Problem 40
Question
Which of the following reactants will react with phenol to give salicylaldehyde after hydrolysis? (a) methyl chloride (b) trichloromethane (c) dichloromethane (d) none of these
Step-by-Step Solution
Verified Answer
The correct reactant is (b) trichloromethane.
1Step 1: Identify the Key Reaction
To solve this problem, we must identify the reaction that forms salicylaldehyde from phenol. This reaction is the Reimer-Tiemann reaction, which involves several stages including the formation of a dichlorocarbene intermediate that converts phenol into salicylaldehyde.
2Step 2: Reactant Identification
In the Reimer-Tiemann reaction, phenol reacts with chloroform (trichloromethane) in the presence of a strong base to form salicylaldehyde. The key reactant here is trichloromethane.
3Step 3: Check Reactant Options
Among the given options, trichloromethane, which is option (b), is the chemical name for chloroform. Methyl chloride and dichloromethane do not produce salicylaldehyde in this reaction.
4Step 4: Choose the Correct Answer
Since trichloromethane is the reactant that leads to the formation of salicylaldehyde when reacted with phenol, option (b) is the correct choice.
Key Concepts
Salicylaldehyde FormationDichlorocarbene IntermediatePhenol Reaction
Salicylaldehyde Formation
Salicylaldehyde formation is a fascinating topic in organic chemistry. It predominantly involves the Reimer-Tiemann reaction, which is integral for converting phenol into salicylaldehyde. This transformation is crucial because salicylaldehyde is a valuable compound utilized in various industrial and pharmaceutical applications. The process starts with using trichloromethane, commonly known as chloroform, as one of the reactants.
When phenol is treated with chloroform in the presence of a strong base, it proceeds through multiple stages to yield salicylaldehyde. The base helps in deprotonating phenol, facilitating the formation of reactive intermediates crucial in this chemical pathway.
When phenol is treated with chloroform in the presence of a strong base, it proceeds through multiple stages to yield salicylaldehyde. The base helps in deprotonating phenol, facilitating the formation of reactive intermediates crucial in this chemical pathway.
- This reaction specifically introduces a formyl group (\(-CHO\)) at the ortho position of phenol, resulting in the formation of salicylaldehyde.
- Thus, the positioning of this formyl group is critical, and chloroform plays a vital role in this precise transformation.
Dichlorocarbene Intermediate
A key intermediate in the Reimer-Tiemann reaction is the dichlorocarbene. This reactive species is important for facilitating the conversion of phenol to salicylaldehyde. Formation of dichlorocarbene is initiated by treating trichloromethane with a strong base, typically sodium hydroxide (NaOH).
The strong base deprotonates trichloromethane, leading to the elimination of a chloride ion, which subsequently results in the creation of the dichlorocarbene intermediate \( (:\text{CCl}_2) \). This dichlorocarbene is a highly reactive species due to its electron-deficient carbon atom, making it a target for further reactions with nucleophiles.
The strong base deprotonates trichloromethane, leading to the elimination of a chloride ion, which subsequently results in the creation of the dichlorocarbene intermediate \( (:\text{CCl}_2) \). This dichlorocarbene is a highly reactive species due to its electron-deficient carbon atom, making it a target for further reactions with nucleophiles.
- Phenoxide ion, the deprotonated form of phenol, acts as the nucleophile in this reaction.
- The dichlorocarbene then attacks this nucleophile at the ortho position, contributing to the final arrangement of atoms seen in salicylaldehyde.
Phenol Reaction
Reactions involving phenol are central in producing valuable chemicals like salicylaldehyde. Phenol, an aromatic compound, features an \(-OH\) group which significantly influences its reactivity. In the Reimer-Tiemann reaction, phenol’s \(-OH\) group is not merely passive. It actively participates by forming a phenoxide ion upon deprotonation by a base.
Phenoxide ions serve as more potent nucleophiles than phenol itself, allowing them to facilitate key reactions with electrophilic species such as dichlorocarbene. In this scenario, the strong \(-CH_2Cl_2\) bond forms with the phenoxide ion, ultimately guiding the reaction toward the synthesis of salicylaldehyde.
Phenoxide ions serve as more potent nucleophiles than phenol itself, allowing them to facilitate key reactions with electrophilic species such as dichlorocarbene. In this scenario, the strong \(-CH_2Cl_2\) bond forms with the phenoxide ion, ultimately guiding the reaction toward the synthesis of salicylaldehyde.
- Unsubstituted phenol is necessary for this reaction to proceed efficiently to achieve the formylation at the ortho position.
- The aromatic structure of phenol allows for effective resonance stabilization, aiding in the stability of the resulting intermediates.
Other exercises in this chapter
Problem 37
Which one of the following sets of reactants is used in Reimer-Tiemann reaction? (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CO}_{2}+\mathrm{KOH}\)
View solution Problem 39
In Williamson synthesis, ethoxyethane is prepared by (a) heating sodium ethoxide with ethyl bromide (b) passing ethanol over heated alumina (c) heating ethanol
View solution Problem 43
Which of the following reactions does not yield an alkyl halide? (a) diethyl ether \(+\mathrm{RCOCl}\) (b) diethyl ether \(+\mathrm{PCl}_{5}\) (c) diethyl ether
View solution Problem 45
When benzenediazonium chloride is treated with water, the compound formed is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (b) \(\mathrm{C}_{6} \mathrm{H}
View solution