Which alkene do you predict will be the major product of each reaction in this  experiment?  For reaction: A, B and C?

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Which alkene do you predict will be the major product of each reaction in this 

experiment? 

For reaction: A, B and C?

### Table: Reaction Details 

- **Reaction A**
  - **Organic Substrate**: 2-bromo-2-methylbutane
  - **Reagent**: Potassium hydroxide in 1-propanol

- **Reaction B**
  - **Organic Substrate**: 2-bromo-2-methylbutane
  - **Reagent**: Potassium t-butoxide in 2-methyl-2-butanol

- **Reaction C**
  - **Organic Substrate**: 2-methyl-2-butanol
  - **Reagent**: Phosphoric acid

### Reaction A: E2 Reaction with a Strong, Small Base

* **Reactant Structure**: 2-bromo-2-methylbutane with a bromine atom attached to a carbon in the chain.
* **Reagent**: KOH in 1-propanol
* **Outcome**: The reaction results in the formation of two possible alkenes, displaying structural isomerism. 

### Reaction B: E2 Reaction with a Strong, Bulky Base

* **Reactant Structure**: 2-bromo-2-methylbutane similar to Reaction A.
* **Reagent**: Potassium t-butoxide in tert-butanol
* **Outcome**: This reaction also yields two potential alkenes. The bulky base influences the product distribution towards less substituted alkenes.

These diagrams illustrate E2 elimination reactions, demonstrating how different bases affect the formation of alkenes from the same organic substrate. Reaction A utilizes a smaller base favoring Zaitsev’s product, while Reaction B employs a bulky base, promoting Hofmann elimination.
Transcribed Image Text:### Table: Reaction Details - **Reaction A** - **Organic Substrate**: 2-bromo-2-methylbutane - **Reagent**: Potassium hydroxide in 1-propanol - **Reaction B** - **Organic Substrate**: 2-bromo-2-methylbutane - **Reagent**: Potassium t-butoxide in 2-methyl-2-butanol - **Reaction C** - **Organic Substrate**: 2-methyl-2-butanol - **Reagent**: Phosphoric acid ### Reaction A: E2 Reaction with a Strong, Small Base * **Reactant Structure**: 2-bromo-2-methylbutane with a bromine atom attached to a carbon in the chain. * **Reagent**: KOH in 1-propanol * **Outcome**: The reaction results in the formation of two possible alkenes, displaying structural isomerism. ### Reaction B: E2 Reaction with a Strong, Bulky Base * **Reactant Structure**: 2-bromo-2-methylbutane similar to Reaction A. * **Reagent**: Potassium t-butoxide in tert-butanol * **Outcome**: This reaction also yields two potential alkenes. The bulky base influences the product distribution towards less substituted alkenes. These diagrams illustrate E2 elimination reactions, demonstrating how different bases affect the formation of alkenes from the same organic substrate. Reaction A utilizes a smaller base favoring Zaitsev’s product, while Reaction B employs a bulky base, promoting Hofmann elimination.
**Reaction C: Acid Catalyzed E1 Reaction with a Tertiary Alcohol**

The diagram depicts the reaction of a tertiary alcohol under acidic conditions (using \( \text{H}_3\text{PO}_4 \)) leading to the formation of two possible alkene products. The structure on the left is a tertiary alcohol, which undergoes an elimination reaction to form alkenes.

- The reactant is a cyclic compound with a hydroxyl group (OH) attached.
- Upon reaction with phosphoric acid (\( \text{H}_3\text{PO}_4 \)), the alcohol is converted into two potential alkenes:
  - The first alkene shown has a single bond between the latter two carbon atoms.
  - The second potential product is an alkene with a double bond between a different set of carbon atoms in the molecule.

**Instructions:**
For each reaction, make a prediction about which alkene will be the major product and which will be the minor product.

**Analysis:**
The ratio of alkene products for each reaction will be determined using gas chromatography. Generally, the alkene with the lower boiling point will have a lower retention time in the chromatography process. This technique helps identify which product is predominant.
Transcribed Image Text:**Reaction C: Acid Catalyzed E1 Reaction with a Tertiary Alcohol** The diagram depicts the reaction of a tertiary alcohol under acidic conditions (using \( \text{H}_3\text{PO}_4 \)) leading to the formation of two possible alkene products. The structure on the left is a tertiary alcohol, which undergoes an elimination reaction to form alkenes. - The reactant is a cyclic compound with a hydroxyl group (OH) attached. - Upon reaction with phosphoric acid (\( \text{H}_3\text{PO}_4 \)), the alcohol is converted into two potential alkenes: - The first alkene shown has a single bond between the latter two carbon atoms. - The second potential product is an alkene with a double bond between a different set of carbon atoms in the molecule. **Instructions:** For each reaction, make a prediction about which alkene will be the major product and which will be the minor product. **Analysis:** The ratio of alkene products for each reaction will be determined using gas chromatography. Generally, the alkene with the lower boiling point will have a lower retention time in the chromatography process. This technique helps identify which product is predominant.
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