Treatment of alkene with hydrogen bromide in the presence of dibenzoyl peroxide

Question

Treatment of alkene with hydrogen bromide in the presence of dibenzoyl peroxide results in the formation of anti-Markovnikov alkyl bromides, as shown below.

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 HCI, 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.) d. Using the enthalpy diagram you drew above, explain why anti-Markovnikov hydrohalogenation occurs readily with HBr but not with HCI or HI. It turns out that this reaction does not work if HBr is replaced by either HCI or Hl. 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, .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 = CI, 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 (delta H) for this step in ail three cases (X = CI, Br, and I) and determine whether it is exothermic or endothermic in each case. Treatment of alkene with hydrogen bromide in the presence of dibenzoyl peroxide results in the formation of anti-Markovnikov alkyl bromides, as shown below.

Explanation / Answer

You need to draw out a table in a and b.

bonds broken- bonds formed= enthalpy change

example

a. c-c pi broken 66 - c-cl formed 81 = -15 negative is exothermic

repeat for all halogens in a and b

c. plug the structures in where relevant for c

d. you'll see that br has very mild changes in enthalpy after you do the grid and plug in the numbers

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