(i) Out of (CH₃)₃C–Br and (CH₃)₃C–I, which one is more reactive towards SN₁ and why?
(ii) Write the product formed when p-nitro chlorobenzene is heated with aqueous NaOH at 443 K followed by acidification.
(iii) Why dextro and laevo – rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation?

(i) Leaving group ability decreases in order as I⁻ > Br⁻ > Cl⁻ > F⁻, therefore I⁻ is better leaving group than Br⁻
Hence,  (CH₃​)₃​C−I, is more reactive towards SN₁.
(ii) The presence of an electron withdrawing group (−NO₂​) at ortho- and para- positions increases the reactivity of haloarenes for nucleophilic substitution reaction.
(iii) Butan-2-ol has four different groups attached to the second tetrahedral carbon. Therefore, it is chiral molecule and the dextro and laevo – rotatory isomers of Butan-2-ol are enantiomers. 

The enantiomers possess identical physical properties like melting point, boiling point, solubility, refractive index etc. They only differ with respect to the rotation of plane polarised light. Therefore, dextro and laevo – rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation.