Heterocyclic Aromatic Compounds

[Chemspiration]

:) I would like to take the time to introduce my self. Hi I'm a witty newbie that enjoys the complexities of our universe, Ok now that that is over. I gonna ask a serious question that is bothering me.

 

The benzene structure or the heterocylcic structure is closed off with Hydrogen in every position around the structure, If you substitute the bottom H, with CH3 you have toluene or methylbenzene or phenylmethane, Furthermore if you substitute that same CH3 side chain with a Nitrogen, thus you have created Pyridine.

 

Question 1. In which manner is the positioning of the structures numbered.. For example the bottom hydrogen in the benzen structure would be classified as the what # position

furthermore were does the numbering start.

 

Question 2. how does one break the bond of the bottom hydrogen without disturbing the others bonds that are in synchronization. I believe that all the carbons have one and a half bonds in this particular structure. so how do you substitute the hydrogen for a nitrogen...

 

thanks Chemspiration :)

[Chemspiration]

somebody please help :)

[Chemspiration]

Am I the only one in here??? :)

[zadojla]

No, there are about a half dozen of us in here now, but probably none of us know the answer. I knew it once, back in 1968, but I've forgotten.

[pgrmdave]

Wait for Tim_Lou, he's the mod here and will probably know the answer.

[Chemspiration]

thanks guys

[Tim_Lou]

ah....

i know answer one....

no matter where the functional group is, you name it so the the sum of these numbers are the lowest.

if there is only one functional group attached to it, it would be just one, if two, you gotta consider where to start so that the sum of these positions are the lowest.

 

question two seems rather complicated to me...

[Tim_Lou]

i found a website that might be helpful:

http://www.cem.msu.edu/~reusch/VirtualText/benzrx1.htm

take a look at the "slide show" in the middle of the page, it shows the mechanics of these reactions...

 

the animated model is kinda cool :hyper:

[Chemspiration]

Wow that link is tight!

this is exactly what sort of information i am after. this is of great use to me. ]

thanks again.

 

:hyper:

[Chemspiration]

so how many functional groups are in benzene???? :hyper:

[maddog]

so how many functional groups are in benzene???? :D

I agree with Tim_Lou on 1 as well. From my physics training... If you have one methyl

group, it is quite arbitrarily simple. The convention is to label it 1. When there are more

branches, you arrange the ring so as to have the second in the lowest position. :(

 

For 2, I could go into QM though I think that might be more in depth than you like. The

simpler version would be depending on the amount of energy, the nearby bonds would be

only affected somewhat. The electrons in the valence shell may get excited enough to

release a photon to give off excess energy. It's kind of a P.Chem question.

 

I do hope the website helped in case I didn't. :hyper:

 

Maddog

[Medicinal_Chemist]

Hi.

 

What you described was not a heterocyclic aromatic compound. To be heterocyclic, it would need to have a heteroatom, usually nitrogen, forming the ring itself, and sharing the pi-electrons in sp2-hybridised orbitals. A typical example would be pyridine.

 

To answer your questions:

1) it depends on what functional groups are present. Generally, the 1-position is that with the most important functional group. By convention, a carboxylic acid or ester will always have priority. So for example if you had a methyl group para to a carboxylic acid, the name would be 4-methylbenzoic acid. To look at some examples to get a feel for it, browse the aldrich catalogue at http://www.aldrich.com

 

2) I'm not quite sure what you mean by this; if you mean how do you break the bond of the para-hydrogen in order to replace it with another functional group, then the main reactions you would use are a Freidel-Crafts alkylation or acylation, a halogenation (usually bromination), nitration (using a mixture of conc. sulphuric and nitric acids at reflux). You then do functional group interchanges on these to get the groups you want. Of course, where the groups initially add is down to what groups are on the ring to start with, and whether they are ortho-para or meta directing. I suggest you read a book called "Introduction to Organic Chemistry" by Streitweiser et al. if you are genuinely interested.

 

Hope that helps.

 

Andy

[UncleAl]

Organic Chemistry, 5th edition, Morrison and Boyd

Advanced Organic Chemistry, 5th Editon, Smith and March

 

Organic nomenclature is by the book. Groups have Cahn-Ingold-Prelog priority. Nucleophilic aromatic substitution is SOP for aromatics' derivatization. Once you get something useful on the ring you can play with diazotization (anilines) or Pd-couplings (good leaving groups). Etc.

[Odin]

:naughty: I would like to take the time to introduce my self. Hi I'm a witty newbie that enjoys the complexities of our universe, Ok now that that is over. I gonna ask a serious question that is bothering me.

 

The benzene structure or the heterocylcic structure is closed off with Hydrogen in every position around the structure, If you substitute the bottom H, with CH3 you have toluene or methylbenzene or phenylmethane, Furthermore if you substitute that same CH3 side chain with a Nitrogen, thus you have created Pyridine.

 

Question 1. In which manner is the positioning of the structures numbered.. For example the bottom hydrogen in the benzen structure would be classified as the what # position

furthermore were does the numbering start.

 

Question 2. how does one break the bond of the bottom hydrogen without disturbing the others bonds that are in synchronization. I believe that all the carbons have one and a half bonds in this particular structure. so how do you substitute the hydrogen for a nitrogen...

 

thanks Chemspiration :xx:

 

Positioning starts in the upper left and works its way to the right.Eg:

 

1-acetyl 3-ethyl benzene

 

There is a CH3 at the 1 position and a CH3CH2 at the three

 

Secondly In order for the pyridine alkaloid to exist it has to be part of the ring structure itself.You have to synthesize it from benzoic acid and nitric oxide

 

So the formulation is C6H5N and not N-C6H5