Draw A Resonance Structure That Places A Pi Bond

Let's dive into the fascinating world of resonance structures and how to strategically draw them to place a pi bond where it's needed. Resonance, at its core, is about depicting electron delocalization within a molecule, a concept crucial for understanding molecular stability, reactivity, and properties. By mastering the art of drawing resonance structures, you'll gain a deeper appreciation for how electrons are distributed and how this distribution impacts chemical behavior.

Understanding Resonance: The Basics

Resonance isn't about molecules vibrating between different forms. Instead, it's a way of representing a single molecule whose true electronic structure is a hybrid of multiple contributing structures. These contributing structures, or resonance structures, differ only in the arrangement of electrons, not the position of atoms. The actual molecule is more stable than any single resonance structure suggests; this increased stability is known as resonance stabilization.

Key Principles of Resonance:

• Atoms Don't Move: Only electrons are rearranged. The sigma bond framework remains constant.
• Valid Lewis Structures: Each resonance structure must be a valid Lewis structure, obeying the octet rule (or duet rule for hydrogen) and having a proper formal charge distribution.
• Same Number of Valence Electrons: Each resonance structure must have the same number of valence electrons.
• Energy Considerations: Resonance structures that contribute most to the overall hybrid structure are those that:
  • Have more filled octets.
  • Place negative charges on more electronegative atoms and positive charges on less electronegative atoms.
  • Have less charge separation.
• Resonance Hybrid: The actual molecule is a blend of all contributing resonance structures, not just a rapid interconversion between them. We often represent the resonance hybrid with dashed lines indicating partial bonds and partial charges.

Drawing Resonance Structures: A Step-by-Step Guide

Here's a systematic approach to drawing resonance structures, with a specific focus on strategically placing pi bonds:

1. Identify Potential Resonance Contributors:

• Look for adjacent pi bonds and lone pairs. These are the prime indicators of possible electron delocalization. Common motifs include:
  • Lone pair adjacent to a pi bond (e.g., enols, amides).
  • Pi bond adjacent to a pi bond (e.g., conjugated systems).
  • Pi bond adjacent to a positive charge (e.g., allylic carbocations).
  • Lone pair adjacent to a positive charge.
  • Pi bond adjacent to an atom with an empty p orbital.
• Consider the overall molecular structure and any patterns that might suggest electron delocalization across multiple atoms.

2. Use Curved Arrows to Show Electron Movement:

• Curved arrows are the language of resonance. They depict the movement of electron pairs, not individual electrons.
• Always start the arrow at a region of high electron density (a lone pair or a pi bond) and point it towards a region of low electron density (an atom that can accept more electrons or a bond that needs to be formed).
• Important Rules for Curved Arrows:
  • An arrow starting from a lone pair forms a new pi bond between the atom bearing the lone pair and the atom the arrow points to.
  • An arrow starting from a pi bond breaks the pi bond. The electrons from the pi bond become a lone pair on the atom the arrow points to.
  • Never exceed the octet rule for second-row elements (C, N, O, F).
• Placing the Pi Bond: The strategic placement of a pi bond is directly dictated by the curved arrow movements. Think about where the electron density needs to shift to create the desired pi bond.

3. Draw the New Resonance Structure:

• Based on the curved arrow movements, redraw the molecule, showing the new bond arrangement and the new positions of lone pairs and formal charges.
• Formal Charge Calculation: Remember to recalculate the formal charge on any atom that gains or loses electrons.
  • Formal Charge = (Valence Electrons) - (Non-bonding Electrons) - (1/2 Bonding Electrons)
• Ensure that the new resonance structure is a valid Lewis structure. All atoms should have a complete octet (or duet for hydrogen) unless there are specific exceptions (like boron compounds).

4. Assess the Relative Contribution of Each Resonance Structure:

• Not all resonance structures contribute equally to the overall resonance hybrid. The most significant contributors are those that:
  • Have more filled octets. Structures with all atoms having an octet are generally more stable.
  • Have negative charges on more electronegative atoms (O, N, Cl) and positive charges on less electronegative atoms (C). This aligns with the electronegativity principle, where electronegative atoms prefer to bear negative charge.
  • Have minimal charge separation. Structures with fewer formal charges are generally more stable than those with more significant charge separation.
• Use resonance arrows (a double-headed arrow) to connect the different resonance structures.

5. Represent the Resonance Hybrid (Optional but Recommended):

• To visually represent the resonance hybrid, draw the sigma bond framework of the molecule.
• Use dashed lines to indicate partial pi bonds where the pi bond is delocalized across multiple atoms in the contributing resonance structures.
• Use δ+ and δ- symbols to indicate partial charges resulting from the delocalization of electrons.

Examples of Drawing Resonance Structures with Pi Bond Placement

Let's illustrate this process with several examples, focusing on how curved arrows strategically place the pi bond.

Example 1: Allylic Carbocation (CH2=CH-CH2+)

1. Identify Potential Resonance Contributors: We have a pi bond adjacent to a positive charge. This is a classic resonance situation.
2. Use Curved Arrows: Draw a curved arrow starting from the pi bond between the first two carbons and pointing towards the third carbon, which bears the positive charge. This arrow indicates that the pi electrons are moving to form a new pi bond between the second and third carbons.
3. Draw the New Resonance Structure: The new resonance structure will have a pi bond between the second and third carbons, and the positive charge will now be on the first carbon. The structure becomes (+CH2-CH=CH2).
4. Assess Contribution: Both resonance structures are equally contributing because the charge is on a carbon atom in both forms and all atoms have a complete octet.
5. Resonance Hybrid: The resonance hybrid would show a partial positive charge on both the first and third carbons and a partial pi bond character between all three carbons.

Example 2: Enol (CH2=CH-OH)

1. Identify Potential Resonance Contributors: We have a lone pair on the oxygen atom adjacent to a pi bond.
2. Use Curved Arrows:
   • Draw a curved arrow starting from one of the lone pairs on the oxygen atom and pointing towards the bond between the two carbons. This will form a new pi bond between the carbon and the oxygen.
   • Since the first carbon already has four bonds, we need to break the existing pi bond between the two carbons. Draw a second curved arrow starting from the pi bond between the two carbons and pointing towards the first carbon. This places a lone pair on the first carbon.
3. Draw the New Resonance Structure: The new resonance structure will have a pi bond between the carbon and the oxygen, a lone pair on the first carbon (giving it a negative formal charge), and a positive formal charge on the oxygen. The structure becomes (-CH2-CH=O+H).
4. Assess Contribution: The original enol structure is the major contributor because all atoms have a complete octet and there is no charge separation. The second resonance structure has charge separation and the oxygen has a positive charge.
5. Resonance Hybrid: The resonance hybrid would show a partial negative charge on the first carbon, a partial double bond character between the carbon and the oxygen, and a partial positive charge on the oxygen.

Example 3: Amide (R-CO-NH2)

1. Identify Potential Resonance Contributors: We have a lone pair on the nitrogen atom adjacent to a pi bond (the carbonyl group).
2. Use Curved Arrows:
   • Draw a curved arrow starting from the lone pair on the nitrogen atom and pointing towards the bond between the carbon and the oxygen. This will form a new pi bond between the carbon and the nitrogen.
   • Since the carbon already has four bonds, we need to break the existing pi bond between the carbon and the oxygen. Draw a second curved arrow starting from the pi bond between the carbon and the oxygen and pointing towards the oxygen. This places a lone pair on the oxygen.
3. Draw the New Resonance Structure: The new resonance structure will have a pi bond between the carbon and the nitrogen, three lone pairs on the oxygen (giving it a negative formal charge), and a positive formal charge on the nitrogen.
4. Assess Contribution: The original amide structure is the major contributor, but the second resonance structure is also significant because it places a full negative charge on the oxygen, which is more electronegative than nitrogen. This resonance contributes to the partial double bond character of the C-N bond in amides, making them less reactive than esters.
5. Resonance Hybrid: The resonance hybrid would show a partial double bond character between the carbon and the nitrogen, a partial negative charge on the oxygen, and a partial positive charge on the nitrogen. This explains the planar geometry around the amide nitrogen.

Example 4: Benzene (C6H6)

1. Identify Potential Resonance Contributors: Benzene is the classic example of resonance. It has alternating single and double bonds in a six-membered ring.
2. Use Curved Arrows: Draw three curved arrows to move the pi bonds. Start from one pi bond and move it to the adjacent single bond. Continue around the ring, moving each pi bond to the next adjacent single bond.
3. Draw the New Resonance Structure: The new resonance structure will have the double bonds in the positions where the single bonds were, and vice versa.
4. Assess Contribution: Both resonance structures are equally contributing because they are identical except for the placement of the double bonds. This equal contribution is what gives benzene its exceptional stability.
5. Resonance Hybrid: The resonance hybrid is represented by a circle inside the hexagon, indicating that the pi electrons are delocalized equally around the entire ring. All carbon-carbon bonds in benzene are equivalent and have a bond order of 1.5.

Example 5: Conjugated Ketone

Consider a conjugated ketone where a double bond is next to the ketone.

1. Identify Potential Resonance Contributors: The double bond conjugated with the ketone offers resonance stabilization.
2. Use Curved Arrows: Start an arrow from the double bond toward the carbon-carbon single bond connecting it to the ketone. From that single bond, draw an arrow to the oxygen of the ketone.
3. Draw the New Resonance Structure: The new structure results in a positive charge on the carbon where the original double bond was, a double bond between the two carbons that were connected by a single bond, and a negative charge on the oxygen of the ketone.
4. Assess Contribution: The original ketone structure usually contributes more, but the resonance structure helps explain the reactivity of conjugated ketones.
5. Resonance Hybrid: This hybrid shows partial positive charge on one carbon, partial double-bond character between two adjacent carbons, and partial negative charge on the oxygen.

Common Mistakes to Avoid

• Moving Atoms: Resonance structures only differ in the arrangement of electrons, not atoms. Never move atoms when drawing resonance structures.
• Exceeding the Octet Rule: Ensure that no second-row element (C, N, O, F) has more than eight electrons in its valence shell. The octet rule is fundamental to Lewis structures and resonance.
• Breaking Sigma Bonds: Only pi bonds and lone pairs are involved in resonance. Sigma bonds remain intact.
• Incorrect Arrow Placement: Make sure the arrows start from a region of high electron density (lone pair or pi bond) and point towards a region of low electron density (an atom that can accept more electrons or a bond that needs to be formed).
• Forgetting Formal Charges: Always recalculate formal charges after moving electrons. Incorrect formal charges invalidate the resonance structure.
• Drawing Impossible Structures: Ensure that all resonance structures are valid Lewis structures.

Advanced Considerations

• Hyperconjugation: While technically not resonance, hyperconjugation involves the interaction of sigma bonding electrons with an adjacent empty or partially filled p orbital or pi* antibonding orbital. This can contribute to stability, especially in carbocations and radicals.
• Resonance and Reactivity: Resonance can significantly influence the reactivity of a molecule. By understanding the electron distribution in the resonance hybrid, you can predict where electrophiles and nucleophiles will attack.
• Resonance in Aromatic Systems: Aromatic compounds, like benzene, owe their exceptional stability to resonance delocalization of pi electrons. The cyclic, planar structure and the presence of (4n+2) pi electrons (Hückel's rule) are crucial for aromaticity.

Conclusion

Drawing resonance structures is a fundamental skill in organic chemistry. By following the steps outlined above and practicing regularly, you can master this technique and gain a deeper understanding of electron delocalization, molecular stability, and chemical reactivity. Remember to pay close attention to curved arrow notation, formal charges, and the relative contributions of different resonance structures. Strategic placement of pi bonds through resonance is key to understanding a molecule's properties and behavior. Embrace the power of resonance, and you'll unlock a new level of insight into the world of chemistry.