Given below are two statements: Statement - I: High concentration...

Statement - I: Rate of ${S}_{N}2 ∝ [ R - X ] \left[{Nu}^{-}\right]<br/>${S}_{N}2 reaction $is favoured by high concentration of nucleophile \left({Nu}^{-}\right)\text{ and } less crowding in the substrate molecule.$

${S}_{N}2$ Reaction Mechanism

This reaction proceeds through a backside attack by the nucleophile on the substrate. The nucleophile approaches the given substrate at an angle of ${180}^{°}$ to the carbon-leaving group bond. The carbon-nucleophile bond forms and carbon-leaving group bond breaks simultaneously through a transition state.$</p><p>Now, the leaving group is pushed out of the transition state on the opposite side of the carbon-nucleophile bond, forming the required product. It is important to note that the <em>product is formed with an inversion of the tetrahedral geometry at the atom in the centre.</em></p><p>Statement - II: Solvolysis follows${S}_{N}1$path.$

he ${S}_{N}1$ reaction is a nucleophilic substitution reaction where the rate-determining step is unimolecular. It is a type of organic substitution reaction. ${S}_{N}1$ stands for substitution nucleophilic unimolecular. Thus, the rate equation (which states that the ${S}_{N}1$ reaction is dependent on the electrophile but not on the nucleophile) holds in situations where the amount of the nucleophile is far greater than the amount of the carbocation intermediate.$</p><p>This reaction involves the formation of a carbocation intermediate. It is generally seen in the reactions of tertiary or secondary alkyl halides with secondary or tertiary alcohols under strongly acidic or strongly basic conditions. The${S}_{N}1$reaction is often referred to as the dissociative mechanism in inorganic chemistry.$

Both are correct statements.