The reactant $\text{CH}_3\text{CH}_2\text{CHCl}_2$ (1,1-dichloropropane) is a geminal dihalide. When treated with a strong base like sodamide ($\text{NaNH}_2$), it undergoes double dehydrohalogenation ($\beta$-elimination). The removal of two molecules of hydrogen chloride ($\text{HCl}$) from adjacent carbon atoms leads to the formation of a carbon-carbon triple bond. Thus, the final product is an alkyne, specifically propyne ($\text{CH}_3-\text{C}\equiv\text{CH}$) . (Note: As the options were missing in the raw data, they have been reconstructed based on the standard AIPMT 2002 question).