MCQ
$C{H_2} = C{H_2} \xrightarrow[KOH / H_2o ]{KMnO_4} \mathop {}\limits_{} X$ . Product ‘$X$’ in above reaction is
- ✓Ethylene glycol
- BGlucose
- CEthanol
- DAll of these
✓
Answer
Correct option: A.
Ethylene glycol
a
$CH _2= CH _2 \underset{ KOH / H _2 O }{\stackrel{ KMnO _4}{\longrightarrow}} HO - CH _2- CH _2- OH + MnO _2$
$CH _2= CH _2 \underset{ KOH / H _2 O }{\stackrel{ KMnO _4}{\longrightarrow}} HO - CH _2- CH _2- OH + MnO _2$
Ethenes react with acidified potassium permanganate. Potassium
Permanganate is a strong oxidant, and will initially convert the double bond to two alcohol ( $OH )$ groups.
Ethene $+$ Acidified Potassium Permanganate $\rightarrow$ Ethan - $1,2$ - diol(Ethylene -
glycol)The purple colour of permanganate will fade as the reaction proceeds.
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