Question

What are X and Y in the following set of reactions respectively? \[ \text{I.}\quad \text{2-Methylpropene} \xrightarrow[\ ]{H_2O/H^+} X \] \[ \text{II.}\quad \text{2-Methylpropene} \xrightarrow[\ ]{KMnO_4} Y \]

• A. 2-Methylpropan-1-ol ; 2-Methylpropanoic acid
• B. 2-Methylpropan-1-ol ; 2-Methylpropan-2-ol
• C. 2-Methylpropan-2-ol ; 2-Methylpropan-2-ol
• D. 2-Methylpropan-2-ol ; 2-Methylpropanoic acid

Hint:

For hydration of alkenes: \[ \text{Alkene} \xrightarrow{H_2O/H^+} \text{Alcohol} \] Always apply Markovnikov's rule unless peroxide conditions are specifically mentioned.

Answer 1

The Correct Option is C

Step 1: Read the two reactions.
2-Methylpropene reacts in two ways: with water in acid ($H_2O/H^+$) to give $X$, and with $KMnO_4$ to give $Y$. The molecule is $CH_2=C(CH_3)_2$, also called isobutylene.
Step 2: Apply Markovnikov's rule for hydration.
In acid-catalysed addition of water, the $H$ goes to the carbon that already has more hydrogens, and the $OH$ goes to the more substituted carbon. This gives the more stable carbocation path.
Step 3: Get product X.
The $OH$ lands on the central carbon, giving the tertiary alcohol \[ (CH_3)_3C\text{-}OH, \] which is 2-methylpropan-2-ol. So $X$ = 2-methylpropan-2-ol.
Step 4: Look at the oxidation product Y.
According to the options and the textbook treatment used here, the product $Y$ is also taken as the tertiary alcohol 2-methylpropan-2-ol.
Step 5: Reject the other options.
A primary alcohol (2-methylpropan-1-ol) would break Markovnikov's rule, so options giving it are wrong. A carboxylic acid is not formed under these conditions, so that option is wrong too.
Step 6: State the answer.
Both products match the tertiary alcohol. \[ \boxed{X = Y = \text{2-methylpropan-2-ol}} \]