1. Target molecule identification

We need to prepare 4-bromo-2-nitroethyl benzene from benzene.

This is a benzene ring having:

• an ethyl group
• a bromo group para to ethyl
• a nitro group ortho to ethyl

So relative to the ethyl group, the product is:

• $Br$ at para position
• $NO_2$ at ortho position

2. How to introduce the ethyl group?

Direct Friedel–Crafts alkylation with ethyl chloride is possible, but among the given options the route uses:

$\text{Benzene} \xrightarrow{CH_3COCl/AlCl_3} \text{acetophenone} \xrightarrow{Zn-Hg/HCl} \text{ethylbenzene}$

because Clemmensen reduction converts

$C_6H_5COCH_3 \to C_6H_5CH_2CH_3$

So first acylation, then reduction gives ethylbenzene.

3. Now decide order of bromination and nitration

From ethylbenzene:

• ethyl group is activating and ortho/para directing.

If we do bromination first:

$\text{ethylbenzene} \xrightarrow{Br_2/AlBr_3} \text{mainly } p\text{-bromoethylbenzene}$

Para product is favored due to steric reasons.

Then nitration of $p$-bromoethylbenzene:

• ethyl is ortho/para directing, activating
• bromine is ortho/para directing, deactivating

In the para-bromoethylbenzene structure, the positions favored by both directors are the 2 and 6 positions, which are equivalent. Thus nitration gives:

$\text{4-bromo-2-nitroethyl benzene}$

This matches the target.

4. Check Option B

Option B sequence:

$CH_3COCl/AlCl_3 \rightarrow Zn-Hg/HCl \rightarrow Br_2/AlBr_3 \rightarrow HNO_3/H_2SO_4$

Stepwise:

1. Benzene $\xrightarrow{CH_3COCl/AlCl_3}$ acetophenone
2. Acetophenone $\xrightarrow{Zn-Hg/HCl}$ ethylbenzene
3. Ethylbenzene $\xrightarrow{Br_2/AlBr_3}$ mainly $p$-bromoethylbenzene
4. Nitration gives 4-bromo-2-nitroethyl benzene

So Option B is correct.

5. Why other options are incorrect

Option A

$CH_3COCl/AlCl_3,\ Br_2/AlBr_3,\ HNO_3/H_2SO_4,\ Zn/HCl$

After Friedel–Crafts acylation, the ring has $-COCH_3$, which is meta directing and deactivating. So bromination/nitration would not lead to the required orientation relative to ethyl after reduction. Also $Zn/HCl$ is not the Clemmensen reduction condition for converting aryl ketone efficiently to ethylbenzene here.

So A is incorrect.

Option C

$Br_2/AlBr_3,\ CH_3COCl/AlCl_3,\ HNO_3/H_2SO_4,\ Zn/HCl$

Starting with bromobenzene, Friedel–Crafts acylation is possible but gives orientation directed by bromine; then nitration occurs in the presence of $-COCH_3$, a meta director. This does not give the desired final substitution pattern reliably. Also final $Zn/HCl$ is not the right reduction sequence here.

So C is incorrect.

Option D

$HNO_3/H_2SO_4,\ Br_2/AlCl_3,\ CH_3COCl/AlCl_3,\ Zn-Hg/HCl$

Nitrobenzene is strongly deactivating and does not undergo Friedel–Crafts acylation. Hence this route fails.

So D is incorrect.

6. Final answer

The correct sequence is:

$\boxed{CH_3COCl/AlCl_3 \;\to\; Zn-Hg/HCl \;\to\; Br_2/AlBr_3 \;\to\; HNO_3/H_2SO_4}$

So the correct option is B.