The UV spectrum of 1-phenylprop-2-en-1-ol shows an intense abs...

Question

The UV spectrum of 1-phenylprop-2-en-1-ol shows an intense absorption at 220 nm and a weaker absorption at 258 nm. When this compound is treated with dilute sulfuric acid, it rearranges to an isomer with an intense absorption at 250 nm and a weaker absorption at 290 nm. Suggest a structure for the isomeric product and propose a mechanism for its formation.

Accepted Answer

All right. Hi everyone. So for this question it says that two different types of absorption are observed in the U. V. Spectrum of three methyl one, fennel butte two in one. Oh That was a lot. But we have a weak absorption at 2:20n and a more intense absorption at 258 nm. So when this compound is reacted with dilute sulfuric acid, an isometric product with a weak absorption at 290 nm and a more intense absorption at 250 nm is obtained. Proposed a plausible mechanism for the reaction described above. And you should include the structure of the reaction product in your proposed mechanism. Alright great. So now let's start by talking about our wavelength in comparison from start to finish of this reaction. Right. So notice how um in our isometric product we have an absorption at 290 nanometers which is greater than both 58 and 2 20 of your original product. Right. It's a longer wavelength. So recall that when you have a molecule that can absorb um UV light at a longer wavelength that means you're going to have a more conjugated system you're going to have a more conjugated system. And what I mean by that is that when you're talking about a conjugated system you have multiple pi bonds that are separated by one sigma bond each. So before we begin let's go ahead and draw out our three method. One fennel beauty too. In one. Oh so the first thing I'm gonna draw is our final group. Or benzene ring on position one here. This is gonna be our benzene ring. We have carbon one over here. Carbon one has an O. H. group. Which I'm gonna go ahead and draw. And then this is carbon too. And now carbon too Actually has a double bond between itself and carbon three. Carbon three has a methyl substitution in addition to the fourth carbon in our chain. Like. So so this is our starting material. And recall that we're reacting the substance with dilute sulfuric acid, which means that we're going to have free floating protons in solution. And in order to start off this mechanism right? Um oxygen here in our starting material from our alcohol is gonna go ahead and take up one of those protons floating around in solution. And so we are going to end up with the following intermediate product. Our federal group is going to stay the same. Let me just draw out the rest of the backbone really quickly Method groups over here and now, what has changed, right, we still have oxygen attached, but this time we're going to have An additional proton attached to our oxygen, which is going to give it a formal charge of one And of positive one, I should say which oxygen is 100% not a fan of. Right? So it's going to go ahead and actually detach itself now from our starting material so that it can become neutral once again. So to draw out the next step, I'm gonna actually scroll down really quickly to give myself some space and then we can proceed with the rest of the mechanism because now that oxygen has left where it was attached to position one, we're actually going to end up with a carbon cat. I'll Right because now carbon one having been bonded to oxygen is now electron deficient. So it's gonna be a carbon cat ion. It's gonna be a carbon cat ion which can be fixed actually by this double bond in between carbons two and three, which is um adjacent to our position one. Right? So what's going to happen in this case is that we're going to see a resonance because this secondary carbon cat eye on that we have here in um in carbon one, it can be improved upon, right? Because we have the possibility of making it a tertiary carbon cat eye on on position three here. So it's going to happen is that our double bond in between positions two and three is gonna go ahead and jump in between positions um positions one and two to get rid of that primary carbon cat ion and generate a tertiary carbon cat ion instead. So let me just draw out what that would look like trying really hard to make the benzene rings look decent here. But now we're going to have a double bond in between Carbons one and 2 instead. Like we talked about before And carbon three here is going to be our tertiary carbon cattle. So this is going to be the end of our residents and our prime tertiary, sorry, carbo cat ion is going to be on position three like we said before, so now um this carbon cat ion must be fixed, right, and must be taken care of. So what's going to happen here is that this tertiary carbon cat ion is going to go ahead and be attacked by water because recalling the question that we have dilute sulfuric acid, so we have acid diluted in water in this case, so we're going to have water molecules floating around in solution which can go ahead and attack our tertiary credible cattle. So once again, I'm gonna give myself some space and I'm gonna draw the next step here. So this is our final group. That doesn't change, This is our double bond in between Carbons one and 2. These are the method groups And now we have water bonded to carbon three instead. So we have O. H two and now oxygen is actually going to have a formal charge of plus one, it's going to have a formal charge of plus one which can actually be resolved um if you recall by yet another water molecule. So if I actually expand on this structure a little bit by drawing out these hydrogen, I can go ahead and show you that another water molecule, you can go ahead and take one of these hydrogen, Thus giving our oxygen a negative charge attached to carbon three. So this last step is actually going to be our final product. This is our double bond. These are method groups And now we're going to have an O. H. group bonded to position three instead. So let me just move this over because I think I moved it a little bit off screen, so let me just go ahead and move all this off to the side. So you can see it better. There we go. And this is going to be our final product. So notice how in this final product we have an extended conjugated system Because this double bond in between Carbons one and can extend the conjugation of our bending ring, right? Because we have double bonds from our bending ring. And this all keen in between carbons one and two separated by a sigma bond. So like we talked about in the beginning of the question um this extends the conjugation of our system which can therefore result in the absorption of UV light at a longer wavelength, which is exactly what we saw at 2 90 nanometers. So this is going to be our mechanism. I'm missing a kind of highlight our answer which in this case was well, the entire mechanism. So let me just finish doing that and then I'm gonna go, I'm gonna do a quick scroll through so that you guys can see everything that um that took place here. Alright, so the first thing that happened was that our oxygen on position one went ahead and took up a free floating proton. That resulted in a positively charged oxygen, which functioned as a leaving group and therefore generating a carbon cat ion at position one, which, through residents, was switched over to position three to make it a tertiary carbon cat ion. Water went ahead and attacked the carbon cat ion at position three, which led to a positively charged oxygen on position three, which was deprived donated by yet another water molecule to generate an O. H. Group At position three with a double bond in between Carbons one and 2. So with that being said um that is going to be it for this question. If you made it to the end. Thank you so much and I hope you found this helpful.

Key Concepts

UV-Vis Spectroscopy

Rearrangement Mechanisms

Isomerism

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