Correct option(s) for the following sequence of reactions is(are)

Let the starting compound (given in the question’s flow-chart) be benzyl chloride, $$Ph-CH_2Cl$$. The sequence shown in the paper is (only the reagents were hidden):

$$Ph-CH_2Cl \xrightarrow{\,Q\,} S \xrightarrow{\,W\,} R \xrightarrow{\,V\,} T$$

We have to decide what the hidden reagents $$Q,\,W,\,V$$ and the intermediates $$R,\,S$$ are, only one of the four option-sets being completely correct.

Step 1 : Nature of product obtained from an alkyl chloride and nitrite reagents
For an alkyl (or benzyl) halide $$RX$$

• With potassium (or sodium) nitrite, $$KNO_2 / NaNO_2$$, the major product is the alkyl nitrite $$R-ONO$$.
• With silver nitrite, $$AgNO_2$$, the major product is the nitroalkane $$R-NO_2$$.

This is a standard JEE fact that arises from the difference in the ambident nucleophile $$NO_2^-$$ and the way silver promotes $$N$$-attack.

Step 2 : Which of the two - $$R-ONO$$ or $$R-NO_2$$ - is needed for the next step?
The next hidden reagent is $$W$$ and, from the given options, the only plausible reducing agent supplied is $$LiAlH_4$$. $$LiAlH_4$$ converts a nitro compound $$R-NO_2$$ to a primary amine $$R-NH_2$$, but it does not convert an alkyl nitrite ($$R-ONO$$) to an amine in one step.

Therefore the intermediate just before $$LiAlH_4$$ must be $$Ph-CH_2NO_2$$, and the required reagent $$Q$$ must be $$AgNO_2$$.

Step 3 : Product after $$LiAlH_4$$ reduction
Reduction of benzyl nitro compound gives benzylamine (phenyl­methanamine):
$$Ph-CH_2NO_2 \xrightarrow{\,LiAlH_4\,} Ph-CH_2NH_2$$

So $$R$$ has to be phenylmethanamine.

Step 4 : Checking the four answer-sets

Option A $$Q = KNO_2,\, W = LiAlH_4$$: KNO2 would give the nitrite $$Ph-CH_2ONO$$, which $$LiAlH_4$$ does not reduce to an amine → wrong.

Option B $$R =$$ benzenamine (aniline): the reduction product is benzylamine, not aniline → wrong.

Option C $$Q = AgNO_2,\, R =$$ phenyl­methanamine: exactly matches the deductions above → correct.

Option D $$W = LiAlH_4,\, V = AgCN$$: while $$W = LiAlH_4$$ is correct, $$AgCN$$ cannot convert a benzylamine to a nitrile in a single step (the Sandmeyer reaction applies to aryl diazonium salts, not to benzylamine) → wrong.

Hence the only correct set is

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