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(C) Friedel-Crafts alkylation is an electrophilic aromatic substitution reaction. It occurs readily in benzene rings that are activated by electron-do

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$2$ and $4$

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(C) Friedel-Crafts alkylation is an electrophilic aromatic substitution reaction. It occurs readily in benzene rings that are activated by electron-donating groups (e.g.,$-OH$,$-CH_2CH_3$).$1.$ $p$-Nitrotoluene: The $-NO_2$ group is strongly electron-withdrawing,deactivating the ring.$2.$ Ethylbenzene: The $-CH_2CH_3$ group is electron-donating,activating the ring.$3.$ Benzoic acid: The $-COOH$ group is strongly electron-withdrawing,deactivating the ring.$4.$ Phenol: The $-OH$ group is strongly electron-donating,activating the ring.Therefore,compounds $2$ and $4$ will undergo Friedel-Crafts alkylation.

Column-$I$ (Reactants)	Column-$II$ (Products)
$A$. Cumene $+ O_2$	$i$. Phenol
$B$. Benzene sulphonic acid $\xrightarrow{NaOH, H^+}$	$ii$. $2$-Methylbutan-$1$-ol
$C$. Chlorobenzene $+ NaOH \xrightarrow{623 \ K, 300 \ atm}$	$iii$. Cumene hydroperoxide
$D$. $2$-Methylbutanal $\xrightarrow{NaBH_4}$	$iv$. Sodium phenoxide

Which of the following compounds will undergo a Friedel-Crafts alkylation reaction?

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