An organic compound ‘A’ with molecular formula CH4O reacts with HI to give an alkyl iodide, which then reacts with potassium cyanide to produce ‘B’. Compound ‘B’ on reduction followed by reaction with HNO2 gives an alcohol, which on oxidation gives ‘C’. Compound ‘C’ reacts with CH3MgBr to give ‘D’, which on hydrolysis gives compound ‘E’, which is also an alcohol.
Nitrous acid (HNO2), used in the reaction sequence of the passage, is prepared by mixing:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → NaNO2 + a mineral acid such as HCl or H2SO4
Nitrous acid, \( \ce{HNO2} \), is made in situ, in the reaction mixture itself, by adding a mineral acid to sodium nitrite. The acid gives its proton to the nitrite ion:
\( \displaystyle \ce{NaNO2 + HCl -> HNO2 + NaCl} \)
In the passage, compound B (\( \ce{CH3CN} \)) is reduced to ethylamine, and the nitrous acid formed in this way converts this primary aliphatic amine into ethanol, with loss of nitrogen gas.
Why the other options are incorrect
Option (1) NaNO2 + NaOH — Incorrect. There is no acid in this mixture to give a proton to the nitrite ion, so no \( \ce{HNO2} \) can form; the solution stays alkaline.
Option (2) HNO3 + OH− — Incorrect.
This is a neutralisation: nitric acid and hydroxide give the nitrate ion and water, \( \ce{HNO3 + OH^- -> NO3^- + H2O} \).
Nitrogen stays in the +5 state of nitrate, while nitrous acid has nitrogen in the +3 state.
NCERT (2026-27 edition):
5. Reaction with nitrous acid – “Three classes of amines react differently with nitrous acid which is prepared in situ from a mineral acid and sodium nitrite.”