Match List I with List II

Choose the correct answers from the options given below:

A-IV, B-III, C-I, D-II

The correct answer is option 2. A-IV, B-III, C-I, D-II

Let us look at each of these matches in detail:

(A) Phenol + \(Br_2\) water: (IV) White precipitate:

When phenol reacts with bromine water, a white precipitate forms. Phenol is an aromatic compound with the chemical formula \(C_6H_5OH\). The hydroxyl group \((-OH)\) attached to the benzene ring makes phenol more reactive towards electrophilic substitution reactions. Bromine water is an aqueous solution of bromine \((Br_2)\) and is used to test for the presence of phenolic compounds.

Reaction Mechanism:

The hydroxyl group \((-OH)\) in phenol is an activating group and is ortho/para-directing, meaning it increases the electron density on the ortho (positions 2 and 6) and para (position 4) positions of the benzene ring. When phenol is treated with bromine water, the bromine molecules undergo electrophilic substitution at the activated positions (ortho and para to the -OH group). Phenol reacts with bromine water to form 2,4,6-tribromophenol, a compound where three bromine atoms are substituted into the ortho and para positions relative to the hydroxyl group.

The product, 2,4,6-tribromophenol, is a white solid that precipitates out of the solution.

Observations:

Color Change: The brownish color of bromine water disappears as it reacts with phenol, indicating the consumption of bromine.

White Precipitate: The formation of the white precipitate (2,4,6-tribromophenol) confirms the reaction.

When phenol reacts with bromine water, it forms a white precipitate of 2,4,6-tribromophenol. This reaction is a classic test for phenol and other activated aromatic compounds that readily undergo electrophilic substitution with bromine.

(B) Phenol + \(CO_2/NaOH\): (III) Kolbe's Reaction :

Kolbe’s reaction, also known as Kolbe Schmitt Reaction, is a type of addition reaction named after Hermann Kolbe and Rudolf Schmitt. When phenol is treated with sodium hydroxide, phenoxide ion is generated. The phenoxide ion generated is more reactive than phenol towards electrophilic aromatic substitution reaction. Hence, it undergoes an electrophilic substitution reaction with carbon dioxide, which is a weak electrophile. Ortho-hydroxybenzoic acid (salicylic acid) is formed as the primary product. This reaction is popularly known as Kolbe’s reaction.

Kolbe’s reaction can be classified as a carboxylation chemical reaction. The reaction occurs when sodium phenolate is allowed to absorb carbon dioxide, and the resulting product is heated at a temperature of a 125-degree Celsius and a pressure of over a hundred atmospheres. An unstable intermediate is now formed.

This unstable intermediate goes through a proton shift, leading to the formation of sodium salicylate. Now, this mixture is treated with sulfuric acid. The acidification of the mixture yields salicylic acid. The illustration for the Kolbe’s reaction mechanism is given below:

(C) Phenol + \(CHCl_3 / KOH\): (I) Reimer-Tiemann Reaction:

Reimer Tiemann reaction is a type of substitution reaction named after chemists Karl Reimer and Ferdinand Tiemann. The reaction is used for the ortho-formylation of \(C_6H_5OH\) (phenols). When phenol, i.e. \(C_6H_5OH\), is treated with \(CHCl_3\) (chloroform) in the presence of \(NaOH\) (sodium hydroxide), an aldehyde group \((-CHO)\) is introduced at the ortho position of the benzene ring leading to the formation of o-hydroxybenzaldehyde. The reaction is popularly known as the Reimer Tiemann reaction.

Mechanism:

Step I: First, a strongly basic hydroxide solution is used to deprotonate the chloroform. Deprotonation is the elimination of the hydrogen atom. The chloroform carbanion is formed by removing the hydrogen atom. The chloroform carbanion will swiftly undergo alpha elimination, yielding dichlorocarbene \((CCl_2)\), the reaction’s major reactive species.

Step II: Base extracts the hydrogen atom from the \(-OH\) group to generate the phenoxide ion. The phenoxide’s negative charge is delocalized into the aromatic ring, making it more nucleophilic.

Step III: The addition of dichlorocarbene results in the formation of an intermediate i.e., dichloromethyl-substituted phenol. The resulting intermediate is then exposed to basic hydrolysis to yield ortho-hydroxybenzaldehyde.

(D) Phenol + Benzene diazonium chloride and \(NaOH\)(II) Azo-dye test:

When phenol reacts with benzene diazonium chloride (\(C_6H_5N_2^+Cl^-\)) in the presence of sodium hydroxide (\(NaOH\)), it undergoes a reaction known as the azo-dye test. Here’s a detailed explanation:

Reagents:

Phenol (\(C_6H_5OH\))

Benzene Diazonium Chloride (\(C_6H_5N_2^+Cl^-\))

Sodium Hydroxide (\(NaOH\))

Reaction Process:

Diazotization Reaction: Benzene diazonium chloride is prepared from aniline (a primary aromatic amine) using sodium nitrite (\(NaNO_2\)) and hydrochloric acid (\(HCl\)).

Coupling Reaction: Phenol reacts with the diazonium salt (\(C_6H_5N_2^+Cl^-\)) to form an azo dye. This reaction is a coupling reaction where the diazonium salt reacts with the phenol to produce an azo compound.

The product is p-hydroxyazobenzene

p-hydroxyazobenzene is an orange colored dye, and its formation indicates a positive test. The reaction typically results in the formation of a brightly colored azo dye.