Hydrocarbons Part 2

Concept:Vicinal dibromides react with excess $\mathrm{NaNH_2}$ to undergo double dehydrohalogenation, forming alkynes.Hydration of terminal alkynes using $\mathrm{HgSO_4}$ / dil $\mathrm{H_2SO_4}$ yields methyl ketones ($\mathrm{RCOCH_3}$), which give a positive iodoform test but a negative Tollens test.Explanation:Compound $\mathrm{P}$ has the molecular formula $\mathrm{C_9H_{10}Br_2}$.Excess sodamide eliminates two molecules of $\mathrm{HBr}$ from the vicinal dibromide, producing an alkyne $\mathrm{Q}$.Upon adding dilute $\mathrm{HCl}$, the alkyne is protonated and obtained as the neutral compound $\mathrm{Q}$.Treatment of $\mathrm{Q}$ with $\mathrm{HgSO_4}$ and dilute $\mathrm{H_2SO_4}$ adds water across the triple bond (Markovnikov addition).For a terminal alkyne ($\mathrm{RC\equiv CH}$), the product is a methyl ketone ($\mathrm{RCOCH_3}$).This methyl ketone gives a positive iodoform test (due to the $\mathrm{CH_3CO}$ group) but a negative Tollens test (it is not an aldehyde).Therefore, $\mathrm{Q}$ must be a terminal alkyne, and $\mathrm{P}$ must be a vicinal dibromide that can eliminate to form $\mathrm{C_6H_5CH_2C\equiv CH}$ (benzylacetylene).Among the given options, only the structure $\mathrm{C_6H_5CH_2CHBrCH_2Br}$ (option B) has two bromine atoms