Compound $\mathrm{P}$ on treatment with ${\mathrm{CH}}_2{\mathrm{N}}_2$ (diazomethane) produces compound $\mathrm{Q}$. Compound $\mathrm{Q}$ on reaction with $\mathrm{HI}$ produces two alkyl iodides $\mathrm{R}$ and $\mathrm{S}$. Alkyl iodide $\mathrm{S}$ with higher number of carbon atoms on reaction with $\mathrm{KCN}$ followed by hydrolysis gives $3$-methylbutanoic acid. The compound $\mathrm{P}$ is: