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(I) Account for the… (Class 12 Chemistry SQP 2026-27, Q31A)

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ChemistryCBSE Class 12 · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 12 Chemistry sample paper 2026-27: question 31(A)

From the official CBSE sample question paper for 2026-27, section E, question 31(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q31(A) · 5 marks · Long answerAldehydes, Ketones and Carboxylic Acids

(I) Account for the following (3x1)
a. The pH should be around 3.5 during the addition of ammonia derivative compounds to aldehyde/ ketone.
b. Alcohols are added to carbonyl compounds in the presence of dry HCl gas.
c. The reaction of aldehydes with sodium hydrogen sulphite is commonly used for the separation and purification of aldehydes.

(II) A student has three colourless samples: methanol, acetaldehyde, and acetic acid. They have only , and 2,4-DNP reagent. Devise a flowchart to identify each compound using only these two reagents. (2)

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

(a) Think about what too much acid does to the nucleophile and what too little does to the carbonyl.

Hint 2

Only one of the three samples is acidic enough to react with NaHCO₃.

Hint 3

2,4-DNP reacts only with compounds containing a C=O group.

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. (I)(a) In strongly acidic medium the ammonia derivative () is protonated to and loses its lone pair, so it is no longer nucleophilic. In basic medium the carbonyl oxygen is not protonated, so the carbonyl carbon is not electrophilic enough. A weakly acidic pH of about 3.5 balances both effects and gives the fastest rate.

    1 mark

  2. (I)(b) Dry HCl protonates the carbonyl oxygen. This increases the electrophilicity of the carbonyl carbon, so the weak nucleophile (alcohol) can attack to form a hemiacetal and then an acetal. The HCl is kept dry because water would reverse the acetal formation.

    1 mark

  3. (I)(c) Aldehydes form crystalline, water-soluble bisulphite addition products that separate from impurities. The adduct is reconverted to the pure aldehyde by treating it with dilute mineral acid or alkali. So the reaction is used for separation and purification.

    1 mark

  4. (II) Flowchart:

    Step 1: Add solution to each sample.

    • Brisk effervescence of → acetic acid ()
    • No effervescence → methanol or acetaldehyde, go to Step 2
    1 mark

  5. Step 2: Add 2,4-DNP reagent to the remaining two samples.

    • Yellow/orange precipitate of the 2,4-dinitrophenylhydrazone → acetaldehyde (has a C=O group)
    • No precipitate → methanol
    1 mark

Answer: (I) (a) optimum pH balances protonation of the carbonyl and the nucleophilicity of NH₂–Z; (b) dry HCl activates the C=O; (c) the reversible, separable bisulphite adduct; (II) NaHCO₃ effervescence → acetic acid; then 2,4-DNP precipitate → acetaldehyde; no precipitate → methanol.

Where the marks are

  • (a) Explains both the too-acidic and the basic extremes (1 mark)
  • (b) Protonation increases the electrophilicity of the carbonyl carbon (1 mark)
  • (c) Adduct separable and regenerable with dilute acid/alkali (1 mark)
  • NaHCO₃ test identifies acetic acid by effervescence (1 mark)
  • 2,4-DNP distinguishes acetaldehyde (precipitate) from methanol (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Aldehydes, Ketones and Carboxylic Acids, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27: 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.