Identify each reagent in List II

• (I) $CH_3MgBr, H_2O$: Grignard reagent, adds a methyl group to aldehydes/ketones, giving alcohols after hydrolysis.
• (II) $Zn(Hg)$ and conc. $HCl$: Clemmensen reduction, reduces carbonyl group $>C=O$ to $-CH_2-$.
• (III) $NaBH_4, H^+$: Reduces aldehydes/ketones to alcohols.
• (IV) $DIBAL-H, H_2O$: Reduces suitable acid derivatives (like nitriles/esters under controlled conditions) to aldehydes.

Match reaction (C)

$C_6H_5CHO \rightarrow C_6H_5CH(OH)CH_3$

Here benzaldehyde is converted to a secondary alcohol with one extra methyl group added.

This is exactly the action of Grignard reagent $CH_3MgBr$ followed by hydrolysis.

So, $(C) \to (I)$

Match reaction (B)

$C_6H_5COC_6H_5 \rightarrow C_6H_5CH_2C_6H_5$

The printed product appears as $C_6H_5CH=C_6H_5$, but from the reagent set and standard transformations, benzophenone is reduced to diphenylmethane by Clemmensen reduction:

$C_6H_5COC_6H_5 \xrightarrow[conc.\ HCl]{Zn(Hg)} C_6H_5CH_2C_6H_5$

Thus, $(B) \to (II)$

Now eliminate options using (B) and (C)

We need:

• $(B) \to (II)$
• $(C) \to (I)$

Among the options, only Option C satisfies both.

Check consistency of remaining matches in Option C

Option C gives:

• $(A) \to (IV)$
• $(D) \to (III)$

This is chemically consistent with the usual pairing:

• one reaction corresponding to selective formation of an aldehyde by DIBAL-H,
• another corresponding to reduction of aldehyde/ketone to alcohol by $NaBH_4$.

Final answer

$\boxed{\text{Option C}}$