A neutral organic compound 'X' (Molecular formula C_2H_6O) on reacting with acidified K_2Cr_2O_7 gives an organic compound 'Y' which is acidic in nature. 'X' reacts with 'Y' on warming in the presence of conc. H_2SO_4 to give a sweet smelling compound 'Z'. (i) Identify 'X', 'Y' and 'Z'. (ii) Write the chemical equations for the reactions in the conversion of (1) 'X' to 'Y' and (2) 'X' to 'Z'. (iii) State the role of (1) acidified K_2Cr_2O_7 in the conversion of 'X' to 'Y' and (2) conc. H_2SO_4 in the reaction of 'X' and 'Y'. (iv) Name the reaction which occurs when 'Z' reacts with an alkali.

ScienceCBSE Class 10 · Board exam 2023, set 31/5/1 · 5 marks · Long answer

CBSE Class 10 Science 2023: question 34(a)

Set in the CBSE Class 10 board exam of 2023, paper 31/5/1, section D, question 34(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(a) · 5 marks · Long answerCarbon and its Compounds

A neutral organic compound 'X' (Molecular formula $C_{2} H_{6} O$ ) on reacting with acidified $K_{2} C r_{2} O_{7}$ gives an organic compound 'Y' which is acidic in nature. 'X' reacts with 'Y' on warming in the presence of conc. $H_{2} S O_{4}$ to give a sweet smelling compound 'Z'.

(i)
Identify 'X', 'Y' and 'Z'.
0 marks
(ii)
Write the chemical equations for the reactions in the conversion of (1) 'X' to 'Y' and (2) 'X' to 'Z'.
0 marks
(iii)
State the role of (1) acidified $K_{2} C r_{2} O_{7}$ in the conversion of 'X' to 'Y' and (2) conc. $H_{2} S O_{4}$ in the reaction of 'X' and 'Y'.
0 marks
(iv)
Name the reaction which occurs when 'Z' reacts with an alkali.
0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 31/5/1.

Hints

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

Hint 1

C₂H₆O that is neutral and has an –OH group is which common alcohol?

Hint 2

Oxidising an alcohol gives which kind of acid?

Hint 3

A sweet smell suggests an ester. What is the reverse reaction with NaOH called?

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) X = ethanol, $C_{2} H_{5} OH$
½ mark
Y = ethanoic acid (acetic acid), $C H_{3} COOH$
½ mark
Z = ethyl ethanoate (ethyl acetate), $C H_{3} COO C_{2} H_{5}$ , a sweet-smelling ester
½ mark
(ii)(1) $C H_{3} C H_{2} OH acidified K_{2} C r_{2} O_{7} C H_{3} COOH$
1 mark
(2) $C H_{3} COOH + C_{2} H_{5} OH conc. H_{2} S O_{4} warm C H_{3} COO C_{2} H_{5} + H_{2} O$
1 mark
(iii)(1) Acidified $K_{2} C r_{2} O_{7}$ acts as an oxidising agent: it supplies oxygen to convert ethanol into ethanoic acid.
½ mark
(2) Conc. $H_{2} S O_{4}$ acts as a dehydrating agent (and catalyst): it removes the water formed and helps the esterification proceed.
½ mark
(iv) When the ester Z reacts with an alkali (e.g. NaOH) it gives back the alcohol and the sodium salt of the acid. This is called saponification: $C H_{3} COO C_{2} H_{5} + NaOH \to C H_{3} COONa + C_{2} H_{5} OH$
½ mark

Answer: (i) X = ethanol, Y = ethanoic acid, Z = ethyl ethanoate; (ii) equations as shown; (iii) K₂Cr₂O₇: oxidising agent; conc. H₂SO₄: dehydrating agent; (iv) saponification

Where the marks are

Identifies X, Y and Z (½ mark each)
Equation for oxidation of ethanol to ethanoic acid (1 mark)
Esterification equation (1 mark)
Role of acidified K₂Cr₂O₇: oxidising agent (½ mark)
Role of conc. H₂SO₄: dehydrating agent (½ mark)
Names saponification (½ mark)

Revise the chapter

Chapter: Carbon and its Compounds (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Carbon and its Compounds, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 31/5/1: every question, by section.

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CBSE Class 10 Science 2023: question 34(a)

Hints

Step-by-step solution

Revise the chapter

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