Question

What is a `"free radical"`?

Answer 1

A free radical species is essentially a reactive intermediate that has a single UNPAIRED electron...........

Explanation:

And in organic chemistry, perhaps the best example of a free radical is the bromine free radical formed by ultraviolet light:

`Br_2(g) + hnu rarr 2dotBr`

Bromine radicals are intrinsically reactive species. Why? Because when they react with another species, typically they form ANOTHER radical, so that a cascade of radical reactions can occur, viz.

`"Radical formation:"`

`Br_2(g) + hnu rarr 2dotBr`

And for the free radical halogenation of toluene, we can write:

`C_6H_5CH_3 + dotBr rarr C_6H_5dotCH_2 +HBr(g)`

And this benzyl radical can continue the chain of reaction:

`C_6H_5dotCH_2 + Br_2 rarr C_6H_5CH_2Br + dotBr`

And thus the formation of benzyl bromide has generated another `dotBr`, which continues the chain. The reaction is (eventually) terminated by the coupling of 2 radicals, viz.:

`C_6H_5dotCH_2 +dotBrrarr C_6H_5CH_2Br`

`2C_6H_5dotCH_2rarrC_6H_5CH_2CH_2C_6H_5`

The presence of diphenylethane in (some) quantity in the product mix is convincing evidence of a radical chain mechanism as proposed.

Answer 2

A species that bears an unpaired electron, i.e. `dotX`......

Explanation:

And free radical species TEND to be highly reactive species. Why so? Because in the very act of reaction they tend to generate ANOTHER free radical that continues the chain of reaction.....

This is best exemplified by free radical halogenation of toluene by bromine...

Now the bromine molecule may be photolyzed by UV light.......this is the so-called initiation step....

`"1. Initiation."`

`Br_2stackrel(hnu)rarr2dotBr`

Normally, the light source is UV; I knew a very good organic chemist who could do brominations that other chemists could not replicate. He managed to trace his success to the fluorescent lights that were used in his lab. When other chemists switched to these lights they were able to reproduce his results. This won him major brownie points in his Ph.D. defence.

`"2. Propagation."`

`dotBr+H_3C-C_6H_5rarrH_2dotC-C_6H_5 + HBr`....

and then the benzyl radical forms the DESIRED `Br-C` bond.......

`Br-Br+H_2dotC-C_6H_5rarrdotBr+Br-CH_2-C_6H_5`

The reaction of the phenyl radical gives (i) the desired benzyl bromide product, and (ii) it gives another equivalent of bromine radical, which can continue to propagate the chain of reaction.

`"3. Termination."`

The radical species react with the reagent in excess. They CAN, however, couple with each other if a radical pair meet......and thus `"terminate"` the chain of reaction.

i.e. `2dotBrrarrBr_2` or.........

`dotBr+dotCH_2PhrarrBrCH_2Ph`

or..........

`Ph-dotCH_2+dotCH_2-PhrarrPh-CH_2-CH_2Ph`

The presence of carbon-carbon coupled products in some (small) quantity is excellent evidence for the intermediacy of a benzyl radical species. `C-C` coupling reactions are notoriously difficult things to do.