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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.

πŸ”‘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/OHβˆ’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 (C6H5Oβˆ’C_6H_5O^-). Electron-withdrawing groups (βˆ’NO2-NO_2) increase acidity, while electron-donating groups (βˆ’CH3-CH_3, βˆ’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 βˆ’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).

πŸ“Formulae

Rβˆ’CH=CH2+H2Oβ†’H+Rβˆ’CH(OH)βˆ’CH3R-CH=CH_2 + H_2O \xrightarrow{H^+} R-CH(OH)-CH_3

C6H5βˆ’CH(CH3)2β†’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+CO2β†’2.H+1.400K,4βˆ’7atmC6H4(OH)COOHC_6H_5ONa + CO_2 \xrightarrow[2. H^+]{1. 400K, 4-7 atm} C_6H_4(OH)COOH

Rβˆ’CH2OHβ†’CrO3/PCCRβˆ’CHOR-CH_2OH \xrightarrow{CrO_3 / PCC} R-CHO

πŸ’‘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 βˆ’OH-OH and βˆ’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 βˆ’NO2-NO_2 group exerts a strong electron-withdrawing effect through resonance (βˆ’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+β‡ŒCH3CH2OH2+CH_3CH_2OH + H^+ \rightleftharpoons CH_3CH_2OH_2^+. (2) Formation of carbocation (slowest step): CH3CH2OH2+β†’CH3CH2++H2OCH_3CH_2OH_2^+ \rightarrow CH_3CH_2^+ + H_2O. (3) Elimination of a proton to form ethene: CH3CH2+β†’CH2=CH2+H+CH_3CH_2^+ \rightarrow CH_2=CH_2 + H^+.

Problem 3:

How do you convert Phenol to Aspirin?

Solution:

Phenol β†’NaOH,CO2,H+\xrightarrow{NaOH, CO_2, H^+} Salicylic acid β†’(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).

Preparation and Properties of Alcohols and Phenols Revision - Class 12 Chemistry CBSE