In the following sets of resonance forms, label the major and minor contributors and state which structures would be of equal energy. Add any missing resonance forms.
`[CH_(3) - overset(-)CH - CH = CH - NO_(2) hArr CH_(3)- CH = CH - overset(-)CH - NO_(2)]`

Answer

To solve the problem, we need to analyze the given resonance forms and determine which are the major and minor contributors, as well as identify any missing resonance forms.

### Step-by-Step Solution:

1. **Identify the Resonance Structures**:
We have two resonance structures provided:
- Structure 1: `[CH3 - CH(-) - CH = CH - NO2]`
- Structure 2: `[CH3 - CH = CH(-) - NO2]`

2. **Draw Possible Resonance Forms**:
We can draw additional resonance forms by moving the negative charge and the double bonds.
- From Structure 1, we can move the negative charge from the carbon to the adjacent double bond, resulting in a new resonance form:
- New Structure 3: `[CH3 - CH = CH - CH(-) - NO2]`
- From Structure 2, we can also move the negative charge to the nitrogen in the nitro group, leading to:
- New Structure 4: `[CH3 - CH = CH - N(+) - O(-) - O]`

3. **Evaluate Stability of Each Structure**:
- **Structure 1**: The negative charge is on a carbon atom, which is less electronegative than oxygen. This makes it less stable.
- **Structure 2**: The negative charge is on a carbon adjacent to a double bond, which can help stabilize it through resonance. However, the nitro group is not directly involved in stabilizing the charge.
- **Structure 3**: The negative charge is now on a carbon that is part of a double bond, which is more stable due to resonance.
- **Structure 4**: The negative charge is on the oxygen of the nitro group, which is highly electronegative, making this structure very stable.

4. **Label Major and Minor Contributors**:
- **Major Contributor**: Structure 4 (due to the negative charge on the electronegative oxygen).
- **Minor Contributor**: Structure 1 (due to the negative charge on carbon).
- Structures 2 and 3 are intermediate in stability and can be considered of equal energy.

5. **Conclusion**:
- Major Contributor: Structure 4
- Minor Contributor: Structure 1
- Structures 2 and 3 are of equal energy.