A compound [X] (Isopropyl phenyl ether) undergoes reactions as given below. Identify compounds [C] and [D] formed in these reactions.
Show Hint
Quick Tip:
• Alkyl aryl ethers cleave at the alkyl side with concentrated HI.
• Phenol is readily oxidized to quinones.
• Alcohols containing the group $\mathrm{CH_3CH(OH)-}$ give positive iodoform test.
% Concept
Concept Used: • Cleavage of ethers using concentrated HI
• Oxidation of phenol
• Iodoform reaction
The given compound [X] is isopropyl phenyl ether:
\[
\mathrm{C_6H_5OCH(CH_3)_2}
\]
This is an alkyl aryl ether. Such ethers undergo cleavage with concentrated HI at the alkyl--oxygen bond because the aryl--oxygen bond has partial double bond character due to resonance and is therefore difficult to break.
% Step 1
Step 1: Cleavage of Ether with Concentrated HI
\[
\mathrm{C_6H_5OCH(CH_3)_2 + HI
\rightarrow
C_6H_5OH + (CH_3)_2CHI}
\]
Hence,
\[
[A] = \mathrm{C_6H_5OH}
\]
which is Phenol
and
\[
[B] = \mathrm{(CH_3)_2CHI}
\]
which is Isopropyl iodide (2-iodopropane)
% Step 2
Step 2: Oxidation of Phenol
Phenol undergoes oxidation in the presence of acidified sodium dichromate:
\[
\mathrm{C_6H_5OH
\xrightarrow{Na_2Cr_2O_7/H^+}
Benzoquinone}
\]
Thus,
\[
[C] = \mathrm{Benzoquinone}
\]
% Step 3
Step 3: Hydrolysis of Isopropyl Iodide
Isopropyl iodide reacts with aqueous KOH to form propan-2-ol:
\[
\mathrm{(CH_3)_2CHI
\xrightarrow{aq.\ KOH}
(CH_3)_2CHOH}
\]
The product formed is propan-2-ol.
% Step 4
Step 4: Iodoform Test
Propan-2-ol contains the group:
\[
\mathrm{CH_3CH(OH)-}
\]
Compounds containing this group give a positive iodoform test.
Hence, with iodine in alkaline medium:
\[
\mathrm{(CH_3)_2CHOH
\xrightarrow{I_2/Na_2CO_3}
CHI_3}
\]
Yellow precipitate of iodoform is produced.
Therefore,
\[
[D] = \mathrm{CHI_3}
\]
that is, Iodoform.
% Final Conclusion
Final Conclusion:
\[
[C] = \mathrm{Benzoquinone}
\]
and
\[
[D] = \mathrm{Iodoform}
\]
Therefore, the correct answer is:
\[
\boxed{(A)}
\]
