The following questions are case - based questions. Read the passage carefully and answer the questions that follow:
Nucleophilic Substitution Nucleophilic Substitution reaction of haloalkane can be conducted according to both ${\mathit{S}}_{\mathit{N}}1$ and ${\mathit{S}}_{\mathit{N}}2$ mechanisms. ${\mathit{S}}_{\mathit{N}}1$ is a two step reaction while ${\mathit{S}}_{\mathit{N}}2$ is a single step reaction. For any haloalkane which mechanism is followed depends on factors such as structure of haloalkane, properties of leaving group, nucleophilic reagent and solvent.
Influences of solvent polarity: In ${\mathit{S}}_{\mathit{N}}1$ reaction, the polarity of the system increases from the reactant to the transition state, because a polar solvent has a greater effect on the transition state than the reactant, thereby reducing activation energy and accelerating the reaction. In ${\mathit{S}}_{\mathit{N}}2$ reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on $\mathit{N}\mathit{u}$ than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate $\left({\mathit{S}}_{\mathit{N}}1\right)$ of tertiary chlorobutane at $25°\mathit{C}$ in water (dielectric constant 79) is 300000 times faster than in ethanol (dielectric constant 24). The reaction rate $\left({\mathit{S}}_{\mathit{N}}2\right)$ of 2-Bromopropane and $\mathit{N}\mathit{a}\mathit{O}\mathit{H}$ in ethanol containing $40\%$ water is twice slower than in absolute ethanol. Hence the level of solvent polarity has influence on both ${\mathit{S}}_{\mathit{N}}1$ and ${\mathit{S}}_{\mathit{N}}2$ reaction, but with different results. Generally speaking weak polar solvent is favourable for ${\mathit{S}}_{\mathit{N}}2$ reaction, while strong polar solvent is favourable for ${\mathit{S}}_{\mathit{N}}1$. Generally speaking the substitution reaction of tertiary haloalkane is based on ${\mathit{S}}_{\mathit{N}}1$ mechanism in solvents with a strong polarity (for example ethanol containing water).
Answer the following questions:
(a) Why racemisation occurs in ${\mathit{S}}_{\mathit{N}}1$ ?
(b) Why is ethanol less polar than water?
(c) Which one of the following in each pair is more reactive towards ${\mathit{S}}_{\mathit{N}}2$ reaction?
(i) ${\mathit{C}\mathit{H}}_3-{\mathit{C}\mathit{H}}_2-\mathit{I}$ or ${\mathit{C}\mathit{H}}_3{\mathit{C}\mathit{H}}_2-\mathit{C}\mathit{l}$
OR
(c) Arrange the following in the increasing order of their reactivity towards ${\mathit{S}}_{\mathit{N}}1$ reactions:
(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2methylbutane, 2-Bromo-3-methylbutane