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Alcohols, Phenols and Ethers - Mechanism of dehydration

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Dehydration of alcohols involves the elimination of a molecule of water to form an alkene or an ether, typically catalyzed by protic acids like concentrated H2SO4H_2SO_4 or H3PO4H_3PO_4.

The reaction follows a carbocation mechanism, and the ease of dehydration follows the order: Tertiary (33^\circ) > Secondary (22^\circ) > Primary (11^\circ) alcohols, due to the relative stability of the intermediate carbocations.

The mechanism consists of three major steps: (1) Formation of protonated alcohol (oxonium ion), (2) Formation of carbocation (slowest, rate-determining step), and (3) Elimination of a proton to form the C=CC=C double bond.

Temperature and reagent concentration determine the product: Ethanol yields ethene at 443 K443\ K with conc. H2SO4H_2SO_4, but yields ethoxyethane (ether) at 413 K413\ K with excess alcohol.

According to Saytzeff's Rule, in the dehydration of unsymmetrical secondary or tertiary alcohols, the more highly substituted alkene (the one with the greater number of alkyl groups attached to the doubly bonded carbon atoms) is the preferred product.

📐Formulae

CnH2n+1OHΔH+CnH2n+H2OC_nH_{2n+1}OH \xrightarrow[\Delta]{H^+} C_nH_{2n} + H_2O

CH3CH2OH+H+CH3CH2O+H2 (Protonated alcohol)CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2\text{O}^+H_2 \text{ (Protonated alcohol)}

CH3CH2O+H2slowCH3C+H2+H2O (Carbocation formation)CH_3CH_2\text{O}^+H_2 \xrightarrow{\text{slow}} CH_3\text{C}^+H_2 + H_2O \text{ (Carbocation formation)}

CH3C+H2fastCH2=CH2+H+ (Formation of ethene)CH_3\text{C}^+H_2 \xrightarrow{\text{fast}} CH_2=CH_2 + H^+ \text{ (Formation of ethene)}

💡Examples

Problem 1:

Explain the dehydration of CH3CH(OH)CH2CH3CH_3CH(OH)CH_2CH_3 (Butan-2-ol) and identify the major product.

Solution:

CH3CH(OH)CH2CH3H2SO4,440KCH3CH=CHCH3 (But-2-ene, 80%)+CH2=CHCH2CH3 (But-1-ene, 20%)CH_3CH(OH)CH_2CH_3 \xrightarrow{H_2SO_4, 440K} CH_3CH=CHCH_3 \text{ (But-2-ene, 80\%)} + CH_2=CHCH_2CH_3 \text{ (But-1-ene, 20\%)}

Explanation:

Butan-2-ol forms a secondary carbocation. Deprotonation can occur from either C1C_1 or C3C_3. According to Saytzeff's rule, the more substituted alkene, But-2-ene, is the major product because it is more stable.

Problem 2:

Write the conditions required for the dehydration of Propan-2-ol to Propene.

Solution:

CH3CH(OH)CH385% H3PO4,440KCH3CH=CH2+H2OCH_3-CH(OH)-CH_3 \xrightarrow{85\% \text{ } H_3PO_4, 440K} CH_3-CH=CH_2 + H_2O

Explanation:

Secondary alcohols like Propan-2-ol require milder conditions (85% H3PO485\% \text{ } H_3PO_4) compared to primary alcohols (Conc. H2SO4Conc. \text{ } H_2SO_4) because the secondary carbocation intermediate is easier to form.

Mechanism of dehydration - Revision Notes & Key Formulas | ICSE Class 12 Chemistry