Complete the given reaction of alkene with `KMnO_(4)` in different conditions
(i) `underset("2-Methylbut-1-ene")(CH_(3)-CH_(2)-underset(CH_(3))underset(|)C=CH_(2)) underset(273 K)overset("Alkaline" KMnO_(4))to ?`
(i) `underset("2-Methylbut-1-ene")(CH_(3)-CH_(2)-underset(CH_(3))underset(|)C=CH_(2)) underset(Delta)overset("Acidified" KMnO_(4))to ?`

Explanation

To complete the reactions of 2-Methylbut-1-ene with KMnO4 under different conditions, we will analyze each scenario step by step.

### Step 1: Reaction with Alkaline KMnO4 at 273 K
1. **Identify the Alkene**: The given alkene is 2-Methylbut-1-ene, which has the structure CH3-CH2-CH(CH3)=CH2.
2. **Understand the Reagent**: Alkaline KMnO4 (also known as Baeyer’s reagent) is used for the syn-dihydroxylation of alkenes. It adds hydroxyl groups (OH) across the double bond.
3. **Mechanism of Reaction**: The reaction leads to the formation of a cis-diol. This means that both hydroxyl groups will be added to the same side of the double bond.
4. **Draw the Product**: The double bond between C1 and C2 will be converted to a single bond, and two hydroxyl groups will be added to the same side:
- The product will be 2-Methylbutane-2,3-diol, represented as:
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