So let's go ahead and actually label these things. So on the left side here, remember our change in free energy, which is looking at our reactants compared to our products-- we're ignoring this little bump. The change in free energy, which I'm going to indicate with the green line, extends between these two points.
And the same on the right side, just again extending between the start and endpoints of our reactions. So I'll go ahead and indicate that this green line in both cases refers to our delta G values. And in a different color, let's say red, I'm going to go ahead and indicate the activation energy, which takes into account the change in energy between the high-energy intermediate and the reactants. So in the case on the left, that change is indicated here with red.
And on the right side, the change between the intermediate and the reactant is a bit longer, so we'll go ahead and indicate that here. Now, activation energy is an important quantity to take into account, because in order for molecules to react, they must have enough energy to overcome this activation energy barrier. Essentially, in the case of a spontaneous reaction for example, I think of it like the energy one needs to get a ball to start rolling down a hill. We all know that gravity will make a ball roll down a hill, which is like a negative delta G value, it's telling us that the reaction is very thermodynamically favorable.
But we need to sometimes give the ball push in order for the reaction to occur. And so that's kind of this little help that it needs to go over before it can actually proceed. For a non-spontaneous reaction, the idea is essentially the same. We still need to have some activation energy. But in addition, because it requires an input of energy, we can think about it as rolling ball up a hill instead of down a hill. Now in general, the idea is that the lower this free energy change, the faster a reaction will occur.
And remember I'm saying faster, so I'm talking about kinetics. I'm talking about the rate of a reaction. So just to write that out, the activation energy, the smaller it is, the faster the reaction will proceed.
Now in biochemistry in particular, it's really important to distinguish between these two terms of thermodynamics and kinetics, which we've drawn out in our diagrams as the change in delta G over the change in activation energy. Because many biochemical reactions in our body are kinetically unfavorable, that is to say they have a very high energy of activation even if they are thermodynamically favorable.
This is why our bodies have enzymes, which essentially lower the activation energy of a reaction. So I went ahead and drew a dotted white line that's a little bit lower, so you can see that when an enzyme is present, the height of the barrier has decreased. And if it's decreased, the reaction will proceed faster.
Now, there's one analogy that my chemistry professor used to tell us all the time that really helped me understand the interplay between kinetics and thermodynamics as they apply to whether or not a reaction will occur. So I'm going to go ahead and scroll down so we can briefly talk about this analogy, which is I think a fun way to think about all of this.
Therefore, P2 is the thermodynamic product the more stable product. We now need to consider how the outcome of this situation changes with the competing reactions of the starting material as we alter the reaction temperature and therefore the average energy of the molecules changes.
At low temperature, the average energy of the molecules is low and more molecules have enough sufficient energy cross activation energy E A 1 than E A 2.
Therefore the reaction preferentially proceeds along the green path to P1. The reaction is not reversible since the molecules lack sufficient energy to reverse to SM , i. Addition of HBr to 2,3-dimethyl-1,3-cyclohexadiene may occur in the absence or presence of peroxides. In each case two isomeric C 8 H 13 Br products are obtained. Which of the following is a common product from both reactions? Addition of the HX to unsubstituted cycloalka-1,3-dienes in either 1,2- or 1,4- manner gives the same product becasuse of symmetry.
The 1,4- product is more thermodynamically stable because there are two alkyl groups on each side of the double bond. This form offers stability to the overall structure. Consider the reaction with 1,3-buta-diene reacting with HCl. Propose a mechanism for the reaction.
Even though the cation would prefer to be in a secondary position in the transition state, the final product is less stable with a terminal alkene. Therefore the major product will be the 1,4 adduct. Objectives After completing this section, you should be able to explain the difference between thermodynamic and kinetic control of a chemical reaction; for example, the reaction of a conjugated diene with one equivalent of hydrogen halide.
Key Terms Make certain that you can define, and use in context, the key terms below. The thermodynamic product: trans bromobutene It is perhaps simple enough to see why 4 is more stable than 3. The kinetic product: 3-bromobutene Several explanations may be proposed to explain the nature of the kinetic product. Experimental Results In , Nordlander et al. Conclusion The reactivity of conjugated dienes hydrocarbons that contain two double bonds varies depending on the location of double bonds and temperature of the reaction.
References Smith, M. March's Advanced Organic Chemistry , 7th ed. This is the origin of some secondary kinetic isotope effects; in our case, it means that 7a is marginally less stable than 7b. What is unusual about the products of 1,2- and 1,4- addition of HX to unsubstituted cyclic 1,3-dienes? Is the 1,2-addition product formed more rapidly at higher temperatures, even though it is the 1,4-addition product that predominates under these conditions?
Why is the 1,4-addition product the thermodynamically more stable product? Out of the following radical cations which one is not a reasonable resonance structure? Which of the following is that isomer: 5,5-dibromohexene 2,5-dibromohexene 2,2-dibromohexene 2,3-dibromohexene 6.
Which of the following will be the kinetically favored product from the depicted reaction? The kinetically controlled product in the above reaction is: 3-ChloroButene 1-ChloroButene 9.
For the reaction in question 8, which one is the result of 1,4-addition? Answers to Problems 1. A Same product for both modes of addition.
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