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Phenols: acidity, substitutions and Reimer-Tiemann reaction: NEET MCQs with answers

50 questions on this topic of Organic Compounds Containing Oxygen (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. What is the total number of valence electrons present in the neutral reactive intermediate :CCl₂ formed in the Reimer-Tiemann reaction?

  1. 4
  2. 6
  3. 8
  4. 10
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Answer: (B) 6

Dichlorocarbene (:CCl₂) has a central carbon atom bonded to two chlorine atoms via single covalent bonds and bears one unshared pair of electrons. Thus, the valence shell of carbon possesses 2 + 2 + 2 = 6 electrons, making it an electron-deficient electrophile with an incomplete octet.

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2. In the Reimer-Tiemann reaction, phenol treated with chloroform and aqueous NaOH, followed by acidification, gives:

  1. Benzoic acid
  2. Picric acid
  3. Salicylaldehyde (o-hydroxybenzaldehyde)
  4. Salicylic acid
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Answer: (C) Salicylaldehyde (o-hydroxybenzaldehyde)

The Reimer-Tiemann reaction introduces a -CHO group at the ortho position of phenol, giving salicylaldehyde.

NCERT Class 12

hard

3. In the Reimer-Tiemann reaction of phenol with chloroform and aqueous alkali, which step is recognized as the rate-determining step (RDS) of the overall transformation?

  1. Generation of dichlorocarbene (:CCl₂) from chloroform
  2. Hydrolysis of the gem-dichloride intermediate to an aldehyde
  3. Electrophilic attack of dichlorocarbene on the phenoxide ring
  4. Deprotonation of the intermediate cyclohexadienone derivative to restore aromaticity
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Answer: (C) Electrophilic attack of dichlorocarbene on the phenoxide ring

The rate-determining step in the Reimer-Tiemann reaction is the electrophilic attack of the neutral, electron-deficient dichlorocarbene (:CCl₂) species onto the ortho-position of the electron-rich phenoxide ring, which disrupts the aromatic sextet.

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4. Match the substituents on the phenol ring in List-I with their relative acidic strength compared to unsubstituted phenol in List-II: List-I: A. -OCH₃ at para position B. -NO₂ at meta position C. -Cl at para position D. -CH₃ at meta position List-II: I. More acidic due to -I only (no resonance with oxygen) II. Less acidic due to strong +R effect dominating over -I effect III. More acidic because -I dominates over +R effect IV. Less acidic due to +I effect only

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

At the para position, -OCH₃ shows a strong +R effect that destabilizes the phenoxide ion, making it less acidic (A-II). At the meta position, -NO₂ acts only via electron-withdrawing -I effect, increasing acidity (B-I). Halogens like -Cl exhibit a stronger -I effect than +R effect in phenols, increasing acidity relative to phenol (C-III). A meta -CH₃ group donates electron density solely through +I effect, slightly decreasing acidity (D-IV).

easy

5. Which of the following substituted phenols has the lowest pKa value?

  1. 2,6-Dichlorophenol
  2. 3-Chlorophenol
  3. 4-Chlorophenol
  4. 2-Chlorophenol
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Answer: (A) 2,6-Dichlorophenol

The lowest pKa corresponds to the strongest acid. Chlorine exhibits a strong -I effect that stabilizes the conjugate phenoxide base. 2,6-Dichlorophenol possesses two chlorine atoms at the ortho positions, very close to the negative oxygen, maximizing stabilization via inductive electron withdrawal and giving it the highest acidity (lowest pKa).

hard

6. Identify the final organic product (Z) in the given sequence of reactions: Phenol + Concentrated H₂SO₄ (heated at 373 K) → X X + Concentrated HNO₃ (heated) → Y Y + PCl₅ → Z

  1. 2,4,6-Trichlorophenol
  2. 2,4-Dinitrochlorobenzene
  3. 4-Chlorobenzenesulphonic acid
  4. Picryl chloride (2,4,6-trinitrochlorobenzene)
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Answer: (D) Picryl chloride (2,4,6-trinitrochlorobenzene)

Heating phenol with concentrated H₂SO₄ gives phenol-2,4-disulphonic acid (X). Subsequent treatment with concentrated HNO₃ replaces the sulphonic acid groups by electrophilic ipso-substitution, forming picric acid (2,4,6-trinitrophenol, Y). Reaction of picric acid with PCl₅ replaces the phenolic OH group with Cl, yielding picryl chloride (Z).

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7. Sodium phenoxide heated with CO₂ under pressure, followed by acidification, gives salicylic acid. This is:

  1. Reimer-Tiemann reaction
  2. Cannizzaro reaction
  3. Williamson synthesis
  4. Kolbe's reaction
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Answer: (D) Kolbe's reaction

In Kolbe's reaction, the phenoxide ion (more reactive than phenol towards electrophiles) reacts with CO₂ to form ortho-hydroxybenzoic acid.

NCERT Class 12

easy

8. When phenol is treated with dilute HNO₃ at 298 K, two mononitro derivatives are obtained. How are these two isomers separated industrially?

  1. Fractional crystallization based on molecular mass
  2. Column chromatography using alumina
  3. Sublimation under reduced pressure
  4. Steam distillation based on differences in intramolecular and intermolecular hydrogen bonding
Show answer and explanation

Answer: (D) Steam distillation based on differences in intramolecular and intermolecular hydrogen bonding

o-Nitrophenol has intramolecular hydrogen bonding (chelation), making it more volatile, whereas p-nitrophenol has intermolecular hydrogen bonding, leading to association and a higher boiling point. Consequently, o-nitrophenol is steam-volatile and can be cleanly separated from p-nitrophenol by steam distillation.

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9. The correct decreasing order of pKa values among the following substituted phenols is:

  1. p-Cresol > Phenol > m-Nitrophenol > p-Nitrophenol
  2. p-Nitrophenol > m-Nitrophenol > Phenol > p-Cresol
  3. Phenol > p-Cresol > p-Nitrophenol > m-Nitrophenol
  4. m-Nitrophenol > p-Nitrophenol > Phenol > p-Cresol
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Answer: (A) p-Cresol > Phenol > m-Nitrophenol > p-Nitrophenol

Acidic strength decreases with increasing pKa. Electron-withdrawing -NO₂ groups increase acidity (lower pKa) more strongly at the para position via resonance (-R and -I) than at the meta position (-I only). Electron-donating methyl group in p-cresol decreases acidity, giving it the highest pKa value.

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10. Consider the following statements regarding the acidic character of substituted phenols: Statement I: Introduction of an electron-withdrawing nitro group at the meta-position increases acidity mainly through inductive effect rather than resonance effect. Statement II: The conjugate base of m-nitrophenol does not have any resonance structure in which the negative charge resides directly on the carbon bearing the nitro group. 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.
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Answer: (A) Both Statement I and Statement II are correct.

The negative charge in phenoxide is delocalized exclusively to the ortho and para positions relative to -O⁻. Thus, a meta-nitro group cannot participate in direct -R (mesomeric) delocalization of the negative charge, stabilizing the phenoxide ion primarily via its electron-withdrawing -I effect. Both statements are factually correct.

easy

11. Which of the following compounds will liberate CO₂ gas with aqueous sodium bicarbonate (NaHCO₃)?

  1. 4-Chlorophenol
  2. 4-Nitrophenol
  3. Phenol
  4. 2,4,6-Trinitrophenol
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Answer: (D) 2,4,6-Trinitrophenol

Only phenols with strong electron-withdrawing groups that make them more acidic than carbonic acid (H₂CO₃, pKa ≈ 6.4) can decompose NaHCO₃ to release CO₂. 2,4,6-Trinitrophenol (picric acid, pKa ≈ 0.38) is much more acidic than carbonic acid and readily liberates carbon dioxide.

hard

12. Consider the following four substituted phenols: (I) 3,5-Dinitrophenol (II) 4-Fluorophenol (III) 4-Chlorophenol (IV) 4-Methoxyphenol Which of the following represents the correct increasing order of their acidic strength?

  1. (IV) < (III) < (II) < (I)
  2. (IV) < (II) < (III) < (I)
  3. (II) < (IV) < (III) < (I)
  4. (IV) < (I) < (III) < (II)
Show answer and explanation

Answer: (B) (IV) < (II) < (III) < (I)

4-Methoxyphenol has a strong +R effect from the -OCH₃ group, making it the least acidic. Between 4-fluorophenol and 4-chlorophenol, fluorine's stronger +R effect relative to chlorine destabilizes the phenoxide more, making 4-chlorophenol slightly more acidic than 4-fluorophenol. 3,5-Dinitrophenol has two strong -I groups at meta positions, making it the most acidic in the set. Thus, the order is (IV) < (II) < (III) < (I).

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