Iodination of alkanes using iodine $(\ce{I2})$ is usually an unfavorable reaction. Tetraiodomethane $(\ce{CI4})$ can be used as the iodine source for iodination, in the presence of a free-radical initiator such as hydrogen peroxide. Propose a mechanism (involving mildly exothermic propagation steps) for the following proposed reaction. Calculate the value of $\Delta H$ for each of the steps in your proposed mechanism ($\ce{cp = cyclopentyl}$). $$\ce{cp-H + CI4 ->[H2O2, heat] cp-I + HCI3}$$

The following bond dissociation enthalpies may be helpful: \begin{array}{lr}\hline \text{bond} & \Delta H / \pu{kJ mol-1}\\\hline \ce{I3C-I} & 188 \\ \ce{HO-OH} & 213 \\ \ce{HO-I} & 234 \\ \ce{cp-H} & 397 \\ \ce{I3C-H} & 418 \\ \ce{cp-I} & 222 \\\hline \end{array}

From the book: L. G. Wade Organic Chemistry; 8th edition; Pearson: 2013; exercise 4-57.

The correct answer is given as: \begin{align} \text{Initiation:}\\\hline \ce{HOOH &-> 2HO*}& \Delta H &= \pu{+213 kJ/mol}\\ \ce{HO* + I-CI3 &-> IOH + CI3}& \Delta H &= \pu{-46 kJ/mol}\\[2ex]\hline \text{Propagation:}\\\hline \ce{cp-H + *CI3 &-> cp* + H-CI3}& \Delta H &= \pu{-21 kJ/mol}\\ \ce{cp* + I-CI3 &-> cp-I + *CI3}& \Delta H &= \pu{-34 kJ/mol}\\\hline \end{align}

The overall enthalpy is $\Delta H = \pu{-55 kJ/mol}$ for the propagation steps.

In the book, there is also an overall table of bond dissociation enthalpies. One important one I used here is the one of water, $\pu{498 kJ mol-1}$.

I have a significantly more exothermic process here: \begin{align} \text{Initiation:}\\\hline \ce{HOOH &-> 2HO*}& \Delta H &= \pu{+213 kJ/mol}\\ \ce{cp-H + OH* &-> cp* + H2O} & \Delta H &= \pu{-101 kJ/mol}\\[2ex]\hline \text{Propagation:}\\\hline \ce{cp* + I3C-I &-> cp-I + I3C*}& \Delta H &= \pu{-135 kJ/mol}\\ \ce{I3C* + cp-H &-> cp* + I3C-H}& \Delta H &= \pu{-21 kJ/mol}\\\hline \end{align}

Here the initiation is less endothermic, and the propagation is significantly more exothermic. So why is this not given as the correct answer? Am I missing something here? I'm quite new at organic chemistry, so I think I must be wrong, not the textbook.


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