(a) Compute the heats of reaction for abstraction of a primary hydrogen and a secondary hydrogen from propane by a fluorine radical.

(b) How selective do you expect free-radical fluorination to be?

(c) What product distribution you expect to obtain from the free-radical fluorination of propane?

An atom or group of atoms containing odd or unpaired electrons is known as free radical. The unpaired electron is represented by a single unpaired dot in the formula. Free radicals are electrically neutral. They are highly reactive species formed by homolytic fission of a covalent bond

In a free-radical chain reaction, free radicals are generally created in the initiation step. A free radical and a reactant are combined to yield a product in the propagation step. Lastly, the number of free radicals generally decreases in the termination step.

BDE is the amount of enthalpy required to break a bond homolytically in such a way that each bonded atom retains one of the bond’s two electrons.

Mathematically,

${\mathrm{\Delta H}}^0=\sum (BDE of bonds broken)-\sum (BDE of bonds formed)$

Primary hydrogen is hydrogen that resides on a carbon which is only attached to one carbon atom.

Secondary hydrogen is hydrogen that resides on a carbon which is only attached to two carbon atoms.

(a)

abstraction of primary hydrogen by a fluorine radical

$\begin{array}{rcl}& & \text{breaking} \text{primary} \text{C}-\text{H} \text{bond:} \text{410} \text{kJ/mol}\\ & & \text{forming} \text{F}-\text{H} \text{bond} \text{:} -\text{569} \text{kJ/mol}\\ & & \mathit{\Delta }\mathbf{\text{H}}\mathbf{ }\mathbf{\text{=}}\mathbf{ }\mathbf{\text{410}}\mathbf{ }\mathbf{\text{kJ/mol + (}}\mathbf{-}\mathbf{\text{569kJ/mol)}}\mathbf{ }\mathbf{\text{=}}\mathbf{ }\mathbf{-}\mathbf{\text{159}}\mathbf{ }\mathbf{ }\mathbf{\text{kJ/mol}}\\ & & \end{array}$

abstraction of secondary hydrogen by fluorine radical

$$\begin{gathered} \text{breaking} \text{secondary} \text{C}-\text{H} \text{bond:} 397 \text{kJ/mol} \\ \text{forming} \text{F}-\text{H} \text{bond} \text{:} -\text{569} \text{kJ/mol} \\ \mathit{\Delta }\mathbf{\text{H}}\mathbf{ }\mathbf{\text{=}}\mathbf{ }\mathbf{397}\mathbf{ }\mathbf{\text{kJ/mol + (}}\mathbf{-}\mathbf{\text{569kJ/mol)}}\mathbf{ }\mathbf{\text{=}}\mathbf{ }\mathbf{-}\mathbf{172}\mathbf{ }\mathbf{ }\mathbf{\text{kJ/mol}} \end{gathered}$$

(b) Free-radical fluorination is nearly non-selective.

(c)

products from free-radical fluorination of propane

The second reaction has more negative change in enthalpy which implies that it is more feasible. Secondary carbon is more easily fluorinated.