Question

Which of the following compound(s) gives yellow precipitate on reaction with $I_2$/NaOH?

(I) $\qquad$ $\qquad$ (II) $\qquad$ CH$_3$CH$_2$OH

(III) $\qquad$ PhCH$_2$CH$_2$OH $\qquad$ (IV) $\qquad$

Choose the correct answer from the options given below:

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

Answer 1

Answer: (C)

I, II, IV only

Answer 2

The iodoform test is used to detect the presence of a methyl ketone group ($\text{CH}_3\text{CO-}$) or a secondary alcohol group with a methyl group at the alpha carbon ($\text{CH}_3\text{CH(OH)-}$), or ethanol ($\text{CH}_3\text{CH}_2\text{OH}$), or acetaldehyde ($\text{CH}_3\text{CHO}$). The reaction produces a yellow precipitate of iodoform ($\text{CHI}_3$).

Let's analyze each given compound:

(I) PhCOCH$_3$ (Acetophenone)

• This compound contains a methyl ketone group ($\text{CH}_3\text{CO-Ph}$).

• It will undergo the iodoform reaction to produce sodium benzoate and iodoform.

  $\text{PhCOCH}_3 + 3\text{I}_2 + 4\text{NaOH} \rightarrow \text{PhCOONa} + \text{CHI}_3 \downarrow + 3\text{NaI} + 3\text{H}_2\text{O}$

• Therefore, (I) gives a yellow precipitate.

(II) CH$_3$CH$_2$OH (Ethanol)

• Ethanol is a primary alcohol that can be oxidized to acetaldehyde ($\text{CH}_3\text{CHO}$) by the $\text{I}_2/\text{NaOH}$ reagent. Acetaldehyde contains the $\text{CH}_3\text{CHO}$ group, which gives a positive iodoform test.

  $\text{CH}_3\text{CH}_2\text{OH} + \text{I}_2 + 2\text{NaOH} \rightarrow \text{CH}_3\text{CHO} + 2\text{NaI} + 2\text{H}_2\text{O}$

  $\text{CH}_3\text{CHO} + 3\text{I}_2 + 4\text{NaOH} \rightarrow \text{HCOONa} + \text{CHI}_3 \downarrow + 3\text{NaI} + 3\text{H}_2\text{O}$

• Therefore, (II) gives a yellow precipitate.

(III) PhCH$_2$CH$_2$OH (2-Phenylethanol)

• This is a primary alcohol. Upon oxidation, it forms phenylacetaldehyde ($\text{PhCH}_2\text{CHO}$).

  $\text{PhCH}_2\text{CH}_2\text{OH} \xrightarrow{\text{I}_2/\text{NaOH}} \text{PhCH}_2\text{CHO}$

• Phenylacetaldehyde does not have a $\text{CH}_3\text{CHO}$ group (it has a $\text{CH}_2\text{CHO}$ group). Only ethanol and acetaldehyde among aldehydes give a positive iodoform test.

• Therefore, (III) does not give a yellow precipitate.

(IV) PhCH(OH)CH$_3$ (1-Phenylethanol)

• This is a secondary alcohol with the $\text{CH}_3\text{CH(OH)-}$ group.

• It will be oxidized to a methyl ketone (acetophenone, PhCOCH$_3$) by the $\text{I}_2/\text{NaOH}$ reagent.

  $\text{PhCH(OH)CH}_3 + \text{I}_2 + 2\text{NaOH} \rightarrow \text{PhCOCH}_3 + 2\text{NaI} + 2\text{H}_2\text{O}$

• Acetophenone (PhCOCH$_3$) contains a methyl ketone group and will then undergo the iodoform reaction.

  $\text{PhCOCH}_3 + 3\text{I}_2 + 4\text{NaOH} \rightarrow \text{PhCOONa} + \text{CHI}_3 \downarrow + 3\text{NaI} + 3\text{H}_2\text{O}$

• Therefore, (IV) gives a yellow precipitate.

Conclusion:

Compounds (I), (II), and (IV) give a yellow precipitate on reaction with $\text{I}_2/\text{NaOH}$.