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Science — CBSE Class 10 board question

Q1. [5]
Attempt either (a) or (b) :
  1. (i) Name the following compounds : (I) CH₃–C(=O)–CH₃ (II) H–C≡C–H [1]
  2. (ii) Name the reaction which is commonly used in the conversion of vegetable oils to fats. [1]
  3. (iii) Why is the melting point of C₄H₁₀ higher than C₃H₈ ? [1]
  4. (iv) What happens when a piece of sodium is dropped into ethanol ? [1]
  5. (v) Write the products in the following reaction : CH₃COOC₂H₅ + NaOH ⎯→ [1]
Previously asked in CBSE board exam
2026 31/3/1 Q29 (b)
Generated by claude-sonnet-4-6 · 2026-06-15 06:47 · grounding rag
Model Answer

(i)
(I) CH₃–C(=O)–CH₃ → Propanone
(II) H–C≡C–H → Ethyne

(ii) The reaction commonly used for the conversion of vegetable oils to fats is Hydrogenation (addition of hydrogen in the presence of a nickel catalyst).

(iii) C₄H₁₀ (butane) has a longer carbon chain and greater molecular mass than C₃H₈ (propane). Greater the molecular mass, stronger the intermolecular forces, and hence higher the melting point.

(iv) When sodium is dropped into ethanol, it reacts vigorously to produce sodium ethoxide and hydrogen gas is liberated:
$$2\text{Na} + 2\text{C}_2\text{H}_5\text{OH} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2\uparrow$$

(v)
$$\text{CH}_3\text{COOC}_2\text{H}_5 + \text{NaOH} \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{CH}_3\text{COONa}$$
Products: Ethanol and Sodium ethanoate

Source: Chapter 4, Sections 4.2.5 and 4.4.2

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Explanation
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