Use the bond-dissociation enthalpies in Table 4-2 (page 167) t...

Question

Use the bond-dissociation enthalpies in Table 4-2 (page 167) to calculate the heats of reaction for the two possible first propagation steps in the chlorination of isobutane. Use this information to draw a reaction-energy diagram like Figure 4-8, comparing the activation energies for formation of the two radicals.

Accepted Answer

All right. Hi everyone. So this question says to calculate the heats of the reaction for the two possible sites for radical formation in the first propagation steps for the chlorination of 23 dimethyl butane label, the reaction coordinate diagram shown below to compare the activation energies of the two radical formations. Here we have an unlabeled reaction coordinate diagram and some relevant bond association energies which we'll discuss later in this video. All right. So first let's start by drawing our starting material. Here we have 23 dimethyl butane that would be a four carbon parrot chain with methyl substituents on carbon number two of the parent chain and carbon number three recall that in the first propagation step, a chlorine radical is going to remove a hydrogen from a carbon to produce an ocular radical. And what I want to point out here is that there is an element of symmetry to the given starting material. This means that two radicals are possible from this one of which is a primary radical because all of our primary carbons here are identical and we can form a tertiary radical because we have two tertiary carbons that are also identical to each other. So from this, we can go ahead and showcase both proton abstractions and compare by using their bond association energies that were provided in the beginning of this video, recall that the delta H standard of a given reaction can be calculated by taking the bond association energies of those that have been broken and subtracting the bond association energies of those that have been formed. So let's go ahead and showcase each propagation step or each first propagation step to go ahead and calculate their individual delta H standard values. So I'm going to take my drawing here of 23 dimethyl butane and I'm going to go ahead and copy paste it just to showcase both proton abstractions here. Alright. All right. So let's start with a primary radical abstraction. Oh sorry, primary a radical formation. So here we have 23 dimethyl butane reacting with a chlorine radical. Now here the chlorine radical is going to take a proton R a primary carpet. In this example, we'll say it's going to remove a proton from the methyl substituent branching off of carbon too. But again, all of the primary carbons in this case would produce the same product. So first, I'm drawing the same carbon chain and now the methyl carbon has one less hydrogen and an unpaired electron HCL is also formed as a by product here. So now let's go ahead and showcase our tertiary a radical formation and then we'll calculate each delta H standard value. So now when showcasing the formation of the tertiary alki radical, I am going to remove a proton from carbon two of the parent chain that creates a radical on position too and HCL as a by product once more. So now let's talk about delta HD and we'll start with the delta H standard for the formation of the primary cul radical. In this particular case, the bond that was broken was the bond between the primary carbon and hydrogen that had a bond association energy of 423 kilojoules per month. So I would subtract that or rather I would subtract the bond association of energy of those formed from this. Now, the bond formed was the new bond between hydrogen and chlorine in HCL that has a bond association energy of 432 kilojoules per month. So 423 kg joules per mole subtracted by 432 kg joules per mole equals negative nine kilojoules per mole. Now recall that since Delta H standard is a negative value, that indicates that this is an exothermic process. But now let's compare our tertiary radical delta H standard or actually before we calculate delta H standard for the tertiary radical formation. Let's recall the bonds that have been broken here. In this case, the bond that broke was the bond between a tertiary carbon and hydrogen that has a bond association energy of 403 kilojoules per month. And once again, the bond that formed is one between hydrogen and chlorine in HCL. So its bond association energy is the same as before. So now Delta H standard is equal to 403 kilojoules per mole subtracted by 432 kilojoules per mole. And this equals negative 29 kilo joules per month. So even though both of these reactions are exothermic, the formation of the tertiary cul radical is lower in energy. So they're both exothermic. However, the tertiary cul radical is lower in energy since the tertiary radical is more stable than the primary one. So now when drawing out or rather labeling the reaction coordinate diagram given on the reactant side, we would draw the structure of 23 dimeo butane and a chlorine radical that would be on the reacting side here on the left. However, when considering the product side, on the right, we have one peak that is lower in energy than the other. So the lower peak would correspond to the formation of the tertiary radical and the higher peak would correspond to the primary radical. So lets go ahead and fill that in. Once again, on the left, I would go ahead and draw are starting material as well as a chlorine radical. So now on the right side, the higher peak or rather the end that is slightly higher up here would correspond to our primary radical. So that's what I'm going to redraw here as well as HCL is a by product. So now on the bottom, on the right side here, I would go ahead and draw the tertiary radical and there you have it here is our completed reaction coordinate diagram. And so with that being said, if you watch this video all the way up to this point, thank you so very much and I hope you found this helpful.

Key Concepts

Bond-Dissociation Enthalpy

Propagation Steps in Radical Reactions

Reaction-Energy Diagram

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