$C{H}_3—C{H}_2 CH=CH—C{H}_3+HCl \to C{H}_2—C{H}_2—CH—C{H}_3$
$|$
$Cl$
$(2-chloropen\tan e)$
$+C{H}_3—C{H}_2—CH—C{H}_2—C{H}_3$
$|$
$(3-Chloropen\tan e)$

The carbocation formed from compound (A) is slightly less stable than carbocation

jeading to the formation of compound (B) therefore the amount of 2-chloropentane (B) will

be little bit more than that of 3- chloropentane (A).