If we moved electrons towards the + charge then it wouldn't be a valid structure because there are two of the same charges. We can't have two of negative charges or two positive charges right next to each other because same charges repel each other. As shown below.

In this case it's kinda like symmetry with the oxygens one resonance structure has one oxygen with a double bond so the second one will have the other oxygen with a double bond. 

This is all because of the exceptions of when to count a lone pair or not. Because we don't count the lone pair on nitrogen it is sp2 hybridized. At 10:30 those are the two main rules for lone pairs to know.  Now when the lone pairs do not fall into the categories shown at 10:30 so no double bond directly on the atom with the lone pair or multiple lone pairs. Then we follow what is shown at  14:54 , and go down the aromatic checklist. If counting the lone pair makes it better then the structure favors that, and the opposite is true.

Farzeen Qadri At timestamp 13:24 you'll see the overall mechanism and explanation of what happens when we have excess of the grignard reagent. Then shortly after you'll see an example of this mechanism. 

Yes, you could also place the lone pair on a solid line and the hydrogens as dashed and wedged. 

Jen - Chemmunity Team  Hussein Murad The new practice problem video is up, you'll find that below:

iso is included as an alphabetical determinant same with "neo" and "cyclo". Others that are not included are "sec" and "tert"

Savannah Hollen  at 10:30 this is why number 7 is A not D, this goes back to which lone pairs count and which do not. And at 4:00 it has to be E for number 9 because structure IV violates one of the rules since the NH2 is sp3 hybridized and not planar. 

Just sent in a request to the team, once they create a video explaining this and other examples I'll link that up here. 👍

Recall that meso compounds have a line of symmetry, this structure doesn't have that so it will not be meso. Instead this gives us 2 products that are enantiomers. 

Oxygen does have a higher electronegativity than sulfur but this is a trick question, it is an exception to the rule. When we are comparing sulfur and oxygen we look at size not electronegativity. 

The larger radius size of the sulfur atom causes the H-S bond to be longer than the H-O bond. The longer bond is weaker and makes it easier to break which makes it easier to deprotonate. Another reason is that the conjugate base of the Sulfur containing structure is more stable than that of the Oxygen because the negative charge is spread out over a larger surface area.