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Phenols undergo… (Class 12 Chemistry 2025, Q30)

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ChemistryCBSE Class 12 · Board exam 2025, set 56/1/1 · 4 marks · Case-based

CBSE Class 12 Chemistry 2025: question 30

Set in the CBSE Class 12 board exam of 2025, paper 56/1/1, section D, question 30, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q30 · 4 marks · Case-basedCompetency-basedAlcohols, Phenols and Ethers

Phenols undergo electrophilic substitution reactions readily due to the strong activating effect of OH group attached to the benzene ring. Since, the OH group increases the electron density more to o– and p– positions therefore OH group is ortho, para-directing. Reimer-Tiemann reaction is one of the examples of aldehyde group being introduced on the aromatic ring of phenol, ortho to the hydroxyl group. This is a general method used for the ortho-formylation of phenols.
Answer the following questions :

  1. (a)

    What happens when phenol reacts with
    (i)
    (ii) Conc.

    2 marks
  2. (b)

    Why phenol does not undergo protonation readily ?

    1 mark
  3. (c)

    Which is a stronger acid – phenol or cresol ? Give reason.

    1 mark

  4. OR

    (c)

    Write the IUPAC name of the product formed in the Reimer-Tiemann reaction.

    1 mark

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 56/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Bromination in a non-polar solvent gives a less vigorous reaction than bromine water.

Hint 2

Concentrated HNO₃ substitutes all activated o/p positions.

Hint 3

An electron-releasing group affects the stability of the phenoxide ion.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. (a)(i) With in (low polarity solvent, low temperature), monobromination occurs and gives a mixture of 2-bromophenol and 4-bromophenol (para is the major product).

    1 mark

  2. (a)(ii) With concentrated , phenol is nitrated at all ortho and para positions to give 2,4,6-trinitrophenol (picric acid).

    1 mark

  3. (b) The lone pair of electrons on the oxygen of phenol is delocalised into the benzene ring by resonance. It is therefore not easily available for accepting a proton, so phenol is not readily protonated.

    1 mark

  4. (c) Phenol is the stronger acid. The group in cresol is electron releasing (+I effect). It intensifies the negative charge on the phenoxide ion and destabilises it, so cresol is less acidic.

    1 mark

  5. (c) OR: The Reimer–Tiemann reaction of phenol (/aq. NaOH, then ) gives salicylaldehyde, IUPAC name 2-hydroxybenzaldehyde. (Alternative to (c) above; carries the same 1 mark.)

Answer: (a)(i) 2-bromophenol + 4-bromophenol; (ii) 2,4,6-trinitrophenol; (b) O lone pair delocalised into the ring; (c) phenol, since CH₃ (+I) destabilises cresoxide / 2-hydroxybenzaldehyde

Where the marks are

  • o- and p-bromophenol formed (1 mark)
  • Picric acid (2,4,6-trinitrophenol) formed (1 mark)
  • Resonance makes the O lone pair unavailable for protonation (1 mark)
  • Phenol stronger (½ mark) with +I reason (½ mark), OR 2-hydroxybenzaldehyde (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Alcohols, Phenols and Ethers, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 56/1/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.