P PM value is set high is basically because these protons should be adjacent to an electron withdrawing group or an electron atom such as bromine. In this case, because we have an alkyl bromide. So we're going to choose ach two group which is bonded to bromine. And this gives us a singlet since there are no neighboring hydrogens, right. And the high P PM value results from the benzene ring as well as the electron bromine. This is our possible structure. If we think about this molecular formula, we can essentially say that the molecular formula of this compound would be 123456 carbon atoms, seven and eight. So that would be C eight, the number of hydros would be four plus two plus three. So we can say that we have a total of 11 and then we have bromine. So that'd be the molecular formula of the alkyl hide. And this is our starting material. Now, according to the problem, it reacts with acetone and the whig reaction. And we have to remember that first of all, we are using an alkyl hide and treating it with trenny phosphate. PPH three followed by the addition of base such as Beyl lithium. What we are forming is essentially our ili. And we can show that it by removing our bromine atom from the structure, adding a negative charge on to carbon and now adding a positively charged phosphorus, we will have a formal positive charge on phosphorus which will be bonded to three panel subsidence. When that elite reacts with acetone, we're going to form our carbon carbon bond. So let's draw the structure of acetone. And let's understand that in order to get the final product, we essentially want to combine our negatively charged carbon and the carbon from the carbon group in a double bond by removing that oxygen atom. So what do we get? Well, essentially we can just redraw the structure, remove our PPH three from the structure. And now the carbon atom that we have will become double bonded, we're forming a double bond between our two carbon atoms. So let's throw a double bond. And now let's add the two methyl subscriptions from Acetone. Simply speaking, we are combining the two carbon atoms into a common double bond by removing that oxygen atom as well as our phosphorus. So we get our product and we can see that the molecular formula of this compound would be c we're adding two more or actually three more carbon atoms. So eight plus 123 would be C 11. Now, the number of hydrogen atoms we can count this carefully. So we have 1234 plus three, that's seven, right? Then we have eight and eight plus six. This gives us 14 hydrogen atoms. This is exactly the molecular formula that we have. So that means our structures are sensible. We are going to label our lky bromide as well as our products from the Wittig reaction. So now let's label them one of them and the other that would be it for today. Thank you for watching and I hope to see you in the next video.

Shown below is the ¹H NMR spectrum of the alkyl bromide used to make the phosphonium ylide that reacts with a ketone in a Wittig reaction to form a compound with molecular formula C₁₁H₁₄. What product is obtained from the Wittig reaction?
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Key Concepts

NMR Spectroscopy

Wittig Reaction

Molecular Formula and Product Prediction

Watch next

Shown below is the 1H NMR spectrum of the alkyl bromide used to make the phosphonium ylide that reacts with a ketone in a Wittig reaction to form a compound with molecular formula C11H14. What product is obtained from the Wittig reaction?<IMAGE>

Key Concepts

NMR Spectroscopy

Wittig Reaction

Molecular Formula and Product Prediction

Watch next

Shown below is the ¹H NMR spectrum of the alkyl bromide used to make the phosphonium ylide that reacts with a ketone in a Wittig reaction to form a compound with molecular formula C₁₁H₁₄. What product is obtained from the Wittig reaction?
<IMAGE>