JEE Challenger
More from Amines
ChemMedAdv 2008 • Paper 2

Match Organic Compounds with Characteristic Tests and Reactions

Match the compounds in Column I with their characteristic test(s)/reaction(s) given in Column II. Indicate your answer by darkening the appropriate bubbles of the 4×44 \times 4 matrix given in the ORS.

Column I

(A)

Diagram 0

(B)

Diagram 1

(C)

Diagram 2

(D)

Diagram 3

Column II

(p) sodium fusion extract of the compound gives Prussian blue colour with FeSO4\text{FeSO}_4

(q) gives positive FeCl3\text{FeCl}_3 test

(r) gives white precipitate with AgNO3\text{AgNO}_3

(s) reacts with aldehydes to form the corresponding hydrazone derivative

Options

A

(A)→(R,S),  (B)→(P,Q),  (C)→(P,Q,R),  (D)→(P,S)(\text{A}) \rightarrow (\text{R}, \text{S}),\; (\text{B}) \rightarrow (\text{P}, \text{Q}),\; (\text{C}) \rightarrow (\text{P}, \text{Q}, \text{R}),\; (\text{D}) \rightarrow (\text{P}, \text{S})

Correct
B

(A)→(R),  (B)→(P,Q),  (C)→(P,Q,R),  (D)→(P,S)(\text{A}) \rightarrow (\text{R}),\; (\text{B}) \rightarrow (\text{P}, \text{Q}),\; (\text{C}) \rightarrow (\text{P}, \text{Q}, \text{R}),\; (\text{D}) \rightarrow (\text{P}, \text{S})

C

(A)→(R,S),  (B)→(P),  (C)→(P,R),  (D)→(P,S)(\text{A}) \rightarrow (\text{R}, \text{S}),\; (\text{B}) \rightarrow (\text{P}),\; (\text{C}) \rightarrow (\text{P}, \text{R}),\; (\text{D}) \rightarrow (\text{P}, \text{S})

D

(A)→(S),  (B)→(P,Q),  (C)→(P,Q),  (D)→(P,R,S)(\text{A}) \rightarrow (\text{S}),\; (\text{B}) \rightarrow (\text{P}, \text{Q}),\; (\text{C}) \rightarrow (\text{P}, \text{Q}),\; (\text{D}) \rightarrow (\text{P}, \text{R}, \text{S})

Topics & Concepts

AminesAminesAlcohols and EthersAldehydes and Ketones

Step-by-Step Solution

To find the correct matches between Column I and Column II, we evaluate the properties and characteristic reactions of each compound:

Analysis of Tests/Reactions in Column II:

  1. (p) Sodium fusion extract gives Prussian blue colour with FeSO4\text{FeSO}_4:

    • This is Lassaigne's test for the detection of nitrogen in organic compounds containing both carbon and nitrogen:
      Na+C+N→ΔNaCN\text{Na} + \text{C} + \text{N} \xrightarrow{\Delta} \text{NaCN}
      6CN−+Fe2+→[Fe(CN)6]4−6\text{CN}^- + \text{Fe}^{2+} \to [\text{Fe(CN)}_6]^{4-}
      4Fe3++3[Fe(CN)6]4−→Fe4[Fe(CN)6]3↓(Prussian blue)4\text{Fe}^{3+} + 3[\text{Fe(CN)}_6]^{4-} \to \text{Fe}_4[\text{Fe(CN)}_6]_3 \downarrow \quad (\text{Prussian blue})
    • Compounds containing nitrogen along with carbon (compounds B, C, and D) give this test.
    • Compound A does not contain carbon, hence it does not form cyanide (CN−\text{CN}^-) and gives a negative test.
    • Matches: (B) →\to p, (C) →\to p, (D) →\to p

  2. (q) Gives positive FeCl3\text{FeCl}_3 test:

    • Neutral ferric chloride (FeCl3\text{FeCl}_3) reacts with compounds having a phenolic −OH-\text{OH} group to form characteristic colored coordination complexes.
    • Compounds B (HO−C6H4−…\text{HO}-\text{C}_6\text{H}_4-\dots) and C (HO−C6H4−…\text{HO}-\text{C}_6\text{H}_4-\dots) possess a phenolic −OH-\text{OH} group, so they give a positive test.
    • Matches: (B) →\to q, (C) →\to q

  3. (r) Gives white precipitate with AgNO3\text{AgNO}_3:

    • Silver nitrate reacts with free halide ions in aqueous solution:
      Ag++Cl−→AgCl↓(white precipitate, soluble in excess NH4OH)\text{Ag}^+ + \text{Cl}^- \to \text{AgCl}\downarrow \quad (\text{white precipitate, soluble in excess } \text{NH}_4\text{OH})
      Ag++Br−→AgBr↓(pale yellow precipitate)\text{Ag}^+ + \text{Br}^- \to \text{AgBr}\downarrow \quad (\text{pale yellow precipitate})
      Ag++I−→AgI↓(yellow precipitate)\text{Ag}^+ + \text{I}^- \to \text{AgI}\downarrow \quad (\text{yellow precipitate})
    • Only chloride salts produce a white precipitate. Thus, compounds A and C (which contain Cl−\text{Cl}^-) give a white precipitate with AgNO3\text{AgNO}_3.
    • Matches: (A) →\to r, (C) →\to r

  4. (s) Reacts with aldehydes to form the corresponding hydrazone derivative:

    • Hydrazine (NH2−NH2\text{NH}_2-\text{NH}_2) and substituted hydrazines (such as 2,4-dinitrophenylhydrazine) react with aldehydes to form hydrazones:
      R−CHO+H2N−NH−R′→H+R−CH=N−NH−R′+H2O\text{R}-\text{CHO} + \text{H}_2\text{N}-\text{NH}-\text{R}' \xrightarrow{\text{H}^+} \text{R}-\text{CH}=\text{N}-\text{NH}-\text{R}' + \text{H}_2\text{O}
    • Compounds A (hydrazinium chloride) and D (2,4-dinitrophenylhydrazinium bromide) can react with aldehydes to form hydrazone derivatives.
    • (Note: In the official JEE Advanced evaluation, both the inclusion and exclusion of (s) for (A) and (D) were accepted).
    • Matches: (A) →\to s, (D) →\to s (or not matching if only free base is considered).

Final Match:

  • (A) →\to r or (r, s)
  • (B) →\to p, q
  • (C) →\to p, q, r
  • (D) →\to p or (p, s)
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