What is the product of the Friedel-Crafts rea | vedclass

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(D) The Friedel-Crafts alkylation of benzene with chloroform $(CHCl_3)$ in the presence of anhydrous $AlCl_3$ proceeds through the formation of a carb

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All of the above

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(D) The Friedel-Crafts alkylation of benzene with chloroform $(CHCl_3)$ in the presence of anhydrous $AlCl_3$ proceeds through the formation of a carbocation intermediate. Initially,benzene reacts with $CHCl_3$ to form benzal chloride $(C_6H_5CHCl_2)$. This intermediate further reacts with more benzene molecules via electrophilic substitution to form triphenylmethane $((C_6H_5)_3CH)$. Since the reaction continues until all chlorine atoms are replaced by phenyl groups,the final major product is triphenylmethane,but the reaction mixture contains intermediate products like $(C_6H_5)_2CHCl$ and $C_6H_5CHCl_2$ depending on the stoichiometry. Thus,all the given options represent products formed during the reaction sequence.

Column-$I$ (Reaction)	Column-$II$ (Mechanism)
$(A)$ $OH^- + CH_3Cl \to HOCH_3 + Cl^-$	$(1)$ $S_N2$ mechanism
$(B)$ $(CH_3)_3CBr + OH^- \to (CH_3)_3COH + Br^-$	$(2)$ $S_E2$ aromatic
$(C)$ $CH_3CH_2Br \xrightarrow[ethanol]{KOH} CH_2 = CH_2$	$(3)$ $S_N1$ mechanism
$(D)$ $C_6H_6 + Br_2 \xrightarrow{Fe} C_6H_5Br + HBr$	$(4)$ $\beta$-elimination (dehydrohalogenation)

List-$I$ (Reaction)	List-$II$ (Major product)
$A. \ CH_3CH=CH_2 \xrightarrow{HI}$	$I. \ 1\text{-bromo-}2\text{-phenylethane}$
$B. \ C_6H_5CH_2CH_3 \xrightarrow{Br_2/\Delta}$	$II. \ 1\text{-iodopropane}$
$C. \ C_6H_5CH=CH_2 \xrightarrow{HBr}$	$III. \ 2\text{-iodopropane}$
	$IV. \ 1\text{-bromo-}1\text{-phenylethane}$

What is the product of the Friedel-Crafts reaction of benzene with chloroform?

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