Question

what is the action of alcoholic KOH on iodoethane​

Answer 1

Explanation:

In presence of alcoholic KOHKOH, ethyl iodide undergoes dehydrohalogenation, to form ethylene \displaystyle C_2H_4C

2

H

4

. Ethylene decolurizes alkaline \displaystyle KMnO_4KMnO

4

.

Alkaline \displaystyle KMnO_4KMnO

4

is called Baeyer's reagent and is reduced to \displaystyle MnO_2MnO

2

by ethylene.

\displaystyle C_2H_5-I \xrightarrow [Dehydrohalogenation]{alc. KOH} C_2H_4C

2

H

5

−I

alc.KOH

Dehydrohalogenation

C

2

H

4

\displaystyle 2KMnO_4 + H_2O + KI \rightarrow 2MnO_2 + 2KOH + KIO_32KMnO

4

+H

2

O+KI→2MnO

2

+2KOH+KIO

3

\displaystyle CH_2=CH_2 + H_2O + [O] \rightarrow CH_2-OH-CH_2-OHCH

2

=CH

2

+H

2

O+[O]→CH

2

−OH−CH

2

−OH

Answer 2

Explanation:

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