What is the major final product in the following sequence of reactions? 1. CH;CH,CI, AICI3 2. ΗΝO3, Η,SO, 3. Cl2, FeCl3

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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Can you explain the reasons and concepts for why these answers are true?

**Question 19:**

What is the major final product in the following sequence of reactions?

**Reaction Sequence:**

1. React benzene with \(\text{CH}_3\text{CH}_2\text{Cl}\) and \(\text{AlCl}_3\).
2. Then treat with \(\text{HNO}_3\) and \(\text{H}_2\text{SO}_4\).
3. Finally, react with \(\text{Cl}_2\) and \(\text{FeCl}_3\).

**Product Options:**

- **A. Compound I**: A benzene derivative with a nitro group (\(\text{NO}_2\)) and a chlorine (\(\text{Cl}\)) on the aromatic ring, and an ethyl group (\(\text{C}_2\text{H}_5\)) on the meta position relative to the nitro group.
  
- **B. Compound II**: A benzene derivative with an amino group (\(\text{NH}_2\)) and a chlorine on the aromatic ring, and an ethyl group on the meta position relative to the amino group.
  
- **C. Compound III**: A benzene derivative with a nitro group and a chlorine on the aromatic ring, and an ethyl group on the para position relative to the chlorine.
  
- **D. Compound IV**: A benzene derivative with a nitro group and a chlorine on the aromatic ring, and an ethyl group on the para position relative to the nitro group.

**Correct Answer:**

- **D. Compound IV**

**Explanation:**

The reaction sequence involves:

1. **Friedel-Crafts Alkylation**: Adding an ethyl group to the benzene ring.
2. **Nitration**: Introducing a nitro group using \(\text{HNO}_3\) and \(\text{H}_2\text{SO}_4\).
3. **Chlorination**: Adding a chlorine atom using \(\text{Cl}_2\) and \(\text{FeCl}_3\).

The final product (Compound IV) reflects the expected positions of substituents based on the directing effects of existing groups during each step. Here, the ethyl and nitro groups direct subsequent substitutions accurately.
Transcribed Image Text:**Question 19:** What is the major final product in the following sequence of reactions? **Reaction Sequence:** 1. React benzene with \(\text{CH}_3\text{CH}_2\text{Cl}\) and \(\text{AlCl}_3\). 2. Then treat with \(\text{HNO}_3\) and \(\text{H}_2\text{SO}_4\). 3. Finally, react with \(\text{Cl}_2\) and \(\text{FeCl}_3\). **Product Options:** - **A. Compound I**: A benzene derivative with a nitro group (\(\text{NO}_2\)) and a chlorine (\(\text{Cl}\)) on the aromatic ring, and an ethyl group (\(\text{C}_2\text{H}_5\)) on the meta position relative to the nitro group. - **B. Compound II**: A benzene derivative with an amino group (\(\text{NH}_2\)) and a chlorine on the aromatic ring, and an ethyl group on the meta position relative to the amino group. - **C. Compound III**: A benzene derivative with a nitro group and a chlorine on the aromatic ring, and an ethyl group on the para position relative to the chlorine. - **D. Compound IV**: A benzene derivative with a nitro group and a chlorine on the aromatic ring, and an ethyl group on the para position relative to the nitro group. **Correct Answer:** - **D. Compound IV** **Explanation:** The reaction sequence involves: 1. **Friedel-Crafts Alkylation**: Adding an ethyl group to the benzene ring. 2. **Nitration**: Introducing a nitro group using \(\text{HNO}_3\) and \(\text{H}_2\text{SO}_4\). 3. **Chlorination**: Adding a chlorine atom using \(\text{Cl}_2\) and \(\text{FeCl}_3\). The final product (Compound IV) reflects the expected positions of substituents based on the directing effects of existing groups during each step. Here, the ethyl and nitro groups direct subsequent substitutions accurately.
**Question 20:** What reagents would be necessary to produce the final disubstituted product from benzene?

**Reaction Sequence Diagram:**
1. **Benzene** is transformed stepwise:
   - First, benzene (hexagonal ring) with alternating double bonds is converted to a nitrobenzene (benzene ring with a nitro group, NO₂).
   - Next, nitrobenzene is converted to the final product, a compound with both a nitro group (NO₂) and a carboxylic acid group (COOH).

**Options:**
- **A.** 1. KMnO₄  2. HNO₃, H₂SO₄  3. CH₃Cl, AlCl₃
- **B.** 1. HNO₃, H₂SO₄  2. CH₃Cl, AlCl₃  3. KMnO₄
- **C.** 1. KMnO₄  2. CH₃Cl, AlCl₃  3. HNO₃, H₂SO₄ 
- **D.** 1. CH₃Cl, AlCl₃  2. HNO₃, H₂SO₄  3. KMnO₄  (correct answer, highlighted)
- **E.** 1. HNO₃, H₂SO₄  2. CH₃COCl, AlCl₃  3. Zn/Hg, HCl

**Explanation of the Correct Option (D):**
- **Step 1**: Use of CH₃Cl and AlCl₃ indicates a Friedel-Crafts alkylation, which attaches a methyl group to the benzene ring.
- **Step 2**: Treatment with HNO₃ and H₂SO₄ is a nitration process, adding a nitro group to the benzene ring.
- **Step 3**: Oxidation using KMnO₄ converts the methyl group to a carboxylic acid group, finalizing the transformation to the disubstituted product with both NO₂ and COOH groups.
Transcribed Image Text:**Question 20:** What reagents would be necessary to produce the final disubstituted product from benzene? **Reaction Sequence Diagram:** 1. **Benzene** is transformed stepwise: - First, benzene (hexagonal ring) with alternating double bonds is converted to a nitrobenzene (benzene ring with a nitro group, NO₂). - Next, nitrobenzene is converted to the final product, a compound with both a nitro group (NO₂) and a carboxylic acid group (COOH). **Options:** - **A.** 1. KMnO₄ 2. HNO₃, H₂SO₄ 3. CH₃Cl, AlCl₃ - **B.** 1. HNO₃, H₂SO₄ 2. CH₃Cl, AlCl₃ 3. KMnO₄ - **C.** 1. KMnO₄ 2. CH₃Cl, AlCl₃ 3. HNO₃, H₂SO₄ - **D.** 1. CH₃Cl, AlCl₃ 2. HNO₃, H₂SO₄ 3. KMnO₄ (correct answer, highlighted) - **E.** 1. HNO₃, H₂SO₄ 2. CH₃COCl, AlCl₃ 3. Zn/Hg, HCl **Explanation of the Correct Option (D):** - **Step 1**: Use of CH₃Cl and AlCl₃ indicates a Friedel-Crafts alkylation, which attaches a methyl group to the benzene ring. - **Step 2**: Treatment with HNO₃ and H₂SO₄ is a nitration process, adding a nitro group to the benzene ring. - **Step 3**: Oxidation using KMnO₄ converts the methyl group to a carboxylic acid group, finalizing the transformation to the disubstituted product with both NO₂ and COOH groups.
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