Choose the correct answer from options given below:

The four reagents that are routinely used to identify and distinguish aldehydes and ketones are Fehling’s solution, the Iodoform reagent, 2,4-DNP (Brady’s reagent) and Schiff’s reagent. Each of them gives a very characteristic observation that the examiner places in List-II. We recall the basic chemistry of every reagent first, then match.

Case A : Fehling’s solution
Fehling’s solution contains $$Cu^{2+}$$ complexed with tartrate ion in an alkaline medium. Only aliphatic aldehydes can reduce $$Cu^{2+}$$ to $$Cu_2O$$, giving a brick-red precipitate. Neither ketones nor aromatic aldehydes respond.
⇒ Observation matched with Fehling’s solution is “formation of a red (brick-red) precipitate”. This statement is numbered IV.

Case B : Iodoform test ($$I_2/NaOH$$)
The haloform reaction is given by carbonyl compounds that contain the $$CH_3CO$$- group (i.e. methyl ketones) and by acetaldehyde (which is readily oxidised to that group). The tell-tale observation is a pale-yellow precipitate having the characteristic smell of iodoform, $$CHI_3$$.
⇒ “Formation of yellow precipitate of $$CHI_3$$ (iodoform)” is statement II.

Case C : 2,4-DNP (Brady’s reagent)
Both aldehydes and ketones add to 2,4-dinitrophenylhydrazine giving intensely coloured yellow to orange crystalline hydrazones. This test is used merely to confirm the presence of the carbonyl (>C=O) group.
⇒ “Formation of yellow/orange crystalline precipitate” is statement I.

Case D : Schiff’s reagent
Schiff’s reagent is a decolourised solution of fuchsin dye in $$SO_2$$. With aldehydes the magenta colour reappears almost instantaneously, whereas ketones do not restore the colour under ordinary conditions.
⇒ “Solution turns colourless to magenta (positive with aldehydes)” is statement III.

Collecting all the matches:
A - IV (B - II) C - I D - III

Therefore the correct choice is
Option B which is: A - IV, B - II, C - I, D - III.