Problem 116
Question
Schotten-Baumann reaction is (a) sodium phenate \(+\) methyl iodide \(\quad{ }_{-\mathrm{HCl}}\) phenyl benzoate (b) benzenediazonium chloride \(+\) phenol \(\longrightarrow\) p-Hydroxyazobenzene (c) phenol + benzoyl chloride \(\stackrel{\mathrm{NaOH}}{\longrightarrow}\) phenyl benzoate (d) phenol + chloroform \(\frac{\mathrm{NaOH}}{\text { salicyladehyde }}\)
Step-by-Step Solution
Verified Answer
Option (c): phenol + benzoyl chloride with NaOH is a Schotten-Baumann reaction.
1Step 1: Understand Schotten-Baumann Reaction
The Schotten-Baumann reaction is a chemical reaction used to synthesize amides from amines and acid chlorides. This reaction typically involves the use of an aqueous base, like NaOH, to facilitate the reaction.
2Step 2: Analyze Option Effects
- (a) Sodium phenate with methyl iodide does not form phenyl benzoate.
- (b) The reaction between benzenediazonium chloride and phenol forms a diazo compound, not phenyl benzoate.
- (c) Phenol with benzoyl chloride in the presence of NaOH is a typical Schotten-Baumann reaction that produces phenyl benzoate.
- (d) Phenol with chloroform under the conditions specified leads to salicylaldehyde, not phenyl benzoate.
3Step 3: Identify the Correct Reaction
The reaction in option (c) describes the synthesis of phenyl benzoate using phenol and benzoyl chloride, which is characteristic of a Schotten-Baumann reaction.
Key Concepts
Synthesis of AmidesPhenyl Benzoate FormationRole of Aqueous Base in Reactions
Synthesis of Amides
Amides are versatile compounds formed in organic chemistry, often synthesized through the Schotten-Baumann reaction. This reaction involves the conversion of an amine and an acid chloride to create an amide. Here, an aqueous base like NaOH is critical. The base not only facilitates the reaction by neutralizing the released HCl but also ensures the reaction environment remains conducive for the amide formation.
The process can be generalized as follows:
- Amines react with acyl chlorides such as benzoyl chloride.
- This reaction occurs in the presence of a suitable base, such as NaOH.
- The aqueous base prevents the acid chloride from being destroyed by water and neutralizes the HCl byproduct.
Phenyl Benzoate Formation
Phenyl benzoate is an ester that can be synthesized via the Schotten-Baumann reaction. Here, phenol and benzoyl chloride are the starting materials, and NaOH is the aqueous base used to control the reaction conditions.
Phenyl benzoate formation can be succinctly summarized:
- Phenol acts as the nucleophile and attacks the carbonyl carbon of the benzoyl chloride.
- NaOH, as an aqueous base, facilitates this nucleophilic attack by deprotonating the phenol to make it more reactive.
- The reaction results in the displacement of the chloride ion and the formation of phenyl benzoate.
Role of Aqueous Base in Reactions
A key component of the Schotten-Baumann reaction is the use of an aqueous base, such as NaOH. This base plays multiple, essential roles in the reaction process.
Firstly, the aqueous base neutralizes the hydrochloric acid (HCl) formed as a byproduct during the reaction between amines (or phenols) and acid chlorides. Without this neutralization step, the acidic conditions would inhibit further reaction and potentially degrade the starting materials or products.
Additionally, the aqueous base helps in the deprotonation of the reactants. For instance, when phenol is the reactant, it is converted to the more reactive phenoxide ion, enhancing its capability to attack the electrophilic carbonyl carbon in benzoyl chloride.
- Neutralizes HCl created in the reaction.
- Facilitates nucleophilic attack on acyl chlorides.
- Improves reaction efficiency and product yield.
Other exercises in this chapter
Problem 114
A compound is soluble in concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\). It does not decolourize bromine in carbon tetrachloride but is oxidized by chromic anh
View solution Problem 115
Vinyl carbinol is (a) \(\mathrm{HO}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}\left(\mathrm{CH}_{2} \mathrm{OH}\right)=\math
View solution Problem 117
An organic compound of molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) does not react with sodium. With excess of HI, it gives only one type of
View solution Problem 118
A compound of the formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) reacts with sodium and undergoes oxidation to give a carbonyl compound which does not re
View solution