Nucleophilic Substitution Nucleophilic Substitution reaction of haloalkane can be conducted according to both S_N 1 and S_N 2 mechanisms. S_N 1 is a…

Nucleophilic Substitution Nucleophilic Substitution reaction of haloalkane can be conducted according to both $\displaystyle \mathrm{S}_{\mathrm{N}} 1$ and $\displaystyle \mathrm{S}_{\mathrm{N}} 2$ mechanisms. $\displaystyle \mathrm{S}_{\mathrm{N}} 1$ is a two step reaction while $\displaystyle \mathrm{S}_{\mathrm{N}} 2$ 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 $\displaystyle \mathrm{S}_{\mathrm{N}} 1$ 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 $\displaystyle \mathrm{S}_{\mathrm{N}} 2$ 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 Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate $\displaystyle \left(\mathrm{S}_{\mathrm{N}} 1\right)$ of tertiary chlorobutane at $\displaystyle 25$ °C in water (dielectric constant $\displaystyle 79$) is $\displaystyle 300000$ times faster than in ethanol (dielectric constant $\displaystyle 24$). The reaction rate $\displaystyle \left(\mathrm{S}_{\mathrm{N}} 2\right)$ of $\displaystyle 2$-Bromopropane and NaOH in ethanol containing $\displaystyle 40$% water is twice slower than in absolute ethanol. Hence the level of solvent polarity has influence on both $\displaystyle \mathrm{S}_{\mathrm{N}} 1$ and $\displaystyle \mathrm{S}_{\mathrm{N}} 2$ reaction, but with different results. Generally speaking weak polar solvent is favourable for $\displaystyle \mathrm{S}_{\mathrm{N}} 2$ reaction, while strong polar solvent is favourable for $\displaystyle \mathrm{S}_{\mathrm{N}} 1$. Generally speaking the substitution reaction of tertiary haloalkane is based on $\displaystyle \mathrm{S}_{\mathrm{N}} 1$ mechanism in solvents with a strong polarity (for example ethanol containing water). Answer the following questions:

Why racemisation occurs in $\displaystyle \mathrm{S}_{\mathrm{N}} 1$?

Why is ethanol less polar than water?