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resonance structures
Started by princessjasmine99
post this in the DAT Discussions Forum
More bonds less charges = more stable -> major contributor.
Specifically, what's the problem? Remember, sometimes H's are hidden, and you can't double bond an sp3 Carbon. Does that help?
What are the options?
I'm not writing this down, so I may very well miss something, but here's how I see it:
1st Resonance Structure
1.) The pi-bond on the oxygen forms a loan pair on the oxygen, making the oxygen negative and leaving behind a carbocation.
2.) The loan pair on the nitrogen forms a pi-bond with the carbocation causing the nitrogen to become positive and making the carbocation neutral.
3.) You have a positively charged nitrogen and a negatively charged oxygen with the same number of bonds as the orginal molecule. Going with what I said earlier in the thread, you have the same number of bonds but you now have charges on atoms which were previously uncharged species.
4.) Because the original structure has no charges and the same number of bonds as the resonance structure I described, the major contributor should be the one with less charges which is the original structure.
2nd Resonance Structure
5.) Another structure which could form would be to repeat step 1 from above then continue to step 6.
6.) Now, the pi-bond from the alkene moves over to form a pi-bond between the carbocation and the second carbon from the left. This leaves you with a carbocation on the far left and a negatively charged oxygen in the molecule which is again less favorable than the original structure meaning that it is also a minor contributor.
If I could draw it out in text I would.
I'm not writing this down, so I may very well miss something, but here's how I see it:
1st Resonance Structure
1.) The pi-bond on the oxygen forms a loan pair on the oxygen, making the oxygen negative and leaving behind a carbocation.
2.) The loan pair on the nitrogen forms a pi-bond with the carbocation causing the nitrogen to become positive and making the carbocation neutral.
3.) You have a positively charged nitrogen and a negatively charged oxygen with the same number of bonds as the orginal molecule. Going with what I said earlier in the thread, you have the same number of bonds but you now have charges on atoms which were previously uncharged species.
4.) Because the original structure has no charges and the same number of bonds as the resonance structure I described, the major contributor should be the one with less charges which is the original structure.
2nd Resonance Structure
5.) Another structure which could form would be to repeat step 1 from above then continue to step 6.
6.) Now, the pi-bond from the alkene moves over to form a pi-bond between the carbocation and the second carbon from the left. This leaves you with a carbocation on the far left and a negatively charged oxygen in the molecule which is again less favorable than the original structure meaning that it is also a minor contributor.
If I could draw it out in text I would.
Last edited:
ahh thank you so much for taking the time to write all that out! Here are the options! this is from bootcamp
I thought the correct answer would be a or c because the negative charge is on the most electronegative atom and positive is on the least! I feel like this should be such an easy question! The correct answer is B!
- B.
- C.
- D.
- E.
I thought the correct answer would be a or c because the negative charge is on the most electronegative atom and positive is on the least! I feel like this should be such an easy question! The correct answer is B!
It's B because it obeys the octet rule!
Yeah, octet rule > electronegativity.
Since the original structure wasn't an option, B has to be the answer because A lacks a complete octet on the carbocation to the far left. 🙂
Since the original structure wasn't an option, B has to be the answer because A lacks a complete octet on the carbocation to the far left. 🙂
Those are all crappy resonance structures and of the ones that could form, form very minor products.
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If N has a + how does it have an octet? 😕
Yikes. E was ruled out immediately due to losing an H. B is the best answer due to conjugation and a non terminal carbocation
thank you everyone 🙂
Which carbocation? We looking at the same B here??
I meant that there is no carbocation on a terminal carbon. All the others, except E, which can easily be ruled out due to the Amine group losing a H, have a carbocation. The terminal carbocation would never be a major resonance structure. It's a test taking strategy, that may or may not make sense to anyone else.
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I was thinking more like B because O is more electronegative and therefore can have the neg charge there more than any other, in addition to the octet rule and the pi bonds can move around in diff configurations in that form. N is more likely to to give up an electron so is more likely to be positively charged. C and H has roughly the same so neither one of them wants to be minus or positive charged.
You always want the neg charge on the most electroneg atom. Haha, I am not that great on O chem tho. I just learned the concepts and looked at the big picture when I was studying and taking the test. Worked for me.
But A, B, and C all have the negative charge on the oxygen. OP wasn't sure which of those to choose from. That's where you look at the structure with the least charges and most bonds and see if one has full octets around each atom.
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