The reagent(s) for the following conversion i | vedclass

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(B) The conversion of $1,2-dibromoethane$ to ethyne involves dehydrohalogenation.First,treatment with alcoholic $KOH$ removes one molecule of $HBr$ to

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alcoholic $KOH$ followed by $NaNH_2$

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(B) The conversion of $1,2-dibromoethane$ to ethyne involves dehydrohalogenation.First,treatment with alcoholic $KOH$ removes one molecule of $HBr$ to form vinyl bromide $(CH_2=CHBr)$.Second,because the $C-Br$ bond in vinyl bromide has partial double bond character due to resonance,it is less reactive towards elimination. Therefore,a stronger base like $NaNH_2$ (sodamide) is required to remove the second molecule of $HBr$ to yield ethyne $(HC \equiv CH)$.Thus,the correct sequence is alcoholic $KOH$ followed by $NaNH_2$.

The reagent$(s)$ for the following conversion is/are:
$Br-CH_2-CH_2-Br \xrightarrow{?} HC \equiv CH$

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The reagent$(s)$ for the following conversion is/are:$Br-CH_2-CH_2-Br \xrightarrow{?} HC \equiv CH$