In a reaction of {C_6H_5Y},the major product | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The reaction involves electrophilic aromatic substitution on a substituted benzene ring ${C_6H_5Y}$.Groups that are electron-withdrawing by resona

Vedclass · Accepted Answer

$-COOH$

Vedclass · Answer

(A) The reaction involves electrophilic aromatic substitution on a substituted benzene ring ${C_6H_5Y}$.Groups that are electron-withdrawing by resonance or induction (deactivating groups) are meta-directing.Among the given options,$-COOH$ is an electron-withdrawing group and acts as a meta-directing group.Conversely,$-NH_2$,$-OH$,and $-Cl$ are electron-donating groups (by resonance) and act as ortho/para-directing groups.Therefore,the correct option is $A$.

In a reaction of ${C_6H_5Y}$,the major product $(> 60\%)$ is the $m$-isomer. Therefore,the group $Y$ is:

Similar Questions

Which of the following are aromatic?

Product $(B)$ in this reaction is

The order of reactivity of the following compounds towards electrophilic substitution will be - [where $\phi = C_6H_5$]:
$I. \phi-CH_3$
$II. \phi-CH_2-CH_3$
$III. \phi-CH(CH_3)_2$
$IV. \phi-C(CH_3)_3$

What type of difference is observed in the hybridization of carbon in benzene during nitration and the formation of the $\sigma$-complex?

Acylation process is preferred over direct alkylation in the Friedel-Crafts reaction because:

Vedclass Test Series

Exam Paper Generator

Online Exam Module