COMEDK · Chemistry · 26. Alcohols Phenols and Ethers
An aliphatic compound [X], Molecular formula (\(\text{C}_4\text{H}_{10}\text{O}\)) can be prepared from Acetone and R– Mg–X. [X] with 20% Phosphoric acid gives an unsaturated compound \(\text{C}_4\text{H}_8\) [Y]. Compound [Y] is also obtained on heating [X] with Copper metal at 573K. [X] shows no reaction with PCC but on heating with acidified \(\text{KMnO}_4\) it forms Acetone + \(\text{CO}_2\) + \(\text{H}_2\text{O}\). Compounds [X] and [Y] are -------------------- and --------------------
- A [X]: 2–Methylpropan–2–ol [Y]: 2–Methylpropene
- B [X]: 2–Methylpropan–1–ol [Y]: But–1–ene
- C [X]: Butan–1–ol [Y]: But–1–ene
- D [X]: Butan–2–ol [Y]: But–2–ene
Answer & Solution
Correct Answer
(A) [X]: 2–Methylpropan–2–ol [Y]: 2–Methylpropene
Step-by-step Solution
Detailed explanation
Acetone (\(\text{CH}_3\text{COCH}_3\)) reacts with methylmagnesium halide (\(\text{CH}_3\text{MgX}\)) followed by hydrolysis to form a tertiary alcohol, 2-methylpropan-2-ol. \(\text{CH}_3\text{COCH}_3 + \text{CH}_3\text{MgX} \rightarrow (\text{CH}_3)_3\text{C-OH}\) Thus,…
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