(a) $\mathrm{CH_3CH_2Br} + \mathrm{HS}^{-}$ → $\mathrm{CH_3CH_2SH} + \mathrm{Br}^{-}$ Here, –SH, a nucleophile displaces Br– from the bromoalkane. Since, the reaction is brought about by a nucleophile and substitution occurs thereafter, the reaction will be termed as a nucleophilic substitution reaction. (b) $\mathrm{(CH_3)_2C= CH_2} + \mathrm{HCl}$ → $\mathrm{(CH_3)_2ClC-CH_3}$ In the given reaction we see that the $\mathrm{H}^{+}$ and $\mathrm{Cl}^{-}$ reacted with the alkene and thus, produced a haloalkane. However, no atom has been displaced. Thus, the reaction is an electrophilic addition reaction. (c) $\mathrm{CH_3CH_2Br} + \mathrm{HO}^{-}$ → $\mathrm{CH_2 =CH_2} + \mathrm{H_2O} + \mathrm{Br}^{-}$ Here there is no atom which is displaced or substituted. Although the reaction is brought about by a nucleophile – OH it is not a nucleophilic substitution. An HBr molecule has been removed from the reactant and hence, it is an elimination reaction. (d) $\mathrm{(CH_3)_3C-CH_2OH} + \mathrm{HBr}$ → $\mathrm{(CH_3)_2CBrCH_2CH_3} + \mathrm{H_2O}$ This is an example of a rearrangement followed by nucleophilic substitution. Initially, a 1° carbocation is formed which rearranges to produce a more stable 3° carbocation. Finally, Br– attacks the carbocation and product is formed.