The correct match between Item-I (starting ma | vedclass

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(C) The correct matches are as follows:$1$. Benzene $(I)$ reacts with $CO, HCl$ and $AlCl_3$ (Gattermann-Koch reaction) to form benzaldehyde. Thus,$I-

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$I-R, II-P$ and $III-Q$

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(C) The correct matches are as follows:$1$. Benzene $(I)$ reacts with $CO, HCl$ and $AlCl_3$ (Gattermann-Koch reaction) to form benzaldehyde. Thus,$I-R$.$2$. Benzonitrile $(II)$ reacts with $SnCl_2, HCl$ followed by $H_3O^{+}$ (Stephen reduction) to form benzaldehyde. Thus,$II-P$.$3$. Benzoyl chloride $(III)$ reacts with $H_2, Pd-BaSO_4, S$ and quinoline (Rosenmund reduction) to form benzaldehyde. Thus,$III-Q$.Therefore,the correct match is $I-R, II-P$ and $III-Q$.

Item-$I$	Item-$II$
$I$. Benzene	$P$. $HCl$ and $SnCl_2, H_3O^{+}$
$II$. Benzonitrile	$Q$. $H_2, Pd-BaSO_4, S$ and quinoline
$III$. Benzoyl Chloride	$R$. $CO, HCl$ and $AlCl_3$

The correct match between Item-$I$ (starting material) and Item-$II$ (reagent) for the preparation of benzaldehyde is :
Item-$I$ Item-$II$
$I$. Benzene $P$. $HCl$ and $SnCl_2, H_3O^{+}$
$II$. Benzonitrile $Q$. $H_2, Pd-BaSO_4, S$ and quinoline
$III$. Benzoyl Chloride $R$. $CO, HCl$ and $AlCl_3$

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The correct match between Item-$I$ (starting material) and Item-$II$ (reagent) for the preparation of benzaldehyde is :Item-$I$Item-$II$$I$. Benzene$P$. $HCl$ and $SnCl_2, H_3O^{+}$$II$. Benzonitrile$Q$. $H_2, Pd-BaSO_4, S$ and quinoline$III$. Benzoyl Chloride$R$. $CO, HCl$ and $AlCl_3$