Using the BDEs in Table 4-2 (page 167),
(c) Sugges...

Question

Using the BDEs in Table 4-2 (page 167),

(c) Suggest two reasons why iodine does not react well with methane.

Accepted Answer

All right. He hello everyone. So this question is asking us to consider the free radical IOD nation of ethane in which ethane reacts with I two in the presence of light to produce ido ethane. And H I provided two reasons why iodine does not react favorably with ethane. All right. So in this particular case, this is a free radical halogen nation reaction which if you recall is characterized by the mechanism proceeding in three distinct steps. The first of which is initiation in which we create our first reactive intermediates, our first radicals which then proceed to the propagation step. Now, in the propagation step are reactive intermediates combined with stable molecules to produce the product. And another reactive intermediate. This causes a chain reaction that eventually leads us into the termination step. That's the final step. After all reactants have been depleted or if the reactive intermediates are somehow eliminated either way. However, the termination step results in the destruction of all radical or yes, all radical intermediates which eventually stops the reaction. So in understanding potential reasons as to why iodine does not react favor favorably with ethane, let's imagine are theoretical steps. If this reaction were to take place starting off with the initiation step. Now, in the initiation step, the bond between the two iodine atoms would break in the presence of the energy provided by light. Here, this is a homoly cleavage. So one electron from the bond itself goes to each iodine atom to produce two iodine radicals. From here we get to the propagation phase. Now, the propagation phase has two individual steps. The first of which is where an iodine radical combines with a molecule of ethane to remove a proton from one of the carbons of ethane. I should actually move this to the bottom. So let me go ahead and do that. There we go. All right. So here the iodine radical is going to go ahead and cause the carbon hydrogen bond to break one of those electrons creates a new bond between hydrogen and iodine and the other creates a new carbon radical. This produces H I and an ethel radical. So in the second propagation step are ethyl radical can combine with another equivalent of I two. This causes the bond between iodine atoms to break with one of them creating a new carbon iodine bond and the other creating an iodine radical. This iodine radical is going to theoretically repeat this process forward in a chain reaction. So that's how we get Ido ethane and the aforementioned iodine radical. So to understand why this particular reaction may not be favorable, we have to understand that the entropy values associated with our propagation steps. So recall here that delta H standard is equal to the bond association energies of the bonds that have been broken subtracted by the bond dissociation energies of those that have been formed. So let's go ahead and consider both propagation steps, calculate their individual delta H standard values and then add those numbers together to find the overall delta H standard value starting off with the first step. The first propagation step I should say in the first propagation step, the hydrogen to iodine bond has been fought and that bond has a delta H standard value equal to 298 kilojoules per mo. On the other hand, the bond between our primary carbon from ethane and hydrogen has broken. And that bond has a bond association energy equal to 423 kilo joules per month. So calculating the delta H standard for this particular step is equal to 423 kettle jewels per mole subtracted by 298 kilo jewels purple. This is equal to 125 kilojoules per ball for the first propagation step. So from here, let's repeat this process for the second propagation step. Now, in the second propagation step, the bond that broke is the bond between the two iodine atoms which has a bond association energy of 149 kg joules per mole on the other hand, the bond that has formed is the bond between carbon and iodine having a bond association energy of 238 KJ pol. So calculating the delta H standard of this particular step is equal to 149 kilo jules per m subtracted by 238 kilojoules per mole. And this equals negative 89 kilojoules per mole. So with this information in mind, we can go ahead and find delta H standard of the reaction overall by adding the delta H standard values of each propagation step. So that's 125 kilo jules per more added to negative 89 kilo jules per m which ultimately equals positive 36 kilojoules per m. And with this information in mind, we begin to understand why free radical IO nation is not favorable because the delta H standard value for the overall reaction is positive, which indicates that this reaction is endothermic. And if it's endothermic, it's going to favor the formation of the reactants as opposed to the products. This is because there is no decrease in energy to drive the reaction to actually form the products. So this leaves us to two reasons as to why this reaction is not favorable. There's a thermodynamic reason and there's therefore a kinetic reason because thermo dynamically, the reaction is endothermic as a result, the first step in propagation or rather part of the reason why the overall reaction is. So endothermic is because of how endothermic the first step or the first propagation step was considering it had a positive delta H standard value. So because the first propagation step happens to be very endothermic, its also going to occur quite slowly. And from there, we've arrived at our final answer. So I'm going to briefly pause the recording so I can go ahead and write it up. All right, now, the recording has resumed and I was actually able to type the final answer. So it reads as follows. The overall reaction is endothermic, making it thermo dynamically unfavorable. The first propagation step is endothermic, making it kinetically unfavorable. These are the two reasons why iodine does not react favorably with ethane and there you have it. So with that being said, thank you so very much for watching and I hope you found this helpful.

Using the BDEs in Table 4-2 (page 167),
(c) Suggest two reasons why iodine does not react well with methane.

Key Concepts

Bond Dissociation Energy (BDE)

Reactivity of Halogens

Hydrocarbon Stability

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