An alkyl halide with molecular formula \(C_5H_{11}Br\) on dehydrohalogenation give two isomeric alkenes \(X\) and \(Y\) with formula \(C_5H_{10}\). On reductive ozonolysis \(X\) and \(Y\) gave four compounds \(CH_3CHO\), \(CH_3-CO-CH_3\), \(CH_3CH_2CH)\), \(HCHO\). The alkyl halide is:

2-Bromo-3-methylbutane

The correct answer is option 2. 2-Bromo-3-methylbutane.

An alkyl halide with molecular formula \(C_5H_{11}Br\) undergoes dehydrohalogenation to give two isomeric alkenes, \(X\) and \(Y\), with the formula \(C_5H_{10}\). This implies that the alkyl halide is a pentyl halide (halogen attached to a five-carbon chain).

On reductive ozonolysis, \(X\) and \(Y\) give four compounds: \(CH_3CHO\) (acetaldehyde), \(CH_3-CO-CH_3\) (acetone), \(CH_3CH_2CHO\) (propanal), and \(HCHO\) (formaldehyde). This implies that \(X\) and \(Y\) are alkenes that can be cleaved into these products.

Given these clues, let's analyze the options:

1. 3-Bromopentane: Upon dehydrohalogenation, it can give either 2-pentene or 1-pentene. On reductive ozonolysis, neither of these alkenes would give the specified products.

2. 2-Bromo-3-methylbutane: Dehydrohalogenation can give 2-methyl-2-butene or 2-methyl-1-butene. These alkenes could potentially give the specified products upon reductive ozonolysis.

3. 2-Bromo-2-methylbutane: Dehydrohalogenation can give 2-methyl-2-butene. This alkene might give some of the specified products, but it wouldn't give acetone upon ozonolysis.

4. 1-Bromo-2,2-dimethylpropane: Dehydrohalogenation can give 2,2-dimethylpropene. This alkene wouldn't give the specified products upon reductive ozonolysis.

Considering the given information, the alkyl halide most likely is 2-Bromo-3-methylbutane, making the answer (2).