Attempt either (a) or (b) :
(a) A carbon compound 'P' of molecular formula C₂H₆O is found to be neutral when tested with red and blue litmus. A gas that burns with a pop sound is produced when sodium metal reacts with compound 'P'.
(i) Identify the compound 'P'.
(ii) The carbon compound 'P' is heated with conc. H₂SO₄ at 443 K to produce compound 'Q'.
(I) Write the chemical equation for the reaction.
(II) What is the role of conc. H₂SO₄ in this reaction ?
(iii) What is likely to be observed on heating compound 'P' with ethanoic acid with an acid as a catalyst ? Write the chemical equation for the reaction.
Generated by claude-sonnet-4-6 · 2026-06-15 06:46 · grounding rag
Model Answer
(i) Compound 'P' is ethanol (C₂H₅OH). It is neutral to litmus and reacts with sodium to produce hydrogen gas (pop sound):
$$2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2\uparrow$$
(ii)
(I) When ethanol is heated with conc. H₂SO₄ at 443 K, it undergoes dehydration to form compound 'Q' — ethene (C₂H₄):
$$C_2H_5OH \xrightarrow{\text{conc. H}_2\text{SO}_4,\ 443\text{K}} CH_2=CH_2 + H_2O$$
(II) Conc. H₂SO₄ acts as a dehydrating agent — it removes water from ethanol to produce ethene.
(iii) A fruity-smelling ester (ethyl ethanoate) will be formed. This reaction is called esterification.
$$C_2H_5OH + CH_3COOH \underset{\Delta}{\overset{\text{conc. H}_2\text{SO}_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O$$
Source: Chapter 4, Section 4.4 — Some Important Carbon Compounds: Ethanol and Ethanoic Acid
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Explanation
- Identification of P is straightforward: C₂H₆O neutral to litmus = ethanol (not an acid/base); reacts with Na to give H₂ (pop sound). Methanol also fits formula but ethanol is the standard NCERT example.
- Dehydration at 443 K is a key fact: conc. H₂SO₄ as dehydrating agent → ethene. Write the balanced equation with conditions above the arrow.
- Esterification: heating alcohol + carboxylic acid with acid catalyst gives ester + water. The reaction is reversible (⇌). Examiners expect the fruity smell observation and the correct equation. Remember ethyl ethanoate = CH₃COOC₂H₅.
- Mark split is roughly: (i) 1 mark, (ii) 2 marks, (iii) 2 marks — match detail to marks.