$2,3-\mathrm{Dimethylbutane}$reacts with bromine in the presence of light to give a mono brominate product. Further reaction gives a good yield of a dibrominated product. Predict the structures of these products, and propose a mechanism for the formation of the monobrominated product.

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

Reaction of $2,3-\mathrm{Dimethylbutane}$with bromine in presence of light

Themonobrominated product formed is $2-\mathrm{bromo}-2,3-\mathrm{dimethylbutane}$and the dibrominated product formed is$2,3-\mathrm{dibromo}-2,3-\mathrm{dimethylbutane}$ .

The mechanism consists of three parts—initiation step, propagation step I, and propagation step II.

Mechanism for the formation of monobrominated product

The mechanism consists of three parts—initiation step, propagation step I, and propagation step II.

Mechanism for the formation of dibrominated product