The image presents a multiple-choice question asking which option correctly shows the movement of electrons. The question is aimed at understanding electrophilic addition reactions in organic chemistry. ### Diagram Details: There are four possible answers, each illustrating a different mechanism of electron movement between an alkene and HCl. Each option shows the initial electron flow and the resulting products: 1. **Option A:** - Shows the curved arrow from the pi bond of the alkene to the hydrogen (H) of HCl, and another arrow from the bond between H and Cl to Cl. - Product: A carbocation where a double bond existed, plus a chloride ion (\(Cl^-\)). 2. **Option B:** - Similar electron movement as option A, but the carbocation is on a different carbon atom in the alkene. - Product: A differently positioned carbocation, plus a chloride ion (\(Cl^-\)). 3. **Option C:** - Electron movement focuses on creating the carbocation at the most substituted carbon atom (Markovnikov's rule). - Product: A more stable, tertiary carbocation, plus a chloride ion (\(Cl^-\)). 4. **Option D:** - The arrow shows electron movement creating a carbocation at a less substituted carbon atom. - Product: A less stable carbocation, plus a positive chloride ion (\(Cl^+\)), which is incorrect as Cl cannot be positive in this context. The diagram is designed to test understanding of the Markovnikov addition and electrophilic attack in the presence of HCl, focusing on the correct flow of electrons leading to the more stable carbocation formation. **Question 41** **Problem Statement:** What are the major stereoisomeric products formed in the following reaction? Reactants: - A branched alkene with the structure shown. - \( \text{Br}_2 \) (bromine) in the presence of \( \text{CH}_3\text{OH} \) (methanol). **Reaction Scheme:** \[ \text{Br}_2 + \text{Alkene} \rightarrow ? \] (with \( \text{CH}_3\text{OH} \) present) **Answer Choices:** 1. **Choice 1** (Selected with a blue dot): - Two stereoisomers shown. - Both have methoxy (\( \text{OCH}_3 \)) groups and bromine (\( \text{Br} \)) attached to the adjacent carbon atoms. 2. **Choice 2**: - A single product with a methoxy group and bromine both on the same side of the molecule. 3. **Choice 3**: - A single product with methoxy and bromine groups attached on opposite sides. 4. **Choice 4**: - A single product with bromine and methoxy groups attached similar to choice 3, but with opposite orientation. 5. **Choice 5**: - Two stereoisomers similar to choice 1 but with bromine and methoxy groups swapped. **Explanation:** The reaction involves the addition of bromine and methanol across a double bond in the alkene, resulting in two major stereoisomeric products due to the formation of a bromonium ion intermediate and subsequent nucleophilic attack by methanol. The correct option represents these possible products.