Show the transition state of each step of the following alkene...

Question

Show the transition state of each step of the following alkene addition reaction. Be sure to indicate whether the transition state in each is reactant-like or product-like.

Diagram showing alkene addition reaction with transition states labeled as reactant-like or product-like.

Accepted Answer

Welcome back everyone to another video, provide the transition states and indicate whether they are reactant like or product. Like for each step of the following addition reaction and Alkin reacts with hydropic acid and step one to produce a carbo cion with the bromide anion. And in step two, we are producing an alkyl hide. First of all, we want to recall the reaction coordinate diagram or an electro pic edition of any HX to an L kin. First of all, we're starting with our reactants, we reach the transition state ts one, then we drop down slightly in energy to produce our intermediate, which is our carbo cion and the anion. In this case, bromide, we then reached transition state two and drop down in energy to produce our alkyl Halide. The way determining step is the first step because it is the highest in energy. First of all, we want to recall Hammond's postulate and essentially, it tells us that the transition state will resemble the species with the highest energy. In other words, when we consider step one and when we take a look at TS one or transition state one, we notice that according to the energy barrier, it is closer to our intermediate than it is to our reaction. Essentially, we can clearly see this if we just visually measure the distances, the distance in energy from the reactants to the transition state is greater than the distance between the transition state and the intermediate form. Because the transition state is closer to the intermediates, that means our first transition state will be product like right, because it's closer to the products. So we can say PL product like and now when we consider transition state two, we noticed that the transition state two is actually closer to that carbo cion than it is to our alky Halide form. Right? The reaction is exothermic. We have a gradual drop in energy until we reach our products. And we noticed that the transition state ts two is closer to the carbo cion, meaning it is reactant like because the carbo cion and the bromide anion, those are our reactants in the second step of reaction. What we can do now is draw the transition states based on this information, we are going to add brackets. First of all, we understand that transition state one is product like and what we're going to do is just draw our LK and we should not hurry adding this pie bond. We understand that our pi bond is going to be broken. We are going to draw that pi bond in a dashed bond to indicate that this is where we are going to break our bond. Now, we clearly noticed that we're adding our hydrogen to the terminal carbon atom. So we're going to continue adding that dashed bond until we reach hydrogen. And finally, we're going to break our hydrogen bromine bond, which essentially means that we want to draw another dashed bond to indicate that we're breaking that hygen roaming bond. We want to close the brackets and add the transition state tumble. Now, based on Hammond's postulate, we want to modify it slightly because we know that the transition state is product like it means that we want to make our hydrogen bromine bond slightly longer. And the reason why we're doing this is simply to indicate that hydrogen is closer to the starting L king, which will essentially mean that the transition state will resemble the products, right? Because we're adding that hydrogen to the alkane to produce our carbo cion. So we want to make sure that the given broom in atom is actually further away, which will essentially show that the transition state is product like. So now we clearly see that when bruin is far away, the part which has our carbon chain represents the actual carbo carion much more, right? So this will be our first transition state. We can also state adjacent to it that it will be product like. And now we can continue with step number two. Now, before we do that, we can also add our partial charges, let's indicate our partial positive charge for carbon, right? Because it will get a positive charge and the partial negative charge for bromine because it will form the bromide anion. And now, for the second step, we are essentially going to form our carbon bromine bond. So we can first of all draw our carbon chain and we can indicate the bond formation, right? Essentially what we want to do is draw a dashed bond and add bromine that we are indicating that there is only one bond formed. We're not breaking any bonds. We can state that bromine will have a partial negative charge and carbon will have a partial positive charge. That's based on the fact that the initial carbon cion is positive and the bromide anion is negative, we're going to close the brackets and the transition state symbol and state that this transition state will be reactant like that would be it. We can label our answers for our two transition states for steps one and two. And we can conclude the video. Thank you for watching and I hope to see you in the next one.

Show the transition state of each step of the following alkene addition reaction. Be sure to indicate whether the transition state in each is reactant-like or product-like.

Key Concepts

Transition State Theory

Alkene Addition Reactions

Reactant-like vs. Product-like Transition States

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