Question

modify the second structure given to draw the new resonance structure. include lone pairs and charges in your structure. use the + and - tools to add/remove charges to an atom, and use the single bond tool to add/remove double bonds.

Explanation:

Step1: Analyze electron movement

The left structure shows electron movement: the lone pair on O moves to form a double bond, and the double bond in the alkene moves to form a carbocation. For the right structure, we reverse this. The double bond between the carbonyl - like carbon and the alkene carbon (in the right structure, the double bond is between the two carbons adjacent to the carbonyl - like carbon and \(CH_2^+\)) will move, and the lone pair on O (with a negative charge) will participate.

Step2: Modify the bonds and charges

- For the double bond between the carbon connected to \(H_3C\) and the adjacent carbon (in the right structure, the double bond is \(C = C\) near \(CH_2^+\)): we need to move the double bond. The \(CH_2^+\) will lose its positive charge as the double bond moves, and the oxygen will change its bond.
- The oxygen in the right structure has a negative charge and lone pairs. We will change the double bond between the carbonyl - like carbon (connected to \(H_3C\)) and the adjacent carbon. The double bond between the two carbons (the one with \(H_3C\) and the one in the double bond with \(CH_2^+\)) will be converted to a single bond, and the double bond in the alkene (between the carbon and \(CH_2^+\)) will be moved to form a double bond with the oxygen. Wait, no, let's correct. The resonance involves the movement of the pi electrons. In the left structure, the carbonyl group (C = O) has electrons moving, and the alkene (C = C) has electrons moving. In the right structure, to get the resonance structure, we need to move the double bond from the C = C (near \(CH_2^+\)) to form a double bond with the oxygen, and the \(CH_2^+\) will become neutral, and the oxygen will have a double bond (losing the negative charge? No, wait, the oxygen has a negative charge in the right structure. Let's draw the correct resonance.

The right structure is \(H_3C - C(=O^-) - C = CH_2^+\)? No, the right structure is \(H_3C - C(=O^-) - C = CH_2^+\) (with the negative charge on O and positive on \(CH_2\)). Wait, no, the right structure as given: \(H_3C - C(=O^-) - C = CH_2^+\) (the O has three lone pairs and a negative charge, the C - C is double, and \(CH_2\) is positive). To draw the resonance structure, we move the double bond from \(C = CH_2^+\) to form a double bond with the O, and the \(CH_2^+\) becomes \(CH_2\) (neutral), and the O loses its negative charge (since it forms a double bond, so O will have two lone pairs and a double bond, so charge 0? No, wait, the formal charge: O with a double bond (C = O) and two lone pairs has formal charge 0, and the carbon that was \(CH_2^+\) will have a positive charge? Wait, no, let's do formal charge.

Formal charge on O: in the right structure, O has 6 valence electrons (3 lone pairs: 6 electrons) and a single bond? No, the right structure has O with a double bond? Wait, the right structure shows O with a double bond? Wait, the right structure: \(H_3C - C(=O^-) - C = CH_2^+\). So O has a double bond (C = O) and a negative charge? No, if O is double - bonded, it has 2 lone pairs (4 electrons) and a double bond (2 electrons), total 6, formal charge = 6 - (4 + 2) = 0. But the right structure shows O with a negative charge, so maybe O is single - bonded. Wait, the left structure has O with a double bond (C = O) and a lone pair? No, the left structure's O has a double bond and a curved arrow, meaning electrons are moving.

Let's start over. The left structure: \(H_3C - C(=O) - C = CH_2\) (with arrows showing electron movement: one arrow from O's lone pair to form a double bond, and one arrow from C = C to form a positive charge on \(CH_2\)). The right structure is the resonance form: \(H_3C - C(=O^-) - C = CH_2^+\) (O has a negative charge, C = C double bond, \(CH_2\) positive). To get the other resonance structure (the one on the left), we need to reverse the electron movement. So in the right structure, we need to:

- Change the double bond between \(C = CH_2^+\) to a single bond, and form a double bond between \(C\) (connected to \(H_3C\)) and \(O\).
- Remove the positive charge from \(CH_2^+\) (since the double bond is moving, the \(CH_2\) will have a double bond, so no positive charge) and remove the negative charge from O (since O will have a double bond, so formal charge 0).

So using the tools:

- Use the single bond tool to remove the double bond between the two carbons (the one with \(H_3C\) and the one in the double bond with \(CH_2^+\))? No, wait, the double bond is between the carbon (let's call it C1: connected to \(H_3C\)), C2: connected to C1 and C3 (C3 is connected to \(CH_2^+\)), and C3: connected to \(CH_2^+\). So C2 - C3 is double bond, C1 - C2 is single bond (with O single - bonded? No, in the right structure, C1 - O is double bond? Wait, the right structure: \(H_3C - C(=O^-) - C = CH_2^+\). So C1 (connected to \(H_3C\)) is double - bonded to O (with O having negative charge), and C1 is single - bonded to C2, and C2 is double - bonded to C3 (\(CH_2^+\)).

To draw the resonance structure, we move the double bond from C2 - C3 to C1 - C2, and the double bond from C1 - O (but O has a negative charge, so C1 - O is a double bond? No, O has a negative charge, so C1 - O is a single bond with O having three lone pairs and a negative charge. Wait, I think I messed up. Let's use the arrow - pushing. The resonance is between the enolate (O^- and C = C) and the carbonyl with a carbocation.

The correct resonance modification:

- The double bond between C2 and C3 (C = CH_2^+) will move to form a double bond between C1 and C2.
- The lone pair on O (O^-) will move to form a double bond between C1 and O, so C1 = O (double bond), O has two lone pairs (formal charge 0), and C3 (CH_2) will lose the positive charge (since the double bond moved, so C3 is now CH_2 (neutral)), and C2 will have a positive charge? Wait, no, let's calculate formal charges.

Original right structure:

- O: 6 valence electrons (3 lone pairs: 6 e^-), single bond to C1, formal charge = 6 - (6 + 1) = - 1 (correct, as shown).
- C1: connected to \(H_3C\), single bond to O, single bond to C2, formal charge = 4 - (3) = + 1? No, C1 has 4 valence electrons. Bonds: single to \(H_3C\), single to O, single to C2: 3 bonds, so 6 electrons, formal charge = 4 - 6 = - 2? No, I'm getting confused. Let's use the drawing tools.

To modify the second structure (right structure) to get the resonance structure:

1. Remove the double bond between C2 and C3 (the C = CH_2^+ bond) using the single bond tool (click on the double bond to make it single).
2. Add a double bond between C1 and C2 (the C connected to \(H_3C\) and C2) using the double bond tool.
3. Remove the negative charge from O (using the - tool? No, the + and - tools: the O has a negative charge, we need to remove it? Wait, no, when we form a double bond with O, the O will have a double bond, so its formal charge will be 0. So use the + tool to remove the negative charge from O (since forming a double bond, O will have 2 lone pairs and a double bond, formal charge 0).
4. Add a positive charge to C3 (CH_2) ? No, wait, the left structure has \(CH_2\) as neutral. Wait, the left structure is \(H_3C - C(=O) - C = CH_2\) (neutral), and the right structure is the resonance form with O^- and \(CH_2^+\). So to get the resonance structure from the right structure, we need to move the double bond back.

So the steps in the drawing tool:

- Click on the double bond between the carbon (connected to \(H_3C\)) and the adjacent carbon (no, the double bond is between the carbon in the middle and \(CH_2^+\)): use the single bond tool to make it single.
- Click on the single bond between the carbon (connected to \(H_3C\)) and the oxygen: use the double bond tool to make it double (so O now has a double bond, losing the negative charge).
- Click on the \(CH_2^+\) to remove the positive charge (using the - tool? No, the + and - tools: the \(CH_2^+\) has a positive charge, we need to remove it (since the double bond moved, the carbon will be neutral). So use the - tool on \(CH_2^+\) to remove the positive charge.

After these modifications, the structure will have a double bond between C (connected to \(H_3C\)) and O, a double bond between C (connected to O) and the adjacent carbon, and \(CH_2\) will be neutral, and O will be neutral (with two lone pairs).

Answer:

The resonance structure is drawn by:

• Converting the double bond between the carbon adjacent to \(CH_2^+\) and \(CH_2^+\) to a single bond.
• Converting the single bond between the carbon (connected to \(H_3C\)) and oxygen to a double bond.
• Removing the positive charge from \(CH_2^+\) and the negative charge from O. The final structure will have \(H_3C - C(=O) - C = CH_2\) (with O having two lone pairs, double - bonded to C, and \(CH_2\) neutral, and the middle C - C double - bonded).