Amines Part 3

Concept:The dipole moment of $\mathrm{R-CN}$ is higher due to greater electronegativity difference. Acidic hydrolysis of $\mathrm{R-NC}$ gives $\mathrm{RNH_2}$ and $\mathrm{HCOOH}$, not a carboxylic acid. Statement II describes a valid multi-step conversion from $\mathrm{R-CN}$ to $\mathrm{R-NC}$.Chemical Equation / Formula:$$\mathrm{R-NC} \xrightarrow{\mathrm{H_3O^+}} \mathrm{RNH_2} + \mathrm{HCOOH}$$$$\mathrm{R-CN} \xrightarrow{\mathrm{H_3O^+}} \mathrm{RCOOH} \xrightarrow{\mathrm{SOCl_2}} \mathrm{RCOCl} \xrightarrow{\mathrm{NH_3}} \mathrm{RCONH_2} \xrightarrow{\mathrm{NaOCl/OH^-}} \mathrm{RNH_2} \xrightarrow{\mathrm{CHCl_3/KOH/\Delta}} \mathrm{R-NC}$$Explanation:• Statement I is false because $\mathrm{R-NC}$ hydrolysis produces $\mathrm{RNH_2}$ and $\mathrm{HCOOH}$, not $\mathrm{RCOOH}$. Also the claim that dipole moment of $\mathrm{R-CN}$ is greater is correct, but the hydrolysis product is wrong.• Statement II is true because $\mathrm{R-CN}$ first hydrolyzes to $\mathrm{RCOOH}$, then converted to $\mathrm{RCOCl}$ with $\mathrm{SOCl_2}$, then to $\mathrm{RCONH_2}$ with $\mathrm{NH_3}$, then to $\mathrm{RNH_2}$ via Hofmann bromamide reaction, and finally to $\mathrm{R-NC}$ via carbylamine reaction (using $\mathrm{CHCl_3/KOH}$).Answer:Statement I is false, Statement II is true. Therefore, option B is correct.