Organic compound ‘A’ of molecular formula C₂H₆O liberates hydrogen gas with sodium
metal. ‘A’ gives ‘B’ of formula C₄H₁₀O, when it reacts with concentrated H₂SO₄ at
413K. At 443K with concentrated H₂SO₄ ‘A’ gives compound ‘C’ of formula C₂H₄. This
compound ‘C’ decolourises bromine water. What are ‘A’, ‘B’ and ‘C’?
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Dear Student,
◆ Answer -
- Compound A => Ethanol
- Compound B => Diethyl ether
- Compound C => Ethene
◆ Reactions -
Ethanol (A) liberates hydrogen gas with sodium
metal.
2C2H5-OH + 2Na --> 2C2H5-ONa + H2
Ethanol (A) reacts with concentrated H₂SO₄ at 413K to give dimethyl ether (B).
2C2H5-OH --> C2H5-O-C2H5 + H2O
Ethanol (A) reacts with concentrated H₂SO₄ at 443K to give ethene (C).
C2H5-OH --> CH2=CH2 + H2O
Ethene (C) reacts with bromine water to form ethyl dibromide which is colourless.
CH2=CH2 + Br2 --> Br-CH2-CH2-Br
Thanks dear...
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