Write the mechanism of the following reaction: $$2\,\mathrm{CH_3CH_2OH}\xrightarrow[413\ K]{\text{conc. }H_2SO_4}\mathrm{CH_3CH_2{-}O{-}CH_2CH_3}$$

Bimolecular dehydration (S_N2 on protonated ethanol): (1) protonation; (2) a second ethanol's O attacks the α-C, displacing H₂O → protonated ether; (3) deprotonation → diethyl ether.

CBSE Class 12 · 2016 · Delhi · Alcohols Phenols and EthersShort Answer

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Write the mechanism of the following reaction:

2 C H_{3} C H_{2} OH conc. H_{2} S O_{4} 413 K C H_{3} C H_{2} - O - C H_{2} C H_{3}

Solution

Step 1 — protonation:

C H_{3} C H_{2} OH + H^{+} \to C H_{3} C H_{2} O + H_{2}

Step 2 — nucleophilic attack by a second ethanol (S

_{N}

2), water leaving:

C H_{3} C H_{2} OH + C H_{3} C H_{2} O + H_{2} \to C H_{3} C H_{2} O + (H) C H_{2} C H_{3} + H_{2} O

Step 3 — loss of H

^{+}

C H_{3} C H_{2} O + (H) C H_{2} C H_{3} \to C H_{3} C H_{2} OC H_{2} C H_{3} + H^{+}

(At 413 K ether forms; at 443 K elimination to ethene wins.)

CBSE Class 12 Board · 2016 · Delhi · Subjective

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