Identify Reactions Yielding Aromatic Compound as Major Product

To determine which of the given reactions produce an aromatic compound as the major product, we analyze each option individually:

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Option (A)

1. First & Second Steps (Dehydrohalogenation): Treatment of $1,2$-dibromopropane ($\text{CH}_3\text{-CH(Br)-CH}_2\text{Br}$) with alcoholic $\text{KOH}$ eliminates one molecule of $\text{HBr}$ to form 2-bromopropene. Subsequent reaction with a stronger base, sodium amide ($\text{NaNH}_2$), eliminates a second molecule of $\text{HBr}$ to give propyne: $\text{CH}_3-\text{CH(Br)}-\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH}} \text{CH}_3-\text{C(Br)}=\text{CH}_2 \xrightarrow{\text{NaNH}_2} \text{CH}_3-\text{C}\equiv\text{CH}$

2. Third Step (Cyclotrimerization): When three molecules of propyne are passed through a red-hot iron tube at $873\text{ K}$, they undergo cyclic trimerization to form $1,3,5$-trimethylbenzene (mesitylene): $3\,\text{CH}_3-\text{C}\equiv\text{CH} \xrightarrow{\text{Red hot Fe tube, } 873\text{ K}} \text{1,3,5-trimethylbenzene (Mesitylene)}$

Since mesitylene is a substituted benzene derivative containing a planar, fully conjugated cyclic ring with $6\pi$ delocalized electrons, it is aromatic. Thus, Option (A) is correct.

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Option (B)

1. Cyclopentadiene has a pair of acidic hydrogen atoms at the $sp^3$ hybridized $-\text{CH}_2-$ position ($pK_a \approx 16$).
2. Reaction with the base sodium methoxide ($\text{NaOMe}$) results in the deprotonation of cyclopentadiene: $\text{Cyclopentadiene} + \text{NaOMe} \longrightarrow \text{Cyclopentadienyl anion } (\text{C}_5\text{H}_5^-) + \text{MeOH}$
3. The cyclopentadienyl anion is cyclic, planar, completely conjugated, and contains $6\pi$ electrons ($4n+2$ where $n=1$). According to Hückel's rule, it is aromatic. Thus, Option (B) is correct.

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Option (C)

1. Treatment of 3-bromocyclobutene with sodium ethoxide ($\text{NaOEt}$) leads to elimination to give cyclobutadiene or substitution products.
2. Cyclobutadiene is a cyclic, planar molecule with $4\pi$ electrons ($4n$ electrons where $n=1$), which makes it anti-aromatic and extremely unstable.
3. Neither the elimination nor substitution product yields an aromatic compound. Thus, Option (C) is incorrect.

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Option (D)

1. In the presence of ultraviolet light ($\text{UV}$) at $500\text{ K}$, benzene undergoes a free-radical addition reaction with excess chlorine ($\text{Cl}_2$): $\text{C}_6\text{H}_6 + 3\text{Cl}_2 \xrightarrow{\text{UV, } 500\text{ K}} \text{C}_6\text{H}_6\text{Cl}_6$
2. The product formed is $1,2,3,4,5,6$-hexachlorocyclohexane (Benzene Hexachloride / BHC).
3. This molecule contains only $sp^3$ hybridized carbons with no delocalized $\pi$-electrons, making it non-aromatic. Thus, Option (D) is incorrect.

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Conclusion

The correct options that produce an aromatic compound as the major product are (A) and (B).