(5 points) Identify the electrophile and the nucleophile in each of the following reaction steps and then draw curved arrows to illustrate the bond-making and bond-breaking processes. C'H;CH CH, + H-Br CH;CH-CH; + Br

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**Identifying Electrophiles and Nucleophiles in Reaction Steps**

In this exercise, we examine the reaction:

\[ \text{CH}_3\text{CH}=\text{CH}_2 + \text{H–Br} \rightarrow \text{CH}_3\text{CH}^+–\text{CH}_3 + \text{Br}^- \]

**Objective:**
- Identify the electrophile and nucleophile in each reaction step.
- Illustrate curved arrows to depict the bond-making and bond-breaking processes.

**Analysis:**

1. **Reactants:**
   - **Propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)):** The double bond indicates areas of high electron density, which can act as a nucleophile.
   - **Hydrogen Bromide (\(\text{H–Br}\)):** Polar molecule where Br is more electronegative, causing a partial positive charge on H and partial negative on Br. The H becomes an electrophile.

2. **Reaction Process:**
   - The \(\text{C=C}\) double bond in propene acts as a nucleophile, attacking the electrophilic hydrogen (H) of \(\text{H–Br}\).
   - A curved arrow should be drawn from the double bond towards the \(\text{H}\) atom in \(\text{H–Br}\) to indicate the formation of a new bond.
   - As the \(\text{C-H}\) bond forms, the \(\text{H-Br}\) bond breaks. A curved arrow from the \(\text{H-Br}\) bond to the \(\text{Br}\) atom indicates bond cleavage, resulting in bromide ion (\(\text{Br}^-\)).
   
3. **Products:**
   - The resultant \(\text{CH}_3\text{CH}^+ – \text{CH}_3\) carbocation and \(\text{Br}^-\) ion are formed.

**Conclusion:**
This step illustrates a key electrophilic addition reaction in organic chemistry, highlighting the roles of electrophiles and nucleophiles, as well as the movement of electrons illustrated by curved arrows. Understanding these concepts is critical for mastering reaction mechanisms.
Transcribed Image Text:**Identifying Electrophiles and Nucleophiles in Reaction Steps** In this exercise, we examine the reaction: \[ \text{CH}_3\text{CH}=\text{CH}_2 + \text{H–Br} \rightarrow \text{CH}_3\text{CH}^+–\text{CH}_3 + \text{Br}^- \] **Objective:** - Identify the electrophile and nucleophile in each reaction step. - Illustrate curved arrows to depict the bond-making and bond-breaking processes. **Analysis:** 1. **Reactants:** - **Propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)):** The double bond indicates areas of high electron density, which can act as a nucleophile. - **Hydrogen Bromide (\(\text{H–Br}\)):** Polar molecule where Br is more electronegative, causing a partial positive charge on H and partial negative on Br. The H becomes an electrophile. 2. **Reaction Process:** - The \(\text{C=C}\) double bond in propene acts as a nucleophile, attacking the electrophilic hydrogen (H) of \(\text{H–Br}\). - A curved arrow should be drawn from the double bond towards the \(\text{H}\) atom in \(\text{H–Br}\) to indicate the formation of a new bond. - As the \(\text{C-H}\) bond forms, the \(\text{H-Br}\) bond breaks. A curved arrow from the \(\text{H-Br}\) bond to the \(\text{Br}\) atom indicates bond cleavage, resulting in bromide ion (\(\text{Br}^-\)). 3. **Products:** - The resultant \(\text{CH}_3\text{CH}^+ – \text{CH}_3\) carbocation and \(\text{Br}^-\) ion are formed. **Conclusion:** This step illustrates a key electrophilic addition reaction in organic chemistry, highlighting the roles of electrophiles and nucleophiles, as well as the movement of electrons illustrated by curved arrows. Understanding these concepts is critical for mastering reaction mechanisms.
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