Match List-I with List-II
List-I | List-II |
(A) \(C_6H_5CHO\) | (I) Precipitation with 2,4-DNP |
(B) \(CH_3C≡CH\) | (II) Precipitation with \(AgNO_3\) (ammoniacal) |
(C) \(HCN\) | (III) Nucleophile |
(D) \(I^-\) | (IV) Cyanohydrin formation |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-I, B-II, C-IV, D-III
Each item is matched by the reaction its functional group gives: the carbonyl group of an aldehyde condenses with ammonia derivatives such as 2,4-DNP and adds HCN; a terminal alkyne has an acidic ≡C–H hydrogen; and an anion with a lone pair, such as \( \ce{I^{-}} \), acts as a nucleophile.
(A) \( \ce{C6H5CHO} \) — This matches (I) Precipitation with 2,4 DNP. Benzaldehyde forms an orange-red 2,4-dinitrophenylhydrazone precipitate, the test for a C=O group.
\( \ce{C6H5CHO + H2N-NH-C6H3(NO2)2 -> C6H5CH=N-NH-C6H3(NO2)2 v + H2O} \)
(B) \( \ce{CH3C} \)≡CH — This matches (II) Precipitation with AgNO3 (ammoniacal). Propyne is a terminal alkyne; its acidic H is replaced by silver, giving a white precipitate of silver propynide.
\( \ce{CH3C#CH + [Ag(NH3)2]^+ -> CH3C#CAg v + NH4^+ + NH3} \)
(C) HCN — This matches (IV) Cyanohydrin formation. HCN adds across C=O of aldehydes and ketones (base-catalysed; \( \ce{CN^{-}} \) is the attacking nucleophile).
\( \ce{CH3CHO + HCN -> CH3CH(OH)CN} \)
(D) \( \ce{I^{-}} \) — This matches (III) Nucleophile. The iodide ion is electron-rich and attacks electron-deficient carbon, as in the Finkelstein reaction (R–Cl + NaI → R–I + NaCl).
Shortcut: \( \ce{I^{-}} \) can only be the nucleophile (D-III); only Option (3) has it.