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

Question

CH$_3$–Br $\xrightarrow{\text{CH$_3$OH/Nu}}$ CH$_3$OH
Correct order of rate of this reaction for given nucleophile:

Answer 1

To solve the problem, we need to evaluate each of the statements and options given:

Statement (A): $\text{NaOCl}$ when reacted with $\text{KI}$ gives $\text{KOI}$.
- Sodium hypochlorite ($\text{NaOCl}$) and potassium iodide ($\text{KI}$) actually react to produce iodine, not potassium hypoiodite ($\text{KOI}$). Therefore, statement (A) is incorrect.
Statement (B): $\text{KOI}$ is the best reducing agent.
- Potassium hypoiodite ($\text{KOI}$) is not commonly considered the most effective reducing agent. Other compounds are much more recognized for this property. Therefore, statement (B) is incorrect.
Statement (C): Methanoic acid gives an iodoform test.
- The iodoform test is given by compounds containing a methyl ketone group ($\text{CH}_3\text{CO}-$) or related alcohols, after their oxidation. Methanoic acid does not contain a methyl ketone group and thus does not give the iodoform test. Therefore, statement (C) is incorrect.
Statement (D): Isopropyl alcohol gives the iodoform test.
- Isopropyl alcohol ($\text{CH}_3\text{CH(OH)CH}_3$) can be oxidized to a methyl ketone, which reacts positively in the iodoform test Producing a yellow precipitate of iodoform. Therefore, statement (D) is correct.
Statement (E): $\text{H}_3\text{C}–\text{CH}=\text{CH}–\text{CO}–\text{CH}_3$ gives the iodoform test.
- This compound contains a methyl ketone group, allowing it to undergo the iodoform reaction, which makes statement (E) correct.

After evaluating each statement, the correct options that hold true are D and E only. Therefore, the correct answer is

D, E only

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