Correct statement is : (A) $\text{NaOCl}$ when reacted with $\text{KI}$ gives $\text{KOI}$ (B) $\text{KOI}$ is best reducing agent (C) Methanoic acid gives iodoform test (D) Isopropyl alcohol gives iodoform test (E) $\text{H}_3\text{C}–\text{CH}=\text{CH}–\text{CO}–\text{CH}_3$ gives iodoform test
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The iodoform test only requires the presence of $\text{R}-\text{CO}-\text{CH}_3$ or $\text{R}-\text{CH}(\text{OH})-\text{CH}_3$. Unsaturation far from the reactive center (E) does not interfere with the test mechanism.
The question asks which statements related to the iodoform test and halogen chemistry are correct.
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Statement (A): $NaOCl + KI \rightarrow KOI$
$NaOCl$ is a strong oxidizing agent and oxidizes iodide ($I^-$).
However, the direct product is not $KOI$ under normal conditions.
$KOI$ (hypoiodite) is unstable and formed only in situ during iodoform reaction.
Hence, Statement (A) is incorrect.
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Statement (B): $KOI$ is best reducing agent
$KOI$ contains iodine in +1 oxidation state.
It acts as an oxidizing agent, not a reducing agent.
Hence, Statement (B) is incorrect.
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Statement (C): Methanoic acid gives iodoform test
Iodoform test requires:
\[
R-CO-CH_3 \quad \text{or} \quad R-CH(OH)-CH_3
\]
Methanoic acid ($HCOOH$) has neither structure.
So it does not give iodoform test.
Statement (C) is incorrect.
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Statement (D): Isopropyl alcohol gives iodoform test
Isopropyl alcohol:
\[
(CH_3)_2CHOH
\]
It contains the required methyl carbinol group $CH(OH)CH_3$.
On oxidation, it forms acetone, which gives yellow precipitate of $CHI_3$.
Hence, Statement (D) is correct.
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Statement (E):
$CH_3-CH=CH-CO-CH_3$ gives iodoform test
This compound contains the methyl ketone group ($COCH_3$).
Presence of double bond does not affect the iodoform reaction.
Hence, Statement (E) is correct.
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Final Conclusion:
Correct statements are (D) and (E).
Correct option is (3).
