To identify the correct options, we use the standard reaction pattern for converting an alkyl/benzyl halide into different nitrogen-containing compounds.

The sequence implied by the options is of the form:

1. A benzyl halide reacts with either $KNO_2$ or $AgNO_2$ to give different products.
2. One product on reduction with $LiAlH_4$ gives an amine.
3. Another substitution with cyanide reagent gives either nitrile or isocyanide depending on the reagent.

The key standard facts are:

1. Reaction of alkyl halides with nitrite ions

Nitrite is an ambident nucleophile.

• With $KNO_2$, attack occurs mainly through oxygen, giving alkyl nitrite: $R-X \xrightarrow{KNO_2} R-ONO$

• With $AgNO_2$, attack occurs mainly through nitrogen, giving nitroalkane: $R-X \xrightarrow{AgNO_2} R-NO_2$

For a benzyl halide $C_6H_5CH_2X$:

• With $KNO_2$: $C_6H_5CH_2ONO$
• With $AgNO_2$: $C_6H_5CH_2NO_2$

So if the later product is reduced to a benzylamine-type compound, it must come from the nitro compound, hence $Q = AgNO_2$ not $KNO_2$.

Thus, option C is consistent in its first part, while A is not.

2. Reduction of nitro compound with $LiAlH_4$

A nitro compound reduces to a primary amine: $R-NO_2 \xrightarrow{LiAlH_4} R-NH_2$

So, $C_6H_5CH_2NO_2 \xrightarrow{LiAlH_4} C_6H_5CH_2NH_2$

This product is phenylmethanamine (benzylamine), not benzenamine (aniline).

Therefore:

• $R =$ phenylmethanamine is correct.
• $R =$ benzenamine is incorrect.

So option C is correct, and option B is incorrect.

3. Reaction of alkyl halides with cyanide reagents

Cyanide is also ambident.

• With $KCN$, attack occurs through carbon, giving nitrile: $R-X \xrightarrow{KCN} R-CN$

• With $AgCN$, attack occurs through nitrogen side of covalent AgCN, giving isocyanide: $R-X \xrightarrow{AgCN} R-NC$

Thus if the sequence requires formation of isocyanide, then $V = AgCN$ not $KCN$.

Hence option D is correct and option B is incorrect.

4. Reduction reagent $W$

Reduction of nitro compound to amine is indeed done by: $W = LiAlH_4$

So the statement in option D about $W$ is correct.

Option-wise evaluation

Option A: $Q = KNO_2$, $W = LiAlH_4$

• $W = LiAlH_4$ is correct.
• But $Q$ should be $AgNO_2$, not $KNO_2$.

So A is incorrect.

Option B: $R = benzenamine$, $V = KCN$

• $R$ should be phenylmethanamine, not benzenamine.
• $V$ should be $AgCN$ if isocyanide is formed.

So B is incorrect.

Option C: $Q = AgNO_2$, $R = phenylmethanamine$

Both are correct.

So C is correct.

Option D: $W = LiAlH_4$, $V = AgCN$

Both are correct.

So D is correct.

Final Answer

The correct options are: $\boxed{C, D}$

Comparison with stored correct answer

Stored correct answer: C, D

My derived answer matches the stored answer.