Process of a substitution reaction

Question

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?

Answer 1

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).

Answer 2

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).