Skip to content

(a) Carry out the… (Class 12 Chemistry 2025, Q33A)

questionbuddy.in/q/cbse-12-chemistry-2025-56-1-1-q33a-carry-out-conversions-ethanal-but

ChemistryCBSE Class 12 · Board exam 2025, set 56/1/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2025: question 33(A)

Set in the CBSE Class 12 board exam of 2025, paper 56/1/1, section E, question 33(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(A) · 5 marks · Long answerAldehydes, Ketones and Carboxylic Acids

(a) Carry out the following conversions : (2 + 3 = 5)
(i) Ethanal to But-2-enal
(ii) Propanoic acid to ethane
(b) An alkene A with molecular formula on ozonolysis gives a mixture of two compounds B and C. Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify the compounds A, B and C.

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

But-2-enal has four carbons, twice the number in ethanal. Which reaction of aldehydes with α-H joins two molecules?

Hint 2

Going from propanoic acid (3 C) to ethane (2 C) means losing CO₂. Think of soda lime.

Hint 3

In (b), Fehling's + iodoform means CH₃CHO; iodoform only means a methyl ketone. Join the two C=O carbons with a double bond to get A.

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) Ethanal → But-2-enal (aldol condensation): (3-hydroxybutanal). On heating, this loses water: (but-2-enal).

    1 mark

  2. (a)(ii) Propanoic acid → Ethane (decarboxylation): . Then .

    1 mark

  3. (b) B gives a positive Fehling test, so it is an aldehyde. It also gives the iodoform test, so it contains a group. Hence B is ethanal, .

    1 mark

  4. C does not give the Fehling test but forms iodoform, so it is a methyl ketone. The carbons left from after removing ethanal's two carbons give C as propanone, .

    1 mark

  5. Joining the two carbonyl carbons by a C=C gives A as 2-methylbut-2-ene, (). Ozonolysis: .

    1 mark

Answer: (a)(i) aldol condensation then dehydration; (ii) sodium salt + soda lime, heat. (b) A = 2-methylbut-2-ene (CH3)2C=CHCH3; B = ethanal CH3CHO; C = propanone CH3COCH3

Where the marks are

  • Aldol reaction of ethanal with dil. NaOH, then heating, to give but-2-enal (1 mark)
  • Sodium propanoate heated with soda lime (NaOH/CaO) to give ethane (1 mark)
  • A identified as 2-methylbut-2-ene (1 mark)
  • B identified as ethanal (1 mark)
  • C identified as propanone (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 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.