Analyse the following reaction and identify the nature of A and B.

Hint: The chemical name of HBr is hydrogen bromide. In the given reaction it should be noted that one reaction is taking place in presence of light and other in the absence of light. The reaction taking place in the presence of light forms free radicals.

Complete Step by Step Solution:
The reaction of a given compound with hydrogen bromide is an electrophilic addition reaction.
In an electrophilic addition reaction, the substrate is first attacked by an electrophile which results in the addition of an atom or molecule across multiple bonds.

In the first step, hydrogen bromide breaks to form hydrogen ion and bromide ion.
The reaction is shown below.
\[HBr \to {H^ + } + B{r^ - }\]
Image: Step 1: Formation of electrophile

In the second step, the hydrogen ion attack the double bond and a carbocation complex is formed.
The reaction is shown below.

Image: Step 2: Formation of carbocation

In the third step, the bromide ion attacks the carbocation to form the compound A.
The reaction is shown below.

Image: Step 3: Formation of compound A
The reaction of a given compound with hydrogen bromide in presence of light is a free radical addition reaction.
Free radical addition reaction involves the addition of free radicals. Free radicals are formed by homolytic cleavage.

The hydrogen bromide undergoes homolysis to form a free radical of hydrogen and bromine.
The reaction is shown below.

Image: Step 1: Free radical formation
Next, bromine radical is added to the compound

Image: Step 2

Image: Step 3: Formation of compound B
Therefore compound A is 1-(1-bromoethylbenzene) and compound B is 1-(2-bromoethyl)benzene.

Note: The anti-Markovnikov free radical addition of haloalkane only takes place in HBr in presence of hydrogen peroxide. Hydrogen peroxide starts the chain reaction in the initiation step..