A compound A with molecular formula C_{10}H_{ | vedclass

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(C) Compound $A$ $(C_{10}H_{13}Cl)$ gives a white precipitate with silver nitrate,indicating it is a tertiary halide or a reactive alkyl halide that f

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$C_6H_5-CH_2-C(Cl)(CH_3)_2$

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(C) Compound $A$ $(C_{10}H_{13}Cl)$ gives a white precipitate with silver nitrate,indicating it is a tertiary halide or a reactive alkyl halide that forms a stable carbocation.$C_6H_5-CH_2-C(Cl)(CH_3)_2 (A) \xrightarrow{alc. KOH} C_6H_5-CH=C(CH_3)_2 (B)$$B$ on ozonolysis:$C_6H_5-CH=C(CH_3)_2 \xrightarrow{O_3/H_2O} C_6H_5CHO (C) + CH_3COCH_3 (D)$$C$ $(C_6H_5CHO)$ has no $\alpha$-hydrogen,so it gives Cannizzaro reaction but not aldol condensation.$D$ $(CH_3COCH_3)$ has $\alpha$-hydrogens,so it gives aldol condensation but not Cannizzaro reaction.Thus,$A$ is $C_6H_5-CH_2-C(Cl)(CH_3)_2$.

List-$I$	List-$II$
$A$. $C_6H_6 + CO + HCl \xrightarrow{AlCl_3} C_6H_5CHO$	$a$. Gattermann-Koch reaction
$B$. $C_6H_5CH_3 \xrightarrow{CrO_2Cl_2} C_6H_5CHO$	$b$. Etard reaction
$C$. $C_6H_5CH_2Br + CH_3Br \xrightarrow{Na/ether} C_6H_5CH_2CH_3$	$c$. Wurtz-Fittig reaction
$D$. $C_6H_6 + (CH_3)_2C = CH_2 \xrightarrow{H_2SO_4} C_6H_5C(CH_3)_3$	$d$. Friedel-Crafts alkylation

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