Question

A gas formed by the action of alcoholic KOH on ethyl iodide, decolorizes alkaline KMnO₄ solution. The gas is

• A. CH₄
• B. C₂H₆
• C. C₂H₄
• D. C₂H₂

Answer 1

The correct option is C

C₂H₄

We know that,

Alkyl halide o reaction with alcoholic KOH or NaOH gives alkene.

$R-C{H}_2-C{H}_2-X+KOH\left(alc\right)⟶R-CH=C{H}_2+KX+H_{2}O$

And

Hydroxylation using $KMn{O}_4:1\%$ cold alkaline $KMn{O}_4$ solution (Baeyer's reagent) or $dil.KMn{O}_4$ gives glycol

Note.

(a) It is an unsaturation test because the pink colour of solution disappears.

(b) It is a example of Syn-addition.

Therefore,

$C{H}_{3}C{H}_{2}l+KOH\left(alc\right)⟶C{H}_2=C{H}_2+Kl+H_{2}O$