CBSE Class 10 Science sample paper 2023-24: question 37
From the official CBSE sample question paper for 2023-24, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.
Q37 · 4 marks · Case-basedCompetency-basedCarbon and its Compounds
The table given below shows the hints given by the quiz master in a quiz.
| S.NO | HINT |
|---|---|
| (i) | Substance 'C' is used as a preservative. |
| (ii) | 'C' has two carbon atoms; 'C' is obtained by the reaction of 'A' in presence of alkaline Potassium permanganate followed by acidification. |
| (iii) | Misuse of 'A' in industries is prevented by adding Methanol, Benzene, and pyridine to 'A'. |
| (iv) | 'F' is formed on heating 'A' in presence of conc Sulphuric acid. |
| (v) | 'F' reacts with Hydrogen gas in presence of Nickel and Palladium catalyst. |
Based on the above hints answer the following questions.
- (a)1 mark
Give the IUPAC names of A and F.
- (b)3 marks
Illustrate with the help of chemical equations the changes taking place. ( and )
- (b)3 marks
Name the chemical reactions which occur in steps 2 and 5. Identify the compounds formed in these steps if 'A' is replaced with its next homologue.
Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2023-24.
Hints
Open one hint at a time and try again before the next.
Hint 1
A two-carbon acid used as a preservative is found in vinegar. Which alcohol is oxidised to it?
Hint 2
Heating an alcohol with conc. sulphuric acid removes water. Which hydrocarbon forms, and why does it react with hydrogen?
Hint 3
For the next homologue, add one unit to each compound.
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.
Reasoning from the hints: C has two carbon atoms and is used as a preservative (in pickles), so C is ethanoic acid. A is oxidised to C, and A is denatured with methanol, benzene and pyridine to stop its misuse, so A is ethanol. Heating A with conc. dehydrates it to F, which takes up hydrogen, so F is an alkene, ethene.
- 1 mark
(a) IUPAC names: A = ethanol (); F = ethene ().
- 1½ marks
(b) (oxidation):
Ethanol is oxidised to ethanoate in the alkaline medium. Acidification gives ethanoic acid (C). - 1½ marks
(b) (dehydration):
Conc. acts as a dehydrating agent. - 1 mark
(b) Alternative, part 1: Step 2 (ethanol to ethanoic acid) is an oxidation reaction. Step 5 (ethene + with Ni/Pd catalyst) is an addition reaction (hydrogenation).
- 1 mark
(b) Alternative, part 2: The next homologue of ethanol is propanol, .
- Step 2: gives propanoic acid.
- 1 mark
- Step 5: propanol dehydrates (step 4) to propene, . Hydrogenation then gives propane:
Answer: (a) A = ethanol, F = ethene. (b) (alk. , then ); (hot conc. ). OR (b): step 2 oxidation, step 5 addition (hydrogenation). With propanol, step 2 gives propanoic acid; propanol gives propene, which is hydrogenated in step 5 to propane.
Where the marks are
- Names A as ethanol and F as ethene (1 mark)
- Correct equation for oxidation of ethanol to ethanoic acid using alkaline followed by acidification (1.5 marks)
- Correct equation for dehydration of ethanol to ethene with hot conc. (1.5 marks)
- OR: names step 2 as oxidation and step 5 as addition/hydrogenation (1 mark)
- OR: identifies the compounds for the next homologue, propanol: propanoic acid in step 2 (1 mark)
- OR: propene, hydrogenated to propane in step 5 (1 mark)
Revise the chapter
- Chapter: Carbon and its Compounds (notes, formulas, where marks are lost and every past question)
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