In the following sequence of reactions, ‘A’ is
\( \text{A} \xrightarrow[95\%\ \text{H}_3\text{PO}_4]{\text{NaI}} \text{B} \xrightarrow{\text{AgNO}_2} \text{C} \xrightarrow[\text{ethanol}]{\text{H}_2/\text{Pd}} \)
\( \text{D} \xrightarrow[\text{(ii) H}_3\text{O}^+]{\text{(i) NaNO}_2 + \text{HCl}} \text{Ethyl alcohol} \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Diethyl ether
Work back from the last step. A primary amine gives an alcohol with nitrous acid (\( \ce{NaNO2} \) + HCl), so D is ethylamine. Each step then fixes the one before it:
\( \ce{C2H5-O-C2H5 + HI -> C2H5I + C2H5OH} \) (A → B)
\( \ce{C2H5I + AgNO2 -> C2H5NO2 + AgI} \) (B → C, nitroethane)
\( \ce{C2H5NO2} \xrightarrow{\ce{\text{H2/Pd, ethanol}}} \ce{C2H5NH2} \) (C → D, ethylamine)
\( \ce{C2H5NH2 + HNO2 -> C2H5OH + N2 + H2O} \) (D → ethyl alcohol)
NaI with 95% \( \ce{H3PO4} \) generates HI, which cleaves the ether into ethyl iodide:
\( \ce{NaI + H3PO4 -> NaH2PO4 + HI} \)
So A is diethyl ether.
Why the other options are incorrect
Option (1) Acetaldehyde — Incorrect. An aldehyde is not cleaved by HI to an alkyl iodide, so B could not be ethyl iodide.
Option (3) Dimethyl ether — Incorrect. It would give \( \ce{CH3I} \), then nitromethane and methylamine, and the final product would be methanol, not ethyl alcohol.
Option (4) Nitroethane — Incorrect. This is compound C in the sequence, not the starting compound A.
NCERT (2026-27 edition):
Cleavage of C–O Bond in Ethers – “Ethers are the least reactive of the functional groups. The cleavage of C-O bond in ethers takes place under drastic conditions with excess of hydrogen halides. The reaction of dialkyl ether gives two alkyl halide molecules.”