Question

Give equations of the following reactions: (i) Oxidation of propan-1-ol with alkaline \(\mathrm{KMnO}_{4}\) solution. (ii) Bromine in \(\mathrm{CS}_{2}\) with phenol. (iii) Dilute \(\mathrm{HNO}_{3}\) with phenol. (iv) Treating phenol wih chloroform in presence of aqueous \(\mathrm{NaOH}\).

Short answer

(i) Propan-1-ol forms propanoic acid. (ii) Phenol with bromine forms 2,4,6-tribromophenol. (iii) Phenol with dilute HNO₃ forms ortho- and para-nitrophenol. (iv) Phenol with chloroform and NaOH forms salicylaldehyde.

Step-by-step solution

Oxidation of Propan-1-ol

Propan-1-ol is oxidized using alkaline \ \(\text{KMnO}_4\) (potassium permanganate). This is a primary alcohol, and upon oxidation, it forms a carboxylic acid. The reaction can be written as: \ \[ \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{COOH} + \text{H}_2\text{O} \] This means that propan-1-ol is converted into propanoic acid.

Reaction of Phenol with Bromine

Phenol reacts with bromine in carbon disulfide \ \(\text{CS}_2\). This results in the formation of 2,4,6-tribromophenol, where three bromine atoms substitute hydrogen atoms at the ortho and para positions: \ \[ \text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr} \] Here, a white precipitate of 2,4,6-tribromophenol forms.

Reaction of Phenol with Dilute Nitric Acid

Phenol reacts with dilute \ \(\text{HNO}_3\) primarily at a moderated temperature to form ortho-nitrophenol and para-nitrophenol: \ \[ \text{C}_6\text{H}_5\text{OH} + \text{HNO}_3 \rightarrow \text{C}_6\text{H}_4\text{(NO}_2)\text{OH} + \text{H}_2\text{O} \] This results in a mix of ortho-nitrophenol and para-nitrophenol as major products.

Chloroform with Phenol in Aqueous NaOH

Treating phenol with chloroform and aqueous \ \(\text{NaOH}\) undergoes the Reimer-Tiemann reaction. This introduces a formyl group onto the benzene ring, primarily at the ortho position, resulting in salicylaldehyde: \ \[ \text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + 3\text{NaOH} \rightarrow \text{C}_6\text{H}_4\text{CHO} + 3\text{NaCl} + 2\text{H}_2\text{O} \] This reaction forms 2-hydroxybenzaldehyde, commonly known as salicylaldehyde.

Key concepts

Oxidation Reactions

Oxidation reactions in organic chemistry involve the transformation of organic molecules by increasing their oxygen content or decreasing their hydrogen content. This usually leads to the conversion of alcohols into aldehydes, ketones, or carboxylic acids. One classic example is the oxidation of propan-1-ol.

In this reaction, propan-1-ol, which is a primary alcohol, is oxidized using an alkaline potassium permanganate (\(\text{KMnO}_4\)) solution. The oxidation leads to the formation of propanoic acid. Potassium permanganate acts as a strong oxidizing agent, and in this context, it acts by supplying the necessary oxygen atoms to convert the alcohol into the acid.

The process can be summarized by the equation: \[\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{COOH} + \text{H}_2\text{O}\]This oxidation reaction is highly useful in preparing carboxylic acids from primary alcohols.

Nitration of Phenol

Nitration is a type of electrophilic aromatic substitution reaction where a nitro group (-NO₂) is introduced to an aromatic ring. When phenol reacts with dilute nitric acid (\(\text{HNO}_3\)), nitration occurs. This reaction primarily takes place under controlled moderate temperatures to avoid over-nitration.

In the case of phenol, the hydroxyl group activates the benzene ring, particularly for ortho and para positions. Hence, the reaction yields a mixture of ortho-nitrophenol and para-nitrophenol. The general reaction can be expressed as follows:

\[\text{C}_6\text{H}_5\text{OH} + \text{HNO}_3 \rightarrow \text{C}_6\text{H}_4\text{(NO}_2)\text{OH} + \text{H}_2\text{O}\]The mixture is often separated due to the different physical properties of ortho and para isomers. While ortho-nitrophenol is slightly more volatile, para-nitrophenol tends to form crystalline solids and is more commonly isolated in labs.

Reimer-Tiemann Reaction

The Reimer-Tiemann reaction is a fascinating method in organic chemistry used to introduce a formyl group (–CHO) into an aromatic ring. This reaction predominantly affects phenols, leading to the formation of salicylaldehyde.

In this transformation, phenol reacts with chloroform (\(\text{CHCl}_3\)) in the presence of aqueous sodium hydroxide (\(\text{NaOH}\)). The active species in this reaction is dichlorocarbene, generated in situ, which then attacks the aromatic ring. The chloroform serves as the source of the carbene, and sodium hydroxide facilitates the removal of hydrogen atoms.

The product of this reaction is 2-hydroxybenzaldehyde, also known as salicylaldehyde. The reaction can be represented by:\[\text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + 3\text{NaOH} \rightarrow \text{C}_6\text{H}_4\text{CHO} + 3\text{NaCl} + 2\text{H}_2\text{O}\]This method is valuable for synthesizing aldehydes directly from phenolic compounds.

Halogenation of Aromatic Compounds

Halogenation is the process whereby a halogen is introduced into an organic molecule. In aromatic compounds, this typically involves substituting hydrogen atoms on the ring with halogens, such as bromine or chlorine.

In the context of phenol, bromination is a classic example. When phenol is treated with bromine in a carbon disulphide (\(\text{CS}_2\)) solvent, it undergoes substitution at the ortho and para positions, due to the directing effect of the hydroxyl group.

The resulting product is 2,4,6-tribromophenol, which precipitates as a white solid:\[\text{C}_6\text{H}_5\text{OH} + 3\text{Br}_2 \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{OH} + 3\text{HBr}\]Halogenation reactions are widely used for modifying aromatic compounds, enhancing their reactivity, and preparing various derivatives for further chemical applications.