Question:
What are X and Y respectively in the following set of reactions?
2-Bromobutane $\xrightarrow{{alc. KOH}}$ C$_4$H$_8$ (major)
C$_4$H$_8$ $\xrightarrow{{KMnO}_4/{H}^+}$ X
C$_4$H$_8$ $\xrightarrow{{Baeyer's reagent}}$ Y
What are X and Y respectively in the following set of reactions?
2-Bromobutane $\xrightarrow{{alc. KOH}}$ C$_4$H$_8$ (major)
C$_4$H$_8$ $\xrightarrow{{KMnO}_4/{H}^+}$ X
C$_4$H$_8$ $\xrightarrow{{Baeyer's reagent}}$ Y
2-Bromobutane $\xrightarrow{{alc. KOH}}$ C$_4$H$_8$ (major)
C$_4$H$_8$ $\xrightarrow{{KMnO}_4/{H}^+}$ X
C$_4$H$_8$ $\xrightarrow{{Baeyer's reagent}}$ Y
Show Hint
KMnO$_4$/H$^+$ oxidatively cleaves alkenes into carboxylic acids or ketones. Baeyer's reagent gives glycols.
Updated On: Jun 3, 2025
- CH3COOH, HO-CH(CH3)-CH2OH
- CH3CH2COOH + CO2, HO-CH2-CH2CH2OH
- CH$_3$COOH, CH$_3$CHO
- CH$_3$CH$_2$COOH + CO$_2$, CH$_3$CHO
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The Correct Option is A
Solution and Explanation
2-Bromobutane on treatment with alc. KOH gives but-2-ene. But-2-ene undergoes oxidative cleavage with KMnO$_4$/H$^+$ to give acetic acid (CH$_3$COOH). Baeyer's reagent reacts with the double bond to give a vicinal diol as the product.
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