A compound on dehydrohalogenation with alcoholic KOH gives alkyne but on dehalogenation with zinc dust gives alkene. The compound is:

C_{2} H_{5} B r

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C H_{3} C H B r_{2}

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C H_{2} B r - C H_{2} B r

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C H B r_{2} - C H B r_{2}

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Solution

The correct option is C $C H_{2} B r - C H_{2} B r$
Reaction with Zn
$C H_{2} B r - C H_{2} B r + Z n \to C H_{2} = C H_{2} + Z n B r_{2}$
Reaction with alc. KOH
$B r - H_{2} C - C H_{2} - B r A l c o h o l i c K O H - -------- \to 1, 2 - d i b r o m o e t h a n e (v i c - d i h a l i d e)$
$H_{2} C = C H - B r N a N H_{2} - ---- \to C H \equiv C H + N a B r + N H_{3}$

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A compound on dehydrohalogenation with alcoholic KOH gives alkyne but on dehalogenation with zinc dust gives alkene. The compound is:

The correct option is C CH2Br−CH2BrReaction with ZnCH2Br−CH2Br+Zn→CH2=CH2+ZnBr2Reaction with alc. KOHBr−H2C−CH2−BrAlcoholicKOH−−−−−−−−−→1,2−dibromo ethane(vic−dihalide)H2C=CH−BrNaNH2−−−−−→CH≡CH+NaBr+NH3