Important Reactions of Alcohols and Phenols

Identify whether the compound is alcohol or phenol, locate the reactive site, note the reagent and conditions, then decide whether substitution, oxidation, dehydration or esterification occurs.

Reaction schemes with products at ortho and para positions help in phenol substitution questions.

Writing benzaldehyde as the Reimer-Tiemann product of phenol is wrong because the -OH group remains on the ring and the formyl group mainly enters the ortho position.

Alcohols undergo reactions such as esterification, dehydration and oxidation, while phenols undergo electrophilic substitution reactions such as bromination, nitration and Reimer-Tiemann reaction.

Alcohols undergo reactions such as esterification, dehydration and oxidation, while phenols undergo electrophilic substitution reactions such as bromination, nitration and Reimer-Tiemann reaction. Key patterns: alcohol + carboxylic acid with acid gives ester; alcohol under acid and heat gives alkene; primary alcohol oxidation gives aldehyde or acid; phenol + Br₂ water gives 2,4,6-tribromophenol; phenol + CHCl3/NaOH then acidification gives salicylaldehyde. Phenol with bromine water gives a white precipitate of 2,4,6-tribromophenol.

Alcohol reactions are controlled by the O-H bond and the carbon bearing -OH. Esterification forms esters with carboxylic acids or acid derivatives. Dehydration removes water to form alkenes under acidic conditions. Oxidation of primary alcohols can give aldehydes and acids, while secondary alcohols give ketones. Phenol activates the benzene ring toward electrophilic substitution, especially at ortho and para positions. With bromine water, phenol gives 2,4,6-tribromophenol. In the Reimer-Tiemann reaction, phenol reacts with chloroform and base followed by acidification to give salicylaldehyde as the major product. Key patterns: alcohol + carboxylic acid with acid gives ester; alcohol under acid and heat gives alkene; primary alcohol oxidation gives aldehyde or acid; phenol + Br₂ water gives 2,4,6-tribromophenol; phenol + CHCl3/NaOH then acidification gives salicylaldehyde. Predict the product when phenol is treated with bromine water. The -OH group strongly activates the ring and directs substitution to ortho and para positions, so three bromine atoms enter at 2, 4 and 6 positions, forming 2,4,6-tribromophenol. Questions ask for product prediction, named reaction identification, observation, conversion, and reagent-condition matching. Writing benzaldehyde as the Reimer-Tiemann product of phenol is wrong because the -OH group remains on the ring and the formyl group mainly enters the ortho position.