Question

Observe the following reaction \[ \mathrm{C_8H_8O} \;(X) \xrightarrow{Y} Z \] Aromatic compound \(X\) gives both iodoform and 2,4-DNP tests. Product \(Z\) liberates \(\mathrm{CO_2}\) with \(\mathrm{NaHCO_3}\) solution. Identify the set(s) in which \(Y\) is correctly represented from the following. \[ \begin{aligned} \text{I. }&\mathrm{Zn-Hg/HCl;\ KMnO_4/OH^-;\ H_3O^+} \text{II. }&\mathrm{KMnO_4/OH^-;\ H_3O^+} \text{III. }&\mathrm{LiAlH_4;\ H_2O} \end{aligned} \] The correct answer is

• A. I, II only
• B. I, III only
• C. II, III only
• D. I only

Hint:

An aromatic methyl ketone \[ \boxed{\mathrm{ArCOCH_3}} \] gives both the \[ \boxed{\text{Iodoform test}} \] and the \[ \boxed{\text{2,4-DNP test}.} \] Strong oxidation with \[ \boxed{\mathrm{KMnO_4}} \] converts the side chain into \[ \boxed{\mathrm{ArCOOH}.} \]

Answer 1

The Correct Option is A

Step 1: Identify compound \(X\). The compound gives both \[ \text{Iodoform test} \] and \[ \text{2,4-DNP test}, \] therefore \(X\) must be a methyl ketone. The aromatic compound with molecular formula \[ \mathrm{C_8H_8O} \] is \[ \boxed{\mathrm{C_6H_5COCH_3}} \] (acetophenone).

Step 2: Check Set-II. Oxidation of acetophenone with alkaline \[ \mathrm{KMnO_4} \] followed by acidification converts the side chain into \[ \boxed{\mathrm{C_6H_5COOH}} \] (benzoic acid), which liberates \[ \mathrm{CO_2} \] with \[ \mathrm{NaHCO_3}. \] Hence, \[ \boxed{\text{Set-II is correct}.} \]

Step 3: Check Set-I. Clemmensen reduction first converts acetophenone into ethylbenzene. Subsequent oxidation with \[ \mathrm{KMnO_4/OH^-} \] followed by acidification again gives \[ \boxed{\mathrm{Benzoic\ acid}.} \] Hence, \[ \boxed{\text{Set-I is also correct}.} \]

Step 4: Check Set-III. Reduction with \[ \mathrm{LiAlH_4} \] produces a secondary alcohol, which does not react with \[ \mathrm{NaHCO_3} \] to evolve \[ \mathrm{CO_2}. \] Hence, \[ \boxed{\text{Set-III is incorrect}.} \] Therefore, \[ \boxed{\text{I and II only}} \] are correct. Thus, \[ \boxed{(A)} \] is the correct answer.