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Practical Organic Chemistry question

2021 · 16 Mar · Shift 2 · Q1
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Practical Organic Chemistry question

2021 · 16 Mar · Shift 2 · Q1

JEE MainChemistryPractical Organic ChemistryMCQ+4 / −1
Match List - I with List - II :

List - I
Test/Reagents/Observation(s)		 List - II
Species detected
(a) Lassaigne's Test (i) Carbon
(b) Cu(II) oxide (ii) Sulphur
(c) Silver nitrate (iii) N, S, P, and halogen
(d) The sodium fusion extract
gives black precipitate with
acetic acid and lead acetate		 (iv) Halogen Specifically


The correct match is :
  1. A
    (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
  2. B
    (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
  3. C
    (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
  4. D
    (a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)
View written solutionFree

Correct answer: C

  1. Identify each test/reagent/observation

    We match each item in List-I with what it detects in List-II.

  1. (a) Lassaigne's Test

    Lassaigne's sodium fusion test is used to detect extra elements present in an organic compound, namely:

    • Nitrogen
    • Sulphur
    • Phosphorus
    • Halogens

    Therefore, (a)→(iii)(a) \to (iii)(a)→(iii)

  1. (b) Cu(II) oxide

    On heating an organic compound with copper(II) oxide, carbon present in the compound gets oxidized to CO2\mathrm{CO_2}CO2​.

    Thus CuO is used for detection of: (b)→(i) Carbon(b) \to (i) \text{ Carbon}(b)→(i) Carbon

  1. (c) Silver nitrate

    Silver nitrate test is used for detection of halogens specifically, because halide ions give precipitates:

    • AgCl\mathrm{AgCl}AgCl : white
    • AgBr\mathrm{AgBr}AgBr : pale yellow
    • AgI\mathrm{AgI}AgI : yellow

    Hence, (c)→(iv)(c) \to (iv)(c)→(iv)

  1. (d) Sodium fusion extract gives black precipitate with acetic acid and lead acetate

    In Lassaigne's extract, sulphur forms Na2S\mathrm{Na_2S}Na2​S. On treatment with lead acetate in acetic acid medium, black precipitate of lead sulphide forms: Na2S+(CH3COO)2Pb→PbS↓+2CH3COONa\mathrm{Na_2S + (CH_3COO)_2Pb \to PbS \downarrow + 2CH_3COONa}Na2​S+(CH3​COO)2​Pb→PbS↓+2CH3​COONa

    Black precipitate of PbS\mathrm{PbS}PbS indicates sulphur.

    Therefore, (d)→(ii)(d) \to (ii)(d)→(ii)

  1. Final matching

    (a)−(iii),(b)−(i),(c)−(iv),(d)−(ii)(a)-(iii),\quad (b)-(i),\quad (c)-(iv),\quad (d)-(ii)(a)−(iii),(b)−(i),(c)−(iv),(d)−(ii)

    This corresponds to Option C.

  1. Comparison with stored answer

    Stored correct answer = C

    Our derived answer = C

    So, the answer agrees with the stored answer.

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