L xrightarrow[(ii) H_3O^{oplus}]{(i) PhMgBr} | vedclass

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(A) The final product is $Ph_2C=CH_2$ ($1$, $1$-diphenylethene). The last step is a Wittig reaction: $N + Ph_3P=CH_2 \rightarrow Ph_2C=CH_2$. This imp

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Benzaldehyde

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(A) The final product is $Ph_2C=CH_2$ ($1$, $1$-diphenylethene). The last step is a Wittig reaction: $N + Ph_3P=CH_2 \rightarrow Ph_2C=CH_2$. This implies $N$ must be benzophenone $(Ph_2C=O)$. The step $M \xrightarrow{CrO_3 / H^{\oplus}} N$ is an oxidation of a secondary alcohol to a ketone. Thus, $M$ is diphenylmethanol $(Ph_2CHOH)$. The first step is the Grignard reaction: $L \xrightarrow[(ii) H_3O^{\oplus}]{(i) PhMgBr} Ph_2CHOH$. Benzaldehyde $(PhCHO)$ reacts with $PhMgBr$ followed by hydrolysis to give diphenylmethanol $(Ph_2CHOH)$. Therefore, $L$ is benzaldehyde.

List-$I$ (Reagents)	List-$II$ (Reaction Name)
$A$. $H_2, Pd-BaSO_4$	$I$. Etard Reaction
$B$. $SnCl_2, HCl$	$II$. Rosenmund Reduction
$C$. $CrO_2Cl_2, CS_2$	$III$. Gatterman-Koch Reaction
$D$. $CO, HCl, \text{Anhyd. } AlCl_3$	$IV$. Stephen Reaction

$L$ $\xrightarrow[(ii) H_3O^{\oplus}]{(i) PhMgBr} M$ $\xrightarrow{CrO_3 / H^{\oplus}} N$ $\xrightarrow{Ph_3P=CH_2} Ph_2C=CH_2$
$L$ in the above sequence of reaction is/are (where $L \neq M \neq N$)

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$L$ $\xrightarrow[(ii) H_3O^{\oplus}]{(i) PhMgBr} M$ $\xrightarrow{CrO_3 / H^{\oplus}} N$ $\xrightarrow{Ph_3P=CH_2} Ph_2C=CH_2$ $L$ in the above sequence of reaction is/are (where $L \neq M \neq N$)