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Alcohols, Phenols and Ethers - Nomenclature and methods of preparation

Grade 12ICSEChemistry

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

πŸ”‘Concepts

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IUPAC Nomenclature: Alcohols are named by replacing the '-e' of the parent alkane with '-ol'. For phenols, the parent structure is C6H5OHC_6H_5OH. Ethers are named as 'alkoxyalkanes' where the smaller alkyl group is the 'alkoxy' part.

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Preparation of Alcohols via Hydration: Acid-catalyzed hydration of alkenes follows Markovnikov's rule, e.g., CH3CH=CH2+H2O→H+CH3CH(OH)CH3CH_3CH=CH_2 + H_2O \xrightarrow{H^+} CH_3CH(OH)CH_3.

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Hydroboration-Oxidation: A method to prepare alcohols from alkenes yielding anti-Markovnikov products. The reagent used is diborane (B2H6B_2H_6) followed by alkaline H2O2H_2O_2.

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Grignard Synthesis: Reaction of RMgXRMgX with formaldehyde gives a primary (1∘1^\circ) alcohol, with other aldehydes gives a secondary (2∘2^\circ) alcohol, and with ketones gives a tertiary (3∘3^\circ) alcohol.

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Preparation of Phenol from Cumene: Isopropylbenzene (C6H5CH(CH3)2C_6H_5CH(CH_3)_2) is oxidized by air to cumene hydroperoxide, which is then hydrolyzed by dilute acid to yield phenol (C6H5OHC_6H_5OH) and acetone (CH3COCH3CH_3COCH_3).

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Williamson Ether Synthesis: An SN2S_N2 reaction between a primary alkyl halide and a sodium alkoxide (Rβˆ’X+Rβ€²Oβˆ’Na+β†’Rβˆ’Oβˆ’Rβ€²+NaXR-X + R'O^-Na^+ \rightarrow R-O-R' + NaX). It is the preferred method for preparing unsymmetrical ethers.

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Dehydration of Alcohols: At 413K413 K in the presence of H2SO4H_2SO_4, primary alcohols undergo intermolecular dehydration to form ethers: 2C2H5OH→H2SO4,413KC2H5OC2H5+H2O2C_2H_5OH \xrightarrow{H_2SO_4, 413K} C_2H_5OC_2H_5 + H_2O.

πŸ“Formulae

Rβˆ’CH=CH2+H2Oβ†’H2SO4Rβˆ’CH(OH)βˆ’CH3R-CH=CH_2 + H_2O \xrightarrow{H_2SO_4} R-CH(OH)-CH_3

3CH3CH=CH2+12B2H6β†’(CH3CH2CH2)3Bβ†’H2O2/OHβˆ’3CH3CH2CH2OH3CH_3CH=CH_2 + \frac{1}{2}B_2H_6 \rightarrow (CH_3CH_2CH_2)_3B \xrightarrow{H_2O_2/OH^-} 3CH_3CH_2CH_2OH

Rβˆ’CHOβ†’2.H2O/H+1.LiAlH4Rβˆ’CH2OHR-CHO \xrightarrow[2. H_2O/H^+]{1. LiAlH_4} R-CH_2OH

C6H5NH2β†’NaNO2+HCl,273βˆ’278KC6H5N2+Clβˆ’β†’H2O,Ξ”C6H5OH+N2+HClC_6H_5NH_2 \xrightarrow{NaNO_2 + HCl, 273-278K} C_6H_5N_2^+Cl^- \xrightarrow{H_2O, \Delta} C_6H_5OH + N_2 + HCl

Rβˆ’X+Rβ€²βˆ’ONaβ†’Rβˆ’Oβˆ’Rβ€²+NaXR-X + R'-ONa \rightarrow R-O-R' + NaX

πŸ’‘Examples

Problem 1:

Predict the product when Propanone reacts with Methylmagnesium bromide followed by hydrolysis.

Solution:

The product is 2-Methylpropan-2-ol (CH3C(CH3)(OH)CH3CH_3C(CH_3)(OH)CH_3).

Explanation:

Propanone (CH3COCH3CH_3COCH_3) is a ketone. Grignard reagents (CH3MgBrCH_3MgBr) react with ketones to form an adduct which, upon hydrolysis, yields a tertiary alcohol: CH3COCH3+CH3MgBr→(CH3)3COMgBr→H2O(CH3)3COH+Mg(OH)BrCH_3COCH_3 + CH_3MgBr \rightarrow (CH_3)_3COMgBr \xrightarrow{H_2O} (CH_3)_3COH + Mg(OH)Br.

Problem 2:

Why is it not possible to prepare tert-butyl ethyl ether by reacting sodium ethoxide with tert-butyl bromide?

Solution:

Reaction will lead to the formation of Isobutylene (CH3C(CH3)=CH2CH_3C(CH_3)=CH_2) instead of the ether.

Explanation:

Tert-butyl bromide is a tertiary (3∘3^\circ) alkyl halide. In the presence of a strong base like sodium ethoxide (C2H5ONaC_2H_5ONa), 3∘3^\circ halides undergo elimination (E2) rather than substitution (SN2S_N2) to form an alkene. To prepare this ether, one should react sodium tert-butoxide with ethyl bromide.

Problem 3:

Identify the major product of the acid-catalyzed hydration of But-1-ene.

Solution:

The major product is Butan-2-ol (CH3CH2CH(OH)CH3CH_3CH_2CH(OH)CH_3).

Explanation:

According to Markovnikov's rule, the hydrogen atom of the water molecule (Hβˆ’OHH-OH) adds to the carbon of the double bond with more hydrogen atoms, and the hydroxyl group adds to the more substituted carbon.

Nomenclature and methods of preparation Revision - Class 12 Chemistry ICSE