Q34 / 180 · COMEDK 2026 (9 May Shift 1) · Chemistry
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 --------------------
- Correct answerA[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
Solution
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, compound [X] is 2-methylpropan-2-ol, which has the molecular formula $\text{C}_4\text{H}_{10}\text{O}$.
Dehydration of [X] with 20% $\text{H}_3\text{PO}_4$ or heating with Cu at 573 K yields 2-methylpropene [Y].
$(\text{CH}_3)_3\text{C-OH} \rightarrow (\text{CH}_3)_2\text{C=CH}_2 + \text{H}_2\text{O}$
Tertiary alcohols do not undergo oxidation with PCC, hence [X] shows no reaction.
Vigorous oxidation of [X] with acidified $\text{KMnO}_4$ involves dehydration to [Y] followed by oxidative cleavage of the double bond to form acetone, $\text{CO}_2$, and $\text{H}_2\text{O}$.
$(\text{CH}_3)_2\text{C=CH}_2 \rightarrow \text{CH}_3\text{COCH}_3 + \text{CO}_2 + \text{H}_2\text{O}$
Therefore, [X] is 2-methylpropan-2-ol and [Y] is 2-methylpropene.
Answer: [X]: 2–Methylpropan–2–ol [Y]: 2–Methylpropene
$\text{CH}_3\text{COCH}_3 + \text{CH}_3\text{MgX} \rightarrow (\text{CH}_3)_3\text{C-OH}$
Thus, compound [X] is 2-methylpropan-2-ol, which has the molecular formula $\text{C}_4\text{H}_{10}\text{O}$.
Dehydration of [X] with 20% $\text{H}_3\text{PO}_4$ or heating with Cu at 573 K yields 2-methylpropene [Y].
$(\text{CH}_3)_3\text{C-OH} \rightarrow (\text{CH}_3)_2\text{C=CH}_2 + \text{H}_2\text{O}$
Tertiary alcohols do not undergo oxidation with PCC, hence [X] shows no reaction.
Vigorous oxidation of [X] with acidified $\text{KMnO}_4$ involves dehydration to [Y] followed by oxidative cleavage of the double bond to form acetone, $\text{CO}_2$, and $\text{H}_2\text{O}$.
$(\text{CH}_3)_2\text{C=CH}_2 \rightarrow \text{CH}_3\text{COCH}_3 + \text{CO}_2 + \text{H}_2\text{O}$
Therefore, [X] is 2-methylpropan-2-ol and [Y] is 2-methylpropene.
Answer: [X]: 2–Methylpropan–2–ol [Y]: 2–Methylpropene
Hence, the correct answer is option: 'A'.