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Do conjugated double bonds take priority over a +1 formal charge?

The first one is what I solved for (ignore the hexagon that has no double bonds attached to it), and the one below it is the answer from the solutions.

Why did they move the conjugated bonds around first? This ends up causing the formal charge to move in the opposite direction than mine. Does it matter for this problem?

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Some may think that mesomeric structures be ‘real’, that molecules ‘resonate back and forth between them, like a pendulum’ or that there be any type of ordering within mesomeric structures.

All of that is not the case. Mesomeric structures are our crutches to understand what is only one state of the molecule. We draw them in a line interconnected by a series of mesomery arrows because we are uncapable of drawing multi-dimensionally. There is no ‘first’ or ‘later’, no ‘full single’ or ‘full double’ bond. You must understand that all those structures are equal and equally far from reality. Reality is much closer to a continuous wave function with electron deficiencies on the benzylic, ortho and para carbons, with varying (but constant for a given bond) bond orders somewhere between 1 and 2.

Your question in its original form is basically moot. However, a marking scheme of your exam may require you to draw five mesomeric structures rather than just four to get full marks. That is up to your teaching staff.

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Given the symmetry of this benzyl cation, the "direction" does not matter. I have put direction in quotes because there actually is no such thing. Electron pushing arrows are a helpful tool (at the very least for accounting purposes), but it is not what is happening in this molecule (or most other cases, either). What you have here is a resonance structure, meaning that the true situation is not well modeled by a single Lewis structure, thus we use several, and the true situation is somewhere in between the structures with varying contributions from the individual ones. So it does not matter how one arrives at these structures, it is important that they be reasonable and do contribute.

One could argue that the answer given is more extensive than your attempt, but I would consider both equal. The reactivity of the molecule is not dominated by the aromatic ring, thus placing extra emphasis on it seems superfluous.

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Our primary motive is to delocalise the charge. More delocalisation of charge implies more stability.

The book solution had moved the conjugated pie bonds inside the ring just to show that they are also in resonance and hence is counted towards the overall structure stabilisation. It doesn't at all matter in which direction your charge being delocalised. What matters is it should be delocalised to maximum extent. More delocalisation => more stability of the overall structure. And don't think the pie conjugation inside the ring gets delocalised first and then the positive charge. There is no such order. As I said what matters is delocalisation. You can delocalise it in any direction. So you are also correct and the book's answer is also correct. The more contributing resonance structure you have, the more will be the stabilisation of the overall molecule. To know which resonance structure (canonical structure) is more contributing and which is least contributing read the article here: http://web.chem.ucla.edu/~harding/tutorials/resonance/imp_res_str.html

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