Which order is correct about electrophilic substitution reaction involved in benzene ring?

Therefore, the correct order of reactivity towards electrophilic substitution is Phenol > benzene > chlorobenzene > benzoic acid.

At what position and on what ring is electrophilic aromatic chlorination of phenyl benzoate expected to occur explain your answer?

At what position, and on what ring, is bromination of phenyl benzoate expected to occur? Explain your answer. ANSWER: Attack occurs in the activated ring and yields ortho and para bromination.

Which position is susceptible for electrophilic substitution in anthracene?

With electrophiles Electrophilic substitution of anthracene occurs at the 9 position.

Which one of the following compounds will be most easily attacked by an electrophile?

Phenol is most easily attacked by an electrophile because presence of -OH group increases electron density at o- and p-positions.

On which position on the ring furan undergoes electrophilic substitution reaction?

Pyrrole, thiophene, and furan gives electrophilic aromatic substitution reaction. These compounds are more reactive compared to benzene. Electrophiles majorly attack on 2nd position rather than 3rd position in these heterocyclic compounds.

What are the ortho meta and para positions on a benzene ring?

They are defined as the following: ortho- (o-): 1,2- (next to each other in a benzene ring) meta- (m): 1,3- (separated by one carbon in a benzene ring) para- (p): 1,4- (across from each other in a benzene ring)

What are aromatic compounds define ortho meta para position of benzene?

Disubstituted benzene rings can be named based on the relative positions of the substituents: the prefix ortho– is used if the substituents occupy adjacent positions on the ring (1,2), meta– is used if the substituents are separated by one ring position (1,3), and para– if they are found on opposite sides of the ring ( …

Which of the following is the most activating in electrophilic aromatic substitution?

Interestingly, fluorine is the most activating of the halogens. The reason is likely that the overlap of the lone pair in the fluorine 2p orbital with the p orbital on carbon is much better (resulting in a stronger pi-bond) than is donation with the 3p (and higher) p orbitals of chlorine, bromine, and iodine.