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(i) Carry out the… (Class 12 Chemistry 2023, Q35a)

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ChemistryCBSE Class 12 · Board exam 2023, set 56/1/1 · 5 marks · Long answer

CBSE Class 12 Chemistry 2023: question 35(a)

Set in the CBSE Class 12 board exam of 2023, paper 56/1/1, section E, question 35(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(a) · 5 marks · Long answerAldehydes, Ketones and Carboxylic Acids

(i) Carry out the following conversions :
(1) Ethanal to But-2-en-1-al
(2) Propanoic acid to 2-chloropropanoic acid
(ii) 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 2023 board paper, set 56/1/1.

Hints

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

Hint 1

For (i)(1), compare the carbon counts: ethanal has 2 carbons and the product has 4. Which reaction joins two aldehyde molecules?

Hint 2

For (i)(2), think of the named reaction that puts a halogen on the α-carbon of a carboxylic acid.

Hint 3

For (ii), a positive Fehling test means an aldehyde, and a positive iodoform test means a group. Find the 2-carbon aldehyde and 3-carbon ketone that fit, then join their carbonyl carbons with a double bond.

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)(1) Ethanal → But-2-en-1-al (aldol condensation)

    Ethanal has -hydrogens, so with dilute NaOH two molecules undergo aldol addition. On heating, the aldol loses water:

    (3-hydroxybutanal)

    (but-2-en-1-al)

    1 mark

  2. (i)(2) Propanoic acid → 2-chloropropanoic acid (Hell–Volhard–Zelinsky reaction)

    A carboxylic acid with an -hydrogen is chlorinated at the -carbon by with red phosphorus, followed by hydrolysis:

    1 mark

  3. (ii) Identifying B

    B gives a positive Fehling test, so it is an aliphatic aldehyde. It also gives the iodoform test with /NaOH, so it has a group. The only aldehyde with both features is ethanal.

    **B = (ethanal)**

    1 mark

  4. Identifying C

    C does not give the Fehling test, so it is a ketone. It forms iodoform, so it is a methyl ketone. A has 5 carbons and B uses 2 of them, so C has 3 carbons.

    **C = (propanone / acetone)**

    1 mark

  5. Identifying A

    Joining the two carbonyl carbons of B and C by a C=C bond gives the alkene:

    **A = (2-methylbut-2-ene)**, formula .

    Check by ozonolysis:

    1 mark

Answer: (i)(1) ; (i)(2) ; (ii) A = (2-methylbut-2-ene), B = , C =

Where the marks are

  • Aldol condensation of ethanal with dil. NaOH, then heating to give (1 mark)
  • HVZ reaction (/red P, then ) on propanoic acid giving (1 mark)
  • B identified as ethanal, (1 mark)
  • C identified as propanone, (1 mark)
  • A identified as 2-methylbut-2-ene, (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 paper2023 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.