Title: Mechanism of Reaction Between Acetone and 1,2-DiaminobenzeneIntroduction:This detailed explanation will help you understand the mechanism of the reaction between acetone ((CH₃)₂CO) and 1,2-diaminobenzene (C₆H₄(NH₂)₂). The steps involve the usage of curved arrows to illustrate the flow of electrons throughout the reaction mechanism.Explanation of the Diagram:The reaction mechanism proceeds as follows:1. **Reaction Initialization:** - The reaction starts with acetone ((CH₃)₂CO) and 1,2-diaminobenzene ((C₆H₄)(NH₂)₂) in the presence of acidic media (shown as H₃O⁺). - Acetone has a carbonyl group where oxygen is more electronegative and pulls electrons towards itself, making the carbonyl carbon electrophilic.2. **Nucleophilic Attack:** - The lone pair of electrons on the nitrogen of the amine group (NH₂) in 1,2-diaminobenzene attacks the electrophilic carbon on the carbonyl group of acetone. - This step forms a tetrahedral intermediate, where the nitrogen from the 1,2-diaminobenzene is now bonded to the carbon from acetone, and the oxygen has an extra pair of electrons making it negatively charged (indicated by the arrow pushing towards the oxygen).3. **Protonation of Intermediate:** - The negatively charged oxygen is protonated by H₃O⁺ in the medium, converting initially formed intermediate to a neutral compound.4. **Formation of Iminium Ion:** - Concurrently, a second NH₂ group from the 1,2-diaminobenzene attacks the newly attached methyl carbon on the tetrahedral intermediate to form a ring structure. - This step is followed by the loss of a water molecule (H₂O) leading to the formation of an iminium ion. This is depicted through multiple steps where electron pairs are repositioned and water is eliminated. 5. **Cyclization:** - The intermediate formed contains a cyclic structure, where electron repositioning and proton removal lead to the final stabilized product, resulting in another aromatic ring. - Loss of water takes place in the successive steps leading to the formation of the final product

The reaction scheme is shown belowWe have to add the curved arrow to this given reaction scheme.

5. Use curved arrows to depict the mechanism of the following reaction: :Ö 0: -NH2 'NH2 업 NH2 HO :0 NH20TH - 0: H 맛으 HT + - C H -NH2 H H H 'NH2 + -OH ÖH H2O HO H H NH2 — NH2 1 I-Z: H H HO® H H CO-H -OH H HO® H H 0④ H

5. Use curved arrows to depict the mechanism of the following reaction: :Ö 0: -NH2 'NH2 업 NH2 HO :0 NH20TH - 0: H 맛으 HT + - C H -NH2 H H H 'NH2 + -OH ÖH H2O HO H H NH2 — NH2 1 I-Z: H H HO® H H CO-H -OH H HO® H H 0④ H

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Title: Mechanism of Reaction Between Acetone and 1,2-Diaminobenzene

Introduction:
This detailed explanation will help you understand the mechanism of the reaction between acetone ((CH₃)₂CO) and 1,2-diaminobenzene (C₆H₄(NH₂)₂). The steps involve the usage of curved arrows to illustrate the flow of electrons throughout the reaction mechanism.

Explanation of the Diagram:

The reaction mechanism proceeds as follows:

1. **Reaction Initialization:**
- The reaction starts with acetone ((CH₃)₂CO) and 1,2-diaminobenzene ((C₆H₄)(NH₂)₂) in the presence of acidic media (shown as H₃O⁺).
- Acetone has a carbonyl group where oxygen is more electronegative and pulls electrons towards itself, making the carbonyl carbon electrophilic.

2. **Nucleophilic Attack:**
- The lone pair of electrons on the nitrogen of the amine group (NH₂) in 1,2-diaminobenzene attacks the electrophilic carbon on the carbonyl group of acetone.
- This step forms a tetrahedral intermediate, where the nitrogen from the 1,2-diaminobenzene is now bonded to the carbon from acetone, and the oxygen has an extra pair of electrons making it negatively charged (indicated by the arrow pushing towards the oxygen).

3. **Protonation of Intermediate:**
- The negatively charged oxygen is protonated by H₃O⁺ in the medium, converting initially formed intermediate to a neutral compound.

4. **Formation of Iminium Ion:**
- Concurrently, a second NH₂ group from the 1,2-diaminobenzene attacks the newly attached methyl carbon on the tetrahedral intermediate to form a ring structure.
- This step is followed by the loss of a water molecule (H₂O) leading to the formation of an iminium ion. This is depicted through multiple steps where electron pairs are repositioned and water is eliminated.

5. **Cyclization:**
- The intermediate formed contains a cyclic structure, where electron repositioning and proton removal lead to the final stabilized product, resulting in another aromatic ring.
- Loss of water takes place in the successive steps leading to the formation of the final product

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