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$C_{2} H_{5} - OH \frac{Conc H_{2} {SO}_{4}}{T_{2}} C_{2} H_{5} - O - C_{2} H_{5}$
$C_{2} H_{5} OH \to_{T_{1}}^{Conc . H_{2} {SO}_{4}} {CH}_{2} = {CH}_{2}$
The values of T₁ and T₂ are respectively.

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$573 K; 613 K$

673 K, 358 K

443 K, 413 K

623 K, 473 K

answer is C.

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Detailed Solution

Excess ethanol in presence of Conc. H₂SO₄ at low temperature (413K) gives ether whereas at higher temperature (443K) gives alkene (elimination product)

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