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NEET Purification and Characterisation of Organic Compounds MCQs with answers

448 questions on Purification and Characterisation of Organic Compounds for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. A compound with molecular formula C₆H₅NO₂ is subjected to the sodium fusion test. Which of the following balanced ionic equations accurately represents the final formation of the coloured complex indicating the presence of nitrogen?

  1. Fe³⁺ + [Fe(CN)₆]⁴⁻ → Fe[Fe(CN)₆]⁻
  2. Fe²⁺ + [Fe(CN)₆]³⁻ → Fe₃[Fe(CN)₆]₂
  3. 4Fe³⁺ + 3[Fe(CN)₆]⁴⁻ → Fe₄[Fe(CN)₆]₃
  4. 3Fe²⁺ + 2[Fe(CN)₆]³⁻ → Fe₃[Fe(CN)₆]₂
Show answer and explanation

Answer: (C) 4Fe³⁺ + 3[Fe(CN)₆]⁴⁻ → Fe₄[Fe(CN)₆]₃

The formation of Prussian blue (ferric ferrocyanide) occurs between Fe³⁺ (ferric) ions and [Fe(CN)₆]⁴⁻ (ferrocyanide) ions: 4Fe³⁺ + 3[Fe(CN)₆]⁴⁻ → Fe₄[Fe(CN)₆]₃ (iron(III) hexacyanoferrate(II)).

medium

2. On complete combustion, 0.2 g of an organic compound gave 0.513 g of CO₂. The percentage of carbon in the compound is:

  1. 70 %
  2. 30 %
  3. 50.9 %
  4. 35 %
Show answer and explanation

Answer: (A) 70 %

%C = (12/44) × (mass of CO₂/mass of compound) × 100 = (12/44) × (0.513/0.2) × 100 = 70%.

hard

3. An aqueous solution (100 mL) contains 10 g of a compound X. It is extracted using an organic solvent for which the partition coefficient K = [X]_organic / [X]_aqueous = 4. What mass of compound X remains in the aqueous layer after two successive extractions using 50 mL of organic solvent each time?

  1. 0.28 g
  2. 3.33 g
  3. 1.11 g
  4. 2.50 g
Show answer and explanation

Answer: (C) 1.11 g

The formula for remaining mass is W_n = W₀ · [V_aq / (V_aq + K · V_org)]ⁿ. Here W₀ = 10 g, V_aq = 100 mL, V_org = 50 mL, and K = 4. For one extraction factor: 100 / (100 + 4 × 50) = 100 / 300 = 1/3. After two extractions: W₂ = 10 × (1/3)² = 10 / 9 ≈ 1.11 g.

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4. Which of the following statements about column chromatography is INCORRECT?

  1. The mobile phase may be a single solvent or a mixture of solvents.
  2. The component with higher affinity for the stationary phase is retained longer in the column.
  3. The stationary phase in column chromatography is always a liquid supported on an inert solid.
  4. The process of washing and extracting the adsorbed components from the column is termed elution.
Show answer and explanation

Answer: (C) The stationary phase in column chromatography is always a liquid supported on an inert solid.

In standard adsorption column chromatography, the stationary phase is a solid adsorbent such as alumina or silica gel, not necessarily a liquid supported on a solid. Thus, statement 3 is incorrect.

easy

5. Given below are two statements: Statement I: The retardation factor (Rf) of a compound on a silica gel plate is an intensive property and is completely independent of the developing solvent system used. Statement II: Amino acids can be detected on a developed chromatogram as coloured spots by spraying with ninhydrin solution. In the light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (C) Statement I is incorrect but Statement II is correct

Statement I is incorrect because the Rf value strongly depends on the nature and polarity of the mobile phase (solvent) as well as the adsorbent. Statement II is correct because ninhydrin is a standard reagent used to detect amino acids, giving characteristic purple or violet spots (Ruhemann's purple).

hard

6. In a modified quantitative estimation of oxygen by the Unterzaucher method, 0.160 g of an organic compound was decomposed thermally. The liberated oxygen was converted quantitatively into carbon monoxide over carbon at 1100 °C, and then oxidized to carbon dioxide using iodine pentoxide (I₂O₅), producing 0.0508 g of iodine (I₂). What is the percentage of oxygen in the compound? [Atomic mass: I = 127 g mol⁻¹, O = 16 g mol⁻¹]

  1. 10.0%
  2. 20.0%
  3. 5.0%
  4. 15.0%
Show answer and explanation

Answer: (A) 10.0%

The sequence of reactions is: Compound → O₂; O₂ + 2C → 2CO (thus 1 mol O₂ yields 2 mol CO; or 1 mol O atom yields 1 mol CO); 5CO + I₂O₅ → 5CO₂ + I₂. Hence, 1 mol of I₂ corresponds to 5 mol of CO, which corresponds to 5 mol of O atoms. Moles of I₂ produced = 0.0508 / (2 × 127) = 0.0508 / 254 = 0.0002 mol. Moles of O atoms = 5 × 0.0002 = 0.0010 mol. Mass of O = 0.0010 × 16 = 0.016 g. Percentage of O = (0.016 / 0.160) × 100 = 10.0%.

medium

7. In Kjeldahl's method, the ammonia evolved from 1 g of an organic compound neutralised 35 mL of 0.1 N H₂SO₄. The percentage of nitrogen in the compound is:

  1. 2.45 %
  2. 9.8 %
  3. 4.9 %
  4. 7.35 %
Show answer and explanation

Answer: (C) 4.9 %

%N = 1.4 × N × V/m = 1.4 × 0.1 × 35/1 = 4.9%.

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8. A chemist tests a sodium fusion extract containing sodium sulphide (Na₂S) by adding sodium nitroprusside solution, but accidentally adds a few drops of concentrated sulphuric acid beforehand. What will be observed?

  1. The purple colour develops much faster and with greater intensity
  2. A blood-red precipitate forms instead of a purple solution
  3. No purple colour appears because the sulphide is converted to volatile H₂S gas and the nitroprusside test requires an alkaline medium
  4. A dark black precipitate of iron(II) sulphide immediately deposits
Show answer and explanation

Answer: (C) No purple colour appears because the sulphide is converted to volatile H₂S gas and the nitroprusside test requires an alkaline medium

The reaction between sulphide ions and sodium nitroprusside to yield the purple complex [Fe(CN)₅NOS]⁴⁻ strictly requires an alkaline medium. In acidic solution, S²⁻ is protonated to weakly dissociated or gaseous H₂S, preventing the formation of the complex anion.

medium

9. In Lassaigne's test for nitrogen, the alkaline sodium fusion extract is first boiled with freshly prepared ferrous sulphate solution. What is the primary role of sodium hydroxide initially present in the extract during this step?

  1. To react with FeSO₄ to form a green precipitate of Fe(OH)₂ which reacts with NaCN
  2. To react with ferrous sulphate to form ferrous hydroxide, which dissolves in NaCN
  3. To oxidise NaCN directly to sodium cyanate (NaCNO)
  4. To convert ferrous ions into insoluble ferric oxide
Show answer and explanation

Answer: (A) To react with FeSO₄ to form a green precipitate of Fe(OH)₂ which reacts with NaCN

The alkaline medium provided by NaOH reacts with FeSO₄ to form a precipitate of ferrous hydroxide, Fe(OH)₂. This freshly precipitated Fe(OH)₂ reacts with sodium cyanide (NaCN) present in the fusion extract to form sodium hexacyanidoferrate(II), Na₄[Fe(CN)₆].

medium

10. In Kjeldahl's method, the ammonia evolved from 0.75 g of an organic compound neutralised 15 mL of 0.5 N H₂SO₄. The percentage of nitrogen in the compound is:

  1. 7 %
  2. 28 %
  3. 14 %
  4. 3.94 %
Show answer and explanation

Answer: (C) 14 %

%N = 1.4 × N × V/m = 1.4 × 0.5 × 15/0.75 = 14%.

medium

11. In Kjeldahl's method, the ammonia evolved from 1 g of an organic compound neutralised 25 mL of 0.2 N H₂SO₄. The percentage of nitrogen in the compound is:

  1. 14 %
  2. 10.5 %
  3. 7 %
  4. 3.5 %
Show answer and explanation

Answer: (C) 7 %

%N = 1.4 × N × V/m = 1.4 × 0.2 × 25/1 = 7%.

easy

12. In the Dumas method, 0.246 g of an organic compound yielded 22.4 mL of dry nitrogen gas measured at 273 K and 1 atm (STP). The percentage of nitrogen in the compound is:

  1. 5.69%
  2. 22.76%
  3. 11.38%
  4. 17.07%
Show answer and explanation

Answer: (C) 11.38%

Volume of N₂ at STP = 22.4 mL. Moles of N₂ = 22.4 / 22400 = 0.001 mol. Mass of N₂ = 0.001 mol × 28 g mol⁻¹ = 0.028 g. Percentage of N = (0.028 / 0.246) × 100 ≈ 11.38%.

hard

13. An organic compound contains nitrogen, sulphur, and bromine. When subjected to sodium fusion with a stoichiometric (limited) amount of sodium metal, it yields an extract containing NaSCN and NaBr. If this extract is directly treated with AgNO₃ solution without boiling with concentrated HNO₃, what precipitate(s) will be formed?

  1. AgBr only (pale yellow precipitate)
  2. Black precipitate of Ag₂S and yellow precipitate of AgI
  3. White precipitate of AgSCN and pale yellow precipitate of AgBr
  4. Curdy white precipitate of AgCl only
Show answer and explanation

Answer: (C) White precipitate of AgSCN and pale yellow precipitate of AgBr

Both thiocyanate ions (SCN⁻) and bromide ions (Br⁻) react with Ag⁺ ions. SCN⁻ gives a white precipitate of AgSCN, while Br⁻ gives a pale yellow precipitate of AgBr. Therefore, failure to decompose SCN⁻ by boiling with HNO₃ results in a mixed precipitate of AgSCN and AgBr.

medium

14. In Carius estimation, 0.25 g of an organic compound gave 0.294 g of AgBr. The percentage of bromine in the compound is:

  1. 67.5 %
  2. 25 %
  3. 37 %
  4. 50 %
Show answer and explanation

Answer: (D) 50 %

%bromine = (80/188) × (0.294/0.25) × 100 = 50%.

medium

15. On complete combustion, 0.2 g of an organic compound gave 0.477 g of CO₂. The percentage of carbon in the compound is:

  1. 35 %
  2. 32.5 %
  3. 47.3 %
  4. 65 %
Show answer and explanation

Answer: (D) 65 %

%C = (12/44) × (mass of CO₂/mass of compound) × 100 = (12/44) × (0.477/0.2) × 100 = 65%.

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