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Which of the following is the deactivating in electrophilic aromatic substitution?

Certain functional groups can accept, rather than donate, a pi bond from the ring, resulting in a new lone pair on a substituent atom. Examples are NO2, carbonyl groups (C=O), sulfonyl, cyano (CN) among others. These groups are universally deactivating, slowing the rate of electrophilic aromatic substitution.

What type of reaction intermediate is generated in the mechanism of electrophilic aromatic substitution reactions?

1. A Mechanism for Electrophilic Substitution Reactions of Benzene. A two-step mechanism has been proposed for these electrophilic substitution reactions. In the first, slow or rate-determining, step the electrophile forms a sigma-bond to the benzene ring, generating a positively charged benzenonium intermediate.

Which of the following is most reactive towards electrophilic substitution reaction?


Which of the following is more reactive than benzene towards electrophilic substitution reaction?


Which of the following is the most reactive towards aromatic electrophilic substitution reaction using Hyperconjugation?

Due to +M-effect of -OH group and hyperconjugation of – CH3 group, the benzene of o-cresol is highly reactive ring towards electrophilic substitution.

Which of the following is the most reactive towards electrophilic aromatic substitution for halogen?

hence, phenol is more readily attacked by an electrophile. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Why Phenol is more reactive than toluene?

Note that the methyl group in toluene increases electron density only by hyperconjugation and inductive effect. Resonance effects are generally far more superior to hyperconjugation. Hence, without doubt, phenol is more activated towards EAS as compared to toluene.

Which of the following is most reactive towards aqueous Naoh?

Solution : Aromatic halides are less reactive than alkyl or aralkyl halides. Hence, C6H5CH2Cl is the most reactive towards aq.