Match the LIST-I with LIST-II 

Choose the correct answer from the options given below :

We recall the characteristic qualitative tests used in practical organic chemistry:

Reagent A : $$NaHCO_3$$ (sodium bicarbonate solution)
  • Carboxylic acids react with $$NaHCO_3$$ producing brisk effervescence of $$CO_2$$.
  • Hence this test identifies the $$-COOH$$ group.
  ⇒ Functional group detected = II (carboxylic acid).

Reagent B : neutral $$FeCl_3$$ solution
  • Phenols form coloured (purple, green, blue) complexes with neutral $$Fe^{3+}$$ ions.
  • Therefore it is used to confirm phenolic $$-OH$$ groups.
  ⇒ Functional group detected = III (phenolic -OH).

Reagent C : ceric ammonium nitrate, $$ (NH_4)_2[Ce(NO_3)_6] $$
  • Alcohols give a wine-red or magenta colour with this reagent due to formation of a Ce(IV) complex.
  • Thus it detects alcoholic $$-OH$$ groups.
  ⇒ Functional group detected = IV (alcoholic -OH).

Reagent D : alkaline $$KMnO_4$$ (Baeyer’s test)
  • Alkaline $$KMnO_4$$ is decolourised by compounds containing $$C=C$$ or $$C \equiv C$$ bonds (unsaturation).
  • Hence it is the standard test for double or triple bonds.
  ⇒ Functional group detected = I (double bond / unsaturation).

Collecting the matches:
  • A → II
  • B → III
  • C → IV
  • D → I

This corresponds to Option A: A-II, B-III, C-IV, D-I.

Therefore, the correct answer is Option A.