When C_6H_5Z undergoes electrophilic substitu | vedclass

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Question

(A) In electrophilic aromatic substitution,the orientation of the incoming group depends on the nature of the substituent already present on the benze

Vedclass · Accepted Answer

$-COOH$

Vedclass · Answer

(A) In electrophilic aromatic substitution,the orientation of the incoming group depends on the nature of the substituent already present on the benzene ring.Groups that are electron-withdrawing by inductive or resonance effects (deactivating groups) are meta-directing.Among the given options,$-COOH$ is an electron-withdrawing group due to the presence of the carbonyl group attached to the benzene ring,which makes it a meta-directing group.$-Cl$ is ortho/para-directing due to the resonance effect.$-NH_2$ and $-CH_3$ are ortho/para-directing due to their electron-donating nature.Therefore,the correct answer is $-COOH$.

When $C_6H_5Z$ undergoes electrophilic substitution,the major product obtained is the meta-isomer. Therefore,the group $Z$ is:

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Identify $A$ and $B$ in the following reaction:
$CH_3-CH_3 \stackrel{B}{\longleftarrow} CH_3COOH \stackrel{A}{\longrightarrow} CH_3CH_2OH$
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The major product formed in the following reaction is
$C_6H_5CH=CH_2$ $\xrightarrow[(ii) H_3O^+]{(i) KMnO_4, KOH, \Delta}$ $\xrightarrow{(iii) Br_2/FeBr_3} \text{Product}$

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