In ${\mathrm{S}}_{\mathrm{N}}2$ reaction if we doubled the concentration of reactant and nucleophile the rate of ${\mathrm{S}}_{\mathrm{N}}2$ reaction increases by:

What is the decreasing order for the rate of ${\mathrm{S}}_{\mathrm{N}}2$ reaction of following compounds?

I.  ${\mathrm{CH}}_3-\mathrm{Cl}$

II 

III 

IV 

$\left(\mathrm{X}\right){\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{Cl}}_2\stackrel{\text{ aq. NaOH }}{\to }\text{ Compound}\left(\mathrm{Y}\right)$

The compound $\left(\mathrm{Y}\right)$ can give a yellow precipitate with both $2, 4-$$\mathrm{DNP}$ and  ${\mathrm{l}}_2/\mathrm{NaOH}$. What is the compound $\left(\mathrm{X}\right)$ ?

What is the correct sequence of reagents for the following conversion?

${\mathrm{CH}}_3-{\mathrm{CH}}_3⟶{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{CN}$

Which configuration will be adopted by the product at carbon atoms marked ($1$) and ($2$), respectively, in the given reaction?

Among the given options, which is the correct option for the above reaction?