The balance chemical equations for the following :
Reaction between 1, 2 - dibromoethane & alcoholic potassium hydroxide.

C H_{2} B r - C H_{2} B r a l c . K O H - ---- \to C H_{3} - C H_{2} B r N a N H_{2} - ---- \to C H \equiv C H

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C H_{2} B r - C H_{2} B r a l c . K O H - ---- \to C H_{2} B r - C H B r N a N H_{2} - ---- \to C H \equiv C H

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C H_{2} B r - C H_{2} B r a l c . K O H - ---- \to C H_{2} = C H B r N a N H_{2} - ---- \to C H \equiv C H

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C H_{3} - C H_{2} B r N a N H_{2} - ---- \to C H_{2} = C H_{2}

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Solution

The correct option is C $C H_{2} B r - C H_{2} B r a l c . K O H - ---- \to C H_{2} = C H B r N a N H_{2} - ---- \to C H \equiv C H$
Reaction:
$C H_{2} B r - C H_{2} B r a l c . K O H - ---- \to C H_{2} = C H B r N a N H_{2} - ---- \to C H \equiv C H$ ethyne
The given reaction is used to form alkyne from alkyl dihalide. It uses reducing agents like KOH and sodamide. It first undergoes $β$ elimination reaction to form vinyl halide which further undergoes dehydrohalogenation to form alkyne.

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Similar questions

{C H}_{3} {C H}_{2} {C H}_{2} B r

{C H}_{3} {C H}_{2} {C H}_{2} {C H}_{2} B r

{({C H}_{3})}_{2} C H {C H}_{2} B r

{({C H}_{3})}_{3} C B r

β

{C H}_{3} {C H}_{2} {C H}_{2} B r (i)

{C H}_{3} - C H {- C H}_{2} B r | {C H}_{3} (i i)

{C H}_{3} - {C H}_{2} - C H - {C H}_{3} | B r (i i i)

(C_{3} H_{4})

B r_{2} / C C l_{4}

C H_{3} - C | C H_{3} = C H_{2} H B r - ---- \to p e r o x i d e C H_{3} - C | C H_{3} H - C H_{2} B r

C H_{3} - C H = C H - C H_{2} B r

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