How to find most stable Resonance Structures

[Alpha Centauri]

So this is definitely one of the topics I seem to struggle in. I have made a post about it before but it didn't get much attention... Can someone please talk me through how to approach these kinds of problems? They are one of the few question types I am consistently getting incorrect and I want that to change 😀

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[kimball]

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[ryeguy95]

Well, first you can go over the structures pretty fast and determine if they even have the same amount of donatable electrons (that is resonating electrons). This specific question has 4 total, 3 in the ring and the Negative charge on the (S)ulfur. You do not count the one on the (N)itrogen because it is locked in the plane of symmetry due to the N being sp2 hybridized *** when you look at choices D and E. D and E only have 3 resonating pairs available so you can get'em outta there. A and C are fairly similar in terms of a LACK of continual resonance. There is a break in the cyclic structure on each ring where the carbon's are not being "covered" by the resonating electrons. Looking at B finally, all carbons are touched by electrons and the negative is on the Electronegative atom = Nitrogen: which allows the negative charge to be more stable on it rather than the less electronegative Carbon as in choice C. Hope this helps. Sorry if I didn't use exactly precise orgo vocab.

 

[flossit]

Well, first you can go over the structures pretty fast and determine if they even have the same amount of donatable electrons (that is resonating electrons). This specific question has 4 total, 3 in the ring and the Negative charge on the (S)ulfur. You do not count the one on the (N)itrogen because it is locked in the plane of symmetry due to the N being sp2 hybridized *** when you look at choices D and E. D and E only have 3 resonating pairs available so you can get'em outta there. A and C are fairly similar in terms of a LACK of continual resonance. There is a break in the cyclic structure on each ring where the carbon's are not being "covered" by the resonating electrons. Looking at B finally, all carbons are touched by electrons and the negative is on the Electronegative atom = Nitrogen: which allows the negative charge to be more stable on it rather than the less electronegative Carbon as in choice C. Hope this helps. Sorry if I didn't use exactly precise orgo vocab.

Can you please show which 3 are resonating in D and E please?

 

[Crithu]

Can you please show which 3 are resonating in D and E please?

D only shows 3 delocalized, although the top of the image may be cut off (The sulfur doesn't have a full octet or a formal charge). The three that are shown would be the double bonds and the pair of electrons on the negatively charged carbon.
In E, the delocalized electrons are the double bonds and the pair of electrons on the negatively charged carbon.

 

[flossit]

So this is definitely one of the topics I seem to struggle in. I have made a post about it before but it didn't get much attention... Can someone please talk me through how to approach these kinds of problems? They are one of the few question types I am consistently getting incorrect and I want that to change 😀

I see a one double bond that can resonate and one lone pair on the negatively charged carbon, but what is the third one in E?

 

[Crithu]

I see a one double bond that can resonate and one lone pair on the negatively charged carbon, but what is the third one in E?

There are 2 sets of double bonds. One c=c and an s=c

 

[Mrhyde]

For me....I simply see which are immediately wrong.


- The first two we can immediately eliminate are Choice D and Choice E
Because a negative charge will never want to be on a carbon atom, that is extremely unstable. Its like a "radical" which is unstable.

-Next we move on to the next one we can eliminate which is Choice C. Nitrogen has 3 bonds and 2 lone pairs (so thats 5 things on one element) which just destroys the octet rule, so its wrong.

-We also eliminate Choice A because of the same reason we just eliminated choice C, but this time, we are looking at the carbon with the double bond connected to the sulfur. That carbon has 5 things connected to it, so it breaks the octet rule.

So the last choice left must be your answer which is Choice B.

---> By seeing why the choices are wrong as i listed above then this question had ZERO to do with Resonance Structures. As this question did not have at least two pretty stable looking compounds where we would then have to see which choice was the better choice despite them both looking pretty "right". We simply quickly skimmed saw the blatant flaws on the wrong choices which all dealt with stability and with breaking the fundamental Octet rule.