Science — CBSE Class 10 board question

(i) When ethanol is heated at 443 K with conc. H₂SO₄, ethene (C₂H₄) is formed by dehydration.

$$\text{C}_2\text{H}_5\text{OH} \xrightarrow{443\text{K, conc. H}_2\text{SO}_4} \text{CH}_2{=}\text{CH}_2 + \text{H}_2\text{O}$$

Electron dot structure of ethene:

$$\text{H}\underset{|}{\overset{|}{\text{ }}} \text{ }$$

$$H : C :: C : H \text{ (each C shares 4 electrons; C=C double bond)}$$

$$\begin{array}{c} H\quad H \\ | \quad\quad | \\ H{:}C{::}C{:}H \end{array}$$

(Two shared pairs between carbon atoms = double bond; each carbon also bonded to two H atoms.)

Role of conc. H₂SO₄: It acts as a dehydrating agent, removing water from ethanol to form ethene.

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(ii) Hydrogenation is the addition of hydrogen to an unsaturated hydrocarbon (alkene or alkyne) in the presence of a catalyst (Ni, Pd, or Pt) to form a saturated compound.

Chemical equation:

$$\text{CH}_2{=}\text{CH}_2 + \text{H}_2 \xrightarrow{\text{Ni, catalyst}} \text{CH}_3{-}\text{CH}_3$$
(Ethene → Ethane)

Industrial role: Hydrogenation is used to convert vegetable oils (unsaturated) into vanaspati ghee (solid fat/saturated) by adding hydrogen in the presence of a nickel catalyst. This increases the shelf life and changes the physical state of the oil.

Source: Carbon and its Compounds, Chapter 4

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• For (i), the key facts are: product = ethene, reaction = dehydration, H₂SO₄ = dehydrating agent. Draw the electron dot structure showing C::C (double bond = two shared electron pairs) with H atoms on each carbon. Don't write just "catalyst" — specify "dehydrating agent."
• For (ii), the textbook exercise (Q12) directly asks this. Always name the catalyst (Ni/Pd/Pt), write a balanced equation, and state the industrial application (vegetable oil → vanaspati). These are the three expected scoring points.