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Detection of hydroxyl, carbonyl, carboxyl and amino functional groups: NEET MCQs with answers

50 questions on this topic of Principles Related to Practical Chemistry (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following organic compounds will evolve brisk effervescence of CO₂ with saturated aqueous NaHCO₃ solution?

  1. Phenol
  2. p-Nitrophenol
  3. 2,4,6-Trinitrophenol
  4. o-Cresol
Show answer and explanation

Answer: (C) 2,4,6-Trinitrophenol

Phenol and mono-nitrophenols are weaker acids than carbonic acid (H₂CO₃) and do not decompose NaHCO₃ to release CO₂. However, 2,4,6-trinitrophenol (picric acid) has three strong electron-withdrawing nitro groups, making it strongly acidic (pKa ≈ 0.38), so it readily displaces CO₂ from sodium bicarbonate.

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2. Given below are two statements: Statement I: Secondary amines react with nitrous acid to yield yellow oily N-nitrosamines which are insoluble in water. Statement II: Tertiary aliphatic amines react with benzenesulphonyl chloride to produce a precipitate soluble in alkali. In the light of the above statements, choose the correct 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
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Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because secondary amines form water-insoluble, yellow oily N-nitrosamines with HNO₂. Statement II is incorrect because tertiary aliphatic amines do not possess any replaceable hydrogen on nitrogen and hence do not react with benzenesulphonyl chloride.

hard

3. An unknown organic compound 'W' gives a positive 2,4-dinitrophenylhydrazine test and forms a silver mirror with Tollens' reagent. When 0.1 mol of 'W' is completely oxidized by Tollens' reagent, how many moles of Ag metal are precipitated?

  1. 0.20 mol
  2. 0.10 mol
  3. 0.05 mol
  4. 0.40 mol
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Answer: (A) 0.20 mol

The oxidation of an aldehyde by Tollens' reagent follows the stoichiometry: RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag(s) + 4NH₃ + 2H₂O. Therefore, 1 mole of aldehyde produces 2 moles of Ag. For 0.1 mol of 'W', 0.20 mol of Ag is precipitated.

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4. Match List-I (Functional group test) with List-II (Characteristic observation): List-I: (A) Biuret test (B) Liebermann's nitroso test (C) Legal's test (D) Tollen's test List-II: (I) Red/violet colour with alkaline nitroprusside (II) Violet-pink colour with dilute CuSO₄ and NaOH (III) Red colour turning blue on addition of alkali (IV) Silver mirror formation Choose the correct answer from the options given below:

  1. A-II, B-III, C-I, D-IV
  2. A-III, B-II, C-IV, D-I
  3. A-II, B-I, C-III, D-IV
  4. A-IV, B-III, C-I, D-II
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Answer: (A) A-II, B-III, C-I, D-IV

Biuret test gives a violet-pink colour for compounds with -CONH- linkages. Liebermann's nitroso test is characteristic of phenols (red turning to blue with alkali). Legal's test detects ketones containing α-hydrogen via alkaline nitroprusside. Tollens' test detects aldehydes via silver mirror.

easy

5. Phthalein dye test is widely used in the laboratory for the detection of:

  1. Aliphatic carboxylic acids
  2. Phenols having an unsubstituted para position
  3. Primary aliphatic amines
  4. Aliphatic aldehydes
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Answer: (B) Phenols having an unsubstituted para position

The phthalein dye test involves heating phenol with phthalic anhydride and concentrated H₂SO₄, followed by addition of excess dilute NaOH. Phenols with a free para position condense to form phenolphthalein-type indicators that give characteristic pink/red/blue colours in alkaline solution.

hard

6. A student performs the sodium nitroprusside test on an unknown organic liquid. Upon adding a few drops of sodium nitroprusside solution followed by dilute aqueous NaOH, an intense blood-red/purple colour develops. This confirms the presence of:

  1. A primary aliphatic amine
  2. A tertiary alcohol
  3. An aliphatic carboxylic acid
  4. A ketone containing a -COCH₃ or active methylene group
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Answer: (D) A ketone containing a -COCH₃ or active methylene group

Ketones containing a -COCH₂- or -COCH₃ group react with nitroprusside ions ([Fe(CN)₅NO]²⁻) in an alkaline medium. The enolate anion formed by NaOH reacts with the nitroprusside ion to form a complex anion [Fe(CN)₅(NO-CH₂COCH₃)]³⁻, producing a characteristic blood-red or purple coloration.

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7. An organic compound 'Z' gives a bright red dye when dissolved in dilute HCl, cooled to 0-5 °C, treated with NaNO₂, and subsequently poured into an alkaline solution of β-naphthol. Compound 'Z' is:

  1. Aniline
  2. N-Methylaniline
  3. Benzylamine
  4. Acetamide
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Answer: (A) Aniline

Aromatic primary amines such as aniline undergo diazotisation at 0-5 °C to form benzenediazonium chloride. When coupled with an alkaline solution of β-naphthol (2-naphthol), an azo dye (1-phenylazo-2-naphthol) with a vivid scarlet/red colour is formed. Benzylamine gives alcohol and nitrogen gas rather than a stable diazonium salt.

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8. In the azo-dye test for aromatic primary amines, the reaction mixture must be kept at a strictly controlled low temperature (0-5 °C). If the temperature rises above 10 °C during diazotisation of aniline, the primary product formed instead of the diazonium salt is:

  1. Nitrobenzene
  2. Chlorobenzene
  3. Phenol
  4. Benzene
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Answer: (C) Phenol

Aromatic diazonium salts are thermally unstable in aqueous medium. If the temperature exceeds 5-10 °C during diazotisation, benzenediazonium chloride undergoes hydrolysis with water to yield phenol and nitrogen gas: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl.

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9. In the qualitative detection of the phenolic hydroxyl group, the ferric chloride reagent must be strictly 'neutral'. If acidic ferric chloride solution is used instead, what is the most likely consequence?

  1. The reaction mixture turns turbid due to immediate chlorination of the aromatic ring
  2. Ferric chloride decomposes rapidly to precipitate brown ferric hydroxide
  3. The phenol is completely oxidised to benzene-1,4-dione without any intermediate complexation
  4. The characteristic violet or coloured complex fails to form because excess H⁺ represses phenoxide ion formation
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Answer: (D) The characteristic violet or coloured complex fails to form because excess H⁺ represses phenoxide ion formation

Phenol reacts with neutral FeCl₃ to form a coloured coordination complex, [Fe(OC₆H₅)₆]³⁻. Because phenoxide ions are required for coordination and phenol is a weak acid, excess acid (H⁺) protonates phenoxide back to phenol, shifting the equilibrium backward and suppressing complex formation. Hence, neutral FeCl₃ is strictly required.

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10. A compound 'M' forms an oxime with hydroxylamine and gives a yellow crystalline precipitate with I₂ and aqueous NaOH. However, it does not reduce Tollen's reagent. Compound 'M' is:

  1. Propanal
  2. Pentanoic acid
  3. Pentan-2-one
  4. Pentan-3-one
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Answer: (C) Pentan-2-one

The formation of an oxime confirms the presence of a carbonyl group (>C=O). Not reducing Tollen's reagent indicates it is a ketone, not an aldehyde. A positive iodoform test (yellow precipitate of CHI₃) confirms the presence of a methyl ketone (CH₃-CO-) group, which corresponds to pentan-2-one.

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11. When acetaldehyde is treated with Schiff's reagent, the restored pink/magenta colour is due to:

  1. Reduction of sulfurous acid to elemental sulfur
  2. Oxidation of acetaldehyde to acetic acid by basic fuchsine
  3. Addition of aldehyde to the decolourised rosaniline hydrochloride complex
  4. Complexation of acetaldehyde with bisulfite ions alone
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Answer: (C) Addition of aldehyde to the decolourised rosaniline hydrochloride complex

Schiff's reagent is prepared by decolourising rosaniline hydrochloride (fuchsine dye) using sulfur dioxide (SO₂). When an aldehyde is added, it reacts with the sulfurous acid/dye complex, restoring the conjugated quinonoid structure of the dye, which brings back the characteristic pink/magenta coloration.

easy

12. During the iodoform test of an unknown compound 'Q', 0.1 mol of 'Q' produces exactly 0.1 mol of CHI₃ upon reaction with iodine and sodium hydroxide. Which of the following cannot be compound 'Q'?

  1. Ethanol
  2. Propan-2-one
  3. Pentan-3-one
  4. Butan-2-ol
Show answer and explanation

Answer: (C) Pentan-3-one

The iodoform test is given by compounds containing the CH₃-C=O or CH₃-CH(OH)- group. Pentan-3-one has the structure CH₃-CH₂-CO-CH₂-CH₃; it is a symmetrical ketone lacking a terminal methyl carbonyl group and therefore does not give a yellow precipitate of iodoform.

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