Question

Question #1cf30

Answer 1

See explanation.

Explanation:

I will try to help as much as I can, since this question requires so much drawings, so I did my best to make it as clear as possible with an OK effort.

In principle, to draw a resonance form, you should have two neighbouring groups: one is rich and the other is poor and they are separated by one single bond only.

Take the first molecule for example, it has two resonance forms only, the one drawn in the question and this one below:

The phenyl group is separated from the negative charge by two single bonds, and therefore, it is not affected by the presence of the negative charge, and thus, it will not change (of course I am not counting the two resonance forms of the phenyl it self, if I have to, therefore, this molecule will have four resonance forms).

However, the double bond `C=O` is separated from the negative charge by one single bond, and therefore, the lone pair of carbene will form a double bond with the carbon of the carbonyl group, and oxygen will acquire a negative charge due to gaining an extra lone pair coming from breaking the `pi` bond with the carbon.

In a similar way, you can find the other resonance forms for the other molecules.

Remember, the Rich gives the Poor.

5 resonance forms (including the original one in question)

4 resonance forms (including the original one in question)

4 resonance forms (including the original one in question)