Tributyltin hydride (Bu₃SnH) is used synthetic...

Question

Tributyltin hydride (Bu₃SnH) is used synthetically to reduce alkyl halides, replacing a halogen atom with hydrogen. Free-radical initiators promote this reaction, and free-radical inhibitors are known to slow or stop it. Your job is to develop a mechanism, using the following reaction as the example.

The following bond-dissociation enthalpies may be helpful:

b. Calculate values of ΔH for your proposed steps to show that they are energetically feasible. (Hint: A trace of Br₂ and light suggests it’s there only as an initiator, to create Br• radicals. Then decide which atom can be abstracted most favorably from the starting materials by the Br• radical. That should complete the initiation. Finally, decide what energetically favored propagation steps will accomplish the reaction.)

Accepted Answer

All right. Hello everyone. So this question says that tribal tin hydride or bu three snh can be used to synthetically reduce all kill halides. In this process, a halogen atom is replaced by a hydrogen atom suggest a mechanism for the initiation and propagation steps for the reaction. Below. Using the following bond association entropies, calculate the values of delta H for the proposed steps and determine whether each step is energetically favorable. And here we have a list of the relevant bond association energies for this reaction. All right. So first, let's start with the mechanism. The fact that we're going to go ahead and showcase an initiation and propagation step means that this is going to be an example of a radical mechanism. And we can further see this when considering the fact that in addition to tribunal and hydride, we have traced BR two and light in solution. Now, the presence of BR two and light is important because this is going to be our radical initiator. Well, let's go ahead and showcase how that happens. So here is the initiation step. Now recall that the bond between the two bromine atoms in BR two is relatively weak because of the high electronegativity of bromine. So what that means is that light provides just enough energy for the bonds to break. And when it does break one electron from the bond is going to go towards each bromine atom. Notice how I using fishhook or single headed arrows to showcase the movement of only one electron. But after the breakage of the bromine to bromine bond, we end up with two bromine radicals. This begins the initiation step because now we have our first radicals in solution. So one bromine radical in the next phase of the initiation step can react with one equivalent of tribunal t hydride. More specifically, it's going to react and it's going to remove that proton from tint. So the bond between tin and hydrogen is going to break one electron creates a new bond between hydrogen and bromine and the other goes towards tin. So at the end of the initiation step, we end up with HBR and a tribunal and radical. And from this, we can proceed on to the propagation step because now that the initiation step has provided us with our first radicals. These radicals, more specifically, the Tretton radical is going to react with our Ocu hide. So here is our cul bromide. And once again, the presence of the tribunal and radical is going to break the bond between carbon and Bromy. So the bond between carbon and bromine is going to break one electron creates a new bond between tin and bromine and the other allows for the formation of an ocular radical. So now here is our all kill radical and I'm going to go ahead and correct my first structure of the oyo Halide because I had placed the two methyl groups on the same carbon as opposed to two different carbons on the rink. But here we have the formation of the new oy radical and the new bond between bromine and tin. So now our cul radical in the next phase of the propagation step is going to remove a proton from another equivalent of Tretton hydride. This breaks the bond between hydrogen and tin allowing for the formation of a new carbon hydrogen bond and the regeneration of the tribunal and radical. So the reason why radical reactions are known as chain reactions is because the newly regenerated Tretton radical is going to go ahead and repeat these same steps until we get to the termination step in which all radicals combine together to create stable molecules. But here we stop at the propagation step. So our mechanism is complete. And so with each step, let's go ahead and calculate their delta H value. Recall that the delta H of a given step or a given reaction overall is equal to the sum of all bond association energies of those that have been broken. Uh In other words, the sum of the bond dissociation energies of the reactants subtracted by the sum of the bond association energies of the products. All right. So starting off with the initiation step, let's go ahead and find the delta H values for each individual step and then combine them together to find the overall delta H value. So starting off in the first initiation step here, we can see that the bond between the two bromine atoms is breaking but no new bonds are forming when considering are bromine radicals as products. So because of this, the delta H value for this particular step is just going to be equal to the bond association energy of the bromine to bromine bond, which is 190 kg joules per month. On the other hand, when considering the second phase of the initiation step, we can see that the bond between tin and hydrogen is breaking on the reactant side here. And that has a bond dissociation energy of 310 keto joules per mole. On the other hand, on the product side, we now have a new hydrogen to Bromma bond with a bond association energy of 366 kilojoules per mole. So the delta H value for this particular step is going to be equal to 310 kg joules per mole subtracted by 366 kg joules per mole, which equals negative 56 kilo jules for more. So now the delta of the initiation step is going to be equal to the sum of these total delta H values. So that's 190 kg joules per mole added to negative 56 keto jules herm, this ultimately equals 134 kg joules per mole. So now let's consider our propagation steps in our first propagation step. We can see that the bond between carbon and bromine on the reactant side is breaking. So the bond between the secondary carbon of the cyclops ring and Bromine had a bond association energy of 309 kilojoules per mole. On the other hand, in the product side, we have a new bond between tin and Bromy that has a bond dissociation energy of 552 kilo jules promo. So delta H of the step is equal to 309 kilojoules per mole subtracted by 552 kg joules per mole, which equals negative 243 kilo joules per month. So now lets consider our second propagation step. Now, in the second propagation step, if we look at the reactant side, once again, the bond between hydrogen and tin is being affected. So its bond association energy is equal to 310 kilo jules peron. On the other hand, on the reacting side, we now have the new formation of a secondary carbon to hydrogen bond with its bond dissociation energy being 413 kilojoules per mole. So delta H of this particular step is equal to 310 glo jules promo subtracted by 413 kg joules per mole, which equals negative 103 kilo joules for more. And so the sum of the delta H values for both of these steps is going to be equal to the delta H value of the propagation step. So scrolling up delta H of propagation is equal to negative 243 kilojoules per more added to negative 103 kilojoules per month. This ultimately equals negative 346 kilo joules per mole. So now, last but not least to find the delta H value of the overall mechanism, we're going to add the delta H values for both the initiation and propagation steps. So delta H overall is equal to 134 kg joules per mole added to negative 346 kilo jules for more. And this is equal to negative 212 kg joules per mole recall that this indicates that the reaction is exothermic due to the fact that delta H is a negative number. So due to the fact that the delta H overall is a negative number and therefore exothermic, this mechanism is energetically favorable. So to go ahead and make space oops for the final answer, I'm going to go ahead and pause this recording so I can make some adjustments. All right. So now the recording has resumed and we have our final answer as for the mechanism, we have the initiation and propagation steps shown. And the last part of the answer reads as follows, the overall entropy of the reaction is highly exothermic, making this mechanism energetically favorable. So with all that being said, thank you so very much for watching. And I hope you found this helpful.

Key Concepts

Free Radical Mechanism

Bond Dissociation Enthalpy (BDE)

Energetic Feasibility and ΔH Calculation

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