Me 1 1. FeBr3, Br2 2. AICI 3, Me Me Me Me N Br Br v o r e Br Br A B D Me I N E Br

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Chapter1: Chemical Foundations
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What is the most likely product from the following sequence?
### Chemical Reaction Problem

**Problem Statement:**
Given the chemical reaction:

1. FeBr₃, Br₂
2. AlCl₃, Acetyl chloride (CH₃COCl)

Determine the product of the reaction.

**Starting Compound:**
An aromatic heterocycle with the structure of a methyl group (Me) attached to one nitrogen in a five-membered ring containing one nitrogen atom.

**Possible Products:**

A. ![Option A](https://i.imgur.com/CvHRnzM.png)
- Structure: Bromine (Br) at position 2, a double-bonded oxygen (carbonyl group) at position 3.

B. ![Option B](https://i.imgur.com/CvHRnzM.png)
- Structure: Bromine (Br) at position 3, a double-bonded oxygen (carbonyl group) at position 2.

C. ![Option C](https://i.imgur.com/CvHRnzM.png)
- Structure: Bromine (Br) at position 2, a double-bonded oxygen (carbonyl group) at position 1.

D. ![Option D](https://i.imgur.com/CvHRnzM.png)
- Structure: Bromine (Br) at position 3, a double-bonded oxygen (carbonyl group) at position 4.

E. ![Option E](https://i.imgur.com/CvHRnzM.png)
- Structure: Bromine (Br) at position 4, a double-bonded oxygen (carbonyl group) at position 3.

**Explanation:**
1. **First Step:** Bromination
   - The FeBr₃ catalyst helps to generate a bromine electrophile (Br⁺) which then brominates the aromatic ring. 
   - Factors such as electron-donating or electron-withdrawing substituents on the ring dictate the position where bromination is likely to occur.
   
2. **Second Step:** Friedel-Crafts Acylation
   - AlCl₃ acts as a Lewis acid, forming a complex with acetyl chloride (CH₃COCl) and facilitating the formation of an acylium ion (CH₃CO⁺), which is the electrophile responsible for the acylation of the aromatic ring.

To solve the problem, consider the electronic and steric factors influencing the positions where bromination and ac
Transcribed Image Text:### Chemical Reaction Problem **Problem Statement:** Given the chemical reaction: 1. FeBr₃, Br₂ 2. AlCl₃, Acetyl chloride (CH₃COCl) Determine the product of the reaction. **Starting Compound:** An aromatic heterocycle with the structure of a methyl group (Me) attached to one nitrogen in a five-membered ring containing one nitrogen atom. **Possible Products:** A. ![Option A](https://i.imgur.com/CvHRnzM.png) - Structure: Bromine (Br) at position 2, a double-bonded oxygen (carbonyl group) at position 3. B. ![Option B](https://i.imgur.com/CvHRnzM.png) - Structure: Bromine (Br) at position 3, a double-bonded oxygen (carbonyl group) at position 2. C. ![Option C](https://i.imgur.com/CvHRnzM.png) - Structure: Bromine (Br) at position 2, a double-bonded oxygen (carbonyl group) at position 1. D. ![Option D](https://i.imgur.com/CvHRnzM.png) - Structure: Bromine (Br) at position 3, a double-bonded oxygen (carbonyl group) at position 4. E. ![Option E](https://i.imgur.com/CvHRnzM.png) - Structure: Bromine (Br) at position 4, a double-bonded oxygen (carbonyl group) at position 3. **Explanation:** 1. **First Step:** Bromination - The FeBr₃ catalyst helps to generate a bromine electrophile (Br⁺) which then brominates the aromatic ring. - Factors such as electron-donating or electron-withdrawing substituents on the ring dictate the position where bromination is likely to occur. 2. **Second Step:** Friedel-Crafts Acylation - AlCl₃ acts as a Lewis acid, forming a complex with acetyl chloride (CH₃COCl) and facilitating the formation of an acylium ion (CH₃CO⁺), which is the electrophile responsible for the acylation of the aromatic ring. To solve the problem, consider the electronic and steric factors influencing the positions where bromination and ac
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