fit2

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Hello I have a question about difference in hybridization and I was wondering if anyone can help me out.

What is difference or better how do you figure out/remember difference in hybridization for something like pyrrole and pyridine.
I read that pyrole has nitrogen that sp2 hybridized but that pyridine's electrons that are in the aromatic system come from a p orbital.

Thanks for the help!
 
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fit2

fit2

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Thanks!
 
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orgoman22

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Hello I have a question about difference in hybridization and I was wondering if anyone can help me out.

What is difference or better how do you figure out/remember difference in hybridization for something like pyrrole and pyridine.
I read that pyrole has nitrogen that sp2 hybridized but that pyridine's electrons that are in the aromatic system come from a p orbital.

Thanks for the help!
If electrons are part of the aromatic system.....they are DELOCALIZED........thus are in a p orbital. We see this situation in pyrrole. In pyrrole, the electrons are in a p orbital, but the N is sp2 since if you examine all the resonance forms it has a double bond associated with it. In Pyridine,,,,,the N is sp2....and the electrons are on the outside,,,,,,not associated with the ring aromaticity...hence in an sp2 orbital. Bottom line......If you can DELOCALIZE the electrons and use them in resonance...they will be in a p orbital. Lets try one ...In an amide.....the Nitrogen is what hybidization ? answer is sp2,,,,,,,,Now,,,,,where are the unshared electrom pair ? Answer.... in a p orbital ...since they can be delocalized !

Hope this helps.

Dr. Romano
 
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fit2

fit2

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Thank you Dr. Romano!
 
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WheatLom

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p for pi. To be aromatic, you need to be single bonded to other atoms in a ring. So... The s and p orbitals are going to be used. S for the connections, p for the aromaticity.

Hybridization even in aromatics, is how many electronic domains attached to the atom. each sigma bond is 1, each double bond is 1, each lone pair is 1, and a triple bond still counts as 1 domain. It doesn't matter if it is delocalized.

The carbons in benzene are all sp2 hybridized and planar. An Ideal aromatic compound is planer.
 
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orgoman22

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p for pi. To be aromatic, you need to be single bonded to other atoms in a ring. So... The s and p orbitals are going to be used.

Hybridization even in aromatics, is how many electronic domains attached to the atom. each sigma bond is 1, each double bond is 1, each lone pair is 1, and a triple bond still counts as 1 domain. It doesn't matter if it is delocalized.
No.....an aromatic system is a continuum of parallel p orbitals. We see this in compounds such as Furan, Pyrrole, Benzene, Azulene, and Thiophene. NO SINGLE bonds are present within these species for aromaticity. This is a very basic concept. A single bond would not have a p orbital vacancy .....LOL......and aromaticity would be broken. Do yourself a favor......get a book written by a PhD in Organic Chemistry and see the numerous examples. Books by Wade, Klein, McMurray, Smith, Feseden, Jones, Solomon, and Carey are my favorites. I think David Klein explains it best.

Dr. Romano
 

WheatLom

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No.....an aromatic system is a continuum of parallel p orbitals. We see this in compounds such as Furan, Pyrrole, Benzene, Azulene, and Thiophene. NO SINGLE bonds are present within these species for aromaticity. This is a very basic concept. A single bond would not have a p orbital vacancy .....LOL......and aromaticity would be broken. Do yourself a favor......get a book written by a PhD in Organic Chemistry and see the numerous examples. Books by Wade, Klein, McMurray, Smith, Feseden, Jones, Solomon, and Carey are my favorites. I think David Klein explains it best.

Dr. Romano

What are you saying? Think before you attack someone.

you need a single bond to connect all the components in a ring. so yes the p and the s orbitals are used in an aromatic system. the s is to maintain the ring structure while the p delocalizes the pi electrons. Why don't you get a book and read up on what connects atoms. You can't have pi electrons without a sigma bond basic GENERAL chemistry.

If you thing you can have an aromatic compound that has no single bonds you are wrong.... don't LOL me. If you figured out a way to have a pi bond without a sigma bond, you'd have a nobel prize

I'm pretty sure benzene is 1,3,5-cyclohexatriene. A sigmas bond connects the 6 carbons, with 3 pi bonds that use p orbitals, cause pi electrons can only use p orbitals.

Janice Smith is my text book. So the source you recommended you are calling wrong.
 
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orgoman22

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What are you saying? Think before you attack someone.

you need a single bond to connect all the components in a ring. so yes the p and the s orbitals are used in an aromatic system. the s is to maintain the ring structure while the p delocalizes the pi electrons. Why don't you get a book and read up on what connects atoms. You can't have pi electrons without a sigma bond basic GENERAL chemistry.

If you thing you can have an aromatic compound that has no single bonds you are wrong.... don't LOL me. If you figured out a way to have a pi bond without a sigma bond, you'd have a nobel prize

I'm pretty sure benzene is 1,3,5-cyclohexatriene. A sigmas bond connects the 6 carbons, with 3 pi bonds that use p orbitals, cause pi electrons can only use p orbitals.

Janice Smith is my text book. So the source you recommended you are calling wrong.
LOL..indeed. I have known Dr. Smith for 35 years. Here is a reference from Wiki......In terms of the electronic nature of the molecule, aromaticity describes a conjugated system often made of alternating single and double bonds in a ring. This configuration allows for the electrons in the molecule's pi system to be delocalized around the ring, increasing the molecule's stability. The molecule cannot be represented by one structure, but rather a resonance hybrid of different structures, such as with the two resonance structures of benzene. These molecules cannot be found in either one of these representations, with the longer single bonds in one location and the shorter double bond in another (see Theory below). Rather, the molecule exhibits bond lengths in between those of single and double bonds.

Benzene is a planar molecule (all the atoms lie in one plane), and that would also be true of the Kekulé structure. The problem is that C-C single and double bonds are different lengths. C-C 0.154 nm C=C 0.134 nm

Note: "nm" means "nanometre", which is 10-9 metre.

That would mean that the hexagon would be irregular if it had the Kekulé structure, with alternating shorter and longer sides. In real benzene all the bonds are exactly the same - intermediate in length between C-C and C=C at 0.139 nm. Real benzene is a perfectly regular hexagon.

Our goal is to �measure� the heat of hydrogenation of this imaginary structure and compare it to the actual heat of hydrogenation of benzene (which can be measured). The idea here is to determine which structure yields the most energy upon saturation of its double bonds with hydrogen. The structure that actually gives up more energy during this process is the one with higher chemical potential energy and therefore less chemical stability.

cyclohexeneSince 1,3,5-cyclohexatriene does not actually exist, a model must to chosen to approximate its heat of hydrogenation. The model we will select will be the substance cyclohexene, C6H10, which is a six carbon ring with one set of double bonds across which the hydrogenation will occur. A schematic picture of a cyclohexene molecule is shown here. In order to approximate the hydrogenation of the three sets of double bonds found in 1,3,5-cyclohexatriene, we will perform the hydrogenation on three separate molecules of cyclohexene and assume that the approximate heat of hydrogenation of 1,3,5-cylochexatriene is equal to three times the heat of hydrogenation of cyclohexene. These assumptions are reasonable.

The measured heat of hydrogenation of benzene (total saturation of all carbons to form cyclohexane) is approximately 50 kilocalories per mole. The heat of hydrogenation of a single cyclohexene molecule is approximately 29 kilocalories per mole. Using the proposed model, the total heat of hydrogenation of the imaginary compound 1,3,5-cylohexatriene is equal to: (3 moles C6H10 ) X (29 kcal/mole) = 87 kcal.

So the heat of hydrogenation of our imaginary model compound is 87-kcal/mole, while the actual measured heat of hydrogenation of benzene is only 50-kcal/mole. The resonance energy can now be quantified by observing the difference in the heat of hydrogenation of the model compound compared to the real benzene molecule:

ΔHH2 Resonance = ΔHH2 �1,3,5-cyclohexatriene� – ΔHH2Benzene = �Resonance energy�

= 87 kcal/mole � 50 kcal/mole = 37 kcal/mole (A significant amount of energy!)

The difference between the ΔHH2 of the two compounds is approximately 37-kcal/mole. This significant energy is the resonance energy and represents the actual increase in chemical stability realized by the benzene molecule. So what does all this have to do with graphite? A single graphene layer is made up of millions of �benzene� ring structures and each of these is resonance stabilized. That means the resonance energy is common to the entire graphite structure. Graphite owes part or most of its �low chemical potential energy� to the fact that it is a pure sp2 carbon, resonance stabilized substance.

You are wrong once again about 1, 3, 5 -hexatriene. Sorry that the facts upset you.

This is my last response to this question.

Dr. Romano
 

WheatLom

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LOL..indeed. I have known Dr. Smith for 35 years. Here is a reference from Wiki......In terms of the electronic nature of the molecule, aromaticity describes a conjugated system often made of alternating single and double bonds in a ring. This configuration allows for the electrons in the molecule's pi system to be delocalized around the ring, increasing the molecule's stability. The molecule cannot be represented by one structure, but rather a resonance hybrid of different structures, such as with the two resonance structures of benzene. These molecules cannot be found in either one of these representations, with the longer single bonds in one location and the shorter double bond in another (see Theory below). Rather, the molecule exhibits bond lengths in between those of single and double bonds.

Benzene is a planar molecule (all the atoms lie in one plane), and that would also be true of the Kekulé structure. The problem is that C-C single and double bonds are different lengths. C-C 0.154 nm C=C 0.134 nm

Note: "nm" means "nanometre", which is 10-9 metre.

That would mean that the hexagon would be irregular if it had the Kekulé structure, with alternating shorter and longer sides. In real benzene all the bonds are exactly the same - intermediate in length between C-C and C=C at 0.139 nm. Real benzene is a perfectly regular hexagon.

I don't know what you are trying to get at. The information that both you and I posted is correct. On an exam, there going to show benzene and when they ask hybridization of any of the carbons its going to be sp2. not some technical answer.

But to say that I am wrong for saying single bonds exist in an aromatic compound is not alright. Im not going to take that lightly. I literally took the information out of Dr. Smith's book which is at my feet and put in on the forum. Im not making information up. If you have a problem, talk to dr. smith cause she is the one that is wrong then. Personally, I think she wrote a hell of a text book that got me a 22 on the DAT.
 

WheatLom

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LOL..indeed. I have known Dr. Smith for 35 years. Here is a reference from Wiki......In terms of the electronic nature of the molecule, aromaticity describes a conjugated system often made of alternating single and double bonds in a ring. This configuration allows for the electrons in the molecule's pi system to be delocalized around the ring, increasing the molecule's stability. The molecule cannot be represented by one structure, but rather a resonance hybrid of different structures, such as with the two resonance structures of benzene. These molecules cannot be found in either one of these representations, with the longer single bonds in one location and the shorter double bond in another (see Theory below). Rather, the molecule exhibits bond lengths in between those of single and double bonds.

Benzene is a planar molecule (all the atoms lie in one plane), and that would also be true of the Kekulé structure. The problem is that C-C single and double bonds are different lengths. C-C 0.154 nm C=C 0.134 nm

Note: "nm" means "nanometre", which is 10-9 metre.

That would mean that the hexagon would be irregular if it had the Kekulé structure, with alternating shorter and longer sides. In real benzene all the bonds are exactly the same - intermediate in length between C-C and C=C at 0.139 nm. Real benzene is a perfectly regular hexagon.

Our goal is to �measure� the heat of hydrogenation of this imaginary structure and compare it to the actual heat of hydrogenation of benzene (which can be measured). The idea here is to determine which structure yields the most energy upon saturation of its double bonds with hydrogen. The structure that actually gives up more energy during this process is the one with higher chemical potential energy and therefore less chemical stability.

cyclohexeneSince 1,3,5-cyclohexatriene does not actually exist, a model must to chosen to approximate its heat of hydrogenation. The model we will select will be the substance cyclohexene, C6H10, which is a six carbon ring with one set of double bonds across which the hydrogenation will occur. A schematic picture of a cyclohexene molecule is shown here. In order to approximate the hydrogenation of the three sets of double bonds found in 1,3,5-cyclohexatriene, we will perform the hydrogenation on three separate molecules of cyclohexene and assume that the approximate heat of hydrogenation of 1,3,5-cylochexatriene is equal to three times the heat of hydrogenation of cyclohexene. These assumptions are reasonable.

The measured heat of hydrogenation of benzene (total saturation of all carbons to form cyclohexane) is approximately 50 kilocalories per mole. The heat of hydrogenation of a single cyclohexene molecule is approximately 29 kilocalories per mole. Using the proposed model, the total heat of hydrogenation of the imaginary compound 1,3,5-cylohexatriene is equal to: (3 moles C6H10 ) X (29 kcal/mole) = 87 kcal.

So the heat of hydrogenation of our imaginary model compound is 87-kcal/mole, while the actual measured heat of hydrogenation of benzene is only 50-kcal/mole. The resonance energy can now be quantified by observing the difference in the heat of hydrogenation of the model compound compared to the real benzene molecule:

ΔHH2 Resonance = ΔHH2 �1,3,5-cyclohexatriene� – ΔHH2Benzene = �Resonance energy�

= 87 kcal/mole � 50 kcal/mole = 37 kcal/mole (A significant amount of energy!)

The difference between the ΔHH2 of the two compounds is approximately 37-kcal/mole. This significant energy is the resonance energy and represents the actual increase in chemical stability realized by the benzene molecule. So what does all this have to do with graphite? A single graphene layer is made up of millions of �benzene� ring structures and each of these is resonance stabilized. That means the resonance energy is common to the entire graphite structure. Graphite owes part or most of its �low chemical potential energy� to the fact that it is a pure sp2 carbon, resonance stabilized substance.

You are wrong once again about 1, 3, 5 -hexatriene. Sorry that the facts upset you.

This is my last response to this question.

Dr. Romano

if you read correctly i put cyclohexatriene so stop, just stop. Its in Janice's book.

also, im not trying to argue. you're trying to be over technical for a simple question and answer.