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Does the cycle in the benzene ring represents double bonds' movement between the carbon atoms or free electrons movement?

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Welcome to Chemistry! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using LATEX syntax. If you receive useful answers, consider accepting one. @IvanNeretin You should write that up into an answer. – bon Feb 25 at 17:48
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The circle, if anything, represents the inability of our everyday physical intuition to cope with the quantum phenomena. See, you would often encounter those two pictures with "double bonds this way" and "double bonds that way", intended to give a vague impression that the molecule switches quickly between the two, but that's not true.

Kekulé benzene

It does not switch; it just stays there, in some sense "halfway between" these structures. There is no movement of double bonds, and in fact there are no double bonds. (Also, there is no movement of electrons either. They just sit there on their molecular orbitals, which by itself is an approximation, though a pretty decent one). All $\ce{C-C}$ bonds in benzene are the same; they are neither double nor ordinary. Read the Wikipedia page on aromaticity, or just look at this picture from it:

"Real" benzene

That's what the circle stands for.

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I found the teaching of this concept is tied far too tightly to the representational aspect. It shouldn't be taught as a 'double bond' where everything exists in a solid state but an electron probability distribution, which, in fact, is what it is. – hownowbrowncow Feb 25 at 21:36
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I read the aromaticity, and didn't anderstand something "instead of being tied to one atom of carbon, each electron is shared by all six in the ring" the free electron shared with all of them or just with the electrons on his side? – studentrr Feb 26 at 7:53
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To begin with, don't call that electron free, because it is not. As for your question, the right answer is closer to the first option. Each electron is delocalized over both sides of the ring. – Ivan Neretin Feb 26 at 8:03
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so the electrons just delocalized on both sides, and when all electrons doing so, it looks like the bonds is moving back and forth? – studentrr Feb 26 at 8:12
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To me, it does not look like that at all. A matter of habit, I guess. Yes, the electrons are just delocalized on both sides, but nothing is moving. – Ivan Neretin Feb 26 at 8:20

This is a typical example of the downfall of the Lewis structures.

$\ce{C_6H_6}$ can be represented by more than one Lewis structure:

Dewar structures:

enter image description here

Kekule structures:

enter image description here

Does anyone of them actually represent benzene?

No!

If it were so, then some of the $\textrm{C-C}$ bonds would be smaller than other same bonds.

But, in reality, it doesn't happen so.


There are two approaches to solve this problem:

  • Resonance theory of Valance Bond Theory

  • Molecular Orbital Theory


A bit intro:

An electron is described by the wavefunction $\psi$ where $$\psi(x)\equiv \langle x|\psi\rangle= \textrm{probability amplitude of finding the electron at} \, x$$ ; the state $|\psi\rangle$ is defined as $$|\psi\rangle = \int_{\textrm{all}\, x} |x\rangle\langle x|\psi\rangle \,\mathrm dx\;.$$

Probability of finding the electron in $x\pm \mathrm dx$ is given by $$\textrm{prob}(x,\mathrm dx)= |\psi(x)|^2\;\mathrm dx$$

Also, from the First Principles of Quantum Mechanics:

When an event can occur in several alternative ways, the probability amplitude for the event is the sum of the probability amplitudes for each way considered separately.$^\dagger$


Resonance theory/ Mesomeric effect:

Consider Hydrochloric acid. How can it be represented by Lewis structure?

$$\ce{H-Cl}\,\,\,\textrm{or}\,\,\, \ce{H^+Cl^-}$$

The former is described by the wavefunction $$\psi_{\ce{H-Cl}}(1,2)= \psi_{\ce{H}}(1)\psi_{\ce{Cl}}(2) + \psi_{\ce{H}}(2)\psi_{\ce{Cl}}(1)$$ and the later by $$\psi_{\ce{H^+Cl^-}}(1,2)= \psi_{\ce {Cl}}(1)\psi_{\ce {Cl}}(2)$$

However, we know neither $\ce{HCl}$ is ionic, nor it is covalent.

To describe the wavefunction of $\ce{HCl}$, quantum superposition is applied viz. $$\psi_{\ce{HCl}}= \psi_{\ce{H^+ Cl^-}}C_1 + \psi_{\ce{H-Cl}}C_2$$ where $C_1$ is the probability-amplitude to find the electron-pair at $|\psi_{\ce{H^+ Cl^-}}\rangle$ provided it was prepared in $|\psi_{\ce{HCl}}\rangle$ that is, $C_1 =\langle \psi_{\ce{H^+ Cl^-}}|\psi_{\ce{HCl}}\rangle;$ also $C_2 = \langle \psi_{\ce{H-Cl}}|\psi_{\ce{HCl}}\rangle$ and $C_1^2 + C_2^2= 1.$

A better description of the wavefunction for the molecule is a superposition of the covalent & ionic descriptions, & we write (with a slightly simplified notation) $$\psi= \psi_{\ce{H_Cl}} + \lambda\psi_{\ce{H^+ Cl^-}}$$ with $\lambda$ some numerical coefficient. In general, we write $$\psi = \psi_\text{covalent} + \lambda \psi_\text{ionic}$$ [...]According to the general rules of quantum mechanics, in which probabilities are related to squares of wavefunctions, we interpret the square of $\lambda$ as the relative proportion of the ionic contribution. If $\lambda^2$ is very small, the covalent description is dominant. If $\lambda^2$ is very large, the ionic description is dominant.$^{\ddagger}$

Resonance is nothing but quantum 'superposition of the wavefunctions representing different electron distributions in the same nuclear framework.'

In the same way, we can describe benzene as the following (un-normalised) wavefunction :

$$\psi = c_1\psi_{\text{Kek}1} + c_2\psi_{\text{Kek}2} + c_3\psi_{\textrm{Dew}1} + c_4\psi_{\textrm{Dew}2} + c_5\psi_{\textrm{Dew}3}$$

[It has been estimated that each Kekule canonical form has a relative contribution of about $0.408$ while the Dewar structures each has a relative contribution of $0.177\;.$ Thus the weight of two Kekule structures each is $0.39$ ie. they contribute to about $39\%$ to the resonance while each Dewar structure has relative weight is $0.073$ ie. they contribute to $7.3\%$ to the resonance hybrid. And thus many times, Dewar structures aren't taken into consideration in creating the resonance hybrid of benzene.]

Remember, the wavefunction on the left is real and all in the right are fictitious.

Variational Principle is used to bring down the expected energy of the wavefunction or orbital.

Resonance thus explains the delocalisation of electrons in unhybridised $p$ orbitals that underwent $\pi $ bonding throughout the whole molecule and thus the circle which represents the delocalisation of $\pi$ bond over the whole molecule.


Molecular Orbital Theory:

Molecular orbitals are one-electron wavefunctions which are formed due to linear combination of atomic or hybrid orbitals. Unlike in VB theory where bonding occurs due to the coupling between atomic orbitals of the participating atoms and thus remain localised between those two atoms, in MO theory, all atomic orbitals apart from those two atoms participate in the combination and thus the MOs are spread over all the nuclei of the molecules.

Delocalisation is thus inherent in the approach of Molecular Orbital Theory. The bonding influence of a single electron pair is spread over the entire molecule and not just the adjacent participating atoms of the molecule.

Now, let's come to the case of benzene.

All the unhybridised $p$ orbitals of $\ce C$ atoms in $\ce{C_6H_6}$ combine to form MO. Since, there are six such orbitals, six MOs would be formed.

The (un-normalised) wavefunction thus can be written as:

$$\psi= c_1\psi_1 + c_2\psi_2 + c_3\psi_3 + c_4\psi_4 + c_5\psi_5 + c_6\psi_6$$

Three bonding MOs and three anti-bonding MOs are formed; the former being the less energetic due to the non-occurrence of internuclear nodes.

enter image description here

There are two pairs of degenerate orbitals;one is lower in energy while the other is in higher level. The electrons occupy the MOs of lower energy level.

Each MO is spreads either all round or partially round the $\ce{C_6}$ ring and this delocalises the $\pi$ bonding. Thus each electron pair helps in binding together several of the $\ce C$ atoms.

Conclusion:

Due to delocalisation, all $\ce{C-C}$ bonds get extra stability which is due to the lowering of energy levels of the electrons. While VB theory explains with imaginary structures, MO theory explains it with the inherent delocalisation in its definition.

But the crux of the point is, the bonded electron pairs never remain confined or localised between two participating atoms; it spreads over the whole molecule; it delocalises.

The circle in that structure you mentioned represents this delocalisation of $\pi$ bond over the entire ring.


References:

$^\dagger$ The Feynman Lectures of Physics Vol. III by Richard P. Feynman, Matthew Sands, Robert B. Leighton.

$^\ddagger$ Elements of Physical Chemistry by Peter Atkins and Julio de Paula.

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Thanks for the explanation, what happens when two rings having these pi bonds above and below them faced close to each other,what type of bonding occur? – studentrr Feb 27 at 8:06
    
@studentrr: huh? Can you elaborate your query? – MAFIA36790 Feb 27 at 8:15
    
@user36790 I believe (s)he is referring to $\pi$-stacking – NewDogOldTricks Feb 27 at 9:17
    
Yes I think it's called pi stacking, but didn't understand how it happen and how strong is it? – studentrr Feb 27 at 10:32
    
@studentrr: You ought ask it as a separate new question here as many people can then view it and may help you. – MAFIA36790 Feb 27 at 10:48

Whilst the circle drawn on a benzene ring is done to represent the resonance stabilised delocalised pi orbital electrons as comprehensively explained by the previous author, there is another more practical reason why a circle is drawn in this manner to represent carbon bonding in benzene. If you consider the case of cyclohextriene [and consider the position double bonds in cyclohextriene occupy] apart from the major fact the 6 carbon atoms in these two compounds are very differently bonded, when they are written in reactions both would actually appear to be identical compounds, [which of course they definately are not]. So benzene's structure is also written this way to differentiate between these two compounds otherwise this situation would cause much confusion in writen chemical reactions separately involving both benzene and cyclohextriene!

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Thanks for the explanation, one question, Why the electrons are stabilised in the ring? – studentrr Feb 27 at 15:01
    
@studentrr: Electrons are stabilised because now they have more space to move about thus lowering the kinetic energy without actually increasing the potential energy . Electrons always want to lower their Coulomb potential energy without making high kinetic energy by confining in a smaller place which can be attributed to Uncertainty Principle. – MAFIA36790 Feb 27 at 20:31
    
Thanks a lot for explaining on energy level, So you mean that electron consume energy when it moves to each atom, and so lowers it's kinetic energy? but didn't understand a bout potential energy? – studentrr Feb 27 at 22:29
    
@studentrr: Okay! I don't encourage long comment session on this. You can come at Chem Chatroom: chat.stackexchange.com/rooms/3229/the-periodic-table where you can make any query and hopefully might get an intuitive chat from the users; so do come there; it is not the right place to make comments after comments. – MAFIA36790 Feb 28 at 2:56
    
@studentrr: Also, in order to ask a specific user, you've to ping him by writing before your comment @userxxxxx. In order to make a call to me, ping me by writing @user36790 before the comment; otherwise I may not get informed about your comment. I could see your above query when I visited the page; had I not seen it; I might never replied! So, do ping the users you want to talk with. – MAFIA36790 Feb 28 at 3:00

The lines present the bonds which are at alternate positions. It basically represent bonds As there are many different structures like electron dot structure and many more.

Benzene has a formula C6H6. Benzene is built from 6 hydrogen atoms ($1s^1$) and 6 carbon atoms ($1s^2\;2s^2\; 2p_x^1\;2p_y^1$).

Each carbon atom has to join to three other atoms (one hydrogen and two carbons) and doesn't have enough unpaired electrons to form the required number of bonds, so it needs to promote one of the $2s^2$ pair into the empty $2p_z$ orbital.

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Do you mean that 2pz orbital will have the free electron to move in it up and down? – studentrr Feb 26 at 8:02
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Why did you write three separate answers? – Nicolau Saker Neto Feb 26 at 8:11
    
Sorry for that first I gave the small info – user153334 Feb 26 at 8:13
    
Cause I want to ask for explanation for that answer specifically – studentrr Feb 26 at 8:18
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You should've edited one of your answers instead. – TIPS Feb 26 at 9:13

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