# Step-by-Step Solution: Electrophilic Addition of HCl to an Alkene Let's break down the **mechanism** of the reaction between an alkene and hydrochloric acid (HCl). We'll use **propene** (\(\ce{CH3-CH=CH2}\)) as a typical example: ## 1. Identify the Reactants - **Alkene:** Propene (\(\ce{CH3-CH=CH2}\)) - **Electrophile:** HCl ## 2. Mechanism Overview Electrophilic addition of HCl to an alkene occurs in two main steps: 1. **Protonation of the alkene** to form a carbocation intermediate. 2. **Nucleophilic attack** by chloride ion (\(\ce{Cl^-}\)). We'll focus on **Step 1** as per your question. --- ## 3. Step 1: Protonation of the Alkene ### **Electron Movement (Arrow Pushing)** - **Double bond** in the alkene acts as a nucleophile (electron-rich). - **Hydrogen atom** in HCl is the electrophile (electron-poor). ### **Curved Arrow Conventions** - **Arrow 1:** Start at the double bond, point to the hydrogen atom of HCl. - **Arrow 2:** Start at the H-Cl bond, point to the chlorine atom. #### **Electron Flow Diagram** ```markdown CH3-CH=CH2 + H-Cl → [CH3-CH(+)-CH3] + Cl^- | | (double (polar bond) bond) ``` ### **Curved Arrows: Explanation** 1. **First Arrow:** - **Origin:** π electrons (double bond) of propene - **Destination:** Hydrogen atom of HCl - **Meaning:** Alkene donates electrons to bond with hydrogen. 2. **Second Arrow:** - **Origin:** H-Cl bond - **Destination:** Chlorine atom - **Meaning:** H-Cl bond breaks; Cl takes both electrons, becoming Cl⁻. **[Diagram Example]** ``` H Cl \ / CH3-C=CH2 || (curved arrow from double bond to H) (curved arrow from H-Cl bond to Cl) ``` - The **double bond attacks H**, forming a new C-H bond. - The **H-Cl bond breaks**, giving Cl both electrons (forming Cl⁻). ### **Regioselectivity (Markovnikov's Rule)** - The **hydrogen** attaches to the carbon with more hydrogens (less substituted). - The **carbocation** forms on the more substituted carbon (more stable). **For propene:** - H adds to terminal carbon (CH2). - Carbocation forms at the central carbon (CH): \[ \ce{CH3-CH=CH2} + \ce{HCl} \xrightarrow{Step\ 1} \ce{CH3-CH^+-CH3} + \ce{Cl^-} \] --- ## 4. **Final Answer: Summarized Mechanism** **Step 1:** 1. Curved arrow from alkene double bond to H of HCl. 2. Curved arrow from H-Cl bond to Cl. **Intermediate:** A carbocation at the more substituted carbon (Markovnikov product), plus Cl⁻. **Image Description:** - The double bond of propene attacks the hydrogen atom of HCl (curved arrow from double bond to H). - The H-Cl bond breaks, forming Cl⁻ (curved arrow from H-Cl bond to Cl). --- ## 5. **Summary Table** | Step | Electron Movement | Product Formed | |-----------|---------------------------------------------------|----------------------------| | 1 | π electrons to H; H-Cl electrons to Cl | Carbocation + Cl⁻ | --- **Final Answer:** The alkene's π electrons attack the hydrogen of HCl, while the H-Cl bond breaks, generating a carbocation (on the more substituted carbon) and Cl⁻. **Draw two curved arrows:** - One from the double bond to H of HCl - One from the H-Cl bond to Cl