I have explained how hyperconjugation leads to the directing properties of alkyl groups below.

Carbon joining substituent now need only single bond. So if we remove one bond between $\ce{H-C}$ along with electron from hydrogen we have a free $\ce{H+}$ ion. The bond which was removed formed a bond with carbon joining the substituent(second picture in second row). From the those two electron (on the left of the second picture in second row) we see one come from electron joining substituent(from hydrogen) and other from the carbon atom at ortho position.

How did one bond between Hydrogen and Carbon move to $\ce{C-C}$ along with one electron from hydrogen?

Hyperconjugation in Toluene

Now we see the electron from hydrogen jumping from ortho to para position to methyl substituent. How does this jump actually happen?

Is it that the electron( from hydrogen of $\ce{CH3}$) on Carbon at ortho jump(Tell me in detail how this jump occur) to meta position due to formation of double bond between the carbon at ortho and meta position. Due to the double bond, electron( from hydrogen of $\ce{CH3}$) on meta came to para position and so on.

How does the double bond between the carbon at ortho and meta position send electron to para position? The explanation given above is what I think. Is it right? In my book the following was written:

Hyperconjugation involves delocalization of $\sigma$-electron through the overlap of $\pi$ orbital of double bond with $\sigma$ orbital of the adjacent single bond.

Please explain the above extract from book using the above example of hyperconjugation of toluene.


2 Answers 2


Why methyl group is 2,4-directing? This question describes in detail why the methyl group is o,p directing.

However, I think the major point of your misunderstanding is in the interpretation of resonance structures. Resonance structures are not stable structures in their own right. A set of resonance structures can be used to represent bonding that cannot be represented using classic Lewis structures. The structures do not exist on their own but the true structure can be viewed as an 'average' of all the contributing resonance structures.

The electrons do not jump between different atoms but are instead delocalized over all of the atoms with a slightly higher chance of being found at the ortho and para positions in the case of toluene and a slightly lower chance of being found at on the methyl carbon. This gives rise to the increased electron density of the ortho and para positions and therefore the increased negative charge and similarly the decreased electron density on the methyl carbon and the positive charge.


The electrons don't "jump" from one position to another:

  1. when the $\ce{H+}$ leaves the methyl group, a full p-orbital is formed on the carbon. this p orbital is on the same plane of the pi bond of the aromatic ring.

  2. when the p-orbital is on the same plane, the electrons can delocalize (move) along the new formed conjugated pi molecular orbital.

  3. so, you get 8 electrons (2 from the methyl group + 6 from the aromatic ring) on 7 carbons all on the same oriented p-orbital (or just $\pi$ bond). that means you have a total of -1 charge over this bond, so where it will be more localized? (or on which carbons there will be a negative charge?)

  4. the picture you posted are resonance structures (they describe the delocalized electrons). There are 3 resonance structures: 1. minus on the carbon of the methyl group 2. minus on orto carbon 3. minus on para carbon The electron "moves" from each structure by moving the double bond. the "minus on the carbon of the methyl group" will contribute to the overall structure the least, because its a primary carbon-anion.

  • 3
    $\begingroup$ @user13186 When drawing resonance structures, you can only move electrons, you cannot move nuclei. When you write, "when the H+ leaves the methyl group, a full p-orbital is formed on the carbon", this would change the H-C-H angle from around 109 to around 120 degrees - you have moved nucelei which is not allowed. $\endgroup$
    – ron
    Feb 4, 2015 at 14:16
  • $\begingroup$ @ron the contributing structures for the resonance exist only after the H+ leaves. thus, the carbon (now carbo-anion) in the methyl group is already in SP2 hybridization meaning it has a near 120 degree angle. this form is stable in this hybridization because it has more stable resonance structures where the methyl group is bonded by double bond with the 6-ring. $\endgroup$
    – user13186
    Feb 5, 2015 at 15:38
  • 1
    $\begingroup$ @user13186 That's not my understanding. The H+ does nor "leave". It stays in its exact position, only the electrons in the C-H bond move and spend more of their time around the ortho and para positions in the aromatic ring. The methyl carbon remains sp3 hybridized. The "double bond" drawn between the aromatic carbon and the methyl carbon involves p-sp3 pi-like overlap. $\endgroup$
    – ron
    Feb 5, 2015 at 16:09

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