(a) $\underset{1-\mathrm{Bromopentane}}{{\mathrm{H}}_3\mathrm{C}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{Br}}$
(b) $\underset{2-\mathrm{Bromopentane}}{{\mathrm{H}}_3\mathrm{C}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\stackrel{\mathrm{Br}}{\stackrel{|}{\mathrm{CH}}}-{\mathrm{CH}}_3}$
(c) $\underset{2-\mathrm{Bromo}-2-\mathrm{methyl}\mathrm{butane}}{{\mathrm{H}}_3\mathrm{C}-{\mathrm{CH}}_2-\underset{{\mathrm{CH}}_3}{\underset{|}{\overset{\mathrm{Br}}{\stackrel{|}{\mathrm{C}}}}}-{\mathrm{CH}}_3}$
(i) These figures show that (a) contains the least steric hindrance so towards the ${\mathrm{S}}_{\mathrm{N}}2$ reaction, 1-bromopentane will be the most reactive.
(ii) Figure b of 2-bromopentane has chiral carbon in it. So, it is optically active.
(iii) Towards the $\mathit{\beta }$-elimination, 2-bromo-2methylbutane will be the most reactive as it will form most stable alkene (on account of the highest number of $\mathit{\alpha }$-hydrogens.)