Confused about arrow heads

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SunTiger
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#1
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#1
Refer to uploaded image. The arrows in the two middle structures should be half headed right? In the image is isn't very clear.
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Pigster
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#2
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Poor IT. Should be double arrow head, otherwise you'd end up with radicals.
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SunTiger
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#3
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(Original post by Pigster)
Poor IT. Should be double arrow head, otherwise you'd end up with radicals.
You sure? Could you explain why, in terms of sigma/pi bonds.
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Pigster
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#4
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Numbering the atoms 1-6 clockwise starting from the N atom.

Step 2: the first equilibrium involves the breaking of the pi bond between C3&4, both e- move to between C4&5. I can see how you'd think that it should be a single arrow head as C4 has no net change in # of e- and I guess in a way a single arrow head would be correct, but I'd think of it is terms of mass movement of the position of the e-, i.e. both e- in the pi bond have moved from one side of C4 to the other side.

Sigma bonds don't get involved, there is no need to mention them.
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SunTiger
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#5
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(Original post by Pigster)
Numbering the atoms 1-6 clockwise starting from the N atom.

Step 2: the first equilibrium involves the breaking of the pi bond between C3&4, both e- move to between C4&5. I can see how you'd think that it should be a single arrow head as C4 has no net change in # of e- and I guess in a way a single arrow head would be correct, but I'd think of it is terms of mass movement of the position of the e-, i.e. both e- in the pi bond have moved from one side of C4 to the other side.

Sigma bonds don't get involved, there is no need to mention them.
How I see it is that the pi electron moves from C4-5 to C3-4, so that is just one electron right? Double bonds are made on one sigma and one pi electron. A half headed arrow shows movement of one electron. Though I may be missing something...
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Pigster
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#6
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(Original post by SunTiger)
How I see it is that the pi electron moves from C4-5 to C3-4, so that is just one electron right? Double bonds are made on one sigma and one pi electron. A half headed arrow shows movement of one electron. Though I may be missing something...
Both the sigma and pi bonds are covalent bonds - a shared pair of e-. A double bond therefore involves two shared pairs i.e. four e-.

Both e- in the pi bond move from one side t't'other.
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SunTiger
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#7
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(Original post by Pigster)
Both the sigma and pi bonds are covalent bonds - a shared pair of e-. A double bond therefore involves two shared pairs i.e. four e-.

Both e- in the pi bond move from one side t't'other.
Ahh yeah, makes sense now. Thanks for the help!
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