The major product formed in the following reaction is:

The reaction is a free-radical chlorination carried out with $$Cl_2$$ under ultraviolet light. Such halogenations proceed through the chain mechanism
  Initiation : $$Cl_2 \xrightarrow{h\nu}\; 2\,Cl^{\bullet}$$
  Propagation : $$Cl^{\bullet}+RH \rightarrow R^{\bullet}+HCl$$ followed by $$R^{\bullet}+Cl_2 \rightarrow RCl+Cl^{\bullet}$$.

In free-radical halogenation, the site from which the hydrogen is abstracted is decided mainly by the stability of the carbon radical that would be formed. The order of radical stability is
$$3^{\circ}\;(\text{tertiary}) \;\gt\; 2^{\circ}\;(\text{secondary}) \;\gt\; 1^{\circ}\;(\text{primary})$$.

The given alkane contains one set of secondary hydrogens and another set of primary hydrogens. Abstraction of a secondary hydrogen produces a secondary radical, which is appreciably more stable than a primary radical. Hence the chain propagation step occurs predominantly at the secondary carbon.

Consequently, chlorine substitutes that secondary hydrogen, giving the monochloride in which the halogen is bonded to the secondary carbon. That compound is shown in Option A, whereas Options B, C and D correspond to products that would arise from less favourable (primary) radical pathways and therefore are obtained only in minor amounts.

Thus the major product is the secondary alkyl chloride represented in Option A.

Answer : Option A which is: the secondary (2-chloro) derivative.