Identify the letters A-E. Letter A= HBr Letter B= tbOK Letter C= H3O+ Letter D= Letter E= H₂C CH3 CH3 A H₂C H₂C Br CH3 E= name B xs NaNH, CI H₂C H₂C CH3 D CI CH3 C CH3 H₂C H₂C OH CH3

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Chapter1: Chemical Foundations
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**Identifying the Letters A-E in the Sequence of Reactions:**

**Reactants and Reagents:**
- **Letter A= HBr**
- **Letter B= tbOK**
- **Letter C= H3O+**

**Reaction Sequence:**
1. **Starting Compound:** The starting molecule is a hydrocarbon chain with five carbons (C5H10), with the following structure:

   ```
   CH3-CH2-CH2-CH3
   ```

2. **Step A:** Addition of HBr to the molecule. This step involves adding hydrogen bromide (HBr) to the hydrocarbon. The product formed is:

   ```
   CH3-CH2-CHBr-CH3
   ```

3. **Step B:** Addition of tbOK. Reacting the bromide-containing molecule with tert-butoxide (tbOK) leads to the elimination of the bromine atom, resulting in an alkene. The product is:

   ```
   CH3-CH2-CH=CH2
   ```

4. **Step C:** Hydration with H3O+. Treating the alkene with hydronium ion (H3O+) results in the Markovnikov addition of water across the double bond, generating an alcohol. The product is:

   ```
   CH3-CH-CH2-CH3
          |
          OH
   ```

5. **Step D:** This step is not explicitly explained in the diagram; it may involve rearrangement or detailed intermediate step not shown in the image.

6. **Step E:** Reaction with excess sodium amide (NaNH2). This step includes treating the chlorinated product with sodium amide, likely leading to deprotonation and substitution by sodium amide. The final product is expected to be:

   ```
   CH3-CH-Cl-CH3
   ```

**Summary:**
In this reaction mechanism summary, HBr leads to the addition of bromine (Step A), tbOK facilitates the elimination reaction to form an alkene (Step B), H3O+ hydrates the alkene to generate an alcohol (Step C), and the final reaction involves treatment with NaNH2, implying further derivatization in Step E.

**Missing Details:**
- **Letter D** is not specified in the text and requires further clarification.
- The final compound’s exact name (Letter E) and full structure are not provided
Transcribed Image Text:**Identifying the Letters A-E in the Sequence of Reactions:** **Reactants and Reagents:** - **Letter A= HBr** - **Letter B= tbOK** - **Letter C= H3O+** **Reaction Sequence:** 1. **Starting Compound:** The starting molecule is a hydrocarbon chain with five carbons (C5H10), with the following structure: ``` CH3-CH2-CH2-CH3 ``` 2. **Step A:** Addition of HBr to the molecule. This step involves adding hydrogen bromide (HBr) to the hydrocarbon. The product formed is: ``` CH3-CH2-CHBr-CH3 ``` 3. **Step B:** Addition of tbOK. Reacting the bromide-containing molecule with tert-butoxide (tbOK) leads to the elimination of the bromine atom, resulting in an alkene. The product is: ``` CH3-CH2-CH=CH2 ``` 4. **Step C:** Hydration with H3O+. Treating the alkene with hydronium ion (H3O+) results in the Markovnikov addition of water across the double bond, generating an alcohol. The product is: ``` CH3-CH-CH2-CH3 | OH ``` 5. **Step D:** This step is not explicitly explained in the diagram; it may involve rearrangement or detailed intermediate step not shown in the image. 6. **Step E:** Reaction with excess sodium amide (NaNH2). This step includes treating the chlorinated product with sodium amide, likely leading to deprotonation and substitution by sodium amide. The final product is expected to be: ``` CH3-CH-Cl-CH3 ``` **Summary:** In this reaction mechanism summary, HBr leads to the addition of bromine (Step A), tbOK facilitates the elimination reaction to form an alkene (Step B), H3O+ hydrates the alkene to generate an alcohol (Step C), and the final reaction involves treatment with NaNH2, implying further derivatization in Step E. **Missing Details:** - **Letter D** is not specified in the text and requires further clarification. - The final compound’s exact name (Letter E) and full structure are not provided
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