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1)2HCHO conc.KOH yields 2)C6H5CHO HNO3/H2SO4 at 273-283K?
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1)2HCHO conc.KOH yields 2)C6H5CHO HNO3/H2SO4 at 273-283K?
2HCHO + conc.KOH → C6H5CHO + HNO3/H2SO4 at 273-283K
Introduction
The given chemical equation represents the reaction between formaldehyde (HCHO) and concentrated potassium hydroxide (KOH) to produce benzaldehyde (C6H5CHO) in the presence of nitric acid (HNO3) and sulfuric acid (H2SO4) at a temperature range of 273-283K. Let's analyze this reaction and its mechanism in detail.
Reaction Mechanism
The reaction can be broken down into two steps:
Step 1: Formation of the Aldol
In the first step, formaldehyde (HCHO) reacts with concentrated potassium hydroxide (KOH) to form an aldol intermediate. This step is known as the aldol condensation reaction.
HCHO + KOH → Aldol Intermediate
The aldol intermediate is an enolate ion, which is stabilized by the presence of the strong base (KOH). The enolate ion can then undergo further reactions to form the desired product.
Step 2: Conversion to Benzaldehyde
In the second step, the aldol intermediate reacts with nitric acid (HNO3) and sulfuric acid (H2SO4) to form benzaldehyde (C6H5CHO). This step involves the oxidation of the aldol intermediate to benzaldehyde.
Aldol Intermediate + HNO3/H2SO4 → C6H5CHO
The presence of nitric acid and sulfuric acid provides the necessary conditions for the oxidation of the aldol intermediate. The reaction proceeds at a temperature range of 273-283K, which ensures the optimal conditions for the reaction to occur.
Overall Reaction
Combining both steps, the overall reaction can be represented as:
2HCHO + conc.KOH + HNO3/H2SO4 → C6H5CHO
In this reaction, formaldehyde (HCHO) is converted into benzaldehyde (C6H5CHO) through the intermediate formation of an aldol compound. The presence of concentrated potassium hydroxide, nitric acid, and sulfuric acid is crucial for the success of this reaction.
Conclusion
The given reaction involves the condensation of formaldehyde with concentrated potassium hydroxide to form an aldol intermediate, which then undergoes oxidation to produce benzaldehyde. The presence of nitric acid and sulfuric acid, along with the specified temperature range, allows for the smooth progression of the reaction.
Introduction
The given chemical equation represents the reaction between formaldehyde (HCHO) and concentrated potassium hydroxide (KOH) to produce benzaldehyde (C6H5CHO) in the presence of nitric acid (HNO3) and sulfuric acid (H2SO4) at a temperature range of 273-283K. Let's analyze this reaction and its mechanism in detail.
Reaction Mechanism
The reaction can be broken down into two steps:
Step 1: Formation of the Aldol
In the first step, formaldehyde (HCHO) reacts with concentrated potassium hydroxide (KOH) to form an aldol intermediate. This step is known as the aldol condensation reaction.
HCHO + KOH → Aldol Intermediate
The aldol intermediate is an enolate ion, which is stabilized by the presence of the strong base (KOH). The enolate ion can then undergo further reactions to form the desired product.
Step 2: Conversion to Benzaldehyde
In the second step, the aldol intermediate reacts with nitric acid (HNO3) and sulfuric acid (H2SO4) to form benzaldehyde (C6H5CHO). This step involves the oxidation of the aldol intermediate to benzaldehyde.
Aldol Intermediate + HNO3/H2SO4 → C6H5CHO
The presence of nitric acid and sulfuric acid provides the necessary conditions for the oxidation of the aldol intermediate. The reaction proceeds at a temperature range of 273-283K, which ensures the optimal conditions for the reaction to occur.
Overall Reaction
Combining both steps, the overall reaction can be represented as:
2HCHO + conc.KOH + HNO3/H2SO4 → C6H5CHO
In this reaction, formaldehyde (HCHO) is converted into benzaldehyde (C6H5CHO) through the intermediate formation of an aldol compound. The presence of concentrated potassium hydroxide, nitric acid, and sulfuric acid is crucial for the success of this reaction.
Conclusion
The given reaction involves the condensation of formaldehyde with concentrated potassium hydroxide to form an aldol intermediate, which then undergoes oxidation to produce benzaldehyde. The presence of nitric acid and sulfuric acid, along with the specified temperature range, allows for the smooth progression of the reaction.
Community Answer
1)2HCHO conc.KOH yields 2)C6H5CHO HNO3/H2SO4 at 273-283K?
1) This is Cannizaro reaction in which oxidation and reduction of the same compound takes place. So, product is HCOOK and CH3OH.
2) Nitration takes place at meta position to give m-NO2C6H4CHO.
2) Nitration takes place at meta position to give m-NO2C6H4CHO.
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1)2HCHO conc.KOH yields 2)C6H5CHO HNO3/H2SO4 at 273-283K?
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1)2HCHO conc.KOH yields 2)C6H5CHO HNO3/H2SO4 at 273-283K? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 1)2HCHO conc.KOH yields 2)C6H5CHO HNO3/H2SO4 at 273-283K? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1)2HCHO conc.KOH yields 2)C6H5CHO HNO3/H2SO4 at 273-283K?.
1)2HCHO conc.KOH yields 2)C6H5CHO HNO3/H2SO4 at 273-283K? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 1)2HCHO conc.KOH yields 2)C6H5CHO HNO3/H2SO4 at 273-283K? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1)2HCHO conc.KOH yields 2)C6H5CHO HNO3/H2SO4 at 273-283K?.
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