You learned in lecture that treatment of an alkene with hydrogen bromide in the

Question

You learned in lecture that treatment of an alkene with hydrogen bromide in the presence of dibenzoyl peroxide results in the formation of anti-Markovnikov alkyl bromides, as shown below. It turns out that this reaction does riot work if HBr is replaced by either HCl or HI. In this question, we will explore the reason why. Bond dissociation energies (BDEs) which you will need to answer this question are provided below: It will also be useful to recall the approximation that, for any given reaction, Delta H = BDE(bonds broken)-BDE(bonds formed). a. The first propagation step for the reaction of a generic halide is illustrated below: Using the bond dissociation energies given above, calculate the enthalpy change (Delta H) for this step in all three cases (X = Cl, Br, and I) and determine whether it is exothermic or endothermic in each case. b. The second propagation step for the reaction of a generic halide is illustrated below: Using the bond dissociation energies given above, calculate the enthalpy change (DeltaH) for this step in all three cases (X = Cl, Br, and I) and determine whether it is exothermic or endothermic in each case. C. Using the numerical values you calculated in parts (a) and (b), construct an enthalpy diagram for the two propagation steps. On the set of axes provided below, start from methylenecyclohexane and X. and draw three separate and dearly-differentiated curves for the addition of HCl, HBr, and HI. Include the structures of all intermediates and products in your diagram. (Note: You do not need to draw the structures of any transition states, and you may assume that all transition states are only slightly higher than the ground state or intermediate which is closest to them In energy.) Reaction coordinate d. Using the enthalpy diagram you drew above, explain why anti-Markovnikov hydrohalogenation occurs readily with HBr but not with HCl or HI.

Explanation / Answer

Anti-Markovnikov Radical Addition of Haloalkane can ONLY happen to HBr and there MUST be presence of Hydrogen Peroxide (H2O2). Hydrogen Peroxide is essential for this process, as it is the chemical which starts off the chain reaction in the initiation step. HI and HCl cannot be used in radical reactions, because in their radical reaction one of the radical reaction steps: Initiation is Endothermic, as recalled from Chem 118A, this means the reaction is unfavorable. To demonstrate the anti-Markovnikov regiochemistry, I will use 2-Methylprop-1-ene as an example below:

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