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Alcohols, Phenols and Ethers - Preparation and Properties of Alcohols and Phenols

Grade 12CBSEChemistry

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

🔑Concepts

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Preparation of Alcohols from Alkenes: Acid catalyzed hydration follows Markovnikov's rule, whereas Hydroboration-Oxidation (B2H6B_2H_6 followed by H2O2/OH−H_2O_2/OH^-) results in Anti-Markovnikov addition of H2OH_2O.

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Preparation of Alcohols from Carbonyl Compounds: Aldehydes are reduced to 1∘1^\circ alcohols and ketones to 2∘2^\circ alcohols using LiAlH4LiAlH_4, NaBH4NaBH_4, or H2/PdH_2/Pd.

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Grignard Reagent Synthesis: Formaldehyde yields 1∘1^\circ alcohols, other aldehydes yield 2∘2^\circ alcohols, and ketones yield 3∘3^\circ alcohols when reacted with RMgXRMgX.

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Industrial Preparation of Phenol: Phenol is manufactured from Cumene (isopropylbenzene). Air oxidation of cumene produces cumene hydroperoxide, which is decomposed by dilute acid to yield Phenol and CH3COCH3CH_3COCH_3 (Acetone) as a valuable byproduct.

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Acidity of Alcohols and Phenols: Phenols are more acidic than alcohols due to the resonance stabilization of the phenoxide ion (C6H5O−C_6H_5O^-). Electron-withdrawing groups (−NO2-NO_2) increase acidity, while electron-donating groups (−CH3-CH_3, −OCH3-OCH_3) decrease it.

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Lucas Test: Used to distinguish between 1∘1^\circ, 2∘2^\circ, and 3∘3^\circ alcohols using Lucas reagent (conc. HCl+ZnCl2HCl + ZnCl_2). 3∘3^\circ alcohols show immediate turbidity, 2∘2^\circ after 5 minutes, and 1∘1^\circ only upon heating.

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Reimer-Tiemann Reaction: Treatment of phenol with CHCl3CHCl_3 and aqueous NaOHNaOH followed by acidification introduces an −CHO-CHO group at the ortho position, forming Salicylaldehyde.

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Kolbe's Reaction: Phenoxide ion reacts with CO2CO_2 followed by acidification to produce 22-hydroxybenzoic acid (Salicylic acid).

📐Formulae

R−CH=CH2+H2O→H+R−CH(OH)−CH3R-CH=CH_2 + H_2O \xrightarrow{H^+} R-CH(OH)-CH_3

C6H5−CH(CH3)2→2.H+/H2O1.O2C6H5OH+CH3COCH3C_6H_5-CH(CH_3)_2 \xrightarrow[2. H^+/H_2O]{1. O_2} C_6H_5OH + CH_3COCH_3

C6H5OH+CHCl3+NaOH→H+C6H4(OH)CHOC_6H_5OH + CHCl_3 + NaOH \xrightarrow{H^+} C_6H_4(OH)CHO

C6H5ONa+CO2→2.H+1.400K,4−7atmC6H4(OH)COOHC_6H_5ONa + CO_2 \xrightarrow[2. H^+]{1. 400K, 4-7 atm} C_6H_4(OH)COOH

R−CH2OH→CrO3/PCCR−CHOR-CH_2OH \xrightarrow{CrO_3 / PCC} R-CHO

💡Examples

Problem 1:

Explain why pp-nitrophenol is more acidic than oo-nitrophenol.

Solution:

pp-nitrophenol has a higher acidity than oo-nitrophenol.

Explanation:

In oo-nitrophenol, there is intramolecular hydrogen bonding between the −OH-OH and −NO2-NO_2 groups, which makes the release of the proton (H+H^+) relatively more difficult compared to pp-nitrophenol, where such intramolecular bonding is absent and the −NO2-NO_2 group exerts a strong electron-withdrawing effect through resonance (−R-R effect) at the para position.

Problem 2:

Write the mechanism for the acid-catalyzed dehydration of Ethanol to Ethene.

Solution:

CH3CH2OH→443KH2SO4CH2=CH2+H2OCH_3CH_2OH \xrightarrow[443K]{H_2SO_4} CH_2=CH_2 + H_2O

Explanation:

The mechanism involves three steps: (1) Formation of protonated alcohol: CH3CH2OH+H+⇌CH3CH2OH2+CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2OH_2^+. (2) Formation of carbocation (slowest step): CH3CH2OH2+→CH3CH2++H2OCH_3CH_2OH_2^+ \rightarrow CH_3CH_2^+ + H_2O. (3) Elimination of a proton to form ethene: CH3CH2+→CH2=CH2+H+CH_3CH_2^+ \rightarrow CH_2=CH_2 + H^+.

Problem 3:

How do you convert Phenol to Aspirin?

Solution:

Phenol →NaOH,CO2,H+\xrightarrow{NaOH, CO_2, H^+} Salicylic acid →(CH3CO)2O,H+\xrightarrow{(CH_3CO)_2O, H^+} Aspirin

Explanation:

First, Phenol is converted to Salicylic acid via Kolbe's reaction. Then, Salicylic acid undergoes acetylation using acetic anhydride in the presence of an acid catalyst to produce Acetylsalicylic acid (Aspirin).