1.BD3 2. H2O2/NAOH

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Chapter1: Chemical Foundations
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For each of the following reactions give their major product 

This image depicts a chemical reaction. The reactant is a methylcyclohexene molecule, which is a cyclohexene ring with a methyl group (CH₃) attached to it. The reaction involves hydrochloric acid (HCl) as the reagent, indicated by an arrow pointing to the right.

In an educational context, this image suggests the study of hydrohalogenation reaction mechanisms, where an alkene reacts with a hydrogen halide to form a haloalkane. Specifically, in this reaction, you might expect the formation of a chlorinated product where the HCl adds across the carbon-carbon double bond.
Transcribed Image Text:This image depicts a chemical reaction. The reactant is a methylcyclohexene molecule, which is a cyclohexene ring with a methyl group (CH₃) attached to it. The reaction involves hydrochloric acid (HCl) as the reagent, indicated by an arrow pointing to the right. In an educational context, this image suggests the study of hydrohalogenation reaction mechanisms, where an alkene reacts with a hydrogen halide to form a haloalkane. Specifically, in this reaction, you might expect the formation of a chlorinated product where the HCl adds across the carbon-carbon double bond.
The image depicts a chemical reaction involving the hydroboration-oxidation process. 

**Molecule Description:**

- The starting molecule is an alkene with a double bond.
- The alkene has a deuterium atom (D) adjacent to the double bond.

**Reaction Steps:**

1. **BD₃ (Borane Deuteride):** 
   - The first step of the reaction involves BD₃, a source of boron and deuterium, which is used for the hydroboration of the alkene. 
   - During this step, BD₃ adds across the double bond in a syn addition, meaning both hydrogen and boron (deuterium here) add to the same face of the alkene.

2. **H₂O₂/NaOH (Hydrogen Peroxide/Sodium Hydroxide):**
   - In the second step, the boron atom is oxidized to replace it with a hydroxyl group (OH) using hydrogen peroxide in the presence of a base (sodium hydroxide).
   - This results in the anti-Markovnikov addition of water, meaning the OH group attaches to the less substituted carbon atom of the former double bond.

**Overall Transformation:**

The process converts the alkene into an alcohol, and the deuterium atom remains attached to the carbon where it was initially present. This reaction is useful for synthesizing alcohols in a regioselective manner while preserving stereochemistry.
Transcribed Image Text:The image depicts a chemical reaction involving the hydroboration-oxidation process. **Molecule Description:** - The starting molecule is an alkene with a double bond. - The alkene has a deuterium atom (D) adjacent to the double bond. **Reaction Steps:** 1. **BD₃ (Borane Deuteride):** - The first step of the reaction involves BD₃, a source of boron and deuterium, which is used for the hydroboration of the alkene. - During this step, BD₃ adds across the double bond in a syn addition, meaning both hydrogen and boron (deuterium here) add to the same face of the alkene. 2. **H₂O₂/NaOH (Hydrogen Peroxide/Sodium Hydroxide):** - In the second step, the boron atom is oxidized to replace it with a hydroxyl group (OH) using hydrogen peroxide in the presence of a base (sodium hydroxide). - This results in the anti-Markovnikov addition of water, meaning the OH group attaches to the less substituted carbon atom of the former double bond. **Overall Transformation:** The process converts the alkene into an alcohol, and the deuterium atom remains attached to the carbon where it was initially present. This reaction is useful for synthesizing alcohols in a regioselective manner while preserving stereochemistry.
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