11.44 Consider the addition of HBr shown here. (a) Draw all four carbocation intermediates possible from protonation of the diene and identify the most stable one. (b) Draw both halogenated products formed by attack of Br? on that carbocation. (c) Which of those products would you expect to be formed in the greatest amount at low temperatures? (d) Which would you expect to be formed in the greatest amount at high temperatures? 11.45 Consider the addition of HBr shown here. (a) There are three carbocation intermediates possible from the protonation of this triene. Draw all three of them and identify the most stable one. (b) Draw all halogenated products formed by attack of Br? on the most stable carbocation. (c) Which of those products would you expect to be formed in the greatest amount at low temperatures? (d) Which would you expect to be formed in the greatest amount at high temperatures? 11.46 The addition of HBr to buta-1,3-diene results in 1,2-addition at cold temperatures and 1,4-addition at warm temperatures. If the 1,2-adduct is first formed at cold temperatures and then warmed up, the 1,4-adduct is formed, as shown here. Draw a mechanism for this isomerization.
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The diene can be protonated at four different positions, resulting in four different carbocations: Show moreā¦
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Electrophilic addition of HBr to alkenes yields a bromoalkane. The reaction begins with an attack on the hydrogen of the electrophilic HBr by the Ļ electrons of the double bond to give a carbocation. This step follows Markovnikov's rule with the electrophilic H atom adding to the sp2 carbon containing the most hydrogens, leading to the formation of the most stable carbocation (1Ā°<2Ā°<3Ā°). If possible, a 1,2-shift of either a neighboring hydride or methyl group can occur prior to the last step in order to form a more stable carbocation. In the final step of the reaction, nucleophilic bromide adds to the carbocation to give the neutral product. Draw curved arrows to show the movement of electrons in this step of the mechanism.
Adi S.
4. When buta-1,3-diene reacts with HBr, two different products form: 3-bromobut-1-ene (A) and 1-bromobut-2-ene (B). At low temperature (0 Ā°C), product A is much more prevalent in the product mixture, even though it is the less thermodynamically favored of the two products. At higher temperature (80 Ā°C), product B predominates instead. a. (i) Write a complete mechanism that explains how the formation of two different products from one starting material is possible. (ii) Show all arrows, charges, and lone electrons. (5 points): b. (i) Draw a complete reaction coordinate diagram (RCD) corresponding to the reaction described above. (ii) Label the axes, all transition states, and all activation energies for both the forward and reverse reactions. (iii) Draw a structure for each reactant, product, intermediate, or transition state on your RCD. (5 points): c. (i) Explain why product B is energetically favored. Then, using an appropriate molecular energy distribution diagram, (ii) explain why the less favored product (A) forms at low temperature instead. (5 points): d. When the above reaction occurs at higher temperature, the more thermodynamically favorable compound (B) is formed in higher proportion. Using your RCD diagram from above, explain this behavior. (5 points):
Sri K.
The reaction of 1,3 -butadiene with HBr is shown below. At $40^{\circ} \mathrm{C}$ the major product is the 1 . 4-addition product; however, at $-80^{\circ} \mathrm{C}$ the major product is the 1,2 -addition product. Why are two products formed? (a) The carbocation intermediate allows delocalisation of the second double bond. (b) There are two double bonds present. (c) The fact that the carbocation is planar allows attack from both sides of the plane. (d) There are 2 moles of $\mathrm{HBr}$.
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Level II
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