CHEM 3410 1st Edition Lecture 2Outline of Current LectureI. Electrophilic Addition to Conjugated Dienes (Continued)II. Kinetic vs Thermodynamic ControlIII. Steric Hindrance RecapOutline of Last LectureI. Introduction to Conjugated DienesA. Conjugated vs NonconjugatedB. Definition of a Conjugated DieneC. Relative StabilityII. Electrophilic Addition RecapIII. Synthesis of Conjugated DienesIV. Electrophilic Addition to Conjugated DienesCurrent LectureI. Electrophilic Addition to Conjugated Dienes (Continued) (1) First, form your resonance stabilized carbocation. When drawing your mechanisms, draw your first intermediate twice. First to show residence, second to show the electrophilic addition.(2) Second, add your Bromine to both structures. The first will be you 1,2 product. The second will be your 1,4 product.RECALL FROM LECTURE 1Resonance gives us a second intermediate and a second product. The products are named “1,2 addition product” and “1,4 addition product” because of where we add things on the molecule. Hydrogen always attaches to carbon one of the alkene as it is pulled on by the electrons from the double bond. These numbers come from the reacting carbons, we are not following the rules of nomenclature. Recall Markovnikov again – The carbocation forms that ismore stable – that is why the Hydrogen attaches to “carbon 1.” The addition to 2 or 4 depends These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.on resonance. The electrophile will attach to carbon 2 of the original alkene (the more stable, available carbon of the alkene from which we began), in the resonance structure, the electrophile will attach to carbon 4. Again, whatever the Hydrogen attacks will always be carbon 1.The mechanism is shown below:Practice with HalogenationRecall from last semester: Halogenation of an alkeneNow, spot the differences in the halogenation of a conjugated dienebelow. Notice that there areno wedges or dashes. Notice that there is not a bromonium ion intermediate. Having a bromonium ion intermediate would not allow resonance because the bromine would be back donating electrons into both carbons it is attached to, preventing formation of an allylic intermediate (an intermediate with resonance.)In this example, we have Br2 reacting with a diene. Remember, the carbocation formed during this conjugate addition will always be allylic. Notice that there are not wedges or dashes. WHEN WE DO BR2 ADDITION TO CONJUGATED DIENES, WE DO NOT GET A BROMONIUM ION INTERMEDIATE. If you are doing Br2 addition to a conjugated diene, you would still get a 1,2 product and a 1,4 product. This means that the reaction is going through an allylic carbocation that is rearranging itself. You cannot get two products if your intermediate is a bromonium ion. With a bromonium ion intermediate, you would only be able to get one product.An allylic intermediate is more stable than a bromonium ion intermediate. This is due to the delocalizedof electrons. You distribute a charge over a much greater area in an allylic carbocation than in a bromonium ion. Entropy is always favored.Also, Bromine is not great at handling a positive charge. Practice with RingsEach double bond will produce its own set of 1,2 and 1,4 products. Though the 1,4 may sometimes be the same, the 1,2 product will always be different. That is because of the difference in resonance forms. Recall that stability the based on substitution of the double bond.Example: Here is the resonance of the reaction if the less stable, less subsituiteddouble bond reacts.Below, you can see the reaction played out using the less stable double bond.Now, see the resonance when the more stable, more substituted double bond is used.Below, you can see the reaction played out using the more stable double bond.Always use the more stable double bond unless otherwise specified because that is how your textbook describes these reactions. It is not a rule. Of course the less stable would be more reactive. (Recall the inverse interaction of stability and reactivity.) More stable, less reactive. Less stable, more reactive.II. Kinetic vs Thermodynamic Controls Temperature Amount 1,2 Produced Amount 1,4 produced Major Product of Reaction Reason“Cold” (0°C)71 % 29% Kinetic Product (always 1,2)Fastest forming (before resonance rearrangement)Added Heat 15% 85% Thermodynamic Product Most stable(40 °C and higher)(can be 1,2 or 1,4) It will most often be the 1,4, but you must always check!The numerical values are specific to the reaction of this example.Temperature can affect reactivity. What the means is that if you want one product over the other,you can control the reaction with temperature: Kinetic vs Thermodynamic Control.When drawing these diagrams, remember than all your transition states will be higher than the energy of your starting materials. Furthermore, no other peaks will be higher than your first transition state (B) because, as you may recall from last semester, the first transition step is the rate determining step or RDS.The slowest part is breaking the double bonds to form the intermediates. Interacting one charged species (intermediates) with another charged species (your halide ion) will be very fast.Also, remember to mark these transition state peaks (B, D) with the double dagger symbol (‡). Your intermediates will be indicated at point C.Why do we always get 1,2 as the kinetic, major product at freezing temperatures?We cooled the reaction, so we took energy out. It requires energy to break the bonds and get over that energy barrier (the blue D on the graph). At 0°C, all processes happen in one direction.There is not enough energy to go backwards and produce an equilibrium, so the reaction wants to do what is easiest. Based on our energy diagram, the 1, 2 intermediate is easiest to get to because it requires less energy. With low temperatures with minimum energy,the 1,2 activation step is more likely to occur.At low temperatures, the intermediate goes to the most readily available product – energy wise. It requires the least effort. It is easiest to get to because we do not have to move charges and rearrange to get to it.Why do we always is it possible to get the 1,2 or 1,4 at elevated temperatures?By heating the system, you are providing enough energy for the reaction to cross that higher energy threshold.