To choose the correct reagent sequence, we identify what overall transformation is most likely being asked in this standard nitrogen-compounds conversion.

The options strongly suggest that compound $X$ is a nitro aromatic compound and product $Y$ is obtained through reactions involving:

• reduction of $-\mathrm{NO_2}$ to $-\mathrm{NH_2}$,
• bromination with bromine water,
• diazotisation with $\mathrm{HNO_2}$,
• and finally replacement of the diazonium group.

We now analyze each option systematically.

1. Key reactions involved

(a) Reduction of nitro group

A nitro group on benzene is reduced by $\mathrm{Fe/H^+}$ to give aniline:

(b) Bromination of aniline with bromine water

Aniline is a very strong activating group and with bromine water gives tribromoaniline:

(c) Diazotisation

A primary aromatic amine reacts with nitrous acid to form diazonium salt:

(d) Reduction of diazonium salt by hypophosphorous acid

$\mathrm{H_3PO_2}$ replaces the diazonium group by hydrogen:

This is a standard way to remove an amino group after using it as a directing/activating handle.

2. Examine each option

Option A

• $Br_2/Fe$ brominates benzene/nitrobenzene type systems.
• $Fe/H^+$ reduces $-NO_2$ to $-NH_2$.
• $LiAlH_4$ is not the standard reagent to convert the resulting aromatic amine into the likely final product here.

This sequence does not naturally lead to the common aromatic substitution/removal pattern implied by the options.

So A is incorrect.

Option B

• $Br_2(aq)$ reacts readily with aniline/phenol, not typically the starting nitro compound.
• $LiAlH_4$ followed by $H_3O^+$ can reduce nitro compounds, but this does not explain selective bromination and final product formation as elegantly as the diazotisation route.

So B is incorrect.

Option C

Let us trace it step by step:

1. Reduction

   $$\mathrm{ArNO_2} \xrightarrow{Fe,H^+} \mathrm{ArNH_2}$$

2. Bromination with bromine water Since $-NH_2$ is strongly activating, bromination occurs at $o,p$ positions, generally giving:

   $$\mathrm{ArNH_2} \xrightarrow{Br_2(aq)} 2,4,6\text{-tribromoaniline}$$

3. Diazotisation

   $$\mathrm{ArNH_2} \xrightarrow{HNO_2} \mathrm{ArN_2^+}$$

   So tribromoaniline becomes the corresponding tribromobenzenediazonium salt.

4. Replacement by hydrogen using $\mathrm{H_3PO_2}$

   $$\mathrm{ArN_2^+} \xrightarrow{H_3PO_2} \mathrm{ArH}$$

   Hence the amino group is removed, leaving the brominated benzene framework.

This is a very standard synthetic route to prepare brominated benzene from nitrobenzene via aniline as an intermediate.

So C is correct.

Option D

The first three steps are same as option C, but the last step is different.

• $CuBr$ in Sandmeyer reaction replaces diazonium group by Br, not by H.
• Therefore, this would introduce an additional bromine at the diazonium position.

So if the intended final product is obtained by removal of amino group, this option is not correct.

Hence D is incorrect.

3. Final answer

The correct sequence is:

Therefore, the correct option is:

4. Comparison with stored correct answer

Stored correct answer: $C$

My derived answer: $C$

So, the derived answer agrees with the stored correct answer.