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Alcohols, Phenols and Ethers - Physical and chemical properties

Grade 12ICSEChemistry

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

πŸ”‘Concepts

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The boiling points of alcohols and phenols are significantly higher than those of hydrocarbons, ethers, and haloalkanes of comparable molecular masses due to the presence of intermolecular HH-bonding.

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Solubility of alcohols in water decreases with the increase in the size of the alkyl or aryl group because the non-polar hydrophobic part dominates.

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Phenols are more acidic than alcohols because the phenoxide ion (C6H5Oβˆ’C_6H_5O^-) is stabilized by resonance, whereas the alkoxide ion (Rβˆ’Oβˆ’R-O^-) is destabilized by the +I+I effect of the alkyl group.

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The Lucas Test distinguishes 1∘,2∘,1^\circ, 2^\circ, and 3∘3^\circ alcohols using anhydrous ZnCl2ZnCl_2 and conc. HClHCl. 3∘3^\circ alcohols give immediate turbidity, 2∘2^\circ within 5 minutes, and 1∘1^\circ only upon heating.

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The Reimer-Tiemann reaction involves the treatment of phenol with chloroform (CHCl3CHCl_3) in the presence of sodium hydroxide to introduce an aldehyde group at the ortho position, forming Salicylaldehyde.

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Kolbe's Reaction involves the treatment of phenoxide ion with CO2CO_2 followed by acidification to produce Salicylic acid (2-hydroxybenzoic acid).

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Ethers have lower boiling points than isomeric alcohols because they cannot form intermolecular HH-bonding with themselves.

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Cleavage of ethers by hydrohalic acids (HXHX): Simple ethers react via SN2S_N2 mechanism, where the halide ion attacks the smaller alkyl group. However, if a 3∘3^\circ alkyl group is present, the reaction proceeds via SN1S_N1 mechanism.

πŸ“Formulae

Rβˆ’OH+HXβ†’ZnCl2Rβˆ’X+H2OR-OH + HX \xrightarrow{ZnCl_2} R-X + H_2O

C6H5OH+CHCl3+3NaOH→343KC6H4(OH)CHO+3NaCl+2H2OC_6H_5OH + CHCl_3 + 3NaOH \xrightarrow{343K} C_6H_4(OH)CHO + 3NaCl + 2H_2O

C6H5ONa+CO2β†’413K,4βˆ’7atmC6H4(OH)COONaβ†’H+C6H4(OH)COOHC_6H_5ONa + CO_2 \xrightarrow{413K, 4-7 atm} C_6H_4(OH)COONa \xrightarrow{H^+} C_6H_4(OH)COOH

Rβˆ’X+Rβ€²βˆ’Oβˆ’Na+β†’Rβˆ’Oβˆ’Rβ€²+NaXR-X + R'-O^-Na^+ \rightarrow R-O-R' + NaX

CH3CH2OH→conc.H2SO4,443KCH2=CH2+H2OCH_3CH_2OH \xrightarrow{conc. H_2SO_4, 443K} CH_2=CH_2 + H_2O

3Rβˆ’OH+PCl3β†’3Rβˆ’Cl+H3PO33R-OH + PCl_3 \rightarrow 3R-Cl + H_3PO_3

πŸ’‘Examples

Problem 1:

Predict the products of the reaction of anisoie (C6H5OCH3C_6H_5OCH_3) with HIHI.

Solution:

C6H5OCH3+HI→C6H5OH+CH3IC_6H_5OCH_3 + HI \rightarrow C_6H_5OH + CH_3I

Explanation:

The reaction occurs via SN2S_N2 mechanism. The nucleophile Iβˆ’I^- attacks the smaller methyl group because the C6H5βˆ’OC_6H_5-O bond has partial double bond character due to resonance, making it stronger and harder to break than the Oβˆ’CH3O-CH_3 bond.

Problem 2:

Write the chemical equation for the nitration of phenol with dilute HNO3HNO_3 at low temperature (298K298 K).

Solution:

C6H5OHβ†’dil.HNO3,298Koβˆ’nitrophenol+pβˆ’nitrophenolC_6H_5OH \xrightarrow{dil. HNO_3, 298K} o-nitrophenol + p-nitrophenol

Explanation:

Phenol is highly reactive towards electrophilic substitution. Dilute HNO3HNO_3 gives a mixture of ortho and para isomers. The ortho isomer can be separated from the para isomer by steam distillation due to intramolecular HH-bonding in oo-nitrophenol.

Problem 3:

How can you distinguish between Ethanol and Phenol using a chemical test?

Solution:

Neutral FeCl3FeCl_3 test.

Explanation:

Phenol reacts with neutral ferric chloride to give a violet-colored complex: 6C6H5OH+FeCl3β†’[Fe(OC6H5)6]3βˆ’+3H++3Clβˆ’6C_6H_5OH + FeCl_3 \rightarrow [Fe(OC_6H_5)_6]^{3-} + 3H^+ + 3Cl^-. Ethanol does not give this color reaction.

Physical and chemical properties - Revision Notes & Key Formulas | ICSE Class 12 Chemistry