The correct option(s) for the following seque | vedclass

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(B) $1$. The reaction sequence starts with $PhCH_3$ (toluene) reacting with $Br_2/light$ to form $P$ $(PhCH_2Br)$.$2$. $P$ $(PhCH_2Br)$ reacts with $A

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$C, D$

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(B) $1$. The reaction sequence starts with $PhCH_3$ (toluene) reacting with $Br_2/light$ to form $P$ $(PhCH_2Br)$.$2$. $P$ $(PhCH_2Br)$ reacts with $AgNO_2$ followed by $H_2, Pd/C$ to give $R$ ($PhCH_2NH_2$,phenylmethanamine). Thus,$Q = AgNO_2$.$3$. $R$ $(PhCH_2NH_2)$ undergoes carbylamine reaction with $CHCl_3/KOH$ to form a foul-smelling isocyanide $(PhCH_2NC)$.$4$. $PhCH_3$ is oxidized to $T$ $(PhCOOH)$,which reacts with $NH_3$ and heat to form $U$ $(PhCONH_2)$.$5$. $U$ $(PhCONH_2)$ is reduced by $W$ $(LiAlH_4)$ to form $R$ $(PhCH_2NH_2)$.$6$. $P$ $(PhCH_2Br)$ reacts with $V$ $(AgCN)$ to form $PhCH_2NC$.$7$. Comparing with the options: $Q = AgNO_2$,$R =$ phenylmethanamine,$W = LiAlH_4$,$V = AgCN$. Therefore,options $(C)$ and $(D)$ are correct.

The correct option$(s)$ for the following sequence of reactions is(are):
$(A)$ $Q = KNO_2, W = LiAlH_4$
$(B)$ $R =$ benzenamine,$V = KCN$
$(C)$ $Q = AgNO_2, R =$ phenylmethanamine
$(D)$ $W = LiAlH_4, V = AgCN$

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The correct option$(s)$ for the following sequence of reactions is(are):$(A)$ $Q = KNO_2, W = LiAlH_4$$(B)$ $R =$ benzenamine,$V = KCN$$(C)$ $Q = AgNO_2, R =$ phenylmethanamine$(D)$ $W = LiAlH_4, V = AgCN$