Kainui
  • Kainui
What is the structure of nitrobenzene complete with charges and electron pairs? @Somy
Organic
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Somy
  • Somy
hmmm u want me kinda to draw all the electrons?
Kainui
  • Kainui
After this it will become clear why groups are ortho/para or meta directors just by looking at them. That's basically what we're going to be doing right now. Just the lone electron pairs. Like if you were drawing water, it would look like this: |dw:1403635731983:dw|
Somy
  • Somy
okaay

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Kainui
  • Kainui
Or draw out the structure of aniline if that's too hard. Nitrobenzene has a tricky structure.
Somy
  • Somy
|dw:1403636011020:dw|
Kainui
  • Kainui
|dw:1403636126168:dw| so would you agree that this is a resonance form? Can you draw another resonance form of this structure? Start out at the electron pair on the ring and try to move it around. This is not the electrophillic substitution part yet.
Kainui
  • Kainui
I can explain anything that makes no sense if you need me to unless you're still wanting to try to figure it out on your own.
Somy
  • Somy
|dw:1403637609728:dw|
Somy
  • Somy
|dw:1403637870549:dw|
Somy
  • Somy
|dw:1403638040951:dw|
Kainui
  • Kainui
Be careful, how my electrons moved is differently and I had to draw two arrows. One from the lone pair on the nitrogen to the bond between the nitrogen and benzene ring. From there that made that carbon have 5 bonds! Not allowed, so this has to kick out a bond, which comes from the pi-bond since that'st he weakest bond to break.|dw:1403638090832:dw| Notice we can also reverse this. This is because these are just resonant structures. The thing to notice is that in electrophillic aromatic substiution, we need an electrophile (something that wants electrons) to come in. IF there are electrons at the ortho position, then that's going to mean we're going to have a more likely chance of reacting there. This lines up with aniline being an ortho/para director so far which is good news! The next resonant form looks like this: |dw:1403638226412:dw||dw:1403638244120:dw| Another thing to keep in mind is that this is a neutral compound overall, so when we moved the electrons around we have a negative charge, however we also have a positive charge. This should make sense I hope? Also, notice the electrons skip the meta position and go straight to the para position. Pretty cool huh? By knowing how to draw resonance structures we can "derive" in a sense whether a compound reacts a certain way.
Kainui
  • Kainui
Now before we move on, see if you can do similar things with the other ortho/para directors. Here's a great picture: http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/Images2/benzdipl.gif The top row is all activators, and remember activators tend to be ortho/para directors. In fact, all of the ones shown in that picture that are called activators are indeed ortho/para directors. However be careful!!! Chlorobenzene is also an ortho/para director, it's a slightly weird exception though in that it is a deactivating group. A quick refresher on activators/deactivators: All that it means is that the stronger you are as an activator, the more influence you have on the location of substitutents put on the ring. For instance, usually we've done reactions where the ortho/para director and meta director both agree on which location to put the third substituent right? Well if there's ever a conflict, you choose the stronger activating group's direction. Why is this? Because activating means it is a good electron donating group. So We see on that link that aniline is the best. This makes sense, it has electrons to donate like we've seen. Now phenol (benzenol) is less. Why might that be? Well it has electrons to donate just like aniline, but remember, oxygen is more electronegative so it tends to not donate its electrons as much. Similarly, this becomes overwhelming for the the halogens and they almost never donate to the ring and are deactivators. Carbon would be better than aniline if only it had an electron pair to donate, which is why it's weaker. So now try to draw the resonance structures of phenol. It should be very similar to aniline. See what you can do, good luck there's a lot here but after you work out several problems you'll see that everything I explained here just makes perfect sense by simply following the rules of drawing resonant structures and electronegativity trends. These are probably the two most important things to know in organic chemistry other than relative pkas.
Somy
  • Somy
IF there are electrons at the ortho position, then that's going to mean we're going to have a more likely chance of reacting there. i didn't quite get what this means Another thing to keep in mind is that this is a neutral compound overall, so when we moved the electrons around we have a negative charge, however we also have a positive charge. Are you talking about C being + ?
Somy
  • Somy
hmmmm okay lets see |dw:1403720962720:dw|
Somy
  • Somy
|dw:1403721158742:dw|
Somy
  • Somy
wait a sec i can't make a double bond between the carbon and OH
Somy
  • Somy
i mean O can only make 2 bonds
Kainui
  • Kainui
|dw:1403747360489:dw| You're right about the resonance arrows. However, oxygen can have more than 2 bonds! It's just that now it's no longer neutral. It's fairly rare for oxygen to give up electrons like this and have a positive charge, so you can see how this would be considered unfavorable and thus would be less of a contributing resonant structure. This is why aniline is a stronger activating group compared to phenol while both are ortho/para directing. =P
Somy
  • Somy
hmmm i didn't know O could make more than 2 bonds again, is it like dative bonding? donating a lone pair to make a bond
anonymous
  • anonymous
|dw:1408213423172:dw|
Somy
  • Somy
wah its still here lol

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