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Alcohols, Phenols and Ethers - Preparation and properties of ethers

Grade 12ICSEChemistry

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

🔑Concepts

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Ethers are organic compounds with the general formula R−O−R′R-O-R', where RR and R′R' can be alkyl or aryl groups. The oxygen atom is sp3sp^3 hybridized with a bond angle of approximately 111.7∘111.7^{\circ} in dimethyl ether.

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Preparation - Dehydration of Alcohols: Primary alcohols undergo intermolecular dehydration in the presence of protic acids like H2SO4H_2SO_4 or H3PO4H_3PO_4 at 413 K413 \, K to form ethers. Higher temperatures (443 K443 \, K) favor the formation of alkenes.

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Preparation - Williamson Ether Synthesis: This is an SN2S_N2 reaction between an alkyl halide and a sodium alkoxide (R−X+R′−ONa→R−O−R′+NaXR-X + R'-ONa \rightarrow R-O-R' + NaX). For better yields, the alkyl halide should be primary; secondary or tertiary halides lead to elimination (alkenes) as the major product.

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Physical Properties: Ethers have lower boiling points than isomeric alcohols because they cannot form intermolecular hydrogen bonds. However, they are soluble in water to a similar extent as alcohols due to hydrogen bonding between water and the ether oxygen.

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Chemical Property - Cleavage of C−OC-O bond: Ethers are relatively unreactive but can be cleaved by concentrated HIHI or HBrHBr at high temperatures. In the case of unsymmetrical ethers with primary/secondary groups, the halide attaches to the smaller alkyl group (SN2S_N2). If a tertiary alkyl group is present, the halide attaches to the tertiary group (SN1S_N1).

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Electrophilic Substitution: In aromatic ethers like anisole (C6H5OCH3C_6H_5OCH_3), the alkoxy group is ortho and para directing and activating due to the +M+M effect. Common reactions include Halogenation, Friedel-Crafts alkylation/acylation, and Nitration.

📐Formulae

2CH3CH2OH→H2SO4,413KCH3CH2OCH2CH3+H2O2CH_3CH_2OH \xrightarrow{H_2SO_4, 413K} CH_3CH_2OCH_2CH_3 + H_2O

R−X+R′−O−Na+→R−O−R′+NaX (Williamson Synthesis)R-X + R'-O^-Na^+ \rightarrow R-O-R' + NaX \text{ (Williamson Synthesis)}

R−O−R′+HX→R−X+R′−OH (where X = I, Br)R-O-R' + HX \rightarrow R-X + R'-OH \text{ (where X = I, Br)}

C6H5OCH3+Br2→CH3COOHp−Br−C6H4OCH3(Major)+o−Br−C6H4OCH3(Minor)C_6H_5OCH_3 + Br_2 \xrightarrow{CH_3COOH} p-Br-C_6H_4OCH_3 (Major) + o-Br-C_6H_4OCH_3 (Minor)

💡Examples

Problem 1:

Predict the products when tert-butyl methyl ether reacts with HIHI.

Solution:

(CH3)3C−O−CH3+HI→(CH3)3C−I+CH3OH(CH_3)_3C-O-CH_3 + HI \rightarrow (CH_3)_3C-I + CH_3OH

Explanation:

This reaction follows the SN1S_N1 mechanism because a stable tertiary carbocation (CH3)3C+(CH_3)_3C^+ is formed upon protonation and cleavage. The iodide ion (I−I^-) then attacks the more stable tertiary carbocation to form tert-butyl iodide.

Problem 2:

What happens when sodium ethoxide (C2H5ONaC_2H_5ONa) reacts with 22-chloro-22-methylpropane?

Solution:

(CH3)3C−Cl+C2H5ONa→(CH3)2C=CH2+C2H5OH+NaCl(CH_3)_3C-Cl + C_2H_5ONa \rightarrow (CH_3)_2C=CH_2 + C_2H_5OH + NaCl

Explanation:

Sodium ethoxide is a strong base. Since the alkyl halide is tertiary, elimination (E2) predominates over substitution (SN2S_N2). Therefore, an alkene (isobutylene) is formed instead of an ether.

Problem 3:

Identify the major product formed during the nitration of Anisole.

Solution:

C6H5OCH3→H2SO4,HNO3p−NitroanisoleC_6H_5OCH_3 \xrightarrow{H_2SO_4, HNO_3} p-Nitroanisole

Explanation:

The methoxy group (−OCH3-OCH_3) is an activating group that increases electron density at ortho and para positions. Due to steric hindrance at the ortho position, the para-isomer (pp-nitroanisole) is the major product.

Preparation and properties of ethers Revision - Class 12 Chemistry ICSE