# Process of a substitution reaction

In this question I am not able to decide by what process this reaction is happening.

I first thought that one of the $$\ce{H}$$ from $$\ce{CH3}$$ will depart $$\ce{Br}$$ from $$\ce{CBrCl3}$$ but I am wrong. Would someone tell me what is the right process for this reaction?

This reaction is already in the literature (peer-reviewed paper), so I don't want to change any outcome already explained by Waylander. But, I like to explore the reaction a bit more for the benefit of novice students of organic chemistry. The reaction seemingly follows the same mechanism as that of a general free radical reaction: Initiation; propagation; and termination. The starting material is active enough to get initiated by light:

$$\ce{Br-CCl3 ->[h\nu] Br^. + ^.CCl3} \tag1$$ $$\ce{Ph-CH3 + Br^. -> Ph-CH2^. + HBr} \tag2$$

Now propagation begins:

$$\ce{Ph-CH2^. + Br-CCl3 -> Ph-CH2-Br + ^.CCl3} \tag3$$ $$\ce{Ph-CH3 + ^.CCl3 -> Ph-CH2^. + HCCl3} \tag4$$

After all limiting reagent is used up, the remaining radicals react with each other to terminate the reaction:

$$\ce{Cl3C^. + ^.CCl3 -> Cl3C-CCl3} \tag5$$

According to the reactions shown in the original paper (Ref.1), the most probable limiting reagent should be toluene. If that's the case, the propagation ends at equation $$(3)$$ after going through many propagation cycles. At the end, what remains are two $$\ce{^.CCl3}$$ radicals from the equation $$(1)$$ and last circle of the equation $$(3)$$. The reaction ends with dimerization of these two radicals.

As pointed out by Waylander, the ratio of $$\ce{HBr:Ph-CH2Br}$$ is ~$$1:20$$. This result supports the mechanism since formation of $$\ce{HBr}$$ is only by the initiation reaction $$(2)$$. The formation of $$\ce{Ph-CH2Br}$$ is by the propagation (chain) reactions $$(23)$$ and $$(4)$$, hence the larger portion. For the same reason, amounts of $$\ce{Ph-CH2Br}$$ and $$\ce{CHCl3}$$ are equimolar.

Note: When the same reaction applied to toluene derivatives with substituted aromatic nucleus, a relatively large polar effect was found in the reaction rates. The effect is depend on $$\sigma^+$$-values of the substituents (Ref.2).

References:

1. Earl S. Huyser, "The Photochemically Induced Reactions of Bromotrichloromethane with Alkyl Aromatics," J. Am. Chem. Soc. 1960, 82(2), 391–393 (https://doi.org/10.1021/ja01487a034).
2. Earl S. Huyser, "Relative Reactivities of Substituted Toluenes Toward Trichloromethyl Radicals," J. Am. Chem. Soc. 1960, 82(2), 394–396 (https://doi.org/10.1021/ja01487a035).

This chemistry was first described in this 1960 JACS paper here.

1. The first step is cleavage of the $$\ce{Br-CCl3}$$ bond to give $$\ce{Br^.}$$ and $$\ce{^.CCl3}$$.
2. The second step is $$\ce{Br^.}$$ abstracting a proton from toluene to give $$\ce{HBr}$$ and a benzyl radical.
3. The third step is the $$\ce{^.CCl3}$$ radical abstracting a proton from toluene to give chloroform and a benzyl radical.
4. The fourth step is a benzyl radical abstracting $$\ce{Br}$$ from $$\ce{Br-CCl3}$$ to give benzyl bromide and $$\ce{^.CCl3}$$.
5. The fifth step is two $$\ce{^.CCl3}$$ radicals quenching to give hexachloroethane $$\ce{Cl3C-CCl3}$$.

So overall the products are $$\ce{HBr}$$, Benzyl bromide, Chloroform, and Hexachloroethane.

Reactions 3 and 4 are the major portion of the reaction so equimolar amounts of benzyl bromide and chloroform are formed along with small amounts of $$\ce{HBr}$$ and hexachloroethane (approx 20:1).