Answer the following questions about the mechanism for the aci...

Question

Answer the following questions about the mechanism for the acid-catalyzed hydration of an alkene:

a. How many transition states are there?

b. How many intermediates are there?

c. Which step in the forward direction has the smallest rate constant?

Accepted Answer

Hi, everybody. Let's take a look at the next problem. It says, consider the acid catalyzed hydration reaction shown below. And we have an alkene that's a trans alkene. The two R groups in opposite sides are both methyl groups. So it's a butane and then it's reacting with H two. So four or hydrosol acid and water and giving us butanol. So there's been a hydrogen added on one side of the double bond and an oh group on the other. And then we have three questions. Number one, determine the number of transition states. Number two, determine the number of intermediates and number three, determine the step in the forward direction with the smallest rate constant. So this is a pretty straightforward reaction. But let's think about what defines what we're looking for here. So transition states are going to be this very uh very slow lives. So they just exist uh or short lived, not. So, excuse me, very short lived states, they just exist, exist sort of very, very short amount of time and they have partial bonds or partial charges, they're sort of on their way to something else as a bond forms or is broken and sort of halfway in between so short life partial bonds and charges, we don't even observe these states. They're sort of theoretical. And then number two, number of intermediates, intermediates have complete bonds or charges. So they live a little bit longer. We can actually observe them sometimes in solution, but they are on the way to the final product. So they are not, you know, they're, they're not as favorable, they're higher energy state than our final product. But they do have complete bonds and charges and do exist long enough to be observed. And then number three, determine the step in the forward direction of the smallest rate constant. That would mean the slowest step and the slowest step will be the one with the highest activation energy put that here too highest, which would be why it's the slowest step because you have to overcome the highest barrier to making this happen. So to be able to define each of these stages, we need to look at our reaction mechanism. So I'm just going to write our two metal groups as R to kind of simplify things. So my initial reactant maybe R on that cc double bond H on the other side got that trans arrangement. So we've got an H and R on each one, I'm just gonna scroll up just a little bit there to have a little more space and that's going to react with a hydro num ion or an acid. So we have that around. So we have hoh two, there's a positive charge there. And the first step will be that the this is an electro pilic addition of proton. So the electrons from the cc double bond will bond to that proton with the electrons from the ho bond going on to the oxygen. So we have this addition of this proton and the breaking of the double bond and that's going to lead to the next step, which will be the formation or which will be the Carbocaine. So as we break the double bond, we end up with a positive charge on the other carbon. So let's draw R carbon with an R and an H and I need to rearrange that it's no longer a 120 degree geometry. So let's redo that there. We have our, our group and then we have two hydrogens because we've added a hydrogen there, then a single bond to the other carbon which has hydrogen and our group, but now has a positive charge on it. It's carbo C and then this Carbocaine, we're gonna have a nucleic attack by water that's going to be very favorable. We've got a positive charge there and our loam pairs on our oxygen. So they'll come on in and attach to that carbon there. So nucleic addition to the carbo cion, but that's going to be an addition of an 02. So we still have one more step that has to happen. So let's redraw that molecule, single bond of the carbon hr So what we have added there is not oh but O H two and we've got a positive charge there. So we just need to deproteinate that hydro num group. So we'll do that with water. So draw 02 water molecule with its lone pairs and those lone pairs grab one of these protons and the electrons from its bond go onto the oxygen. So can't forget about that final step there. And that will lead us to our final product, our alcohol cc single bond hr and oh so that's our product that will have the lowest energy level if we do an energy level diagram, and we started with our reactant or alkene. So every step along the way, in this case, we have two stages along the way, the Carbocaine and then the hydro ion, these are the intermediates. So we have, we put I one and I too, they have complete bonds, complete charges, but they're sort of something along the way, they have a higher energy state than the product. There are these intermediate steps along the way, but they're observable species. So if we have two intermediates here, then we're going to end up with transition states going from one thing to the next. Well, we'd have a transition state between our reactant and intermediate one, between intermediate one and intermediate two and between intermediate two and the product. So by definition, we'll have three transition states. So I'm gonna put the answer at number one will be three transition states. Two intermediates go ahead to highlight those and we'll go ahead and sort of draw those just so we can think about the difference there. So our transition state between one and two, we have a lot of things going on here. We have this double bond breaking and grabbing the proton. So we have a bond breaking and a bond forming. So we indicate that with those dash lines showing those bonds that are sort of halfway in between. So I'm going to draw that I could say about multiple things going on here. So I've drawn a dashed line where the double, the pi bond had been between the two carbons, a dashed line from the carbon to the hydro or excuse me. Yeah, the hydro ion hydrogen and then a dash line from the hydrogen to the oxygen. And the 02, these are all bombs that are in the process of breaking and then forming. And then we'll go ahead and put, we know we have sort of a partial positive charge. A positive charge is forming on the second carbon that's going to be the Carbocaine and a positive charge is going away on that oxygen as we deproteinate it. So we'll indicate that with a partial positive charge on the carbon that will be a Carbocaine and a partial positive charge on our oxygen from the hydro light. So we put these brackets around this and put this called a double dagger looks like a little t with two lines indicating it's a transition state. And then we have another transition state where we have our water molecule adding to the carbo Cron. So again, I've showed a dashed line from the carbon to the 02 since that's a forming bond. And then I have a partial positive on the carbon and a partial positive on the oxygen. So there's that transition state transition state number two. And then I have transition state number three as we dep proton the 02 plus that's added on to the carbo C iron. So now I have a dashed line between the oxygen though it was the water molecule added to the carbon as that bond is breaking. And then a dash line from that hydrogen to an water molecule. And again, we have a posi partial positive charge on the both of the oxygens as that proton transfers between them. And that would be transition state number three. So we'll say TS one, ts two NT S3. So we have the number of transition states, the number of intermediates. And we just need to know which is the slowest step here. Let's think about what's going on in each of our processes. In the first step, we have the double pi bond breaking the C bond forming and the ho bond breaking. So we've got multiple bonds being broken and formed. So this step will definitely have our highest activation energy. We looked at an energy level diagram here for this reaction. This will be the highest hill as we go from reactant to transition state. Number one. So that's gonna be this peak up here, ts one reactant down here. Then we'll have a little valley where a Carbocaine has formed. That will be I one intermediate one. And then we have to bump up a bit but not much to transition state two. All that's happening there is that addition to the carbo cion. So you're just forming a single bond and it's pretty favorable. You have this um electron oxygen atom adding to a positively charged carbon. So that's not a very high activation energy and should be a pretty fast step. And then we're going to drop a whole lot. And then we just have this down to another valley which is intermediate to and then up just a wee bit to deproteinate that oxygen swapping hydrogens between either oxygen or nitrogen as is happening. There is a pretty easy, pretty quick reaction. So just a little hill to transition state three, and we're running out of space and then drop way down to our product. We'll have to make our hill a little shorter here because we're running out of space at the end of the screen. So make that a small hill and then down to our product. So we could see that clearly, the slowest step is going to be that first one because that's going to take the highest activation energy you're forming and breaking multiple bonds, that's much more difficult than adding onto a carbon Canion or dep proton an oxygen. So our first step, it's getting a little crowded in here, but we'll highlight it. So it stands out. Our first step will be the step in the forward direction with the smallest rate constant as it's the slowest step. So for acid catalyzed reaction, there were three transition states. For question number one, there were two intermediates, question number two. And the step in the forward direction with the smallest rate constant is the first step. The electro pilic addition of the hydrogen to the double bond, hope to see you in the next video.

Answer the following questions about the mechanism for the acid-catalyzed hydration of an alkene:
a. How many transition states are there?
b. How many intermediates are there?
c. Which step in the forward direction has the smallest rate constant?

Key Concepts

Acid-Catalyzed Hydration Mechanism

Transition States and Intermediates

Rate Constants and Reaction Steps

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