Electronic Effects
The effect of electrons that are located in the chemical bonds within the atoms of the molecule is termed an electronic effect. The electronic effect is also explained as the effect through which the reactivity of the compound in one portion is controlled by the electron repulsion or attraction producing in another portion of the molecule.
Drawing Resonance Forms
In organic chemistry, resonance may be a mental exercise that illustrates the delocalization of electrons inside molecules within the valence bond theory of octet bonding. It entails creating several Lewis structures that, when combined, reflect the molecule's entire electronic structure. One Lewis diagram cannot explain the bonding (lone pair, double bond, octet) elaborately. A hybrid describes a combination of possible resonance structures that represents the entire delocalization of electrons within the molecule.
Using Molecular Structure To Predict Equilibrium
Equilibrium does not always imply an equal presence of reactants and products. This signifies that the reaction reaches a point when reactant and product quantities remain constant as the rate of forward and backward reaction is the same. Molecular structures of various compounds can help in predicting equilibrium.
![### Monobromination Reaction Products: Major and Minor
**Objective:** Draw the major and minor monobromination products of the given reaction.
**Reaction:**
The substrate is a hydrocarbon with a zigzag chain structure, which likely indicates a short-chain alkane.
In the presence of molecular bromine (Br₂) and light (indicated as hv, which stands for ultraviolet light):
\[ \text{Br}_2 \ (1 \ \text{equiv}) \ \xrightarrow{hv} \]
The expected reactions lead to bromine radicals that will substitute a hydrogen atom in the hydrocarbon chain.
#### Task:
1. **Draw the Major Product:**
- This typically forms when the bromine radical attaches to the most stable intermediate position, like a tertiary or secondary carbon (most stable carbon-radical intermediates).
2. **Draw the Minor Product:**
- This occurs when the bromine radical attaches to a less favorable position, such as a primary carbon.
*Note: You should focus on the likely positions of bromination based on the stability of the formed carbon radicals.*
### Visual Explanation:
1. **Starting Material:**
- A hydrocarbon skeleton represented as a simple zigzag line indicating carbon atoms and implied hydrogen atoms.
2. **Reaction Mechanism:**
- Br₂ (1 equivalent) reacts under UV light (hv), leading to the formation of bromine radicals.
3. **Expected Products:**
- **Major Product:** The diagram indicates a space for drawing the major brominated product.
- **Minor Product:** Another diagram space for sketching the less dominant product of the reaction.
**Summary:**
By analyzing the carbon positions, you can predict and draw the products. Remember, the major product will have bromine at the most stable carbon position, while the minor product will have bromine at a less stable position.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44572439-aee9-40a4-8ec4-b1e043de89f6%2F2ddd9a84-aaf3-4772-b3bc-f245e133fc68%2Fnaipjzc_processed.jpeg&w=3840&q=75)
![### Reaction Mechanism - Exercises
**Problem Statement:**
*Draw the major product of this reaction. Ignore any inorganic byproducts.*
**Reactant Structure:**
![Reactant](mock-image-of-chemical-structure)
The chemical structure depicted shows a hydrocarbon chain with a terminal alkene.
**Reagents:**
- Hydrogen Bromide (HBr) - 1 equivalent
- Hydrogen Peroxide (H₂O₂)
**Reaction Mechanism:**
The arrow indicates the reactants reacting together to form the major product.
![Reaction-Arrow](mock-vertical-arrow)
**Interactive Exercise:**
[Select to Draw](mock-image-of-drawing-area): Draw the major product of the reaction based on the given reactant and reagents.
### Explanation
When reacting with HBr in the presence of H₂O₂, the reaction follows an anti-Markovnikov addition mechanism. This means the bromine will add to the less substituted carbon in the double bond due to the radical initiation by H₂O₂.
1. **Step 1:** Initiation:
- Radical formation initiated by H₂O₂.
2. **Step 2:** Propagation:
- Formation of a bromine radical and its subsequent addition to the hydrocarbon chain.
3. **Step 3:** Termination:
- Formation of the final stable product.
Use the interactive drawing tool to sketch the product structure, ensuring you follow the anti-Markovnikov rule.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44572439-aee9-40a4-8ec4-b1e043de89f6%2F2ddd9a84-aaf3-4772-b3bc-f245e133fc68%2Foq009ef_processed.jpeg&w=3840&q=75)
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