Alkyl halide is converted into alkyl isocyanide by reaction with

Solution

Chemistry ma’am · 4 steps

Cyanide has two ends to attack with; the metal decides which end is free, so look at the bond.

1. Ambident nucleophile. The cyanide ion ⁻C≡N can bond to carbon through either end. Attack through C gives an alkyl cyanide, R-C≡N. Attack through N gives an alkyl isocyanide, R-N≡C.
2. Ionic cyanides. KCN, NaCN and NH₄CN are mainly ionic, so the free CN⁻ ion attacks. Carbon is the better nucleophilic end, because the C–C bond formed is stronger than a C–N bond. So these give cyanides.
3. Covalent AgCN. AgCN is mainly covalent, Ag-C≡N, so the carbon end is already tied to silver. Only the lone pair on nitrogen is free to attack, so the alkyl group bonds to N.

   R-X + AgCN → R-N≡C + AgX

4. So. The alkyl halide becomes an alkyl isocyanide with AgCN, option (d).

The idea to keep. For ambident ions, ionic salts (KCN, KNO2) attack through the stronger nucleophilic atom; covalent silver salts (AgCN, AgNO2) attack through the other atom, giving isocyanides and nitroalkanes.

Answer from the NTA final answer key. Solution by PrepWiser.