How does hyperconjugation lead to the directing properties of alkyl group?

Question

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?

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.

Answer 1

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.

Answer 2

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.