Why electrophile substitution in C6H5OCH3 occurs at ortho and para?

Answer 1

Because there is an extra resonance structure that is possible through donation of the lone pair of electrons of oxygen to the benzene ring (leaving a C=O+CH3 form), making the ortho and para center succeptable to electrophillic attack. With meta attack this resonance structure is not possible.

Answer 2

Because of the interaction of the unshared electron pairs of the oxygen with the aromatic ring, there are resonance forms which will have a negative charge in the ortho and para positions. The oxygen atom is ring-activating, and is an ortho-para director. Substituents which can donate electrons into the ring are activators for electrophilic substitution and are ortho-para directing. Substituents which withdraw electrons from the ring deactivate the ring for electrophilic substitution, but are meta directors.

Answer 3

It occurs at ortho and para because OCH3 group is an activating, electron releasing group.

According to Markovnikov's rule, more abundant products are formed by pathways which go through lower energy intermediates.

If you draw the resonance structures you'll see that in the intermediates leading to the ortho and para products, one of the contributing structures is tertiary.

Electrons from the oxygen can directly stabilize the electron deficient carbon.

Lower energy intermediates = faster reaction = more product forms.

That is why it is ortho/para directing.

Answer 4

so we are talking electrophillic substitution on methoxy benzene.
u have to figure, whether the substituent donates or subtracts electron density from the aromatic ring. in our case oxygen has an unshared electron pair, meaning it is willing to donate it to the aromatic ring. If u now draw the aromatinc ring in the old fashioned way with conjugated double bonds, and try to push the electron pair of O along the bond connecting it to the Ph, forming thus a double bond there, u can see, that in order to avoid a 5-valent carbon, one of the adjacent double bonds in the Ph ring will have to break, leaving a negative charge at o-position. well electrophiles looooooooooove negative charges, so that´s where ur E+ will go.
if u carry on with this little arrow/charge/double-bond-shift game, u will see, that u also can get a negative charge at the p-position, but never at meta. that´s why the attack is always o- or p-.

Answer 5

If you draw the resonance structures of C6H5OCH3 (anisole) ie delocalize an electron pair on the oxygen down into the ring you will see that the electron density will reside on the ortho and para positions.

Answer 6

We need more information. Is the CH3 and O in either the ortho or para position? Most of the time, the electrophile subsitution will occur adjacent to the O group. So where is it?