JEE Main 2022 · 29 July · morning · Single correct · +4 / −1
Considering the above reactions, the compound 'A' and compound 'B' respectively are :
- (a) CH₃CH₂-N(+)≡C(-) , CH₃CH₂-N(+)≡C(-)
- (b)
- (c)
- (d)
Answer and solution
Answer (c)
Solution
Chemistry ma’am · 3 steps
Cyanide has two ends that can attack. The metal decides which end is free to bond with carbon.
- Ambident nucleophile. The cyanide group can attack through carbon or through nitrogen, because both atoms carry a lone pair. Attack through C gives a nitrile (R-C≡N). Attack through N gives an isocyanide (R-N≡C).
Reagent Nature of bond Free end that attacks Product from CH₃CH₂Cl AgCN mainly covalent, Ag-C≡N N (the carbon is held by Ag) A = ethyl isocyanide, CH₃CH₂NC NaCN ionic, free CN⁻ ion C (forms the stronger C–C bond) B = propanenitrile, CH₃CH₂CN - Why it differs. In NaCN, the carbon end of the free ion is available, and a C–C bond is more stable than a C–N bond, so the nitrile forms. In AgCN, carbon is already bonded to silver, so only the nitrogen lone pair can reach the ethyl carbon.
- So. A is ethyl isocyanide and B is propanenitrile (ethyl cyanide). That matches option (c).
The idea to keep. Ionic KCN or NaCN gives the nitrile, because carbon attacks. Covalent AgCN gives the isocyanide, because silver holds the carbon and nitrogen attacks. The same rule holds for KNO2 versus AgNO2.
Answer from the NTA final answer key. Solution by PrepWiser.