Organic Chemistry Question

['Hamas' Jenkins]

This one has me totally vexed. Any help you could offer on where to put the arrows would be greatly appreciated.




It looks like it's already done. I'm not sure if I should approach it as conjugated pi bonds or not.

[Ming the Merciless]

I found this question on another test of the same company




with no curved lines

maybe u got instructors manual???? NICE

[Halfcan]

B B B B B B B B B B B B B B

Always pick B in a multiple choice test.

[Don Corlemahomes]

Originally Posted by 'Hamas' Jenkins

Doesn't the first resonance arrow suggest that you are moving the pi bond in the upper right hand of the benzene ring to the C-N bond, and won't that gobble up that lone pair?

I think the instructions in the lower righthand box are suggesting you indicate the movement of electrons in that last structure from the double bond between C1 and nitrogen back onto the nitrogen as a lone pair.

****, been 5 years and a near total dump of the concepts from that awful class.

[Don Corlemahomes]

****, no. Maybe you have to complete each by adding arrows to show the movement of electrons onto each carbon in the ring, and also show the movement of electrons back onto the nitrogen in the fourth, a la this one:

[Ming the Merciless]

Originally Posted by Cunning Linguist

****, no. Maybe you have to complete each by adding arrows to show the movement of electrons onto each carbon in the ring, and also show the movement of electrons back onto the nitrogen in the fourth, a la this one:

man crush

[Buehler445]

I have no idea. But I have a buddy that has a doctorate and some post doc work in chemistry. PM me if you want me to call him.

[Zebedee DuBois]

aniline

[DrunkBassGuitar]

So in that first picture, if you moved the double bond to the amine, the nitrogen would have a negative charge, because it's going to have a double bond, plus 2 hydrogen, plus it's electron pair. nitrogen doesn't like that (octet rule). also the carbon that loses that double bond is going to be positively charged since it will be only bound to 2 carbons and a hydrogen and is missing a valence electron. It's been years since I've done something like that example. But you have a conjugated system.

The electron pair moves from the N to the carbon closest to it to form the double bond, and an electron pair moves from that carbon to the one adjacent to it. I'm a little confused by how they drew it though. but it's been a long time for me and we only went over ortho, para, and meta for one session

here's a decent picture of what I mean


basically the nitrogen has a partial positive charge and all of the carbons have a partial negative if that makes sense

[-King-]

Four.

[Buzz]

Two single bonds and two lone pairs on N

[Hog's Gone Fishin]

You dipwad, That's already answered . I jerk off friggin pigs and know that !

[Simply Red]

When you all get this sorted - I have a few Algebra II questions.

Thanks!

[Buzz]

Originally Posted by Simply Red

When you all get this sorted - I have a few Algebra II questions.

Thanks!

All Alegbra questions ends in Q.

[Don Corlemahomes]

Originally Posted by DrunkBassGuitar

So in that first picture, if you moved the double bond to the amine, the nitrogen would have a negative positive charge, because it's going to have a double bond, plus 2 hydrogen, plus it's electron pair electron pair is lost. nitrogen doesn't like that (octet rule). also the carbon that loses that double bond is going to be positively chargedNo since it will be only bound to 2 carbons and a hydrogen and is missing a valence electron. It's been years since I've done something like that example. But you have a conjugated system.

The electron pair moves from the N to the carbon closest to it to form the double bond, and an electron pair moves from that carbon to the one adjacent to it. (Yes, which is why N isn't negative and C isn't positive, but the exact opposite) I'm a little confused by how they drew it though. but it's been a long time for me and we only went over ortho, para, and meta for one session

here's a decent picture of what I mean


basically the nitrogen has a partial positive charge and all of the carbons have a partial negative if that makes sense

You haven't really provided any solution, other than describing what you see. OP can do that, but he needs a solution other than "it's already done, and this is happening" offered by others.

['Hamas' Jenkins]

Ok, so moving the lone pair on the last benzene ring on the carbon with the negative formal charge to form a double bond in the upper left hand corner is correct, but my attempts for the other three were not.