Halogenation of which one of the following wi | vedclass

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Question

(C) In electrophilic aromatic substitution,the incoming electrophile attacks the position that is ortho or para to the most strongly activating group

Vedclass · Accepted Answer

$2-$methylphenol

Vedclass · Answer

(C) In electrophilic aromatic substitution,the incoming electrophile attacks the position that is ortho or para to the most strongly activating group (the group with the strongest electron-releasing effect).For $2$-methylphenol ($o$-cresol),the $-OH$ group is a much stronger activating group than the $-CH_3$ group.The $-OH$ group directs the electrophile to its ortho and para positions.The para position with respect to the $-OH$ group is the meta position with respect to the $-CH_3$ group.Therefore,halogenation of $2$-methylphenol yields the $m$-substituted product with respect to the methyl group as the major product.

List-$I$ (Compound)	List-$II$ $(pK_a)$
$A$. $p-$Nitrophenol	$I$. $7.1$
$B$. Phenol	$II$. $10.0$
$C$. Ethanol	$III$. $15.9$
$D$. $p-$Cresol	$IV$. $10.2$

Halogenation of which one of the following will yield $m$-substituted product with respect to the methyl group as a major product?

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What will be the product of the reaction shown below?

Phenol $\xrightarrow[\Delta ]{Zn} A$ $\xrightarrow[\text{Anhydrous } AlCl_3]{CH_3Cl} B$ $\xrightarrow[\Delta ]{KMnO_4/OH^{-}} C$
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Match the following:
List-$I$ (Compound) List-$II$ $(pK_a)$
$A$. $p-$Nitrophenol $I$. $7.1$
$B$. Phenol $II$. $10.0$
$C$. Ethanol $III$. $15.9$
$D$. $p-$Cresol $IV$. $10.2$
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$C_6H_5OH \xrightarrow[(ii) NaOH, (iii) H^+]{(i) CHCl_3, NaOH(aq)} ?$

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