ChemistryCBSE Class 12 · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 12 Chemistry sample paper 2025-26: question 31(A)

From the official CBSE sample question paper for 2025-26, 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

Answer the following questions:
I. Write the structure of expected product of Cannizarro reaction of 2-chlorobenzaldehyde.
II. How would the presence of $- S O_{3} H$ group effect the basic strength of aniline.
III. Convert acetic acid to ethanamine.
IV. Write the steps to prepare Benzoic acid from Benzoyl chloride.
V. Give a chemical test to distinguish between: propanal and propanone

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

Cannizzaro reaction: one aldehyde molecule is reduced and the other oxidised.

Hint 2

Electron-withdrawing groups decrease lone-pair availability on N.

Hint 3

Aldehydes reduce Tollens' reagent; ketones do not.

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.

I. $2 (2 -ClC_{6} H_{4} CHO) conc. NaOH$ 2-chlorobenzyl alcohol ( $2 -ClC_{6} H_{4} CH_{2} OH$ ) + sodium 2-chlorobenzoate ( $2 -ClC_{6} H_{4} COO^{-} Na^{+}$ ). Acidifying the salt gives 2-chlorobenzoic acid.
1 mark
II. –SO $_{3}$ H is a strong electron-withdrawing group (–I, –R). It lowers the electron density on the nitrogen of –NH $_{2}$ , so the lone pair is less available for donation. The basic strength of aniline decreases.
1 mark
III. $CH_{3} COOH LiAlH_{4} CH_{3} CH_{2} OH SOCl_{2} CH_{3} CH_{2} Cl excess NH_{3} CH_{3} CH_{2} NH_{2}$
(Alternatively: CH $_{3}$ COOH → SOCl $_{2}$ → CH $_{3}$ COCl → NH $_{3}$ → CH $_{3}$ CONH $_{2}$ → LiAlH $_{4}$ → CH $_{3}$ CH $_{2}$ NH $_{2}$ .)
1 mark
IV. $C_{6} H_{5} COCl H_{2}, Pd/BaSO_{4}$ C $_{6}$ H $_{5}$ CHO (Rosenmund reduction) $alk. KMnO_{4} then H^{+}$ C $_{6}$ H $_{5}$ COOH. (Benzoyl chloride also gives benzoic acid directly by hydrolysis: C $_{6}$ H $_{5}$ COCl + H $_{2}$ O → C $_{6}$ H $_{5}$ COOH + HCl.)
1 mark
V. Tollens' test: warm each compound with ammoniacal silver nitrate on a water bath. Propanal gives a silver mirror: CH $_{3}$ CH $_{2}$ CHO + 2[Ag(NH $_{3}$ ) $_{2}$ ] $^{+}$ + 3OH $^{-}$ → CH $_{3}$ CH $_{2}$ COO $^{-}$ + 2Ag↓ + 2H $_{2}$ O + 4NH $_{3}$ . Propanone gives no silver mirror.
1 mark

Answer: I. 2-chlorobenzyl alcohol + sodium 2-chlorobenzoate. II. Basicity decreases (electron-withdrawing –SO₃H). III. CH₃COOH → LiAlH₄ → C₂H₅OH → SOCl₂ → C₂H₅Cl → NH₃ → C₂H₅NH₂. IV. Rosenmund reduction then oxidation (or direct hydrolysis). V. Tollens' test: propanal gives a silver mirror, propanone does not.

Where the marks are

Correct Cannizzaro products (1 mark)
Electron-withdrawing –SO₃H decreases basicity (1 mark)
Correct conversion sequence acetic acid → ethanamine (1 mark)
Correct steps benzoyl chloride → benzoic acid (1 mark)
Tollens' (or Fehling's) test with both observations (1 mark)

Revise the chapter

Chapter: Aldehydes, Ketones and Carboxylic Acids (notes, formulas, where marks are lost and every past question)

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 2025-26: every question, by section.

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CBSE Class 12 Chemistry sample paper 2025-26: question 31(A)

Hints

Step-by-step solution

Revise the chapter

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