SuperSaiyan3

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Does this even make sense? Wouldn't treating a subtrate that already has an amine group with a nitro group just put the nitro group elsewhere on that molecule instead of transforming the amine group into a diazo group (-NN)?
 

loveoforganic

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What reaction exactly are you talking about? Give an example.
 
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SuperSaiyan3

SuperSaiyan3

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May 13, 2009
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Okay so this is the reaction that Kaplan was asking me to figure out, which wasn't all too hard.

The Sn (not sure if they meant to say Sn or Zn.. because from my common knowledge, you use Zn & HCl to reduce) is used to reduce the nitro group to an amine.

Then, the NaNO2 comes in. This was the question I was asking. Why wouldn't the nitro group just add in ortho or para (due to the presence of amine group on benzene ring)? When would you know that it would become a diazo group or do an electrophilic substitution elsewhere on the ring??

I quote Kaplan:
"The reaction sequence is an example of aromatic substitution using a diazonium ion. The first step is a reduction; compound W is m-chloroaniline. The second step converts the amine to the corresponding diazonium salt (compound X)."

:confused:
 

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loveoforganic

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The amine has a lone pair of electrons to donate to the diazonium ion. For the aromatic ring to donate electrons to the diazonium ion, it would have to break aromaticity. The lone pair is in a much higher energy state than the bonding aromatic electrons.

edit: Tin in hydrochloric acid is a common reducing agent for aromatic nitro groups.
 

sleepy425

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Okay so this is the reaction that Kaplan was asking me to figure out, which wasn't all too hard.

The Sn (not sure if they meant to say Sn or Zn.. because from my common knowledge, you use Zn & HCl to reduce) is used to reduce the nitro group to an amine.

Then, the NaNO2 comes in. This was the question I was asking. Why wouldn't the nitro group just add in ortho or para (due to the presence of amine group on benzene ring)? When would you know that it would become a diazo group or do an electrophilic substitution elsewhere on the ring??

I quote Kaplan:
"The reaction sequence is an example of aromatic substitution using a diazonium ion. The first step is a reduction; compound W is m-chloroaniline. The second step converts the amine to the corresponding diazonium salt (compound X)."

:confused:
The problem is that NaNO2 cannot be used to nitrate (put a nitro group on) an aromatic ring. You're thinking of nitrate (like sodium nitrate, NaNO3, or nitric acid, HNO3). So lemme summarize nitration, then I'll explain why sodium nitrite (NaNO2) cannot serve as a nitration agent.

Nitration requires some form of the nitro group that can function as an electrophile. You use nitric acid and sulfuric acid to make the electrophile. The sulfuric acid protonates the nitric acid a second time at the same oxygen that is already protonated, so you form a water leaving group. This leaving group leaves and you're left with a positive charge on the nitrogen (this is called a nitronium ion). The nitronium ion is the electrophile that gets attacked by the ring. Since you have a full positive charge on the nitrogen, it is a great electrophile so the weakly nucleophilic aromatic ring can attack it pretty easily.

In the case of the nitrite ion, the nitrogen is neutral, so it cannot function as the electrophile in a nitration of benzene.

Diazotization is a very different process. Nitrite does function as an electrophile in diazotization (in its protonated form), but unlike nitration, the nucleophile in diazotization is an amino group, not the pi electrons of an aromatic ring. Amino groups (even aromatic amino groups), are much better nucleophiles than aromatic pi electrons, so you don't need the electrophile to be that good.

So, the way it works is, you use sodium nitrite and HCl. The HCl is used to generate nitrous acid (HNO2) in situ because nitrous acid is not shelf stable. Nitrous acid is a better electrophile than the nitrite ion because the extra proton ties up the negative charge onto one of the oxygens, leaving an N=O double bond on the other side. The N=O double bond has similar reactivity to a carbonyl (it's less electrophilic though). The amine lone pair attacks the N=O double bond and kicks the electrons up onto the oxygen. After that, there are a few proton transfer steps and eliminations to get rid of the two oxygens and form the diazonium salt. If you want, I can describe those in detail, but that's the gist of it.

Hope that helps.